Lectures on Analysis II. Math 502 with homeworks

Lectures on Analysis II Math 502 John Roe SPRING 2006 1 Lecture 1 The complex plane Remember that a complex number is...

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Lectures on Analysis II Math 502 John Roe SPRING 2006

1

Lecture 1 The complex plane Remember that a complex number is a formal expression of the sort z = x + yi where x and y are real numbers and i2 = −1. (We call x the real part of z and y the imaginary part, and we use the notaion x = Re z, y = Im z.) For algebraists, the ring of complex numbers is the quotient of the polynomial ring R[T ] in one indeterminate by the principal ideal generated by T 2 + 1. We identify each real number x with the complex number x + 0i. There is no problem in adding, subtracting or multiplying complex numbers by the usual rules. However, the following is a nontrivial fact. Theorem 1.1. The complex numbers form a field, i.e. every non-zero complex number has a multiplicative inverse. (Equivalently, for the algebraists again, the ideal generated by T 2 + 1 is maximal.) Proof. We write an explicit formula for the inverse. If z = x+yi is a complex number, then |z| is the positive real number defined by |z|2 = x2 + y 2 . The complex conjugate of z is z¯ = x − yi. One computes z z¯ = x2 + y 2 = |z|2 . Thus, if z 6= 0, one has |z| > 0 and z¯ x y = 2 − 2 i 2 2 |z| x +y x + y2 is the multiplicative inverse of z. It’s often useful to represent the complex number z = x + iy in polar coordinate form as z = reiθ = r cos θ + ir sin θ. Here r = |z|, and θ is a residue class modulo 2π (an element of the quotient group R/2πZ). One calls r the modulus of z and θ the argument. In polar 2

coordinates the law for multiplication of complex numbers takes the simple form (r1 eiθ1 )(r2 eiθ2 ) = r1 r2 ei(θ1 +θ2 ) ; you multiply the moduli and add the arguments. The complex plane is the set of all complex numbers x + yi, thought of as points of a Euclidean plane (a 2-dimensional real inner product space). What do complex number operations look like as transformations of this plane? • The operation of adding (or subtracting) a fixed complex number corresponds to a translation of the plane, and all translations arise in this way. • The operation of multiplying by a fixed complex number corresponds to a linear transformation of the plane. However, not all linear transformations arise in this way. Definition 1.2. An invertible linear transformation T : R2 → R2 is conformal if it preserves angles: that is, for all pairs of vectors u, v, the angle between T u and T v is the same as the angle between u and v. Recall that an invertible linear transformation is orientation-preserving if it can be represented by a matrix with positive determinant. Proposition 1.3. The operation of multiplication by a fixed nonzero complex number is an orientation-preserving conformal linear transformation, and every orientation-preserving conformal linear transformation of the plane is given by such a multiplication. Proof. That multiplication by a complex number z = reiθ is conformal follows from the polar representation of complex multiplication. In the other direction, suppose that the matrix a b T = c d represents a conformal transformation. Look at the following table of vectors and their images under T :

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Vector 1 0 1 1 0 1

Image a c a+b c+d b d

Since the first and third rows must be perpendicular, ab + cd = 0, whence there is a scalar λ with b = −λc and d = λa. Since the first and second rows make an angle of π/4, we must have 1 1 a(a + b) + c(c + d) √ =p =√ . 2 2 2 2 2 1 + λ2 (a + c )((a + b) + (c + d) ) Thus λ = ±1, and the orientation-preserving condition forces λ = 1. Now we see that T is exactly the matrix of multiplication by the complex number a + ci. Exercise 1.1. Show that the modulus obeys the triangle inequality |z ± w| 6 |z| + |w|. This allows us to make the complex plane into a metric space, and thus to introduce topological notions such as open and closed sets, continuity. etc. Exercise 1.2. Show that the dot product of two complex numbers z and w (considered as vectors in R2 ) is the real part of z w. ¯ Use this to give another proof that multiplication by a fixed complex number is a conformal linear transformation.

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Lecture 2 Holomorphic Maps The chief objects of study in complex analysis are differentiable functions (of a complex variable). The definition is the natural one. Definition 2.1. Let Ω be an open subset of C. A map f : Ω → C is differentiable at a point z ∈ Ω if there is a complex number f 0 (z) such that f (z + h) = f (z) + f 0 (z)h + o(h) as h → 0. Equivalently, the limit f (z + h) − f (z) h→0 h lim

exists and equals f 0 (z). Definition 2.2. The function f is holomorphic on the open set Ω if it is differentiable at every point of Ω. The usual proofs of real-variable theory work without change to show that the sum, difference, product, and quotient (where defined) of holomorphic functions are again holomorphic, with the usual expressions for their derivatives. The composite of holomorphic functions is also holomorphic, and the chain rule holds: if h(z) = g(f (z)), then h0 (z) = g 0 (f (z))f 0 (z). Last semester we proved the following important fact (Proposition 7.1 of last semester). Theorem 2.3. A power series f (z) =

∞ X

an z n

n=0

represents a holomorphic function within its radius of convergence. Its derivative is given by the power series f 0 (z) =

∞ X

nan z n−1

n=1

which has the same radius of convergence. This theorem shows that familiar functions like polynomials, ez , sin z, cos z are holomorphic on the whole complex plane.

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It is important to understand that complex differentiability is a much stronger condition than real-variable differentiability. To say that f is differentiable as a real function would say that there is a 2 × 2 real matrix Df (z) such that f (z + h) = f (z) + Df (z) · h + o(h). Thus the extra condition for complex differentiability is that the 2×2 matrix Df (z) is the matrix of multiplication by a complex number, or in other words (thanks to Proposition 1.3) is either zero or an orientation-preserving conformal linear transformation. According to the proof of Proposition 1.3, the matrix of such a linear a −c . On the other hand, if z = transformation must be of the form c a x + iy and f (z) = u(x, y) + iv(x, y), then ∂u/∂x ∂u/∂y . Df = ∂v/∂x ∂v/∂y Comparing these expressions we see that f = u+iv is a complex-differentiable function of z = x + iy if and only if it is real-differentiable and, in addition, satisfies the Cauchy-Riemann equations ∂u ∂v = , ∂x ∂y

∂u ∂v =− . ∂y ∂x

Remark 2.4. It follows from the Cauchy-Riemann equations and the symmetry of mixed derivatives that both the real and imaginary parts of a holomorphic function satisfy Laplace’s equation 1 ∂2u ∂2u + 2 = 0. ∂x2 ∂y Laplace’s equation is a fundamental one in physics and this is one reason for the close connection between complex analysis and many physical problems. A mapping between open subsets of the plane is called a conformal map if it is differentiable (in the real-variable sense) and its derivative at each point is a conformal linear transformation. Geometrically, this says that if two smooth curves meet at a certain angle2 then their images under the map must always meet at the same angle. A conformal map is directly conformal if the derivatives are orientation-preserving. From our discussion we have 1 At least if they are twice differentiable; later we shall see that this assumption is always satisfied. 2 The angle between two curves at their point of intersection is defined to be the angle between their tangent vectors.

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Proposition 2.5. A directly conformal map is a holomorphic function with non-vanishing derivative, and conversely. Example 2.6. Consider the function f (z) = z 2 , whose derivative is 2z. Everywhere except at the origin, this map is conformal. However, it is not conformal at the origin; in fact, it doubles angles between curves that intersect there. Example 2.7. The simplest non-trivial examples of conformal maps are the M¨ obius transformations z 7→

az + b cz + d

where ad − bc 6= 0. The derivative is ad − bc (cz + d)2 which is never zero, so M¨obius transformations are conformal where defined. The composite of two M¨obius transformations is again a M¨obius transformation. If one associates the matrix ( ac db ) to the M¨obius transformation z 7→ (az + b)/(cz + d), then composition of M¨obius transformations corresponds to multiplication of matrices. The correct explanation of this fact involves projective geometry. Definition 2.8. The Riemann sphere CP 1 is the complex projective line; that is, the collection of all 1-dimensional subspaces of the complex vector space C2 . For every complex number z we can define a point of the Riemann sphere, namely the 1-dimensional subspace spanned by the vector (z, 1) (for brevity, we refer to this as the point with homogeneous coordinates (z, 1)) . All points of the Riemann sphere, except one, arise in this way. The exceptional point is the one with homogeneous coordinates (1, 0); let us denote this point by ∞. Thus we can regard the Riemann sphere as the union of C with a single ‘point at infinity’. Now the invertible linear transformation ( ac db ) of C2 induces a map on CP 1 , and on C ⊆ CP 1 this is the M¨obius transformation z 7→ (az + b)/(cz + d) that we started with. Remark 2.9. Notice that the expressions −d/c 7→ ∞,

∞ 7→ a/c

make sense (and are true) on the Riemann sphere. Later we shall see that the Riemann sphere is a simple example of a Riemann surface; that is a domain, more general than a subset of C, on which conformal mapping can be considered. 7

Definition 2.10. A circline (horrible but useful word) means a circle or a straight line in C. From the point of view of the Riemann sphere, a straight line is just a ‘circle’ which happens to go through ∞. Proposition 2.11. The image of a circline under a M¨ obius transformation is another circline. Proof. There are several ways to prove this. Here is one that uses the Riemann sphere. The equation of a circline must be of the form α|z|2 + βz + γ z¯ + δ = 0 for real constants α, β, γ, δ. (Exercise: why?) In terms of homogeneous coordinates (z, w) on the Riemann sphere this equation becomes α β z¯ z w = 0. w ¯ γ δ

a c Applying a M¨obius transformation — that is, replacing (z, w) by (z, w) b d — produces another equation of the same sort where the middle matrix is replaced by a ¯ ¯b α β a c , c¯ d¯ γ δ b d This is therefore the equation of another circline. Exercise 2.1. Show that every equation of the form |z − p| = λ|z − q| defines a circline, and that every circline can be expressed (not uniquely) in this way. Use this fact to give another proof that M¨obius transformations map circlines to circlines.

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Lecture 3 Some examples of holomorphic functions In this lecture we are going to review some examples of holomorphic functions, and mention some objectives that we want to achieve for the course. Power series The first examples are those defined by power series. Definition 3.1. A function that is holomorphic on the whole complex plane is called an entire function. The most important example of an entire function is the exponential function ∞ X zn ez = . n! n=0

We have the exponential law ea eb = ea+b and thus ez never takes the value zero. We shall see that it takes all other complex values. Remark 3.2. The exponential function is equal to its own derivative. Conversely, if f is an entire function which is equal to its own derivative, then by the chain rule d (f (z)e−z ) = (f 0 (z) − f (z))e−z = 0. dz Therefore f (z)e−z is a constant (this follows for instance from the realvariable version of the Mean Value Theorem, proved in Proposition 11.5 of last semester) and so f is a constant multiple of ez . Since the power series for the exponential function has real coefficients, ez¯ = ez . It follows that |ez |2 = ez ez = ez+¯z = e2 Re z so |ez | = eRe z , for all complex numbers z. The sine and cosine functions are defined in terms of the exponential by cos z =

eiz + e−iz , 2

sin z =

9

eiz − e−iz . 2i

These are entire functions also and the identity eiz = cos z + i sin z is valid for all z. In particular ex+iy = ex (cos y + i sin y) which shows that the complex number ex+iy has modulus ex and argument y. It follows that for any nonzero complex number w one can find a complex number z such that ez = w. Indeed, one can find infinitely many such z, differing by integer multiples of 2πi. If a is a positive real number we can define az = ez log a where log a is the ordinary (real) logarithm of a; this is obviously an entire function. We leave aside for now the question of defining az for other values of a. Functions defined by integrals A power series is one way of taking an ‘infinite linear combination’ of elementary functions, in this case powers of z. Another way of taking such a ‘linear combination’ is by an integral. Here the most famous example is the gamma function, which is defined by the integral Z ∞ Γ(z) = tz−1 e−t dt. 0

The integral here is a Lebesgue integral (of a complex-valued function) over the positive half-line. If Re z > a > 0, then the absolute value of the integrand is dominated by the integrable function ta e−t , so that the integral converges. Moreover, the derivative of the integrand with respect to z, that is (log t)tz−1 e−t is also dominated by an integrable function (for instance a multiple of ta/2 e−t/2 ), and thus by the theorem about differentiating under the integration sign (last semester, Exercise 19.2) the function Γ(z) is differentiable. Our final conclusion is that Γ is a holomorphic function in the right-hand half plane, Re z > 0.

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Proposition 3.3. We have Γ(z + 1) = zΓ(z) whenever Re z > 0. We have Γ(1) = 1, and consequently Γ(n) = (n − 1)! for n = 1, 2, . . .. Proof. Integrate by parts to get Z Z ∞ z −t t e dt = Γ(z + 1) =

∞

ztz−1 e−t dt = zΓ(z)

0

0

as required. We can rewrite the proposition as Γ(z) =

Γ(z + 1) Γ(z + 2) = = .... z z(z + 1)

Notice that whatever z is, there is an integer n such that Γ(z + n) has positive real part. Thus we can take this equation as defining Γ for all values of z, except for nonpositive integers where one tries to divide by zero on the right. We say that there is an analytic continuation of Γ to the whole complex plane, with singularities at 0, −1, −2, . . .. The gamma function satisfies a number of important identities, for example: π • Reflection formula: Γ(z)Γ(1 − z) = . sin πz • Duplication formula: Γ(2z) = π −1/2 22z−1 Γ(z)Γ(z + 12 ). √ • Stirling’s asymptotic formula: Γ(x) 2πxx−1/2 e−x . We shall prove these later in the course. Dirichlet series A series such as ∞ X 1 ζ(s) = ns n=1

is called a Dirichlet series; this particular example (the most important) is the Riemann zeta function. It converges for Re s > 1, but (as with the gamma function) it has an analytic continuation to the whole complex plane. This analytic continuation satisfies the reflection law ζ(1 − s) = 2(2π)−s Γ(s) cos(πs/2)ζ(s) 11

(proved by Riemann). A more elementary fact is Euler’s formula for the even integer values of the zeta function, ζ(2k) =

(−1)k 22k−1 B2k π 2k , (2k)!

where the Bernoulli numbers Bn are defined to be the coefficients in the power series expansion ∞

X z = Bn z n /n!. ez − 1 n=0

We’ll probably prove these results in class. It follows that ζ(2k) is an irrational number. The odd integer values are much more inaccessible. I remember the sensation caused by Ap´ery’s proof (1979) that ζ(3) is irrational. There is no other odd integer 2k + 1 for which even the question of the rationality of ζ(2k + 1) is settled. Exercise 3.1. Show that the zeta function can also be defined by the integral Z ∞ s−1 1 t ζ(s) = dt, Γ(s) 0 et − 1 provided that Re s > 1. Infinite products We an infinite prodQ met infinite products last semester: remember thatP uct (1+an ) converges (absolutely) if and only if the sum |an | converges. In particular an infinite product expression such as z

∞ Y n=1

z2 1− 2 n

defines a convergent infinite product which is a function of z. This function clearly has zeroes at all integer points. In fact, as a consequence of our work on the Γ function we shall prove the surprising identity sin πz = πz

∞ Y n=1

1−

z2 n2

.

There are similar expressions (discovered by Euler) for other trigonometrical functions too.

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Here is one interesting Euler argument (at present this is purely formal, but we shall see later how to make it rigorous). Take the logarithm of the infinite product above X z2 log sin πz = log π + log z + log 1 − 2 . n Differentiate with respect to z to get π cot πz =

N ∞ X 1 X 2z 1 = lim + . 2 2 N →∞ z z −n z+n n=1

n=−N

Thus the simplest imaginable expression for a periodic function with singularities at the integer points gives the trigonometric function π cot πz. It is now a small step to consider doubly periodic or elliptic functions such as ℘(z) = lim

N →∞

N X

N X

m=−N n=−N

1 . (z + m + in)2

These functions were of central importance in 19th-century analysis.

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Lecture 4 Plane topology and complex analysis This lecture is mostly devoted to the ramifications of a simple question: √ is there a holomorphic function z? Lemma 4.1. Let Ω and Ω0 be open subsets of C and f : Ω → Ω0 is a holomorphic map. Suppose that f is a one-to-one correspondence (bijection) and that the derivative f 0 (z) is continuous3 and is never zero for any z ∈ Ω. Then the inverse function to f , f −1 : Ω0 → Ω, is a holomorphic map also. Proof. The real-variable version of the Inverse Function Theorem (see lecture 12 of last semester) shows that f −1 is differentiable in the real-variable sense, and that the derivative of f −1 is the inverse of the derivative of f . But the inverse of a conformal linear transformation is clearly conformal, so f −1 is real-differentiable with derivative a conformal transformation, hence it is complex-differentiable. Let’s apply this lemma to the function f (z) = z 2 on various domains. First of all, every nonzero complex number has two ‘square roots’: if w = reiθ , then the two numbers z = ±r1/2 eiθ/2 both satisfy z 2 = w. (Notice that eiπ = −1, so the sign ambiguity is implicit in the fact that θ is only unique modulo 2πZ.) Moreover, we have Lemma 4.2. Every nonzero w0 ∈ C has a neighborhood U on which there is defined a ‘holomorphic branch of the square root’, that is a holomorphic function g : U → C such that g(w)2 = w for all w ∈ U . Proof. Let w0 = r0 eiθ0 and define Ω to be the open half-plane Ω = {reiθ : r > 0, (θ0 − π)/2 < θ < (θ0 + π)/2. The function f (z) = z 2 maps Ω bijectively onto U = {ρeiψ : ρ > 0, θ0 − π < ψ < θ0 + π} which is an open set (a “cut plane”) containing w0 . The result therefore follows from Lemma 4.1. Now of course, there is also a sign ambiguity in taking the square root of a (positive) real number. However, in the real case this ambiguity can be resolved by a once-and-for-all executive decision: to always take the positive square root. In the complex world such a once-and-for-all decision is not possible. 3

Later we shall see that this is automatic for holomorphic functions.

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Proposition 4.3. There is no holomorphic function g(w) defined on all of C, or even on C \ {0}, such that g(w)2 = w. Proof. For those who know some algebraic topology, we can express this very shortly. Take the supposed function g and restrict it to the unit circle T in the complex plane. Then the composite g

T

/T

f

/ T,

where f (z) = z 2 , is the identity. Look at the diagram of homomorphisms induced on π1 (T) = Z, g∗

Z

/Z

f∗

/ Z.

But f∗ is multiplication by 2, and this is ridiculous: there is no homomorphism Z → Z which, when composed with multiplication by 2, yields the identity. Without algebraic topology, let’s take a more pedestrian course. Let h : T → T be the function that sends eiθ , 0 6 θ < 2π, to eiθ/2 ; of course, although this expression is continuous as a function of θ ∈ [0, 2π), it is not continuous at 1 when considered as a function on T. For each θ ∈ [0, 2π) we must have g(eiθ ) = ±h(eiθ ) and, since both sides are continuous functions on the interval [0, 2π), the intermediate-value theorem tells us that the sign must be constant. Thus, globally, either g = h or g = −h; in either case it is not continuous on T. Exercise 4.1. Prove the facts corresponding to those above for the logarithm function. That is, show that every nonzero w0 has a neighborhood on which there is defined a holomorphic function g with eg(w) = w, but that there is no function defined on all of C \ {0} that has this property. Let’s develop these topological ideas. Definition 4.4. A path in the complex plane is a continuous map γ of an interval [a, b] ⊂ R to C. The trace γ ∗ of the path is the set of values {γ(t) : t ∈ [a, b]}; it’s a compact subset of C. The path is closed if γ(a) = γ(b). Let p ∈ C. To each closed path in C \ {p} we are going to associate an integer, the winding number of the path with respect to p . For topologists, this is just the element of π1 (C \ {p}) ∼ = Z defined by the path γ. However, let’s interpret it in terms of analysis. Lemma 4.5. Let γ : [a, b] → C \ {p} be a path. Let w0 ∈ C have the property that ew0 = γ(a) − p. Then there is a unique continuous function w : [a, b] → C such that w(a) = w0 and ew(t) = γ(t) − p. 15

Proof. According to Exercise 4.1, there is an open cover of C\{p} consisting of sets U on each of which a ‘holomorphic logarithm of w − p’ is defined. Pull back this cover via the continuous map γ to an open cover U of [a, b]. By the Lebesgue covering theorem, the cover U admits a positive Lebesgue number. Therefore, we can subdivide [a, b] into finitely many equal-length subintervals [aj , aj+1 ] in such a way that on each of these subintervals there is defined a function gj (t) with egj (t) = γ(t) − p, t ∈ [aj , aj+1 ]. Since eg0 (a) = ew0 , g0 (a) − w0 = 2πik0 for some integer k0 . Replace g0 by g˜0 = g0 − 2πik0 . Inductively replace gj+1 by g˜j+1 = gj+1 − 2πikj+1 , where kj+1 is the integer defined by g˜j (aj+1 ) − gj+1 (aj+1 ) = 2πikj+1 . Now the functions g˜j agree at the endpoints of their subintervals so they fit together to yield a continuous function w such that ew(t) = γ(t) − p. The uniqueness is clear because two possible choices for w would have to differ by a continuous 2πiZ-valued function, which must be constant, and hence zero because it’s zero at t = a. Suppose now that γ is a closed path. Then ew(b) = ew(a) and so w(b) − w(a) = 2πik for some integer k, which is independent of the choice of initial value w0 . Definition 4.6. This integer k is called the winding number of the closed path γ around p, or the index of p with respect to γ. We write it as n(γ, p). Example 4.7. The winding number of a circular path γ(t) = keit , t ∈ [0, 2π], about the origin is equal to 1. Finally, a key definition. Definition 4.8. A connected open subset Ω of C is simply-connected if, for every closed path γ in Ω and every p ∈ C \ Ω, the winding number n(γ, p) is zero. Remark 4.9. Two closed paths γ0 and γ1 are said to be homotopic if there is a “continuous 1-parameter family of closed paths” γs that interpolates between them. It is proved in an algebraic topology course that if two closed paths are homotopic in C \ {p} then they have the same winding number around p. Moreover, a region Ω as above is simply-connected if and only if every closed path in Ω is homotopic to a constant path (this is a non-trivial theorem). However, I don’t think we are going to need any of this. Remark 4.10. Another fact that I hope we won’t need is the Jordan curve theorem. A Jordan curve is a homeomorphism of the circle into C, or in other words a closed path that is a one-to-one map (except for joining the 16

ends). It is a theorem that every Jordan curve divides the plane into exactly two connected open pieces the exterior comprising points of index 0, and the interior comprising points of index ±1. This statement is quite tricky to prove rigorously. Despite its prominence in some treatments of complex analysis, we shall do without it.

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Lecture 5 Path integrals and Cauchy’s theorem Recall that a path is a continuous map γ from some interval [a, b] to C. We’ll say that it is smooth if it is differentiable as a function from (a, b) to C = R2 and its derivative is continuous and bounded. More generally we will consider piecewise smooth paths which have continuous, bounded derivatives except at finitely many points in [a, b]; this allows us to consider paths like the boundary of a triangle, which have finitely many corners. Definition 5.1. Let γ : [a, b] → C be a smooth (or piecewise smooth) path and let f be a continuous function defined on γ ∗ . Then the path integral of f along γ is defined to be Z Z b f (z) dz := f (γ(t)) γ 0 (t)dt. γ

a

Formally, we obtained this expression by ‘substituting z = γ(t)’. We will need some simple estimates for path integrals. The basic one is Proposition 5.2. Let γ be a smooth path and let f be a function defined and continuous on the image of γ. Then Z f (z) dz 6 M Length(γ), γ

where M is an upper bound for |f (z)| along γ. Proof. Use the definition Z

b

Z f (z) dz =

γ

f (γ(t))γ 0 (t)dt.

a

By real-variable theory Z Z b f (z) dz 6 M |γ 0 (t)|dt = M Length(γ) γ

a

as required. Example 5.3. Let R γn be the circular path around the origin of radius r, and let us compute γ z dz for various values of z. We can parameterize the path as z = reit , t ∈ [0, 2π]. Thus ( Z 2π Z 2πi (n = −1) n n int it z dz = r e ire dt = 0 (n 6= −1) γ 0 18

The fundamental theorem of calculus is valid for path integrals in the following form. Proposition 5.4. Let γ be a piecewise smooth path and let F be a holomorphic function defined on some open set containing γ ∗ . Put F 0 (z) = f (z). Then Z f (z)dz = F (γ(b)) − F (γ(a)). γ

Proof. We have f (γ(t))γ 0 (t) = F 0 (γ(t))γ 0 (t) =

d F (γ(t)) , dt

so the result follows from the ordinary form of the fundamental theorem of calculus. However in complex analysis there is an additional complication: even very regular functions may not have antiderivatives for global (topological) reasons. Indeed, it’s obvious that if a function has an antiderivative then its integral around any closed path must vanish. Thus Example 5.3 shows that the function 1/z has no antiderivative on C \ {0}, even though it’s holomorphic (and as smooth as you like) there. This is related to what we studied last time. Proposition 5.5. Let γ be a closed, piecewise smooth path and let p ∈ C\γ ∗ . Then Z dz = 2πin(γ, a) z −p γ where n(γ, p) denotes the winding number of γ around p. Proof. Fix w0 such that ew0 = γ(a) − p. Now define Z s 0 γ (t)dt w(s) = w0 + . γ(t) −p a By the chain rule d −w(s) e (γ(s) − p) = e−w(s) (γ 0 (s) − γ 0 (s)) = 0. ds So ew(s) = γ(s)−p and w(a) = w0 . Thus w is the unique continuous function of Lemma 4.5 that satisfies these conditions, and the result follows from the definition of winding number when we put s = b.

19

Corollary 5.6. The winding number n(γ, p) (considered as a function of p) is constant on the connected components of C \ γ ∗ , and is zero on the unbounded component. Proof. The integral formula and the DCT show that n(γ, p) is a continuous integer-valued function and tends to zero at ∞. Proposition 5.7 (Cauchy’s theorem for a triangle). Let Ω be an open subset of C, and let T be a triangle that is contained (interior and boundary) within Ω. Let ∂T denote the closed path obtained by traversing the three sides of T counterclockwise. For any holomorphic function f on Ω, one has I f (z) dz = 0. ∂T

Proof. Proof by contradiction, so suppose the result false. Define the ‘badness’ B(T ) of a triangle T to be I B(T ) = f (z) dz diam(T )2 . ∂T

By hypothesis there is some triangle T with B(T ) > 0, say B(T ) = 5. Subdivide T = T0 into four sub-triangles by bisecting each side. Each Hsub-triangle has half the diameter of T . On the other hand, the integrals f (z)dz around the boundaries of the four sub-triangles total the integral around the boundary of T . Consequently, one such integral must have absolute value at least 14 of the absolute value of the integral around the boundary of T . We conclude that one sub-triangle, call it T1 , has B(T1 ) > B(T ). Repeat this process obtaining a nested sequence of triangles T0 ⊃ T1 ⊃ T2 ⊃ · · · , each half the diameter of the preceding one, and all with badness B(Tn ) > B(T ) = 5. T A compactness argument shows that Tn = {z0 } for some point z0 . Now since f is differentiable at z0 we have f (z) = f (z0 ) + (z − z0 )f 0 (z0 ) + e(z), with a ‘small’ error term e(z); in particular, there is δ > 0 such that if |z − z0 | < δ then |e(z)| < |z − z0 |. The function z 7→ f (z0 ) + (z − z0 )f 0 (z0 )

20

is linear and thus has an antiderivative; so its integral around any closed contour is zero. Consequently I I f (z) dz = e(z)dz ∂Tn

∂Tn

for each triangle Tn . Pick n large enough that the diameter d of Tn is less than δ. Then by the estimate 5.2, I 6 d · 3d = 3d2 , e(z)dz ∂Tn

so B(Tn ) 6 3 and this is a contradiction. Exercise 5.1. Show that Cauchy’s theorem for a triangle remains true if f is continuous in Ω and differentiable everywhere except at a single point. (Approximate a triangle with the ‘bad’ point as vertex by nearby triangles that don’t contain the bad point.)

21

Lecture 6 Consequences of Cauchy’s theorem We are going to ‘bootstrap’ the Cauchy theorem to successively greater generality. Recall Definition 6.1. A subset Ω of C is convex if, whenever a, a0 ∈ Ω, all the points λa + (1 − λ)a0 , λ ∈ [0, 1], also lie in Ω. Theorem 6.2 (Cauchy’s theorem for a convex set). Let Ω be a convex open subset of C and let f be holomorphic on Ω. Then Rthere is a holomorphic function F on Ω such that f = F 0 . Consequently, γ f (z)dz = 0 for every closed path in Ω. Proof. Fix a ∈ Ω. For each z ∈ Ω let [a, z] denote the straight line path from a to z; by convexity this lies entirely in Ω. Define Z F (z) = f (w)dw. [a,z]

If z + h ∈ Ω also then by Cauchy for a triangle (5.7), Z Z Z F (z + h) − F (z) = f (w)dw − f (w)dw = [a,z+h]

[a,z]

f (w)dw.

[z,z+h]

Therefore Z F (z + h) − F (z) 1 = − f (z) (f (w) − f (z))dw h h [z,z+h] 6 sup{|f (w) − f (z)| : w ∈ [z, z + h]}, and this tends to zero as h → 0 by continuity of f . WeR have shown that F is differentiable and its derivative is f . The equality γ f (z)dz = 0 now follows from the fundamental theorem of calculus. Remark 6.3. Using Exercise 5.1, we obtain the same conclusion under the weaker condition f continuous in Ω, holomorphic except at a single point. We’ll use that in a moment. Corollary 6.4. A convex open set is simply connected. Proof. Let Ω be a convex open set. According to Proposition 5.5, for any path γ ∈ Ω and any p ∈ / Ω, Z 1 dz . n(γ, p) = 2πi γ z − p 22

The integrand is holomorphic in Ω, so the integral vanishes by Cauchy’s theorem. The following amazing consequence shows that the values of a holomorphic function at all points inside a closed path can be recovered from its values on that path. Proposition 6.5 (Cauchy integral formula). Let f be holomorphic on a convex open set Ω and let γ be a closed path in Ω. Let w ∈ Ω \ γ ∗ ; then Z 1 f (z)dz n(γ, w)f (w) = . 2πi γ z − w (Usually we are interested in n(γ, w) = 1.) Proof. Define g on Ω \ {w} by g(z) = (f (z) − f (w))/(z − w). This function is holomorphic on Ω \ {w}. Moreover, it extends to a continuous function on all of Ω, whose value at w is f 0 (w). Thus Remark 6.3 applies, and Z Z f (z)dz 0 = g(z)dz = − 2πif (w)n(γ, w) γ γ z−w as required. Theorem 6.6. Let f be holomorphic in an open subset Ω of C. Suppose that the closed disk D(a; r) is contained in Ω. Then there is a power series P ∞ n n=0 cn (z − a) which converges to f (z) in the open disk D(a; r) (uniformly on compact subsets). Moreover the coefficients are given by I 1 f (w) dw cn = 2πi ∂D(a;r) (w − a)n+1 The power series here is called the Taylor series for f around a. Proof. Without loss of generality take a = 0. For fixed z ∈ D(a; r) write ! I I X ∞ 1 f (w)dw 1 zn f (z) = = f (w)dw, 2πi w−z 2πi wn+1 n=0

using Cauchy’s integral formula. (The integrals are taken around ∂D(a; r).) The series converges uniformly in w for z fixed (by Weierstrass’ M-test) and so we can interchange sum and integral to get I ∞ X f (w)dw 1 f (z) = zn 2πi wn+1 n=0

as asserted. 23

Corollary 6.7. A holomorphic function is infinitely differentiable, and we have Cauchy’s formula for derivatives I n! f (w)dw f (n) (a) = . 2πi ∂D(a;r) (w − a)n+1 Proof. Use the differentiability of power series. Corollary 6.8 (Morera’s Theorem). If a continuous function f on an open R subset Ω of C has ∂T f (z)dz = 0 for all triangles T in Ω, then f is holomorphic. Proof. Without loss of generality take Ω to be a disk. Then the proof of theorem 6.2 shows that f (z) = F 0 (z) for some holomorphic F . But now F is infinitely differentiable; so f is infinitely differentiable too — in particular it is holomorphic. Corollary 6.9. If fn is a sequence of holomorphic functions and fn → f uniformly on compact sets, then f is holomorphic too. R R Proof. For any triangle T , ∂T f (z)dz = lim ∂T fn (z)dz = 0. The result follows from Morera’s theorem. Corollary 6.10 (Removability of Singularities). Let f be continuous in an open set Ω and holomorphic in Ω except possibly at one point a. Then f is, in fact, holomorphic at a too. Proof. Use Cauchy’s and Morera’s theorems. Exercise 6.1. (The reflection principle) Let f be continuous on the closed upper half-plane, holomorphic on the open upper half-plane, and real-valued on the real axis. Show that, if we extend f to the whole of C by defining f (z) = f (¯ z) when z is in the lower half-plane, then the extended function is holomorphic on the entire complex plane.

24

Lecture 7 The rigidity of holomorphic functions Lemma 7.1. Let f be holomorphic in a disc D(a; r), but not identically zero there. Then there is a smaller disc D(a; ρ) within which f (z) = 0 has no solution except possibly z = a. P Proof. Expand f in a Taylor series f (z) = cn (z − a)n about z = a. Since f does not vanish identically, some Taylor coefficient cn is nonzero; let cN be the first such coefficient (so that cN 6= 0 but ck = 0 for k < N ). Then f (z) =

∞ X

cn (z − a)n = (z − a)N (cN + (z − a)g(z)),

n=N

where g(z) =

∞ X

ck+N +1 (z − a)k

k=0

is continuous at a. It follows that |(z − a)g(z)| < 21 |cN | for z sufficiently close to a, and thus cN + (z − a)g(z) does not vanish for z sufficiently close to a. Terminology: A zero of f is a point where f (z) = 0. The exponent of the first nonvanishing term in the Taylor series of f about a zero (i.e. the number N above) is the order of the zero. Let S be a subset of C. Recall that a limit point of S is the limit of a sequence of distinct points of S. A set is closed if and only if it contains all of its limit points. Any limit point of limit points is itself a limit point; in particular, the set of limit points of any set is itself closed. Proposition 7.2 (The principle of isolated zeroes). Let Ω ⊆ C be open and connected. If f : Ω → C is holomorphic and not identically zero, then the set of zeroes of f has no limit points in Ω. Proof. Let A ⊆ Ω be the set of limit points of zeroes of f . Then A is closed. But by lemma 7.1, if a ∈ A then some disc D(a; ρ) is also contained in A. Thus A is open. Connectedness now gives that A is either empty or equal to Ω. It follows that if f and g are holomorphic functions on Ω, and f (z) = g(z) for all z belonging to some set having a limit point in Ω, then f = g everywhere; this is the uniqueness of analytic continuation. (To see it, apply

25

the previous result to f − g.) For example, supposing that we know that a trigonometric identity like sin(z + w) = sin(z) cos(w) + cos(z) sin(w) holds for all real z, w, we can deduce that it holds for complex z, w too. Lemma 7.3. Let f be holomorphic in a disc D(a; r) and not constant there. Then there are points z ∈ D arbitrarily close to a at which |f (z)| > |f (a)|. Proof. Take a = 0 wlog. Using the Taylor series write f (z) = c0 +

∞ X

cn z n = c0 + z N (cN + zg(z))

n=N

with g continuous, cN 6= 0. By replacing f (z) by eiφ f (eiψ z) for suitable constants φ, ψ we may assume wlog that c0 and cN are real and positive. But now for x > 0 real and small enough that |xg(x)| < 21 |cN |, Re f (x) > c0 + 21 cN xN > c0 and so |f (x)| > c0 = |f (0)| also. Theorem 7.4 (Maximum principle). Let f be a nonconstant holomorphic function on a connected open set Ω. Then |f | does not attain a local maximum (even non-strict) anywhere on Ω. Proof. If it did, then according to Lemma 7.3 it would be constant on a disk around the maximum point, hence everywhere in Ω by the uniqueness of analytic continuation. The minimum principle states that |f | attains a local minimum only at the zeroes of f . To prove it, apply the maximum principle to 1/f . Corollary 7.5 (Fundamental theorem of algebra). Every nonconstant polynomial p(z) has a root in C. Proof. A nonconstant polynomial is holomorphic and has the property that |p(z)| → ∞ as |z| → ∞. Thus |p(z)| must attain a (global!) minimum somewhere. By the minimum principle, the point where it does so is a root. There are several ways to use complex analysis in proving the fundamental theorem of algebra. Another one makes use of the Cauchy estimates for a holomorphic function and its derivatives. 26

Proposition 7.6 (Cauchy Estimates). Suppose that f is holomorphic in an open set containing the closed disk D(a; r). Then |f (n) (a)| 6 n!r−n sup{|f (z)| : |z − a| = r}. Proof. This follows immediately from the integral formula I f (w)dw n! (n) f (a) = 2πi ∂D(a;r) (w − a)n+1 of Corollary 6.7. Corollary 7.7 (Liouville’s Theorem). Let f : C → C be an entire function (that is, holomorphic on the whole complex plane). If f is bounded, then it is constant. Proof. Say f is bounded by M . Cauchy estimates give |f 0 (a)| 6 M r−1 for any r. Letting r → ∞ we find that f 0 = 0, so f is constant. Now we can prove the fundamental theorem of algebra again. If p is a nonconstant polynomial, then 1/p is holomorphic everywhere except at the roots of p, and it tends to zero at infinity. Thus, if p had no roots, 1/p would be a constant, so p would be a constant too; a contradiction. Exercise 7.1. Show that if f is an entire function and |f (z)| 6 C(|z|n + 1) for some constant C, then f is in fact a polynomial (of degree at most n).

27

Lecture 8 The Global Cauchy Theorem In this lecture we shall prove the final and most general form of Cauchy’s theorem, removing the restriction to convex regions Ω. First, we introduce some jargon which belongs to the realm of homology theory. Let Ω be an open subset of C. Definition 8.1. An 0-chain in Ω is a finite formal linear combination of points of Ω, with integer coefficients. (Formally speaking, this is an element of the free abelian group generated by the points of Ω.) Similarly, a 1-chain is a finite formal linear combination of piecewise smooth paths in Ω. P If Γ = nj=1 kj γj is a 1-chain, we define Z f (z)dz = Γ

n X

Z f (z)dz

kj γj

j=1

S whenever f is defined and continuous on Γ∗ = j γj∗ . If Γ and Γ0 are two 1-chains such that Z Z f (z)dz = f (z)dz Γ0

Γ Γ∗ ∪(Γ0 )∗ ,

for every continuous f on we shall say that Γ and Γ0 are equivalent. Definition 8.2. The boundary map from 1-chains to 0-chains is the linear map which takes each path γ : [a, b] → Ω to the formal difference of its ending and starting points, [γ(b)] − [γ(a)]. A 1-chain is a 1-cycle if its boundary is the zero 0-chain. For example, a single closed path defines a 1-cycle. More generally, one has the following simple combinatorial fact. Exercise 8.1. Show that every 1-cycle is equivalent to a formal linear combination of closed paths. This implies that if Γ is a 1-cycle and p ∈ / Γ∗ , the number Z 1 dz n(Γ, p) = 2πi Γ z − p is an integer; we call it the winding number of Γ about p. So far, these are standard definitions from singular homology theory. The following is not standard (it is equivalent with the usual definition because of Poincar´e-Lefschetz duality, but that does not matter here): Definition 8.3. Let Γ and Γ0 be 1-cycles in the open set Ω. We say that they are homologous (in Ω) if n(Γ, p) = n(Γ0 , p) for all p ∈ / Ω. Notice that this is an equivalence relation. 28

Example 8.4. In a simply-connected Ω every cycle is homologous to the zero cycle. In fact, this is our definition of ‘simply connected’. Exercise 8.2. Give an example of a cycle (in fact a closed path) in Ω = C \ {0, 1} that is homologous to zero but is not homotopic to zero (i.e., cannot be continuously deformed to a constant path). That is enough topological machinery: now for Cauchy’s theorem. Theorem 8.5. Let f be holomorphic on an open set Ω and let Γ be a 1-cycle in Ω that is homologous to zero. Then for every w ∈ Ω \ Γ∗ , we have the integral formula Z 1 f (z)dz n(Γ, w)f (w) = . 2πi Γ z − w Corollary 8.6. Under the above circumstances, if Γ0 and Γ00 are homologous 1-cycles, then Z Z f (z)dz =

f (z)dz.

Γ0

Γ00

In particular, if Γ is homologous to zero, then Z f (z)dz = 0. Γ

Proof. As in our earlier proof (Proposition 6.5), define ( f (z)−f (w) (z 6= w) z−w g(w, z) = 0 f (z) (z = w) and then put Z g(z, w)dz.

h(w) = Γ

The formula we want is equivalent to proving that h(w) = 0 for all w ∈ Ω\Γ∗ . We shall do that by proving three things: (a) h(w) is a holomorphic function of w (defined on all of Ω), (b) h extends to a holomorphic function defined on all of C (an entire function), and (c) the extended function h tends to zero at infinity. It will then follow from Liouville’s theorem that h ≡ 0. To prove (a), note that we already know that g(z, w) is holomorphic as a function of w for each fixed z (even at z = w the apparent singularity 29

is removable). Now let T be a small triangle. By Fubini’s and Cauchy’s theorems, Z Z Z h(w)dw = g(z, w)dwdz = 0. ∂T

Γ

∂T

Thus h is holomorphic by Morera’s theorem. To prove (b), let Ω0 be the open set of all complex numbers w for which n(Γ, w) = 0. The formula Z f (z) h(w) = dz Γ z−w defines a holomorphic function on Ω0 which agrees with h on Ω ∩ Ω0 . Thus the two formulas together define h on the union Ω ∪ Ω0 . But our hypothesis that Γ is homologous to zero says that Ω ∪ Ω0 is the entire complex plane. To prove (c), notice that for |w| large, w belongs to Ω0 and the second formula for h applies and shows that h(w) → 0 as |w| → ∞. Proof of Corollary. If Γ is homologous to zero, choose w 6∈ Γ∗ and apply the previous result to (z − w)f (z). To get the second statement, notice that Γ0 and Γ00 are homologous if and only if Γ0 − Γ00 is homologous to zero. Using the global Cauchy theorem we can extend some of our previous results which we had derived from Cauchy’s theorem for convex sets. For example, consider Cauchy’s formula for derivatives, Corollary 6.7. Let f be holomorphic on Ω, let Γ be a cycle that is homologous to zero in Ω, and let a ∈ Ω \ Γ∗ . Then I n! f (w)dw n(Γ, a)f (n) (a) = . 2πi Γ (w − a)n+1 For the proof, just observe that Γ is homologous in Ω \ {a} to a small circle around a traversed n(Γ, a) times. The integrand is holomorphic on Ω \ {a}, so the global Cauchy theorem reduces the problem to the case of the circular path, which was handled by Corollary 6.7.

30

Lecture 9 Laurent series and the residue theorem Terminology: Let Ω be an open subset of C and let a be a point of Ω. If f is a holomorphic function defined on Ω \ {a}, we sometimes say that f has an isolated singularity at a. We are going to give a counterpart to the Taylor series expansion valid for functions with a singularity. Theorem 9.1. Let f be holomorphic in Ω0 = D(a; r) \ {a}. Then f has a Laurent expansion convergent in Ω0 , uniformly on compact subsets, of the form ∞ X f (z) = cn (z − a)n , n=−∞

where

1 cn = 2πi

I

f (w)dw , (w − a)n+1

the integral being taken around a circular contour in Ω0 with center a. Proof. Assume without loss of generality that a = 0. Use Cauchy’s integral formula (Theorem 8.5) to write Z f (w)dw 1 f (z) = , 2πi Γ w − z where the cycle Γ is the union of two circles, one of radius R larger than |z| and traversed counterclockwise, one of radius r smaller than |z| and traversed clockwise. (Notice that Γ is homologous to zero in Ω0 .) Now on the larger circle expand ∞ X zk 1 = w−z wk+1 k=0

with uniform convergence in w; on the smaller circle expand ∞

X wk 1 =− w−z z k+1 k=0

again with uniform convergence in w. Substitute these expansions into the Cauchy integral formula and interchange summation and integration (justified by uniform convergence). This gives I I ∞ ∞ X X 1 f (w)dw 1 k k f (z) = z + w f (w)dw z −k−1 ; 2πi 2πi wk+1 k=0

k=0

31

here the integrals in the first sum are around the circle of radius R, the ones in the second sum are around the circle of radius r, but both these circles are homologous to each other and indeed to any circle around a in Ω0 . Rearranging the displayed equation gives Laurent’s theorem as stated. Definition 9.2. If f has an isolated singularity at a, as above, the residue of f at a is the coefficient c−1 in its Laurent expansion. We write it Res(f, a). From the formula in Theorem 9.1 for the Laurent coefficients it follows that I 1 Res(f, a) = f (w)dw 2πi where the integral is taken around a small circular contour enclosing a. Thus the residue measures the ‘local failure of Cauchy’s theorem’ at a. Theorem 9.3 (The residue theorem). Let Ω be an open subset of C and let Γ be a cycle homologous to zero in Ω. Let f be a function holomorphic on Ω except for isolated singularities at points a1 , . . . , am ∈ Ω \ Γ∗ . Then 1 2πi

Z f (z)dz = Γ

m X

n(Γ, aj ) Res(f, aj ).

j=1

Proof. Let γj be a closed circular contour in Ω around aj of radius smaller than 21 min{|aj − ak | : j 6= k}. Then the cycle Γ is homologous in Ω \ {a1 , . . . , am } to m X Γ0 = n(Γ, aj )γj . j=1

Thus by the homology form of Cauchy’s theorem 1 2πi

m

Z

1 X f (z)dz = n(Γ, aj ) 2πi Γ

Z

j=1

f (z)dz γj

as required. There is some standard terminology about isolated singularities. If f has an isolated singularity at a then there is a Laurent series expansion f (z) =

∞ X

cn (z − a)n

n=−∞

about a. If there are only finitely many negative values of n for which cn 6= 0, one says that f has a pole at a; in that case the order of the pole is the 32

number N such that c−N 6= 0 but cn = 0 for all n < −N . Otherwise, if there are infinitely many negative n for which cn 6= 0, one says that f has an essential singularity at a. Example 9.4. The function cot z has a pole of order 1 (a.k.a. a simple pole) at zero. The function 1/(cos z − 1) has a pole of order 2 (double pole) at zero. The function sin(1/z) has an essential singularity at zero. A special case is a pole of order zero, or removable singularity. Here the Laurent series contains no negative powers of z, so it actually converges at z = a. In other words, f can be extended to a holomorphic function on Ω; the singularity can be ‘removed’. It is sometimes convenient to make the convention4 that a ‘zero of order −N ’ means a pole of order N , and a ‘pole of order −N ’ means a zero of order N . With this convention, the order of the zero of f at a is the unique N (positive or negative) such that we can write f (z) = (z − a)N g(z) near a, with g holomorphic near a and g(a) 6= 0. From this, it follows that if f1 and f2 have zeroes of orders N1 and N2 at a, then f1 f2 has a zero of order N1 + N2 and f1 /f2 has a zero of order N1 − N2 . Therefore the quotient of two holomorphic functions has (at worst) isolated poles as singularities. Definition 9.5. A function whose singularities are (at worst) isolated poles is called meromorphic. Exercise 9.1. Suppose that f1 and f2 are as above, possibly having zeroes or poles at a. What, if anything, can you say about the order of the zero or pole of f1 + f2 at a, in terms of the orders of f1 and f2 ? One can classify isolated singularities in terms of the limiting behavior of the function near the singularity. Proposition 9.6. Let a be an isolated singularity of the holomorphic function f . Then (i) the point a is a removable singularity of f iff |f (z)| is bounded in some neighborhood of a; (ii) the point a is a pole of f iff |f (z)| tends to ∞ as z tends to a; (iii) the point a is an essential singularity iff f (z) approaches every value as z approaches a (this means that, given any b ∈ C, we can find a sequence zn → a such that f (zn ) → b). Proof. First look at (i). Obviously, f is bounded near a removable singularity since it extends continuously across the singularity. Conversely, if f 4 See the discussion after Lemma 7.1 to understand what we mean by the “order” of a zero.

33

is bounded by M at a singularity, look at the Laurent expansion. Estimate the coefficients by integrating around a path of radius to get |cn | 6 M −n . Letting → 0 we find that cn = 0 for n = −1, −2, . . ., so the Laurent series is a Taylor series and f has a holomorphic extension. Now for (ii), if f has a pole at a of order N then f (z) = (z − a)−N g(z), where g is holomorphic and g(a) 6= 0. It certainly follows that |f | → ∞ at a. The converse statement will follow from (iii) — if f had an essential singularity than f (z) would approach every value, and in particular could not tend to ∞ — once we have proved that. As for (iii), if f (z) approaches every value as z → a, then a can’t be removable or a pole (by what we have already proved) so it must be an essential singularity. The converse statement (sometimes called the CasoratiWeierstrass theorem) is the meat of the result. Suppose that f does not approach every value and let b be a value which is not approached, so that there exist , δ > 0 with |f (z) − b| > δ whenever |z − a| < . Then g(z) = 1/(f (z) − b) is bounded near a, so it has a removable singularity. Extend g to a function holomorphic in a disk around a by removing the singularity. Writing f (z) = b + 1/g(z), we now see that the singularity of f at a is at worst a pole, of order equal to the order of the zero of g at a.

34

Lecture 10 Counting zeroes and poles In this lecture we are going to use contour integration to count the zeroes and poles of a holomorphic (or meromorphic) function. Lemma 10.1. Let f be holomorphic or meromorphic on an open set Ω. Then the quotient f 0 /f is meromorphic; its only singularities are poles, and these occur at the zeroes and the poles of f . Moreover, the poles of f 0 /f are all simple, and the residue of f 0 /f at a point is equal to the order of the zero (or minus the order of the pole) of f at that point. Proof. If f is holomorphic and nonzero near a, then f 0 (z)/f (z) is holomorphic near a. Thus the only possible singularities of f 0 /f are the zeroes and poles of f . Suppose that such a point (zero or pole) occurs at z = a. Then from the Taylor or Laurent series, f (z) = (z − a)m g(z) near z = a, where g(z) is holomorphic and non-vanishing at a and m is the order of the zero (or minus the order of the pole). Thus f 0 (z) m(z − a)m−1 g(z) + (z − a)m g 0 (z) m g 0 (z) = = + . f (z) (z − a)m g(z) z−a g(z) The second term is holomorphic near 0, so the pole is simple and has residue m. From the residue theorem we deduce Proposition 10.2. Let f be meromorphic in Ω, and let Γ be a cycle homologous to zero in Ω and not passing through any zeroes or poles of f . Then Z 0 X 1 f (z) dz = mj n(Γ, aj ), 2πi Γ f (z) j

where a1 , a2 , . . . are the zeroes and poles of f in Ω and mj is the order of the zero (or minus the order of the pole) of f at aj . Usually we apply this result where Γ is a simple closed curve dividing C into two regions: an interior region consisting of points a such that n(Γ, a) = 1, and an exterior region consisting of points such that n(Γ, a) = 0 In this case the right hand side of Proposition 10.2 is the total number of zeroes minus the total number of poles of f inside Γ (both totals counted according to multiplicity). 35

Remark 10.3. The left hand side of Proposition 10.2 is the winding number n(f∗ Γ, 0), where f∗ Γ is the cycle got by composing each of the individual paths that make up Γ with the function f . More generally, if f is holomorphic then n(f∗ Γ, b) gives the number of times that f takes the value b (rather than 0) inside Γ. Proposition 10.4 (Rouch´es Theorem). Let f and g be holomorphic in an open set Ω, and let Γ be a cycle homologous to zero in Ω. Suppose that neither f nor g has a zero on Γ∗ . If |f (z) − g(z)| < |f (z)| for all z ∈ Γ∗ , then f and g have the same number of zeroes inside Γ. The “number of zeroes inside γ” is, by definition, the number computed in Proposition 10.2. Proof. Consider the family of holomorphic functions ft (z) = tg(z) + (1 − t)f (z) for t ∈ [0, 1]. We have for z ∈ Γ∗ , |ft (z)| = |f (z) + t(g(z) − f (z))| > |f (z)| − t|f (z) − g(z)| > 0 so the functions ft have no zeroes on Γ∗ . Thus the integral Z 0 1 ft (z) dz 2πi Γ ft (z) is a continuous integer-valued function of t, hence constant. For t = 0 we get the number of zeroes of f , for t = 1 the number of zeroes of g. We can use this to give yet another proof of the Fundamental Theorem of Algebra. Corollary 10.5. A polynomial equation of degree n must have n roots. Proof. In Rouch´es theorem, let g(z) = a0 + · · · + an z n be our polynomial equation and let f (z) = an z n . Let Γ be a circle, center the origin. If the radius of Γ is large then |f − g| < |f | on Γ. Therefore, f and g have the same number of roots inside Γ. But f obviously has n roots counted with multiplicity (namely, an n-tuple root at the origin); so g does too. Proposition 10.6 (Open mapping principle). A non-constant holomorphic map on an open set Ω in C is an open map; it takes open subsets (of Ω) to open subsets of C.

36

Proof. It is enough to show that, given any a ∈ Ω, there is a neighborhood W of a in Ω such that f (W ) is a neighborhood of b = f (a). Since f is non-constant there is a disk D(a; ) on which f (z) does not take the value b except at z = a. Let Γ be a circular path around a of radius /2. Then n(f∗ Γ; b) > 0 and so (by continuity) n(f∗ Γ; b0 ) > 0 for b0 close enough to b, say |b0 − b| < δ. Thus f takes the value b0 within Γ. But this is to say that if W = D(a; /2), then f (W ) contains D(b; δ), and so is a neighborhood of b, as required. We say that a map f is locally injective at a point a if there is a neighborhood of a on which f is an injective (that is, one-to-one) map. Proposition 10.7. A holomorphic map f is locally injective at a point a if and only if f 0 (a) 6= 0. Proof. Let f (a) = b. First, suppose that f 0 (a) 6= 0. Argue as in Proposition 10.6. Now n(f∗ Γ, b) = 1 (since f (z) − b has a simple zero at a) and so n(f∗ Γ, b0 ) = 1 also for all b0 belonging to some disk D(b; δ). Let U = f −1 (D(b; δ)) ∩ D(a; /2); then U is open and contains a, and f takes each value in D(b; δ) precisely once on U — in other words, f is injective on U. Now, suppose that f 0 (a) = 0. The same reasoning shows that n(f∗ Γ, b) > 1 and so, for any > 0, there is δ > 0 such that the restriction of f to D(a; ) takes each value b0 ∈ D(b; δ) at least twice, counting multiplicity. Since the zero of f 0 at a is isolated, f (z) − b0 must have simple zeroes in D(a; ) for sufficiently small and b0 6= b. Thus there must be at least two such zeroes, contradicting local injectivity. This allows us to give the final form of the inverse function theorem. Proposition 10.8. A bijective holomorphic map between open subsets of C has an inverse which is also holomorphic. Proof. We now know that a holomorphic map has a continuous derivative, which does not vanish if the map is injective. This allows us to remove the extra conditions in Lemma 4.1.

37

Lecture 11 Calculations with the residue theorem It is necessary to be able to evaluate residues efficiently. Often, the simplest way to proceed is to manipulate Laurent series. Example 11.1. Find the residue of the function 1 − z2 1 + z2 at z = i. Answer. We must find a Laurent expansion in powers of (z − i). We can write 1 − z2 2 + O(z − i) 1 = = + O(1). 2 1+z (z − i)(2i + (z − i)) i(z − i) So the residue is 1/i = −i. Example 11.2. Find the residue of the function ez ez − 1 − z at z = 0. Answer. The function ez − z − 1 has a Taylor series beginning with z 2 /2, so it has a double zero at z = 0. The function ez is nonzero at z = 0. Thus the quotient has a double pole. To find the residue, we need to work out the beginning of the Laurent series: ez 1 + z + O(z 2 ) = = ez − z − 1 (z 2 /2)(1 + z/3 + O(z 2 )) 2 z 2 4 1 − 1 + z + O(z 2 ) = 2 + + O(1). 2 z 3 z 3z Thus the residue is 4/3. There are some general results that can be used to calculate residues. Lemma 11.3. Let f be holomorphic near z = a; then the residue of f (z)/(z− a)n at a is f (n−1) (a)/(n − 1)!. Proof. From the Taylor series for f . Lemma 11.4. Let f and g be holomorphic near z = a, and suppose that g has a simple zero at a. Then the residue of f (z)/g(z) at a is f (a)/g 0 (a).

38

Proof. Write g(z) = (z − a)h(z), where h is holomorphic and nonvanishing at a, and note that g 0 (a) = h(a) (by elementary calculus). Then f (z)/g(z) = (f (z)/h(z))/(z − a), where f (z)/h(z) is holomorphic near a; applying the previous lemma gives the result. One can use this lemma to shorten Example 11.1, for instance. There are analogous formulas for multiple poles, but they’re not worth trying to remember. Exercise 11.1. Derive the formula for the residue at a double pole: if g has a double zero at a, then the residue of f /g at a is 0 f (a) f (a)g 000 (a) 2 − . g 00 (a) 3g 00 (a)2 Verify that this is consistent with our solution to Example 11.2. Combined with the residue theorem, these techniques give a computational method for evaluating certain definite integrals, which is known as contour integration. The idea of the method is to consider a suitable meromorphic function f (z), and evaluate its integral over the cycles in some homology class by using the residue theorem. Then, usually, one lets the cycles approach some limit in which it can be seen that the integral around the cycle tends to some definite integral that we want to evaluate. It is easier to give examples than to try to explain this in general. Example 11.5. Evaluate Z ∞ 1 I= dx. 1 + x2 −∞ Of course we know from elementary calculus that the answer is π, but let’s pretend for a moment that we don’t know this. Let f (z) = 1/(1 + z 2 ), which is a meromorphic function with simple poles at ±i. Let ΓR be the closed path which proceeds from −R to R along the real axis, and then returns via a semicircle in the upper half plane. From the residue theorem we have Z f (z)dz = 2πi Res(f, i) = π ΓR

so long as R > 1. Now consider what happens as R approaches ∞. The integral along the straight portion of ΓR approaches I. The integral around the semicircular portion is bounded by πR ·

R2 39

1 −1

by the R estimate 5.2, and this tends to zero as R → ∞. Thus the integral ΓR f (z)dz approaches I; but on the other hand, it is the constant π. Therefore, I = π. Remark 11.6. We could equally well have used a semicircular “return contour” around the lower half plane. We would have obtained the same result. Generalizing this remark, one can show that if f (z) = p(z)/q(z) is a rational function, the quotient of two polynomials, and if deg(q) > deg(p) + 2, then the sum of all the residues of f is zero. Example 11.7. Evaluate Z ∞ cos x I= dx. 2 −∞ 1 + x This is very similar to the previous example; we use the same semicircular contour. Put eiz . f (z) = 1 + z2 (It is important that we use the function eiz rather than the tempting cos z! The reason is that eiz is bounded in the upper half plane.) Now from the residue theorem Z π f (z)dz = 2πi Res(f, i) = . e ΓR The real part of the integral along the straight portion of Γ approaches I as R → ∞. Around the semicircular portion, the integral tends to zero as before (since |eiz | 6 1 there). Thus, I = π/e. Remark 11.8. In this example we could not have integrated using the lower half plane instead; the function eiz is not bounded there. Example 11.9. Show that Z ∞ at e dt π = , t sin πa −∞ 1 + e for 0 < a < 1. Let I denote the integral we want to evaluate. Let f (z) = eaz /(1 + ez ), and notice that • f (z + 2πi) = e2πia f (z). • f has a simple pole at z = πi, with residue −eiπa . • If x = Re(z) > 1, then |f (z)| 6 eax /(ex − 1), which tends to 0 as x → ∞. 40

• If x = Re(z) < −1 then |f (z)| 6 2eax , which tends to 0 as x → −∞. R Consider the integral ΓR f (z)dz, where ΓR is now a rectangular contour whose vertices are at −R, R, R + 2πi, and −R + 2πi. By the residue theorem, this integral equals −2πieiπa . On the other hand, as R → ∞ the integral along the lower long side of the rectangle tends to I, and the integral along the upper long side tends to −e2πia I (by point 1 above). The integrals along the short sides tend to zero (by points 3 and 4). We conclude (1 − e2πia )I = −2πieiπa . Therefore I=

2πi π = . eiπa − e−iπa sin πa

41

Lecture 12 The Gamma Function We will develop the properties of the Γ function. Earlier we defined this function by Z ∞ tz−1 e−t dt Γ(z) = 0

when Re(z) > 0. The iteration formula Γ(z + 1) = zΓ(z) allows us to find an analytic continuation of Γ(z) to C \ {0, −1, −2, . . .}, and this continuation has simple poles at the non-positive integers. In homework we proved the limit formula n!nz−1 , n→∞ z(z + 1) · · · (z + n − 1)

Γ(z) = lim

again valid for Re(z) > 0. We’ll begin our investigation by giving a different infinite product formula for Γ. Definition 12.1. Euler’s constant γ is the limit γ = lim

n X

n→∞

k −1 − log n .

k=1

1 n

1 Since 6 log(n)−log(n−1) 6 n−1 one easily checks that the expression under the limit is decreasing in n and bounded below, so convergent. The numerical value of Euler’s constant is approximately 0 · 577215665. Proposition 12.2. The Γ function Γ(z) satisfies ∞ h Y 1 z −z/n i . = zeγz 1+ e Γ(z) n n=1

The infinite product converges to an entire function on C. In particular, Γ(z) never vanishes. The convergence of an infinite product is defined in the obvious Q way — as convergence of the partial products. It is easy to check that (1 + an ) P converges if |an | converges. This can be used to show that the product above converges. The convergence is locally uniform hence to a holomorphic function. Proof. Because of the uniqueness of analytic continuation it suffices to check the result for z real and positive. From homework we have the limit formula z(z + 1) · · · (z + n) 1 = lim . Γ(z) n→∞ n!nz 42

Rewrite the expression under the limit as −z log n

ze

n Y k=1

z 1+ = zez( k

P

n 1 k=1 k −log n)

n h Y k=1

1+

z −z/k i e . k

Letting n → ∞ we obtain the result. Remark 12.3. Taking the log of the infinite product gives an infinite series for log Γ(x), x real. Differentiating we find that ∞

1 d2 log Γ(x) X = > 0. dx2 (x + k)2 k=0

Thus Γ is log-convex. It can be proved that it is the only log-convex function which has Γ(1) = 1 and xΓ(x) = Γ(x + 1) (Bohr-Mollerup theorem). Now the reflection formula. Theorem 12.4. We have Γ(z)Γ(1 − z) = π/ sin(πz). Proof. By analytic continuation it is enough to check this for z = a, 0 < a < 1. Write Z ∞Z ∞ Γ(1 − a)Γ(a) = s−a ta−1 e−(s+t) dtds. s=0

t=0

In the t-integral substitute t = us to obtain Z ∞Z ∞ ua−1 e−s(1+u) duds. s=0

u=0

Reverse the order of integration (Fubini’s theorem!) and perform the inner s-integral to obtain Z ∞ a−1 u du. 1 u=0 + u The substitution u = et gives Z ∞ −∞

eat dt π = , 1 + et sin πa

as we worked out in Example 11.9.

43

This has many interesting consequences. For instance, writing it as sin πz = −π/(zΓ(z)Γ(−z)) and combining with the infinite product for 1/Γ(z) we get ∞ Y z2 sin πz = πz 1− 2 . n n=1

By logarithmic differentiation, ∞ n X 1 X 2z 1 π cot πz = + = lim . 2 2 n→∞ z z −n z−k n=1

k=−n

This last result can be used to find the value of Riemann’s zeta function at even Recall that the Bernoulli numbers are defined by z/(ez −1) = P integers. 1 1 n Bn z /n!; we have B2 = 61 , B4 = − 30 , B6 = 42 and so on (the odd ones vanish after B1 ). Then we have Proposition 12.5. For each even integer 2m, ∞ X 1 (−1)m+1 22m−1 B2m π 2m ζ(2m) = = . n2m (2m)! n=1

Proof. Write ∞ X eiz + e−iz 2iz 22m z 2m m+1 z cot z = iz iz = iz + = 1 − (−1) B . 2m e − e−iz e2iz − 1 (2m)! m=1

On the other hand, using our infinite sum for the cotangent, z cot z = 1 + 2

∞ X n=1

z2 . z 2 − n2 π 2

Expand the summand in a binomial series and rearrange as a power series in z (justified by absolute convergence). The coefficient of z 2m is −

∞ X n=1

2 π 2m n2m

.

Comparing coefficients with those appearing in the previous expansion of z cot z (justified by uniqueness of Taylor coefficients) we get the result. The final fact about Γ that we will prove is Stirling’s formula, which gives the asymptotics of Γ(z) for z real and large (actually for Re z tending to +∞ with Im z bounded). 44

Proposition 12.6. As n → ∞ we have √ 1 n! ∼ 2πnn+ 2 e−n ; √ 1 equivalently, as x → ∞, Γ(x) ∼ 2πxx− 2 e−x . The ∼ notation means that the ratio of the two sides tends to 1. It is easy to check that the two forms are equivalent (at least as x tends to infinity through integers; an additional argument is needed to handle real but non-integral x). Proof. Note that Z

k+ 21

Z log x dx = log k +

k− 12

1 2

− 21

t log 1 + k

dt = log k + O(k −2 ),

integrating the Taylor series expansion for log(1 + u). Summing from k = 1 to n we find that Z n+ 1 n X 2 log x dx − log(n!) = O(k −2 ); 1 2

k=1

theR series converges so we may let n → ∞. Substituting the explicit value of log x dx = x log x − x, there is a constant c such that log(n!) − (n + 21 ) log n − n → c, c as √ n → ∞. This gives Stirling’s formula with a constant C = e in place of 2π. To determine the constant we consider Γ( 21 ). From the reflection formula √ we see immediately that Γ( 12 ) = π. On the other hand

Γ( 12 ) = lim

n→∞

√ n! n 22n (n!)2 √ . = lim 1 1 1 n→∞ (2n)! n 2 ( 2 + 1) · · · ( 2 + n)

(This is Wallis’ formula for π.) Substituting in Stirling’s formula we see √

so C =

√

22n C 2 n2n+1 e−2n C =√ , 1 √ n→∞ 2 C(2n)2n+ 2 e−2n n

π = lim

2π as asserted.

45

Lecture 13 More examples of contour integration (This lecture will be presented by Prof. Anna Mazzucato) Z ∞ sin x π dx = x 2 0 Evaluating this example by contour integration requires more delicate estimates of integrals over both large and small arcs than we have used so far. For the large arc we make use of Lemma 13.1 (Jordan’s Lemma). The integral Z π e−R sin θ dθ 0

tends to zero as R → ∞. Proof. The integrand is dominated by 1 and tends to 0 almost everywhere as R → ∞, so this follows from the dominated convergence theorem. It is traditional to give the following elementary proof instead: on the range θ ∈ [0, π/2] one has sin θ > 2θ/π (by convexity), and thus Z

π

−R sin θ

e 0

π/2

Z dθ = 2

−R sin θ

e 0

Z

π/2

dθ 6 2

e−2Rθ/π dθ 6

0

π 2R

This shows that the integral is in fact of order 1/R. On the small side we use Lemma 13.2 (Small arc lemma). Let f (z) be meromorphic near a, with a simple pole at a. Let γ denote the arc segment γ (t) = a + eit , θ0 6 t 6 θ1 . R Then as → 0, the integral γ f (z)dz tends to i(θ1 − θ0 ) Res(f, a). We emphasize that this lemma works only for simple poles. Also, note the consistency with the residue theorem when θ0 = 0 and θ1 = 2π. Proof. Because f has a simple pole we can write f (z) =

r + h(z) z−a

46

where r is the residue and h(z) is holomorphic (and hence bounded) near z. Thus Z Z Z r f (z)dz = dz + h(z)dz. γ γ z − a γ The second integral on the right vanishes as → 0, by our standard estimate 5.2, and the first can be explicitly computed and yields i(θ1 − θ0 )r. Z 0

Z 0

2π

∞

sin mx π dx = 2 (1 − e−ma ). 2 2 x(x + a ) 2a

dx 2π =√ , a > b > 0. a + b cos x a2 − b2 Z ∞ sin x 3 3π dx = . x 8 0 Z ∞ sin ax π aπ dx = tanh . sinh bx 2b 2b 0

47

Lecture 14 More about the Riemann Sphere Remember (Definition 2.8) that the Riemann sphere S is defined to be the complex projective line. That is to say, a point of the Riemann sphere is defined by a pair (z, w) of complex numbers, not both of which are zero, and two pairs (z, w) and (z 0 , w0 ) define the same point of the Riemann sphere if zw0 = z 0 w. The Riemann sphere is given the quotient topology, according to which a subset U of S is open in S if and only if the set of all pairs (z, w) that represent points of U is open in C2 . Lemma 14.1. The Riemann sphere is compact. Proof. By rescaling, every point on the Riemann sphere can be represented by a pair (z, w) with |z|2 + |w|2 = 1. The set of such pairs (z, w) is a compact space (the 3-sphere S 3 ; it’s a closed and bounded subset of C2 ). Thus there is a continuous surjection form a compact space to S; so S itself is compact. Lemma 14.2. The Riemann sphere is Hausdorff. Proof. Define a positive real-valued function on C2 \ {0} × C2 \ {0} by φ((z1 , w1 ), (z2 , w2 )) =

|z1 w2 − z2 w1 | . |z1 w2 − z2 w1 | + |z1 z¯2 + w1 w ¯2 |

The denominator never vanishes (check this!) and thus φ is well-defined and continuous. Moreover, φ respects the equivalence relation defining the Riemann sphere so it passes to a continuous function Φ : S × S → R+ , which has the property that Φ(p, q) = 0 if and only if p = q ∈ S. (Reason: z1 w2 − z2 w1 = 0 implies that (z1 , w1 ) and (z2 , w2 ) represent the same point of S.) The existence of such a function implies that S is Hausdorff, since if p, q are distinct points of S and Φ(p, q) = r > 0, then the sets U = {x : Φ(p, x) < r/2},

V = {x : Φ(p, x) > r/2}

are disjoint open sets containing p and q respectively.

48

We can define a homeomorphism from C to an open subset of S by sending z ∈ C to the point (z, 1) ∈ S. The image of this homeomorphism (sometimes called a chart) covers all of S except for the single point (1, 0), which we may denote by ∞. Thus, topologically, S is the one point compactification of C. We can define another chart for S by sending w ∈ C to the point (1, w) ∈ S. The two charts together cover all of S. Moreover, on their intersection, we have wz = 1 so w = 1/z. This mapping, expressing w in terms of z, is the transition function for the atlas consisting of our two charts. Because the transition function is a holomorphic bijection, the following definitions make sense. Definition 14.3. A continuous function f defined on S (or on an open subset of S) is holomorphic near a point p ∈ S if the composite function f ◦ φ is holomorphic near φ−1 (p), where φ : C → S is a chart containing p in its range. Definition 14.4. A continuous function g with values in S is holomorphic near a point q in its domain if the composite ψ −1 ◦ g is holomorphic near q, where ψ : C → S is a chart containing f (q) in its range. We can of course combine both definitions and speak of holomorphic maps from S to S. Proposition 14.5. The only holomorphic maps from S to C are constant functions. Proof. Since S is compact, any continuous function f : S → C is bounded. Now, for any chart φ, f ◦ φ is a bounded holomorphic function, hence constant by Liouville’s theorem (7.7). Because the charts overlap, f is globally constant. Proposition 14.6. Let Ω be an open subset of C. The holomorphic maps from Ω to S are precisely the meromorphic functions on Ω. Proof. By definition, to give a holomorphic function from Ω to S is the same thing as to give a cover of Ω by two open sets U and V and a function g : Ω → C ∪ {∞} such that g is holomorphic on U and 1/g is holomorphic on V . But this implies that the points where g takes the value ∞ are isolated (by the principle of isolated zeroes for 1/g) and that at each such point g has a pole singularity (because g is the reciprocal of a holomorphic function). Thus g is meromorphic. Let f (z) = p(z)/q(z) be a rational function (the quotient of two polynomials). Then f is meromorphic on C. Moreover, f (1/z) is rational (and 49

hence meromorphic on C) as well. These facts together tell us that f extends to a holomorphic map from S to S. Theorem 14.7. Every holomorphic map from S to S is a rational function. This is an example of the so-called “GAGA principle”: analytic objects defined on compact Riemann surfaces (such as the sphere) turn out in fact to be algebraic objects.5 Proof. Let f be such a map. Notice that f −1 ({∞}) is a closed, discrete subset of S, hence finite. Thus the restriction of f to C is a meromorphic function with finitely many poles. Let the poles in C be a1 , . . . , an with orders m1 , . . . , mn . Then g(z) = (z − a1 )m1 · · · (z − an )mn f (z) has only removable singularities in C; on removing them, we obtain an entire function. Moreover, f (1/w) has at worst a pole at w = 0, which implies that there are constants C and N such that |f (z)| 6 C|z|N for large |z|. Then |g(z)| 6 C 0 |z|N

0

for large |z|, where N 0 = N + m1 + · · · + mn . By Exercise 7.1, g is a polynomial, so f is rational. The Riemann sphere is an example of a Riemann surface. Definition 14.8. Let X be a topological space. An analytic atlas for X is an open cover U = {Uα } of X, together with homeomorphisms φα from Uα to open subsets Vα of C. The homeomorphisms φα (called charts) are required to satisfy the transition condition: whenever Uα ∩ Uβ 6= ∅, the map φα ◦ φ−1 β : φβ (Uα ∩ Uβ ) → φα (Uα ∩ Uβ ) (between two open subsets of C) is a holomorphic bijection. Definition 14.9. A Riemann surface is a Hausdorff topological space equipped with an analytic atlas. Definitions 14.3 and 14.4 generalize in an obvious manner to Riemann surfaces. Thus, holomorphic functions between Riemann surfaces can be defined; in fact, Riemann surfaces give the most general context in which it is possible to speak of holomorphic functions (of a single complex variable). Exercise 14.1. Let S be a compact connected Riemann surface. Show that the only holomorphic functions S → C are the constant functions. 5

GAGA stands for the French: “g´eometrie analytique, g´eometrie alg´ebrique”.

50

Lecture 15 Automorphisms Let S and S 0 be Riemann surfaces (the only examples that we shall consider in this lecture are open subsets of C and the Riemann sphere S). Definition 15.1. A biholomorphic equivalence (other authors often say conformal equivalence) from S to S 0 is a bijective holomorphic map S → S 0 . A holomorphic automorphism of S is a biholomorphic equivalence S → S. The inverse of a biholomorphic equivalence is also a biholomorphic equivalence, by Proposition 10.8. Thus, in particular, the holomorphic automorphisms of S form a group, denoted Aut(S). This is the symmetry group for the ‘holomorphic geometry’ of the Riemann surface. We are going to calculate some of these symmetry groups. Proposition 15.2. The automorphism group of the Riemann sphere is exactly the group of all M¨ obius transformations (denoted P GL(2, C)). Proof. We proved last time that any holomorphic map S → S is a rational function p(z)/q(z), where p and q are polynomials. Thus all we have to do is to prove that a rational function is bijective as a map S → S only if deg(p), deg(q) 6 1. But this is easy: the equation p(z)/q(z) = w translates to the polynomial equation p(z) − wq(z) = 0 and this will in general have as many roots (for z) as its degree max{deg(p), deg(q)}. A M¨obius transformation that fixes the point ∞ (that is, maps ∞ to ∞) is simply an affine transformation of the form z 7→ az + b, a 6= 0. The affine transformations form a subgroup of the group of all M¨obius transformations. Proposition 15.3. The automorphism group of the complex plane is exactly the group of all affine transformations. To put it another way, every automorphism of C extends to an automorphism of S. Proof. Let f : C → C be an automorphism. In particular, it is a continuous map with continuous inverse (a homeomorphism). Thus, if z → ∞, f (z) → ∞ also. Consider f (1/w), which has a singularity at w = 0. Since |f (1/w)| → ∞ as w → 0, this singularity is a pole by Proposition 9.6. Thus f extends to a meromorphic function on S, i.e. a holomorphic function S → S. The extended function is bijective, hence an automorphism of S; and now the classification of automorphisms of S completes the proof. 51

Remark 15.4. Essentially the same argument shows that if S is a compact Riemann surface and o ∈ S, then every automorphism of S \ {p} extends to an automorphism of S. The third example that we shall consider is the open unit disk, denoted U. First, notice that this example is genuinely different from the others. Lemma 15.5. U is not biholomorphically equivalent either to C or to S. Proof. Since S is compact and U is not, they can’t even be homeomorphic. U and C are homeomorphic, but they can’t be biholomorphically equivalent because a holomorphic bijection C → U would in particular be a bounded entire function, and hence constant by Liouville’s theorem. The rotations z 7→ eiφ z, φ ∈ R, are obvious examples of automorphisms of U. Further examples are supplied by Lemma 15.6. Let a ∈ U. Then the M¨ obius transformation z 7→

z−a 1−a ¯z

is an automorphism of U and maps a to the origin. Proof. Let Ta (z) = (z − a)/(1 − a ¯z). It is easy to check that the inverse of the transformation Ta is T−a . Thus we need to prove that Ta maps U to itself. To see this, note that |1 − a ¯z|2 − |z − a|2 = 1 + |a|2 |z|2 − |a|2 − |z|2 = (1 − |z|2 )(1 − |a|2 ) > 0. So |1 − a ¯z| > |z − a|, which gives |Ta (z)| < 1. Corollary 15.7. The group Aut(U) is transitive. By definition, Aut(S) is transitive if for any two points p, q ∈ S there is an automorphism sending p to q. Exercise 15.1. Show that the automorphism groups of S and C are transitive. Theorem 15.8 (Schwarz’ Lemma). Let f : U → U be a holomorphic map (not necessarily bijective) and suppose that f (0) = 0. Then |f (z)| 6 |z| for all z, and |f 0 (0)| 6 1. Moreover, equality occurs6 only if f (z) = eiφ (z) for some real φ. 6

I mean, |f (z)| = |z| for some nonzero z, or |f 0 (z)| = 1.

52

Proof. Let g(z) = f (z)/z; since f has a zero at the origin, g extends to a holomorphic function on U. For each r < 1, g is holomorphic on D(0; r) and |g(z)| 6 1/r

for z ∈ ∂D(0; r).

By the maximum principle (Theorem 7.4), |g(z)| 6 1/r for all z ∈ D(0; r). Let r % 1 to find |g(z)| 6 1 which gives the inequality. If equality occurs, then |g| attains its maximum value 1 at some interior point of U, so g is a constant (by the maximum principle again). Proposition 15.9. Every automorphism of U is of the form z 7→ eiφ Ta (z) where Ta is defined as above. Proof. Let f ∈ Aut(U) and suppose that f −1 (0) = a. Then g = f ◦ (Ta )−1 is an automorphism mapping 0 to 0. By Schwarz’ Lemma the only such automorphisms are of the form w 7→ eiφ w (proof: |g(z)| 6 |z| by Schwarz’ lemma, and |z| 6 |g(z)| by Schwarz’ lemma applied to g −1 . Thus we have equality in Schwarz’ lemma and the result follows). Thus f ◦ (Ta )−1 (w) = eiφ w. Put z = Ta (w); we get f (z) = eiφ Ta (z) as required. Remark 15.10. The automorphisms of U are therefore those M¨obius transformations of the form λ + µz z 7→ ¯ . µ ¯ + λz These are exactly the M¨obius transformations that map U to itself. The group of such M¨obius transformations is denoted P SU (1, 1).

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Lecture 16 Hyperbolic Geometry Let U be the unit disk, as before. It is a consequence of our earlier discussion that if a, b ∈ U the quantity a−b 1 − a ¯b is preserved by every automorphism of U. For it is equal to |φ(b)| where φ is any automorphism that maps a to 0. We can think of this quantity as a sort of ‘distance’ from a to b. Definition 16.1. If a, b ∈ D then the hyperbolic distance between them is −1 a − b d(a, b) = 2 tanh . 1−a ¯b By construction, the hyperbolic distance is invariant, that is, d(f (a), f (b)) = d(a, b) for any automorphism f . Lemma 16.2. If a, 0, and b are collinear (meaning that a is a positive real multiple of −b) then d(a, 0) + d(0, b) = d(a, b). Proof. Use the formula tanh(x + y) =

tanh(x)+tanh(y) 1+tanh(x) tanh(y) .

We are going to show that the hyperbolic distance is a metric, i.e. satisfies the triangle inequality. To see this we need to study some hyperbolic trigonometry. Definition 16.3. A hyperbolic line in U is a circline that meets the unit circle at right angles. Thus the diameters of U are hyperbolic lines; indeed, they are the only hyperbolic lines through the origin. Since the automorphisms of U are conformal (angle-preserving) transformations and map the unit circle to itself, they map hyperbolic lines to hyperbolic lines. Lemma 16.4. There is exactly one hyperbolic line through two given points of U. Proof. Because of transitivity we may assume that one of the points is 0, and then the result is obvious.

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We define the (hyperbolic) angle between two hyperbolic lines to be the ordinary (Euclidean) angle between them. Note that these angles are preserved by automorphisms (since the automorphisms are conformal transformations). Now consider a hyperbolic triangle ABC consisting of three points A, B, C ∈ D and the hyperbolic line segments between them. Let the sides have hyperbolic length AB, BC, CA and the angles at the vertices be γ, α, β. Proposition 16.5 (Hyperbolic cosine law). In the above situation cosh AB = cosh AC cosh BC − sinh AC sinh BC cos γ. Proof. Because of transitivity there is no loss of generality in assuming that C is the origin. Let A, B be represented by the complex numbers a, b respectively. Remember some more trigonometric identities: if t = tanh x/2 then cosh(x) =

1 + t2 , 1 − t2

sinh(x) =

2t . 1 − t2

Thus a−b 2 1 + 1− ¯ba cosh(AB) = a−b 2 1 − 1− ¯ba =

|1 − ¯ba|2 + |a − b|2 |1 − ¯ba|2 − |a − b|2

=

(1 + |a|2 )(1 + |b|2 ) − 2(¯ ab + a¯b) . 2 2 (1 − |a| )(1 − |b| )

But a ¯b+a¯b = 2|a||b| cos γ (see exercise 1.2). Substituting into the expression above we obtain cosh AC cosh BC − sinh AC sinh BC cos γ as required. Corollary 16.6. If A, B, C are three points in U we have d(A, C) 6 d(A, B) + d(B, C) with strict inequality unless A, B, C lie on the same hyperbolic line (with B between A and C). 55

Proof. This follows from the cosine law and the formula cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y). Exercise 16.1. Prove the hyperbolic sine law: in a hyperbolic triangle as discussed above, one has sinh BC sinh CA sinh AB = = . sin α sin β sin γ The Schwarz lemma has a beautiful invariant form relative to the hyperbolic metric. Proposition 16.7 (Ahlfors). Any holomorphic map f : U → U decreases all hyperbolic distances: the hyperbolic distance from f (a) to f (b) is no greater than the hyperbolic distance from a to b. Proof. If a = f (a) = 0 this is the ordinary Schwarz lemma 15.8. The general case can be reduced to this one by replacing f by g = Tf (a) ◦ f ◦ (Ta )−1 which is a holomorphic map sending 0 to 0. Thus |g(Ta (b))| 6 |Ta (b)|. But |Ta (b) is the hyperbolic distance from a to b, and (by construction) |g(Ta (b))| = |Tf (a) (f (b))| is the hyperbolic distance from f (a) to f (b). Corollary 16.8. Let F : U → U be any holomorphic map. Then |F 0 (0)| 6 1, with strict inequality unless F is a rotation. This is a corollary because, by the invariant form of the Schwarz lemma, the hyperbolic distance between 0 and dz is greater than the hyperbolic distance between a = f (0) and f (0 + dz). But the hyperbolic distance between these two points is in turn at least as large as the euclidean distance. Rather than try to make this precise (which would lead us in the direction of Riemannian geometry) we write out an independent proof based on the same idea. Proof. Let F (0) = a. Then G = Ta ◦ F is a holomorphic map U → U and G(0) = 0. By the Schwarz lemma, |G0 (0)| 6 1. But G0 (0) = Ta0 (a)F 0 (0) by the chain rule. Calculation shows that 1 Ta0 (a) = 1 − |a|2 so |F 0 (0)| = (1 − |a|2 )G0 (0) 6 1, with equality only if a = 0 and F = G is a rotation. 56

Remark 16.9. One can show that hyperbolic geometry obeys all the axioms of Euclidean geometry except for the parallel axiom. In hyperbolic geometry there are many lines through a given point that do not meet a given line.

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Lecture 17 The Riemann mapping theorem There are many examples of open subsets of C that are conformally (a.k.a. biholomorphically) equivalent to U. For example, the M¨obius transformation z−i z 7→ z+i is a conformal equivalence from the upper half plane {z : Im(z) > 0} to U. Such a conformal equivalence Ω → U gives a 1 : 1 correspondence between holomorphic functions on Ω and holomorphic functions on U. Theorem 17.1 (Riemann Mapping Theorem). Any proper, simply-connected subset of C is conformally equivalent to U. We are going to prove this following an argument of Koebe. This proof depends only on the following property of simply-connected regions. Lemma 17.2. Let Ω be a simply connected open subset of C. If f : Ω → C is holomorphic and nowhere zero, then there is a holomorphic function g : Ω → C such that g(z)2 = f√(z) for all z ∈ Ω. We say g is a “branch of f ” on Ω. Proof. Fix a point z0 ∈ Ω and define Z z 0 f (w) dw, h(z) = z0 f (w) where the integral is taken along any path in Ω from z0 to z. Since the integrand is holomorphic and Ω is simply connected, Cauchy’s theorem tells us that the choice of path does not matter. Then h0 (z) = f 0 (z)/f (z), which gives d −h(z) e f (z) = 0. dz It follows that there is a constant C such that f (z) = Ceh(z) on Ω. Choose D such that D2 = C and define g(z) = Deh(z)/2 . Now let Ω be a simply connected open subset of C and let a ∈ / Ω. Let 7 S be the class of injective holomorphic functions Ω → U. We are going to construct the Riemann mapping as a certain extremal element of S. Lemma 17.3. The class S is nonempty. 7

In this context it is traditional to say univalent, but it means the same thing.

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√ Proof. Let g(z) be a branch of z − a on Ω. Clearly g is injective Ω → C. Moreover, it is impossible that both w and −w should belong to the image of g. Now, by the open mapping theorem, the image of g contains some disk D, so it must omit the disk −D. Let the disk −D have center b and radius r; then r z 7→ g(z) − b is an injective map sending Ω into U. Lemma 17.4. The set S is precompact in the topology of uniform convergence on compact subsets of Ω. That is, given any sequence of functions fn ∈ S one can find a subsequence that converges uniformly on compact subsets of Ω. Proof. It suffices to show that for some fixed compact subset K ⊆ Ω, the restriction of S to C(K) forms a precompact subset of C(K). But, given K, there is r > 0 such that the closed disk D(z; r) is contained in Ω for all z ∈ K. Now if z, z 0 ∈ K with |z − z 0 | < r/2, I z − z0 1 0 f (w)dw f (z) − f (z ) = 2πi ∂D(z;r) (w − z)(w − z 0 ) by Cauchy’s integral formula. Remembering that |f | 6 1, this gives |f (z) − f (z 0 )| 6 (2/r)|z − z 0 | for every f ∈ S. Thus the restrictions of the functions f to K form an equicontinuous subset of C(K), which is then precompact by the Arz´elaAscoli theorem. Fix a point p ∈ Ω. Lemma 17.5. The derivatives |f 0 (p)| are bounded above for f ∈ S, and the least upper bound is actually attained by some f ∈ S. Proof. Let fn ∈ S be a sequence such that |fn (p)| tends to its least upper bound (possibly +∞). By the previous lemma, we may assume by passing to a subsequence that fn converges uniformly on compact sets to a holomorphic function f . From Cauchy’s formula for derivatives, fn0 converges uniformly on compact sets also. In particular, |fn0 (p)| converges to |f 0 (p)|, which is the least upper bound. (Notice in particular that f 0 (p) 6= 0, so f is not constant.)

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It remains to show that f ∈ S. Since each fn maps Ω into U, f maps Ω into U. But f , being holomorphic and nonconstant, is an open map, so it actually maps Ω into the interior of U, which is U. To show that f is injective, we use Rouch´e’s theorem. Suppose that there are distinct points q, q 0 ∈ Ω such that f (q) = f (q 0 ) = c. Let the cycle Γ be the union of two small disjoint circles around q and q 0 in Ω on which f does not take the value c. Then sup{|f (z) − c| : z ∈ Γ∗ } = δ > 0. There is an n such that |fn − f | < δ on Γ∗ . By Rouch´e’s theorem, fn − c and f − c have the same number of zeroes inside Γ∗ . Thus fn − c has two zeroes, contradicting its injectivity. Completion of the proof of Theorem 17.1. Let f be the extremal element of S constructed by Lemma 17.5. We shall show that f maps Ω onto U. Suppose not, and say c ∈ U does not belong to the image of f . Then Tc ◦ f is in S and does not take the value 0, so by Lemma 17.2 there is g ∈ S such that Tc ◦ f = g 2 . Put g(p) = d and let h = Td ◦ g ∈ S. By construction we have f = F ◦ h, where F (w) = T−c [T−d (w)]2 . By the chain rule we have f 0 (p) = F 0 (0)h0 (p) since h(p) = 0. But since F is a holomorphic map U → U and is not an automorphism, Corollary 16.8 shows that |F 0 (0)| < 1. Thus |h0 (p)| > |f 0 (p)|, contradicting the extremality of f .

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Lecture 18 Multi-valued functions As our proof of the Riemann mapping theorem indicates, when constructing conformal maps we often need to make use of “branches” of “multivalued functions” like the square root function. A full treatment of multivalued functions involves studying Riemann surfaces; but in this lecture we shall take an ad hoc approach. Definition 18.1. Let f be holomorphic (or meromorphic) on an open set Ω. A holomorphic (or meromorphic) function g on Ω is a branch of log f if eg(z) = f (z) for all z ∈ Ω. Similarly, if α = p/q ∈ Q, then g is a branch of f α if g(z)q = f (z)p for all z ∈ Ω. √ This is consistent with our previous definition of a branch of f . Proposition 18.2. If f is holomorphic on a simply connected region Ω, and f (z) 6= 0 for all z ∈ Ω, then there exist branches of log f and of f α on Ω. Proof. We proved this in Lemma 17.2. In the opposite direction: Proposition 18.3. There is no branch of log f on any open set that contains a zero or a pole of f . . Proof. Let a be a zero or a pole of f and let g be a branch of log f defined near a. Then f 0 /f = g 0 . Thus I 0 I f (z) 1 1 dz = g 0 (z)dz = 0 2πi f (z) 2πi by the fundamental theorem of calculus (where the integrals are taken around a small circle with center a). However, the integral gives the order of the zero or pole of f at a (Proposition 10.2). Thus the order is 0, i.e. f has neither a zero nor a pole. Proposition 18.4. Let α = p/q in lowest terms. There is no branch of f α on any open set that contains a zero or pole of f of order that is not a multiple of q. In particular, there is no branch of f α on any open set that contains a simple zero or pole of f (unless α is an integer).

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Proof. Let a be a zero or pole of f and let g be a branch of f α defined near a. We have g(z)q = f (z)p and both sides are meromorphic functions near a. The order of the zero or pole of the left side at z = a is a multiple of q, and of the right side is a is a multiple of p. Since p and q are coprime, the order is a multiple of pq, and this means that f has a zero or pole of order a multiple of q. Now suppose that f is holomorphic (or meromorphic) on Ω, and we want to define a branch of log f or of f α . If there are zeroes or poles of f in Ω of the orders described in the previous propositions, this will not be possible. Such points are called branch points of log f or of f α , and it is necessary to remove them from Ω in order for us to have any chance of defining the branch that we want. However, removing the branch points is usually not sufficient to be able to define a branch of the function. The trouble is that after we remove a finite number of points from an open set Ω, the part that is left will not be simply connected, so that Proposition 18.2 will not apply. Definition 18.5. A cut is a piecewise smooth curve joining two branch points, or joining a branch point to a point outside Ω. It is not hard to see that if Ω contains a finite number of branch points, we can find a finite number of cuts such that the result of removing the branch points and the cuts from Ω is simply connected. (We shan’t prove this formally because it is more important to understand how to compute in specific examples.) Thus, it is always possible to remove cuts from Ω in such a way that a branch of log f of f α can be defined on the complement. However, there is no unique choice of cuts to remove: we must adapt our choice to the problem at hand. Example 18.6. Let Ω = C and consider the problem of defining a branch of log z. There are branch points at 0 and ∞. As we know, there is no branch of log z on C \ {0}. However, if we cut the plane, say along the negative real axis (joining 0 and ∞), then we can define a branch of log z on C \ R− . For instance, each such z can be represented as reiθ where −π < θ < π and r > 0; then log z = log r + iθ defines one of the infinitely many branches of log z on the cut plane. Note that the need for a cut is apparent from the fact that our branch is ‘discontinuous across the cut’— it takes very different values just above and just below a point on the cut. 62

Example 18.7. Again let Ω = C and now let’s consider defining a branch of log(z 2 − 1). This time there are branch points at ±1 and ∞. If we cut along the real axis from −1 to 1 and again from 1 to +∞, the remaining region is simply-connected and so admits a branch of log(z 2 − 1). To describe this branch explicitly, note that on the cut region we can write z − 1 = r1 eiθ1 , z + 1 = r2 eiθ2 with r1 , r2 > 0 and 0 < θ1 , θ2 < 2π. Then log(z 2 − 1) = log(r1 r2 ) + i(θ1 + θ2 ) gives a branch. √ Example 18.8. Now consider the function z 2 − 1. Here there are branch points at ±1 but not at ∞ (because ∞ is a pole of order 2 for the function z 2 − 1). This suggests that we may not need a cut extending to ∞, and that indeed is the case. To see this start with the branch g(z) of√log(z 2 −1) defined in the previous example. Taking eg(z)/2 gives a branch of z 2 − 1, which we may explicitly write as √ r1 r2 · ei(θ1 +θ2 )/2 . Notice now that this function can be extended continuously to the larger region C \ [−1, 1]; in other words, the cut from 1 to ∞ is unnecessary. For as z passes from a point just below to a point just above this cut, both θ1 and θ2 change by 2π. Their sum therefore changes by 4π and so the exponential term ei(θ1 +θ2 )/2 does not change at all. There is therefore no discontinuity across this cut, and it can be removed. Exercise 18.1. Find a region on which a branch of the inverse cosine function is defined. (Start by proving that p cos−1 (z) = i log z + z 2 − 1 and then look for a branch of the function on the right.)

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Lecture 19 Constructing conformal maps In this lecture we shall discuss a cook-book of techniques for constructing explicit conformal maps from various regions of the plane to the unit disk U. Example 19.1. The interior of a disk. This can always be moved onto the unit disk by a suitable affine transformation z 7→ az + b. For some purposes it may be useful to use a more general M¨obius transformation. Recall that points A and B are inverse relative to a circle C of center P and radius r if P, A, B are collinear and d(P, A).d(P, B) = r2 . If the complex numbers a and b represent inverse points with respect to the circle C then the equation |z−a| = λ|z−b|, where λ is the positive constant such that (a−p) = λ2 (b−p), represents C. See Exercise 2.1. Thus the M¨obius transformation z 7→

z−a z−b

send the given circle to the circle center the origin and radius λ. Example 19.2. Any two circles in the plane, one of which is entirely contained in the other, can be mapped to two concentric circles by a M¨obius transformation. One need only find a common pair of inverse points on the line joining the centers of the two circles, which necessitates solving a quadratic equation. There is an application to Steiner’s porism, see http: //www.maths.gla.ac.uk/∼wws/cabripages/inversive/steiner.html. Example 19.3. A half-plane can be mapped to the unit disk by finding a pair of inverse points as in the example above. In this case, λ = 1, and the points a and b are inverse with respect to a line L if a is the reflection of b in L. Example 19.4. Consider a wedge-shaped region. After translation and rotation one can consider only the region {z = reiθ : r > 0, 0 < θ < α}. The mapping z 7→ z π/α (which may be a multiple-valued function; if so, define it on the plane cut along the positive real axis) will take this wedge to a half-plane, which can then be mapped to the disk by a M¨obius transformation. Example 19.5. A region bounded by two circlines can be mapped to a wedge-shaped region by a M¨obius transformation. Indeed, the two circlines will meet at points a and b; the transformation z 7→

z−a z−b

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will send one point of intersection to 0 and the other to ∞, so the enclosed region will be mapped to the wedge between two rays from 0 to ∞. Example 19.6. A strip such as {z : 0 < Im z < c} can be mapped to a half-plane by the complex exponential function. In this case the required functions is z 7→ eπz/c . A related case is that of the region bounded by two circlines tangent at a point. In this case, any M¨obius transformation that sends the point of tangency to ∞ will map our region to a strip (or to the complement of a strip). Example 19.7. Some half-strips can be handled by the same technique. For instance, consider {z : 0 < Im z < c, Re z > 0}. The exponential eπz/c maps this to the set {w : Im w > 0, |w| > 1}. This is a region bounded by two circlines and the techniques of Exercise 19.5 apply. Example 19.8. Find a conformal mapping of U \ [ 12 , 1) onto U. The next stage in complexity is to consider the Schwarz-Christoffel transformations which map the upper half-plane conformally onto a polygon. If the polygon has exterior angles α1 , α2 , . . . , αn then a Schwarz-Christoffel transformation is of the form Z z dw α /π 1 · · · (w − xn )αn /π z0 (w − x1 ) where x1 , . . . , xn are points on the real axis (which will map to the vertices of the polygon). These transformations are of real practical use in solving potential theory problems — whole books have been written about them — but we shall not consider any problems requiring them in this course.

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Lecture 20 Contour integrals with multi-valued functions We can expand our repertoire of contour integration techniques by using suitably chosen branches of multiple-valued functions as integrands. I will work these examples at the chalkboard. Example 20.1. Z ∞ a−1 π x dx = , 0 < a < 1. 1+x sin aπ 0 We have encountered this integral before in the proof of theorem 12.4, the reflection theorem for the gamma function. There, we made a substitution which avoided the multiple-valued function z a−1 . This time, we will tackle the multi-function directly. Outline of argument: cut the plane along the positive real axis, and define a branch of f (z) = z a−1 /(1 + z) in the cut plane. Integrate around a “keyhole” contour (a large and a small circle, and two straight segments just above and just below the cut). The two straight segments give integrals related to the desired one, the large and small circles vanish in the limit, and there is a simple pole at z = −1. Note that the substitution z = ew transforms not merely the integral, but the contour and our entire argument, into Example 11.9. Example 20.2. Z ∞ π2 log x dx = − √ . 4 1+x 8 2 0 Outline of argument: cut the plane along the positive real axis and define a branch of f (z) = (log z)/(1 + z 4 ) in the cut plane. Integrate this function around a semi-circular contour with bottom on the real axis. Over the positive part of the real axis we get the integral I we want; over the negative part we get Z ∞ dx I + iπ ; 1 + x4 0 the integral over the semi-circle vanishes √ in the limit. There are √ two poles inside the contour, at eiπ/4 = (1 + i)/ 2 and e3iπ/4 = (1 − i)/ 2, and we can compute their residues and apply the residue theorem. R ∞ Consider real parts to evaluate I. As a bonus, we find the value of 0 dx/(1 + x4 ) by considering imaginary parts. A tricky alternative contour is to use a quarter -circle with sides on the real and imaginary axes. This works because z 4 = (iz)4 in this case. 66

Example 20.3. Z 1 0

dx π p =p , (x − a) x(1 − x) a(a − 1)

(a > 1).

Outline of argument:pcut the plane between 0 and 1 and define a singlevalued branch of (z − a) z(1 − z) on this cut plane, compare Example 18.8. We apply the residue theorem to the cycle Γ = [γ1 ] − [γ2 ], where γ1 is a ‘dumb-bell’ contour encircling the cut and γ2 is a circle of large radius. R There is one singularity enclosed by Γ, the pole at a. On the one hand, γ2 R tends to 0 as the radius tends to ∞; on the other, γ1 approaches −2 times the integral we want, because the sign change across the cut means that the integrals along the long sides of the dumb-bell are equal. Exercise 20.1. Evaluate the integral of homework 6, problem 1, √ Z ∞ x dx, 2 + 5x + 6 x 0 by means of a contour integral involving a suitably defined branch of the square root function.

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Lecture 21 Analytic Continuation Suppose that a holomorphic function f is defined on some open disk D(a; r) ⊆ C. The classical idea of analytic continuation is to extend f to a holomorphic function defined on some larger open set. Because of the principle of isolated zeroes (otherwise known as the uniqueness of analytic continuation) such an extension is unique if it exists. However, our studies of multi-valued functions show that there need not be a largest domain to which a given function can be analytically continued. Classically, then, one considers analytic continuations along a curve γ; meaning a chain of disks D(ai ; ri ), each one intersecting the one before and together covering γ ∗ , together with holomorphic functions on these disks which agree on the intersections. The discussion becomes much cleaner if we use some of the language of topology. Consider triples (U, f, a), where a is a point of C, U is an open subset of C containing a, and f : U → C is a holomorphic function. We impose an equivalence relation on the set of such pairs by saying that (U, f, a) is equivalent to (V, g, b) if a = b and also f = g on some open set W ⊆ U ∩ V that contains a. Definition 21.1. An equivalence class for this equivalence relation is called the germ of a holomorphic function on C. We define the germ of a meromorphic function in a similar way. Let M denote the collection of all germs of meromorphic functions on C. The functions (U, f, a) 7→ a, (U, f, a) 7→ f (a) respect the equivalence relation described above, and they therefore pass to functions z : M → C and w : M → S. We are going to make M into a topological space, and indeed into a Riemann surface. Let Ω be an open subset of C and let f be a meromorphic function on Ω. Then f defines a map f˜: Ω → M which sends each z ∈ Ω to the germ of f at z — that is, the equivalence class of the triple (Ω, f, z). The image f˜(Ω) is a subset of M: we shall call any subset of M of this kind a basic set. Lemma 21.2. The intersection of two basic sets is a basic set.

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Proof. Consider a basic set of the form B = f˜(Ω). The function z restricts to a bijection B → Ω, and w ◦ z −1 = f : Ω → S. It follows that if two basic sets f˜1 (Ω1 ) and f˜2 (Ω2 ) intersect nontrivially, then f1 = f2 = f say on the intersection Ω1 ∩ Ω2 = Ω, and f˜1 (Ω1 ) ∩ f˜2 (Ω2 ) = f˜(Ω). Because of this property, we can define a topology for M as follows: we declare that a subset U of M is open if, for every point p ∈ U , there is a basic set B containing p and contained in U . (This is the standard definition of the topology generated by a basis.) Lemma 21.3. M is a Hausdorff space. Proof. Let p and q be distinct points of M. Each of these ‘points’ is really a germ, that is an equivalence class of triples. Let (U, f, a) be a triple defining the germ p, and let (V, g, b) be a triple defining the germ q. If a and b are distinct points of C, then let U 0 and V 0 be disjoint open sets with a ∈ U 0 ⊆ U and b ∈ V 0 ⊆ V . Then f˜(U 0 ) and g˜(V 0 ) are disjoint basic (hence, open) subsets of M containing p and q respectively. If a and b are the same point, let W be a small disk D(a; ) contained in U ∩ V . I claim that the basic sets f˜(W ) and g˜(W ), which contain p and q respectively, do not intersect. Suppose the contrary: then there is some point c ∈ W such that the germ of f around c is equal to the germ of g around c — that is, f = g on some neighborhood of c. But then, by the uniqueness of analytic continuation, f = g on the whole disk W , which implies that p = q, contrary to hypothesis. Lemma 21.4. The standard functions z and w on M are continuous. Proof. Let Ω ⊆ C be open and let p ∈ z −1 (Ω). Let the germ p be represented by (U, f, a), with a ∈ Ω. Then f˜(U ∩ Ω) is a basic neighborhood of p and is contained in z −1 (Ω). Thus z −1 (Ω) is open, so z is continuous. The argument for w is similar (see below). Exercise 21.1. Complete the proof that w is continuous. Lemma 21.5. The map z : M → C is a local homeomorphism. What this means is that given any p ∈ M, there is a neighborhood U of p such that z restricts to a homeomorphism of U onto z(U ) ⊆ C. Proof. Let p be defined by the triple (U, f, a). Then B = f˜(U ) is a basic neighborhood of p, and, as we already noted, z restricts to a continuous bijection from B to U . But z maps any basic set to an open set (by definition) so it is an open map. An open continuous bijection is a homeomorphism.

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Lemma 21.6. M is a Riemann surface, and z : M → C and w : M → S are holomorphic functions. Proof. This is now easy. Each point of M has a neighborhood U on which the function z gives a homeomorphism into C, by the previous lemma. We take these homeomorphisms as our charts. The transition functions between the charts are just the identity function on C, which is certainly holomorphic! So M is a Riemann surface. To check whether a function on a Riemann surface is holomorphic, we must look at its representation in a chart. Consider the chart defined by identifying the basic neighborhood B = f˜(U ) of the previous lemma with U . In this chart, z becomes the identity map U → C, and w becomes the map f : U → S. These are both holomorphic. Definition 21.7. We call M the sheaf of germs of meromorphic functions on C. In terms of all this machinery we can finally give the definition that we have been aiming for, that of analytic continuation. Definition 21.8. Let p be the germ of a meromorphic function at a ∈ C, and let γ : [0, 1] → C be a path starting at γ(0) = a. An analytic continuation of p along γ is a lifting of γ to M: that is, a path γ˜ : [0, 1] → M such that the diagram =M || || | γ ˜ || z || | | | | || γ /C [0, 1]

commutes. Example 21.9. Suppose that p is the germ of an entire function f at a ∈ C. Then p can be analytically continued along any path in C, and the analytic continuation along γ sends t to the germ of f at γ(t). Our language of ‘the’ analytic continuation is legitimate because we shall shortly prove that analytic continuations are unique when they exist. Example 21.10. Let p be the germ at 1 ∈ C defined by the branch f (z) of √ z, defined on a small disk with center 1, that has f (1) = 1. Then p can be analytically continued along the path z = eiπt , t ∈ [0, 1], and the value of this analytic continuation at γ(1) = −1 is i. Also, p can be analytically continued along the path z = e−iπt , t ∈ [0, 1], and the value of this analytic

70

continuation at γ(1) = −1 is −i. There is no analytic continuation of p along the path γ(t) = 1 − 2t, t ∈ [0, 1]. Example 21.11. One can find a function f which is holomorphic on U and continuous on the closed disk U, but whose restriction to the boundary circle ∂U is nowhere differentiable. An explicit formula for such a function is f (z) =

n ∞ X z 10

n=0

2n

,

but one can also modify the Baire category argument that we used last semester to construct continuous nowhere differentiable functions. Then the function f cannot be analytically continued along any path that leaves U.

71

Lecture 22 The monodromy theorem As we have seen, analytic continuations may or may not exist. But if they do exist, they are unique: Proposition 22.1. There is at most one analytic continuation of a given germ p along a given path γ. Proof. Let γ1 and γ2 be two analytic continuations of p along γ, and consider the set A = {t : γ1 (t) = γ2 (t)}. Since M is Hausdorff, A is closed. Let t be a point of A and let q = γ1 (t) = γ2 (t). There is a neighborhood B of Q such that z : B → z(B) ⊆ C is a homeomorphism. There is then a neighborhood I = (t − , t + ) of t such that γ1 (s), γ2 (s) ∈ B for s ∈ I. Since z ◦ γ1 = z ◦ γ2 and z is bijective on B, it follows that γ1 (s) = γ2 (s), i.e. I ⊆ A. So A is open. By connectedness, A = [0, 1] which gives the desired uniqueness. If Ω is an open subset of C, we define M(Ω) to be the space of germs of meromorphic functions at points of Ω. It is an open subset of M. Definition 22.2. Let Ω ∈ C and let p be the germ of a meromorphic function defined at a point of Ω. The Riemann surface of p over Ω, Mp (Ω), is the connected component of M(Ω) that contains p. As we saw in Example 21.10, if we analytically continue a germ p along a path γ from a to b, the germ q at b that we obtain by continuation typically depends on the whole path γ, and not just on b. We are now going to investigate conditions under which analytic continuation does depend only on the end-point. Let Ω be a connected open subset of C and p the germ of a meromorphic function at some point a ∈ Ω. If p can be analytically continued along every path in Ω, we shall say that Ω is a regular region for p. Lemma 22.3. Suppose that p is a germ defined at a, and that the disk D(a; r) is a regular region for p. Let g(z) be the function on D(a; r) defined by analytic continuation of p along the straight line segment [a, z]. Then g is meromorphic on D(a; r) and its germ is p. Sketch proof. Define g as indicated and let ρ be the largest radius such that g is meromorphic on D(a; ρ). If ρ < r, use a compactness argument to cover the circle of radius ρ by finitely many domains of germs of analytic continuations; then use the principle of isolated zeroes to argue that g is meromorphic on a slightly larger disk, contrary to the definition of ρ. 72

Proposition 22.4. The connected open set Ω is regular for p if and only if z : Mp (Ω) → Ω is a regular covering map in the sense of topology. Proof. By definition of Mp (Ω), every germ q ∈ Mp (Ω) can be analytically continued along every path in Ω. Now let a ∈ Ω and suppose that the disk D(a; r) ⊆ Ω. Let Γ = {pj } be z −1 {a}, the germs in Mp (Ω) at a. By the lemma, each of these germs is the germ of a meromorphic function gj defined on D(a; r). Each gj defines g˜j : D(a; r) → Uj ⊆ Mp (Ω), which is a homeomorphism of D(a; r) onto a basic open set, with inverse z. Thus z −1 (D(a; r)) ∼ = D(a; r) × Γ where Γ is a discrete set. This is the definition of a regular covering map. Theorem 22.5 (Monodromy theorem). Let Ω be a regular region for a germ p at a ∈ Ω. If γ0 and γ1 are paths in Ω, starting from a and ending at b, that are homotopic relative to their endpoints, then the germ obtained by analytically continuing p along γ0 to b is the same as the germ obtained by analytically continuing along γ1 . In particular, if Ω is simply connected, then the analytic continuation of p along any path in Ω depends only on the endpoint of the path. Proof. Immediate from Proposition 22.4 and the covering homotopy lifting theorem from algebraic topology. To conclude, we shall sketch how these ideas can be applied to the proof of Picard’s theorem, one of the most beautiful results in complex analysis. We need a certain special function, the elliptic modular function J. The explicit construction of J belongs to the theory of elliptic curves, and we shan’t have time to discuss that in this course. Fortunately, however, all that we need about J are some key properties. The function J(τ ) (the notation τ is traditional) is defined on the upper half plane H = {τ : Im(τ ) > 0}, and it maps H onto Ω = C \ {0, 1}. One can show that (a) J 0 (τ ) 6= 0 for τ ∈ H. (b) J is invariant under the modular group Γ, which consists of M¨obius transformations aτ + b τ 7→ cτ + d with a, b, c, d ∈ Z, ad − bc = 1. 73

(c) Two points τ1 , τ2 ∈ H belong to the same Γ-orbit if and only if J(τ1 ) = J(τ2 ). (d) The action of Γ on H if free and proper, which is to say that each τ ∈ H has a neighborhood W such that the sets γW , γ ∈ Γ, are all (pairwise) disjoint. From these properties it follows that Lemma 22.6. The map J : H → Ω = C \ {0, 1} is a regular covering map. Proof. Let a ∈ Ω and choose τ such that J(τ ) = a. Since J 0 (τ ) 6= 0 (property (a)), the inverse function theorem gives us a neighborhood U of τ and a neighborhood V of a such that J restricts to a conformal equivalence U → V . By (d) there is no loss of generality in assuming that the sets γU are all disjoint. By (c) we now have [ J −1 (V ) = γU ∼ = Γ × V. γ∈Γ

Thus each point of Ω has a neighborhood U such that J −1 (U ) can be identified with the product of U and a discrete set; this is the definition of a regular covering map. Theorem 22.7 (Picard’s theorem). A nonconstant entire function takes every value with at most one exception. Proof. Let f be a nonconstant entire function omitting two values; without loss of generality, we may assume that it omits 0 and 1. We obtain a diagram

~

~

~

~

~

~

f

C

H ~? J

/Ω

Since C is simply connected, the covering homotopy theorem (aka the monodromy theorem in this context) says that we can lift f to obtain a function g : C → H such that J ◦ g = f . But now g−i g+i is an entire function from C to U, hence it is constant by Liouville’s theorem. Thus g, and hence f , are constant as well.

74

Lecture 23 Basics of Banach Spaces This lecture reviews material from last semester’s course, lecture 10. Definition 23.1. Let V be a vector space (real or complex). A norm on V is a function V → R, denoted v 7→ kvk, such that (a) kvk > 0 for all v, and kvk = 0 iff v = 0. (b) kλvk = |λ|kvk, where λ is a scalar (real or complex). (c) kv + v 0 k 6 kvk + kv 0 k. A vector space equipped with a norm is called a normed vector space. Clearly, d(v, v 0 ) = kv − v 0 k is then a metric. Because of (b), this metric has an ‘affine’ structure not present in a general metric. Definition 23.2. If a normed vector space is complete in its metric, it is called a Banach space. Example 23.3. The space C(X) of continuous complex-valued functions on a compact Hausdorff space X is a Banach space under the norm kf k = sup{|f (x)| : x ∈ X}. (We proved the completeness last semester.) Lemma 23.4. Let a and b be positive real numbers and let p > 1. Then inf{t1−p ap + (1 − t)1−p bp : t ∈ (0, 1)} = (a + b)p . Proof. The function x 7→ xp , p > 1, is convex (the graph always lies below its chords) and therefore b p a p (a + b) = t + (1 − t) t 1−t p a p b 6t + (1 − t) = t1−p ap + (1 − t)1−p bp t 1−t with equality at t = a/(a + b). Let (X, µ) be a σ-finite measure space. For 1 6 p < ∞ we define Lp (X, µ) to be the space of (equivalence classes moduloR equality almost everywhere of) measurable functions f : X → C for which kf kp dµ < ∞. The Lp norm kf kp is then defined by Z 1/p p kf kp = |f | dµ . 75

The notation is consistent with our previous definitions of the normed vector spaces L1 and L2 . Proposition 23.5. The Lp norm is indeed a norm. Proof. The only nontrivial thing is the triangle inequality. Let f, g ∈ Lp (X, µ). We have for t ∈ (0, 1) Z p kf + gkp 6 (|f (x)| + |g(x)|)p dµ(x) Z t1−p |f (x)|p + (1 − t)1−p |g(x)|p dµ(x) 6 6 t1−p kf kpp + (1 − t)1−p kgkpp . Taking the infimum over all t we find that kf + gkpp 6 (kf kp + kgkp )p , and taking p’th roots gives the result. In the next lecture we shall prove that Lp (X, µ) is complete, and therefore a Banach space. (We already proved this for L1 and L2 last semester.) Remark 23.6. We let L∞ (X, µ) denote the space of essentially bounded measurable functions on X, that is, functions that are bounded off a set of measure zero. This space can be given the essential supremum norm kf k∞ = inf sup{|f (x)| : x ∈ X \ N } N

where the infimum is taken over null sets N . If X is a finite measure space then we have L∞ ⊆ Lq ⊆ Lp ⊆ L1 when p < q. Exercise 23.1. Show that L∞ is a normed vector space. It is natural to consider linear mappings between normed spaces. Proposition 23.7. A linear mapping T : V → W between normed vector spaces is continuous if and only if there is a constant k such that kT vk 6 kkvk (one then says that T is bounded). Proof. If T is bounded then kT u − T vk 6 kku − vk so T is continuous. Conversely, if T is continuous then there is some δ > 0 such that kuk < δ implies kT uk < 1. Take k = 1/δ. 76

Definition 23.8. The best constant k in the above proposition, that is the quantity sup{kT vk : kvk 6 1} is called the norm of T and denoted kT k. Proposition 23.9. With the above norm, the collection B(V, W ) of bounded linear maps from V to W is a normed vector space, and it is a Banach space if W is a Banach space. Proof. It is straightforward to check that the expression in Definition 23.8 is a norm. Let us prove completeness: suppose then that Tn is a Cauchy sequence in B(V, W ). For each v ∈ V we have kTn v − Tm vk 6 kTn − Tm kkvk, so Tn v is a Cauchy sequence in W . Since W is assumed to be complete, this Cauchy sequence converges: denote its limit by T v. A limiting argument shows that T is a linear map and that kT k 6 sup kTn k is bounded. Finally, let > 0. There is an N such that if m, n > N then kTn − Tm k < . Then for any v, and n > N , kTn v − T vk = lim kTn v − Tm vk 6 kvk, m→∞

so we have shown kTn − T k 6 and thus Tn converges to T in the normed space B(V, W ). Definition 23.10. A linear functional on a vector space V over a field k is a linear map V → k. Last semester we proved: Proposition 23.11. A linear functional on a normed vector space is continuous if and only if its kernel is closed. Definition 23.12. Let V be a normed vector space. The dual space V ∗ is the space of continuous linear functionals on V (in other words, it is the space B(V, k)). We consider V ∗ as a normed vector space, with the norm coming from Definition 23.8. From proposition 23.9, V ∗ is always a Banach space (even if V itself is not complete). It is often important to identify explicitly the dual of some given normed space. For example, we proved last semester that the dual of C(X) is the space of Borel measures on X; the dual of a Hilbert space is the same Hilbert space (self-duality).

77

Lecture 24 The Spaces Lp (X, µ) Let (X, µ) be a σ-finite measure space. Recall from last time the definition of the normed vector spaces Lp (X, µ), 1 6 p 6 ∞. Lemma 24.1 (H¨older’s inequality). Suppose that 1 < p, q < ∞ with 1/p + 1/q = 1 (we then say that p, q are conjugate indices). Then for f ∈ Lp (X, µ) and g ∈ Lq (X, µ) we have f g ∈ L1 (X, µ), and Z f gdµ 6 kf kp kgkq . Proof. This can be proved in the same way as Proposition 23.5. Namely, for any nonnegative a, b one can show that (ta)p (t−1 b)q ab = inf + . t>0 p q (To see this, use elementary calculus to discover that the unique minimum of the right-hand side occurs at t = b1/p a−1/q .) Applying this inequality pointwise we find Z Z p Z t−q p f g 6 t |f | + |g|q , p q and now taking the infimum over t yields H¨older’s inequality. Remark 24.2. An important consequence of H¨older’s inequality is that, if p, q are conjugate indices, then there is a map Φ : Lp → (Lq )∗ which sends f ∈ Lp to the linear functional Z φf : g 7→ f gdµ. We are going to prove that if p > 1 then Φ identifies Lp and (Lq )∗ as normed vector spaces. In particular, because dual spaces are complete, this will show that all the spaces Lp are complete. Lemma 24.3. Let (X, µ) be a finite measure space, and let 1 6 p 6 ∞. Then the simple functions are dense in Lp (X, µ). Recall that a simple function is a finite linear combination of characteristic functions of measurable sets (of finite measure).

78

Proof. Let f ∈ Lp (X, µ) and suppose that f > 0. (Every function in Lp can be written as a linear combination of positive real-valued functions, so this is enough.) Define a sequence of simple functions {fn } as follows: let Xk,n = {x ∈ X : k2−n 6 f (x) < (k + 1)2−n }, and put n

fn =

4 X

k2−n χXk,n .

k=0

This is an increasing sequence of simple functions converging pointwise to f . If f is bounded, then the convergence is uniform: this proves the case p = ∞. Otherwise, argue as follows. Since 0 6 fn 6 f , we have |fn − f |p 6 |f |p . The right-hand side here is integrable, so the DCT shows that f |p dµ → 0 as required.

R

|fn −

Theorem 24.4. Let X be a σ-finite measure space and let p, q > 1 be conjugate indices. The map Φ : Lp (X, µ) → Lq (X, µ)∗ , defined above, is an isometric isomorphism. That is, it is a bijective linear transformation that preserves the norm. Notice that Proposition 24.6 from last semester can be thought of as the limiting case of this result where p = ∞, q = 1. It is not true that (L∞ )∗ = L1 in general, i.e. the limiting case p = 1, q = ∞ is false. Proof. For simplicity we shall assume where convenient that X is a finite measure space (the extension to the σ-finite case is left as an exercise). By H¨older’s inequality, kΦ(f )k(Lq )∗ 6 kf kp . On the other hand, let f ∈ Lp and write f = u|f | where u has absolute value 1 everywhere. Now p/q let g = u ¯|f |p/q ; then kgkq = kf kp . We have Z Z p/q Φ(f )(g) = |f ||f | = |f |p = kf kpp . Thus kΦ(f )k(Lq )∗ >

kf kpp p/q

kf kp

= kf kp

and we have shown that Φ is isometric. In particular, it is injective. To show that Φ is surjective we use the same argument based on the Radon-Nikodym theorem as we did in the case q = 1 (in our discussion last 79

semester). Namely, let φ be a continuous linear functional on Lq and consider the set function E 7→ φ(χE ). This set function is a measure, absolutely continuous with respect to µ; so by the Radon-Nikodym theorem there is an integrable function f on X such that Z f dµ φ(χE ) = E

for all measurable sets E. Therefore Z φ(g) =

f gdµ

for all g ∈ L∞ , since both sides are continuous functionals of g ∈ L∞ and agree for simple functions. Once we know that f ∈ Lp , the same argument will show that the identity holds for all g ∈ Lq and will therefore complete the proof. It remains therefore only to prove that f ∈ Lp . Write f = u|f | as above and for each n let fn = u min{|f |, n}. Let gn = u ¯|fn |p/q . Then fn and gn are bounded and, just as above, Z Z φ(gn ) = f gn dµ > fn gn dµ = kfn kp kgn kq . We conclude that kfn kp 6 kφk for all n and now since |fn | → |f | monotonically the MCT tells us that f ∈ Lp with kf kp 6 kφk. Corollary 24.5. The normed vector space Lp (X, µ) is complete, and therefore a Banach space. It follows from the above arguments that the Banach space Lp for 1 < p < ∞ is isomorphic to the dual of its dual. A space with this property is called reflexive. When we discuss the Hahn-Banach theorem in the next lecture we shall see how to ‘prove’ that L1 is not in general reflexive. The word ‘prove’ is in quotes because the construction depends on the axiom of choice.

80

Lecture 25 The Hahn-Banach Theorem Let E be a normed vector space over k = R or C. Recall that the dual space of E is the space E ∗ of continuous linear functionals E → k. It is a Banach space, with norm kφk = sup{|φ(v)| : kvk 6 1}. The Hahn-Banach theorem tells us that the dual space of E is always ‘large’. (It is not obvious that there exist any nonzero continuous linear functionals at all!) Lemma 25.1. Let E be a real normed vector space and let F be a subspace of codimension one, so that E is the direct sum of F and a one-dimensional subspace. Then every continuous linear functional on F can be extended to a continuous linear functional on E that has the same norm. To put this another way, the natural restriction map from E ∗ to F ∗ is surjective. Proof. Let E = F ⊕ hvi, where v is a unit vector. Let φ : F → R be the given linear functional. The extensions of φ are all of the form x + λv 7→ φ(x) + λα

(x ∈ F )

where α is a constant. Our task is to show that α can be picked to ensure that this extension of φ has the same norm as φ itself. Without loss of generality, assume that kφk = 1. In order for our extension to have norm 1, it is necessary and sufficient that |φ(x) + α| 6 kx + vk for all x ∈ F . (A simple rescaling shows that this condition implies the apparently more general |φ(x) + λα| 6 kx + λvk.) The displayed inequality is equivalent to α ∈ [mx , Mx ], where mx = −kx + vk − φ(x),

Mx = kx + vk − φ(x).

Thus in order that a suitable α can be chosen we need to know that the intersection of all the intervals [mx , Mx ] is nonempty, and this will be true if and only if my 6 Mx for all x, y ∈ F . To prove this, start from the fact that φ has norm one. Write φ(x) − φ(y) 6 kx − yk = k(x + v) − (y + v)k 6 kx + vk + ky + vk 81

using the triangle inequality. Rearranging to get all the x terms on the right and the y terms on the left, we get −ky + vk − φ(y) 6 kx + vk − φ(x) which is the desired conclusion my 6 Mx . Exactly the same result holds in the complex case. Lemma 25.2. Let E be a complex normed vector space and let F be a subspace of codimension one, so that E is the direct sum of F and a onedimensional subspace. Then every continuous linear functional on F can be extended to a continuous linear functional on E that has the same norm. Proof. The strategy is to reduce to the real case via complexification. Any complex normed vector space E has an underlying real normed vector space ER . If φ is a continuous complex-linear functional on E, its real part is a continuous real-linear functional on ER , and it has the same norm. Conversely, suppose that ψ is a continuous real-linear functional on ER . Define φ(x) = ψ(x) − iψ(ix) ∈ C. Clearly φ is a real-linear map E → C, and since φ(ix) = ψ(ix) + iψ(x) = iφ(x) it is actually complex-linear and has ψ as its real part. Thus we have shown that there is a norm-preserving 1 : 1 correspondence between the dual of the complex vector space E and the dual of the real vector space ER . Now the complex version of the lemma can be reduced to the real version (one just applies Lemma 25.1 to the underlying real vector spaces). Lemma 25.3. Let E be a normed vector space (real or complex) and let F be a dense subspace. Then every continuous linear functional on F can be extended to a continuous linear functional on E that has the same norm. Proof. Let φ be the given functional. For any x ∈ E there is a sequence {xn } in F converging to x. The inequality |φ(xn ) − φ(xm )| 6 kφkkxn − xm k shows that {φ(xn )} is a Cauchy sequence (of real or complex numbers). Define φ(x) to be its limit. It is easy to prove that this does not depend on the choice of sequence {xn } converging to F and defines a continuous linear functional extending φ. 82

Theorem 25.4 (Hahn-Banach). Let E be a normed vector space and let F be a subspace of E. Then any continuous linear functional on F extends to a continuous linear functional on E with the same norm. Proof. If E is separable, this can be proved by induction. Let {xn } be a countable dense subset of E and let F0 = F and Fn be the subspace spanned by F and {x1 , . . . , xn }. Then each Fn is of codimension zero or one in Fn+1 , and therefore Lemmas 25.1 or 25.2 can be applied inductively to extend any continuous linear functional on F to a continuous linear functional (with the same norm) defined on the union of all the Fn . But this union is dense in E, so Lemma 25.3 completes the proof. If E is not separable, it is necessary to use some transfinite technology. To mimic the above proof as closely as possible, use the axiom of choice to well-order E, say as {xα } where α < Υ, for some ordinal Υ. Let Fα = span (F ∪ {xβ : β < α}) so that F0 = F and FΥ = E. Suppose that the given continuous linear functional on F has already been extended to Fβ for all β < α. If α is not a limit ordinal, then Lemmas 25.1 and 25.2 provide an extension to Fα ; if α is a limit ordinal, then Lemma 25.3 does the same thing. By transfinite induction, the proof is complete. Remark 25.5. In the textbooks you will probably find this argument presented using Zorn’s lemma or Hausdorff’s maximality principle. These are equivalent ways of formulating the transfinite induction but do not require explicit reference to ordinals. Corollary 25.6. The dual space of a (real or complex) normed vector space E separates points of E. That is, given any two distinct points x, x0 ∈ E, there is φ ∈ E ∗ such that φ(x) 6= φ(x0 ). Proof. Consider the one-dimensional subspace spanned by x − x0 . Take any nonzero linear functional on this subspace (necessarily continuous!), and extend it to E by the Hahn-Banach theorem. Exercise 25.1. For any Banach space E construct an isometric injection E → E ∗∗ .

83

Lecture 26 Applications of the Hahn-Banach Theorem (This lecture will be delivered by Prof Mazzucato) Many important applications of the Hahn-Banach theorem come from the following corollary. Proposition 26.1. Let E be a normed vector space and S a subset of E. The closed linear span of S is the intersection of the kernels of all continuous linear functionals annihilating S. Corollary 26.2. A subspace S of E is dense in E if and only if the only continuous linear functional that annihilates it is the zero functional. Proof. Let N be the closed linear span of S. By definition, N is the intersection of all the closed subspaces of E that contain S. Now, if φ ∈ E ∗ annihilates S, then Ker(φ) is such a subspace. Consequently \ N ⊆ {Ker(φ) : φ ∈ E ∗ , φ(S) = {0}}. To prove the converse, let y ∈ E \ N . Let F be the subspace spanned by N and y. Each element of F can be written uniquely as x + λy, where x ∈ N and λ ∈ C. Define a linear functional φ on F by φ(x + λy) = λ. The kernel of this linear functional is N , which is a closed subspace of F . Thus φ is continuous (according to Proposition 10.8 from last semester, a linear functional on a normed vector space is continuous iff its kernel is closed). By the Hahn-Banach theorem, we can extend φ to a continuous linear functional on E. Denoting this extension still by φ we have φ(y) = 1 and so y ∈ / Ker(φ). But φ(S) = {0} since S ⊆ N , so we have shown that \ N ⊇ {Ker(φ) : φ ∈ E ∗ , φ(S) = {0}} which completes the proof. We are going to apply this technique to the proof of Runge’s theorem, which deals with the approximation of functions defined on a subset of C by polynomials or rational functions. The simplest version of Runge’s theorem is the following. Proposition 26.3. Let K be a compact subset of C whose complement is connected. Let f be a function holomorphic on a neighborhood of K. Then there exists a sequence of polynomials converging uniformly to f on K. 84

Remark 26.4. The example of the function 1/z on the circle S 1 shows that the topological hypothesis (connectedness of the complement) cannot be omitted. Proof. Consider the Banach space E = C(K). We want to prove that f ∈ E belongs to the closed linear span of the set S = {1, z, z 2 , . . .}. According to Proposition 26.1, it is equivalent to prove that every continuous linear functional φ on E that annihilates every member of S also annihilates f . By the Riesz representation theorem, any such functional φ is of the form Z gdµ φ(g) = K

for some Borel measure µ on K. Let f be holomorphic on Ω ⊇ K. We will need the following topological Claim 26.5. There is a cycle Γ in Ω that is homologous to zero in Ω and does not meet K, such that the winding number n(Γ, a) = 1 for all a ∈ K. In other words, the cycle Γ ‘encircles K’. We shall prove the claim in a moment. Granted it for now, we can write Z 1 f (z)dz f (w) = 2πi Γ z − w for all w ∈ K, by Cauchy’s integral formula. Thus, by Fubini’s theorem, Z Z Z 1 dµ(w) φ(f ) = f (w)dµ(w) = f (z)dz. 2πi Γ K K z−w Denote the inner R parenthesized expression by h(z), so that our formula is φ(f ) = (2πi)−1 Γ f (z)h(z)dz. We shall now show that h(z) ≡ 0 and this will complete the proof. Note that Z dµ(w) h(z) = K z−w is a holomorphic function of z (by differentiating under the integral sign), and it is defined on the set U = C \ K which is connected (by hypothesis). When |z| is large (greater than sup{|w| : w ∈ K}) we can expand the expression under the integration sign in a uniformly convergent power series: ∞

X wk 1 = . z−w z k+1 k=0

85

Thus for such z h(z) =

∞ X k=0

Z

1 z k+1

wk dµ(w) = 0,

K

since φ annihilates powers of w by assumption. We have shown that h(z) = 0 for all sufficiently large z. Since U is connected (here is where we use that critical assumption) it follows by analytic continuation that h ≡ 0 on U . It remains to prove Claim 26.5. Given K and Ω there is > 0 such that the distance from any point of K to any point not in Ω ia at least 2. Now cover C by a grid of squares of side . Finitely many of these squares will contain a point of K in their interior or boundary (we say it “meets K”, for short). Each such square has a bounding cycle made up of its four sides. Let Γ1 , . . . , ΓN be the bounding cycles of the squares Σj meeting K. PN I claim that the sum Γ = j=1 Γj has the properties claimed. By construction, Γ is a cycle in Ω. To see that it does not meet K, suppose that γ is an edge in the grid that intersects K. Then γ appears as a component in two of the cycles Γj , and it appears with opposite orientations. Thus it cancels from the sum defining Γ. We conclude that only edges that do not intersect K can appear in Γ with a non-zero coefficient. By construction, Γj has winding number 1 about any point in the interior of the square Σj and winding number 0 about any point outside the closure of Σj . This immediately implies that Γ has winding number 0 about any point outside Ω, i.e. Γ is homologous to zero. Any point a ∈ K is either in the interior of exactly one Σj , or is a limit point of a sequence {ak } of such interior points. In the first case we get n(Γ, a) = n(Γj , a) = 1; in the second we have n(Γ, ak ) = 1 by the same argument, and it follows that n(Γ, a) = 1 since ak and a are in the same connected component of Ω \ Γ∗ . The proof of Claim 26.5 is now completed.

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Lecture 27 Convexity and the Hahn-Banach Theorem Let E be a vector space over R. A mapping p : E → R+ is called a sublinear functional if (a) p(x + y) 6 p(x) + p(y) for all x, y ∈ E, and (b) p(λx) = λp(x) for nonnegative scalars λ. Clearly a norm is a sublinear functional, but there may be many others. Moreover, if we examine the proof of the Hahn-Banach theorem, we see that it really made use only of the sublinear properties of the norm (the triangle inequality and multiplication by positive scalars). Thus the very same proof that we gave for the Hahn-Banach theorem will also prove the following slightly more general result. Theorem 27.1. Let E be a real vector space and F a subspace. Let p be a sublinear functional on E. Suppose that ψ : F → R is a linear map such that ψ(x) 6 p(x) for all x ∈ F . Then ψ can be extended to a linear map φ : E → R such that φ(x) 6 p(x) for all x ∈ E. There is a close relationship between sublinear functionals and convex sets. Let p be sublinear and let A be the set {x : p(x) 6 1}. If x, y ∈ A and t ∈ [0, 1] then p(tx + (1 − t)y) = tp(x) + (1 − t)p(y) 6 t + (1 − t) = 1. Thus tx + (1 − t)y ∈ A also. That is to say, A is convex. S As well as convexity, A has a further property; the union s>0 sA is the whole space E. (This is easy to see: any x ∈ E belongs to p(x)A.) A subset of a vector space which has this property is called absorbing. We have seen that a sublinear functional gives a convex, absorbing set. Conversely Proposition 27.2. Let A be a convex, absorbing set in a vector space E. Then the Minkowski functional of A, defined as pA (x) = inf{s > 0 : x ∈ sA} is sublinear. 87

Proof. The hypothesis that A is absorbing shows that the infimum exists (i.e., the set over which it is taken is not empty). It is clear that pA (tx) = tpA (x) for t > 0. To prove the triangle inequality, suppose that pA (x) < r and pA (y) < s. Then r−1 x and s−1 y belong to A. By convexity, so does (r + s)−1 (x + y) =

r −1 s −1 r x+ s y. r+s r+s

Thus p(x + y) 6 r + s. Taking the infimum over r and s now gives the result. Remark 27.3. If the vector space E is normed, and C is an open convex set containing the origin, then C is absorbing. In this case the associated sublinear functional pC is continuous (because there is a constant k such that pC (x) 6 kkxk). Moreover, if x ∈ C then pC (x) < 1 (strict inequality). To see this, note that because C is open, if x ∈ C then there is some λ > 1 such that λx ∈ C also; then pC (λx) 6 1 so pC (x) 6 λ−1 < 1. Exercise 27.1. Give an example of an absorbing subset in a normed vector space which does not contain any neighborhood of the origin. Now we can reformulate the Hahn-Banach theorem as a separation theorem for convex sets. Proposition 27.4. Let A and B be disjoint nonempty convex sets in a normed vector space E, one of which (say A) is open. Then there is a continuous linear functional φ on E such that there is a constant c with φ(x) < c ∀x ∈ A,

φ(x) > c ∀x ∈ B.

One says that φ−1 (c) is a separating hyperplane for the convex sets A and B. Proof. Without loss of generality we may assume that A contains the origin. Let b0 ∈ B and let C = A − B + b0 = {a − b + b0 : a ∈ A, b ∈ B}. Then C is an open, convex set which contains the origin. By Remark 27.3 above, its Minkowski functional pC is defined and satisfies pC (x) 6 kkxk for some constant k.

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Notice that b0 ∈ / C (because A and B do not intersect) and therefore pC (b0 ) > 1. Let F be the 1-dimensional subspace spanned by b0 and let ψ : F → R be the linear functional that sends b0 to 1. Then ψ 6 pC on F . According to the Hahn-Banach theorem, then, ψ can be extended to a linear functional φ on E which is dominated by pC . I claim that φ has the required properties. First, we have φ(x) 6 pC (x) 6 kkxk and so φ is continuous. The set φ(A) is an open, convex subset of R containing 0; the set φ(B) is a convex subset of R containing 1. I claim that they are disjoint. If not, then there exist a ∈ A, b ∈ B such that φ(a) = φ(b). But then 1 = φ(a − b + b0 ) 6 pC (a − b + b0 ) which contradicts Remark 27.3, since a − b + b0 ∈ C. Now since φ(A) and φ(B) are disjoint intervals, with φ(A) open and to the left of φ(B), taking c = sup φ(A) gives us the theorem. Here is another version of the same idea. Proposition 27.5. Let E be a normed vector space, and let A and B be disjoint convex subsets with A compact and B closed. Then there are a continuous linear functional φ on E and constants c, c0 such that φ(a) 6 c < c0 6 φ(b) for all a ∈ A, b ∈ B. Proof. Consider the continuous function x 7→ d(x, B) = inf{kx − bk : b ∈ B}. Since B is closed, this function is zero only on B; so it is strictly positive on A, and therefore has a positive lower bound δ > 0, since A is compact. Then [ A0 = B(a; δ/2) a∈A

is open, convex and disjoint from B. Apply the previous theorem to the sets A0 and B, and then note that φ(A) is a compact subinterval of the open interval φ(A0 ).

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Lecture 28 Weak topologies A normed vector space E has a natural topology that it acquires from its norm. However, it’s frequently useful to consider other topologies — other notions of convergence — on E as well. We have seen this in our study of measure theory. The space L1 (X, µ) of integrable functions has a norm which gives rise to a notion of convergence; but it is frequently useful to consider also the weaker notion of pointwise convergence. Remark 28.1. If a set X is equipped with two topologies T1 and T2 , we’ll say that T1 is weaker than T2 , or that T2 is stronger than T1 , if every set that is open for T1 is also open for T2 . Let X be a set and let {fα }α∈A be a family of maps from X to topological spaces Xα . Definition 28.2. The topology induced by the family {fα } is the topology generated by the sets fα−1 (U ), where U is open in Xα (and α runs over A). In other words, it is the weakest topology on X that makes all the functions fα continuous. Now let E be a Banach space. Definition 28.3. The weak topology on E is the topology induced by the family E ∗ of maps E → C (or E → R). In more concrete terms, a set U is a neighborhood of 0 for the weak topology if and only if there are finitely many linear functionals φ1 , . . . , φn ∈ E ∗ such that |φj (x)| < 1 ∀j = 1, . . . , n =⇒ x ∈ U. The weak topology has the characteristic property that if xn is a sequence (or net) in E that converges weakly to x, then φ(xn ) → φ(x) for every φ ∈ E∗. Example 28.4. Let H be a Hilbert space and {en } an orthonormal sequence. For any x ∈ H, Bessel’s inequality states X |hx, en i|2 6 kxk2 , and it follows in particular that hx, en i → 0. Since every continuous linear functional on H is given by inner product with some x, it follows that en → 0 weakly. Since ken k = 1 for each n, the sequence en does not tend to 0 in norm. Example 28.5. Let E = C(X), where X is compact. Let {fn } ∈ E be a sequence such that

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(a) There is a constant C such that kfn k 6 C for all n. (b) There is f ∈ C(X) such that fn (x) → f (x) (pointwise) for each x ∈ X. Then fn → f weakly. To see this, remember that according to the Riesz representation Rtheorem, every continuous linear functional on E is of the form φ(f ) = f dµ for some Borel measure µ on X. The desired conclusion Z Z fn dµ → f dµ follows from the dominated convergence theorem (with the constant function C as dominating function). Proposition 28.6. A linear map f : E → C (or R) is weakly continuous if and only if it is norm continuous. Proof. A norm continuous functional is weakly continuous by definition of the weak topology. A weakly continuous functional is norm continuous because every weakly open set is norm open. Proposition 28.7. A convex subset of E is weakly closed if and only if it is norm closed. Proof. Any weakly closed set is norm closed. Conversely, let A be convex and norm closed and let b ∈ / A. According to the Hahn-Banach theorem 27.5, ∗ there are φ ∈ E and a constant c such that φ(x) < c < φ(b) for all x ∈ A. Then Fb := φ−1 ((−∞, c]) is a weakly closed set containing A and not containing b. It follows that \ A= Fb b∈A /

is weakly closed. Exercise 28.1. The argument above, as stated, works for real vector spaces only. Work out how to generalize it to the complex case. Corollary 28.8. Weak limits cannot increase norms: if {xn } is a weakly convergent sequence (or net) with limit x, and kxn k 6 c for all n, then kxk 6 c also. 91

Proof. The closed ball B(0; c) is weakly closed (by the proposition above) and so contains the limit of any weakly convergent sequence in it. In the Hilbert space case one can say a little more. Proposition 28.9. Suppose that {xn } is a sequence in a Hilbert space H such that xn → x weakly and kxn k → kxk. Then xn → x in norm. Proof. Consider kxn − xk2 = kxn k2 − 2 Rehx, xn i + kxk2 . Because xn → x weakly, and inner product with x gives a linear functional, hxn , xi → hx, xi = kxk2 . Thus the displayed expression tends to zero. In advanced functional analysis many different weak topologies appear. We mention only one. Definition 28.10. Let E be a Banach space. The weak-∗ topology on E ∗ is the topology induced by the family of maps ex (φ) = φ(x) as x runs over E. The maps ex are elements of E ∗∗ , but unless E is reflexive they are not all the elements of E ∗∗ . Thus the weak-∗ topology on E ∗ is in general weaker than its weak topology. An important property of the weak-∗ topology is the Banach-Alaoglu theorem: the (norm) closed unit ball of E ∗ is compact in the weak-∗ topology. We won’t prove this here.

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Lecture 29 Applications of the Baire category theorem Recall the Baire category theorem: a complete metric space cannot be the countable union of closed nowhere dense subsets. This has some striking applications to functional analysis, which we will consider in the next couple of lectures. Definition 29.1. A subset A of a Banach space is said to be balanced if λa ∈ A for all a ∈ A and scalars λ with |λ| = 1. Lemma 29.2. In a Banach space, every closed convex balanced absorbing set contains a neighborhood of 0. Proof. Let A be closed convex balanced and absorbing. (Such a set is called a barrel.) Because A is absorbing, ∞ [

nA = E.

n=1

By the Baire category theorem, some nA has nonempty interior. It follows that nA − nA = {x − y : x, y ∈ nA} contains a neighborhood of the origin. But nA − nA ⊆ 2nA since A is convex and balanced. It follows that 2nA contains a neighborhood of the origin, so A does too. The following application is called the uniform boundedness principle (or sometimes, in a more general context, the Banach-Steinhaus theorem). Theorem 29.3. Let E be a Banach space and let S be a subset of E ∗ . Suppose that, for every x ∈ E, the set {φ(x) : φ ∈ S} is bounded (in R or C). Then S itself is a bounded set in E ∗ (i.e., it is contained in some ball). One expresses the hypothesis of the theorem by saying that S is pointwise bounded, the conclusion by saying it is uniformly bounded. Proof. Let A = {x ∈ E : |φ(x)| 6 1 ∀φ ∈ S}. Clearly A is closed, convex and balanced. Moreover, the hypothesis of the theorem shows that A is absorbing: if r = sup{|φ(x)| : x ∈ S}, then x ∈ 2rA. By the lemma, A contains a neighborhood of the origin, say B(0; ). Then for all φ ∈ S, kxk < =⇒ kφ(x)k 6 1 93

and so S is contained in the closed ball in E ∗ of radius 1/. Corollary 29.4. Let E be a Banach space and let φn be a sequence in E ∗ . Suppose that, for every x ∈ E, the sequence φn (x) is convergent, say to the limit φ(x). Then φ ∈ E ∗ also. The point of this theorem is that it is not obvious that the limit functional φ is continuous. Proof. Let S be the set {φn }. Because every convergent sequence of real or complex numbers is bounded, S satisfies the hypothesis of the uniform boundedness theorem. Thus, S is a bounded set in E ∗ . That is to say, there is a constant C such that |φn (x)| 6 Ckxk for all n, x. It follows that |φ(x)| = lim |φn (x)| 6 Ckxk n→∞

so φ is bounded. Linearity of φ follows easily from that of φn . Example 29.5. Towards the end of last semester we proved that there exists a continuous function whose Fourier series diverges at a point. Our proof in fact made use of the uniform boundedness theorem. Let C(T) be the Banach space of continuous functions on the circle T and let φn : C(T) → C be the functional that takes the sum of the first n terms of the Fourier series of a given function f and then evaluates at a fixed point of T. We showed, by calculations with the Dirichlet kernel, that kφn k → ∞ — that is, the family {φn } is not uniformly bounded. By the uniform boundedness theorem it can’t be point bounded either: there must be f ∈ C(T) such that φn (f ) is unbounded, which certainly implies that the Fourier series of f diverges at the given point. Our last example involves bilinear maps. Let X, Y, Z be topological spaces. A map B : X × Y → Z is separately continuous if, for each fixed x, the map y 7→ B(x, y) is continuous (from Y to Z) and for each fixed y, the map x 7→ B(x, y) is continuous (from X to Z). In this context, we say that B is jointly continuous if it is continuous from X × Y (with the product topology) to Z. Exercise 29.1. Give an example of a separately continuous map that is not jointly continuous. Theorem 29.6. Let E and F be Banach spaces. Every separately continuous bilinear map E × F → C is jointly continuous.

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Actually one only needs the completeness of E (or F ); the other space need not be complete. Proof. Let B be a separately continuous bilinear map. To show that B is jointly continuous it is enough to prove that if {xn }, {yn } are sequences tending to x and y in E and F respectively, then B(xn , yn ) → B(x, y). Let βn ∈ E ∗ be the linear functional x 7→ B(x, yn ). Because yn → y and B is separately continuous, βn (x) → B(x, y) as n → ∞. In particular, {βn } is a pointwise bounded subset of E ∗ . By the uniform boundedness theorem, there is a constant C such that kβn k 6 C for all n. In particular then |B(xn , yn ) − B(x, y)| 6 |B(xn − x, yn )| + |B(x, yn − y)| 6 Ckxn − xk + |B(x, yn − y)| and both terms tend to 0 as n → ∞. This completes the proof.

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Lecture 30 The open mapping and closed graph theorems These are important applications of the Baire category theorem to functional analysis. Typically, they allow one to deduce the continuity of a map from some apparently much weaker assumptions. Definition 30.1. Let X and Y be topological spaces. A map φ : X → Y is an open map if φ(U ) is open in Y whenever U is open in X. For example, a constant map is (usually!) not open. We proved in Proposition 10.6 that a nonconstant holomorphic function is open. We have already made use, in our discussion of the geometric Hahn-Banach theorem, of the fact that a nonconstant linear functional on a normed space is an open map. The map x 7→ x3 − x from R to R is surjective and continuous, but it is not open. We are going to show that the last phenomenon cannot occur for continuous linear maps between Banach spaces. This is the content of the open mapping theorem: Theorem 30.2. Let E, F be Banach spaces and T : E → F a bounded linear map. Suppose that T is surjective. Then T is open. Both E and F must be complete for the theorem to hold. Proof. Let V be an open ball in E whose center is the origin. Consider the subset B = T (V ) of F . It is closed (by construction), convex, and balanced. Because T is surjective, B is also absorbing: every point of Y has a preimage which is in some multiple of V , so the point itself must be in some multiple of T (V ). By Lemma 29.2, then, T (V ) contains a neighborhood of 0. Let U be an open ball around the origin in E, of radius r. It suffices to show that T (U ) contains a neighborhood of the origin in F . (We know that the closure of T (U ) contains such a neighborhood.) Let U = U0 and let U1 , U2 , . . . be the sequence of balls of center the origin and radius r2−n . For each of these balls we know that T (Un ) contains a neighborhood of the origin. We are going to show that T (U1 ) ⊆ T (U0 ), which will complete the proof. Pick a point y = y1 ∈ T (U1 ). There exist points x ∈ U1 such that kT x − yk is as small as we please; choose x = x1 so that y2 = y1 − T x1 belongs to the neighborhood T (U2 ) of the identity. Then choose x2 ∈ U2 such that y3 = T x2 − y2 belongs to the neighborhood T (U3 ), and continue in this way by induction. We obtain sequences {xn } in E and {yn } in F such that xn ∈ Un , yn = yn−1 − T xn−1 ∈ T (Un ). 96

−n In particular P the norms kxn k and kyn k decrease like some multiple of 2 . Thus x = xn converges in the Banach space E, and we have X X Tx = T xn = (yn−1 − yn ) = y1 = y.

Moreover kxk <

P∞

−n r n=1 2

= r. Thus x ∈ U and the theorem is proved.

Corollary 30.3. A bounded linear bijection between Banach spaces is a homeomorphism. Proof. A bijective continuous map is a homeomorphism precisely when it is open. Corollary 30.4. Let T : E → F be a bounded linear map between Banach spaces. If T is injective and has closed range, then there is a constant C > 0 such that kT xk > Ckxk for all x ∈ E. Proof. Let R denote the range of T . It is a closed subspace of F , hence a Banach space in its own right. Considered as a map from E to R, T is a bounded linear bijection, hence a homeomorphism. By the previous corollary, it has a bounded inverse (from R to E); the constant C is then kT −1 k−1 . Note that in general the range of a bounded linear map will not be closed. For example, consider the map T : L2 [0, 1] → L2 [0, 1] that sends the function f (x) to xf (x). This is a bounded linear map and its range is dense, but not closed. Exercise 30.1. Prove the above statements. Yet another reformulation is the closed graph theorem. Theorem 30.5. Let T : E → F be a linear map between Banach spaces. Suppose that, whenever {xn } is a sequence in E such that both xn and T xn converge, to say x and y respectively, we have T x = y. Then T is continuous. The property in the theorem is weaker than continuity because the convergence of T xn is part of the hypothesis, not part of the conclusion. Proof. Let E ⊕ F denote the set of pairs (x, y) with x ∈ E, y ∈ F , and with norm k(x, y)k = max{kxk, kyk}. It is easy to check that this is a Banach space. The hypothesis of the theorem says exactly that the graph of T , that is the set G = {(x, T x) : x ∈ E} is a closed subspace of E ⊕ F and therefore a Banach space in its own right. Let πE , πF denote the coordinate projections from G to E and F 97

respectively. They are bounded and linear and πE is a bijection. By the open mapping theorem, πE has an inverse which is also a bounded linear map. But now T = πF ◦ (πE )−1 is bounded as well.

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Lecture 31 Operators on Hilbert Space (Given by Professor Nistor) Recall that a Hilbert space H is a (usually complex) inner product space that is complete in its norm. The inner product is a scalar-valued function on H × H, written h·, ·i, which has the properties that hu, λ1 v1 + λ2 v2 i = λ1 hu, v1 i + λ2 hu, v2 i, that hu, vi = hv, ui, and that hu, ui is a nonnegative real number, zero only when u = 0. These properties are modeled on those of the dot product in Euclidean space. The norm associated to an inner product is defined by kvk2 = hv, vi. The proof that this is indeed a norm depends on the Cauchy-Schwarz inequality |hu, vi| 6 kukkvk. All these matters were discussed in Math 501. Example 31.1. The most important example of a Hilbert space is the space L2 (X, µ) of (equivalence classes of) square summable functions on a σ-finite measure space (X, µ). The inner product is Z hf, gi = f¯(x)g(x)dµ(x). A special case (where X is the integers with counting measure) is the space `2 of square summable sequences. Definition 31.2. An operator on a Hilbert space H is a bounded linear map from H to itself. The norm of an operator is defined in the usual way for linear maps between Banach spaces, namely kT k = sup{kT uk : kuk 6 1}. Equipped with this norm, the space B(H) of operators on H becomes a Banach space. (It is not a Hilbert space!) Definition 31.3. An operator T ∈ B(H) is invertible if there is S ∈ B(H) such that T S = ST = I, the identity operator. In infinite dimensions it is necessary to specify two-sided invertibility: ST = I does not imply T S = I. The definition insists that an inverse be a (bounded) operator. However, it follows from Corollary 30.3 that if a

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bounded operator T is bijective (that is, invertible as a set map), then its inverse is also a bounded operator. As a consequence of the Riesz representation theorem for linear functionals on Hilbert space (every linear functional is given by an inner product), we have Proposition 31.4. For every operator T there is a unique operator T ∗ (the adjoint of T ) such that hT u, vi = hu, T ∗ vi for all u, v ∈ H. One has kT k = kT ∗ k. Moreover, the operation T 7→ T ∗ is antilinear, and T ∗∗ = T . ¯ ∗+µ ‘Antilinear’ means that (λS + µT )∗ = λS ¯T ∗ . The proposition was proved in last semester’s course. Example 31.5. Let (X, µ) be a σ-finite measure space and let H be the Hilbert space L2 (X, µ). Let f ∈ L∞ (X, µ). The multiplication operator Mf on H is defined by Mf g(x) = f (x)g(x). Since f ∈ L∞ there is some constant C such that |f (x)| 6 C for almost all x. The estimate Z Z 2 2 2 2 kMf gk = |f (x)| |g(x)| dµ(x) 6 C |g(x)|2 dµ(x) = kgk2 shows that Mf is bounded, with norm less than or equal to C. It is easy to check that the adjoint of Mf is Mf¯, the operator of multiplication by the complex conjugate of f . Exercise 31.1. Show that in the above situation the norm of Mf is exactly equal to the L∞ norm of f . Example 31.6. Again let (X, µ) be a σ-finite measure space, and let k be an L2 function on X × X (with respect to the product measure). The formula Z T f (x) = k(x, y)f (y)dµ(y) defines an operator on H = with kernel k. To see that T Z 2 kT f k =

L2 (X, µ), called the Hilbert-Schmidt operator is bounded, write Z 2 k(x, y)f (y)dµ(y) dµ(x).

By the Cauchy-Schwarz inequality Z 2 Z Z 2 2 k(x, y)f (y)dµ(y) 6 |f (y)| dµ(y) . |k(x, y)| dµ(y) 100

Thus we get 2

Z Z

kT f k 6

|k(x, y)| dµ(y)dµ(x) kf k2 . 2

The expression in parentheses is finite because k ∈ L2 . Thus T is a bounded operator and its norm is at most the L2 norm of k. In general, one does not have equality (the operator norm of T is usually strictly less than the L2 norm of k). Let us calculate the adjoint of the Hilbert-Schmidt operator T . We have ZZ hT f, gi = k(x, y)f (y)g(x)dµ(y)dµ(x) ZZ ¯ x)g(y)f¯(x)dµ(y)dµ(x) = hf, T ∗ gi = k(y, Thus T ∗ is also a Hilbert-Schmidt operator, with kernel function (x, y) 7→ ¯ x). The interchange of order of integration is justified by Fubini’s thek(y, orem. Example 31.7. The unilateral shift on `2 is the operator V which sends the sequence (a1 , a2 , . . .) to the sequence (0, a1 , a2 , . . .). It is clear that V is bounded — in fact it is an isometry, that is kV ak2 = kak2 . The adjoint V ∗ sends (b1 , b2 , b3 , . . .) to (b2 , b3 , . . .). Note that V ∗ V = I,

V V ∗ 6= I.

This example therefore shows that, on Hilbert space, a one-sided inverse need not be a two-sided inverse. Proposition 31.8. For an operator T , kT ∗ T k = kT k2 . This proposition (called the C ∗ -identity) is the key fact that characterizes operators on a Hilbert space. Proof. Write kT xk2 = hT x, T xi = hT ∗ T x, xi 6 kT ∗ T kkxk2 using Cauchy-Schwarz at the last step. Taking the supremum over x in the unit ball gives kT k2 6 kT ∗ T k. On the other hand kT ∗ T k 6 kT ∗ kkT k = kT k2 so we have equality all through, which is what is needed.

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Lecture 32 More about operators The adjoint operation allows us to describe some special kinds of operators on a Hilbert space. Definition 32.1. An operator T is self-adjoint (or Hermitian) if T = T ∗ . For instance, the multiplication operator Mf is self-adjoint when f is a real -valued function. A Hilbert-Schmidt integral operator is self-adjoint ¯ x). The shift operator V is not selfwhen its kernel satisfies k(x, y) = k(y, adjoint. Let T be self-adjoint. For v ∈ H we have hT v, vi = hv, T vi = hT v, vi using the definition of self-adjointness and the symmetry of the inner product. Thus q(v) = hT v, vi is a real number (when considered as a function of v it is called the quadratic form associated to T ). In the next proof, and elsewhere, we shall use the polarization identities 4 RehT u, vi =q(u + v) − q(u − v) 4 ImhT u, vi =q(u + iv) − q(u − iv) which can be proved by expanding the quadratic forms in terms of inner products. Proposition 32.2. If T is self-adjoint, then kT k = sup{|hT v, vi| : kvk 6 1}.

Proof. By Cauchy-Schwarz, |hT v, vi| 6 kT vkkvk 6 kT kkvk2 . This shows that the supremum in the proposition exists and is less than or equal to kT k. To prove the opposite inequality, let c be the supremum in question, that is c = sup{|q(v)| : kvk 6 1} where q is the quadratic form associated to T . Then by the polarization identity 4 RehT u, vi = q(u + v) − q(u − v) 6 c(ku + vk2 + ku − vk2 ) = 2c(kuk2 + kvk2 ). 102

Suppose for a moment that T u 6= 0, and take v = kuk/kT uk T u in the displayed identity. We get 4kT ukkuk 6 2c · 2kuk2 so kT uk 6 ckuk. Since this identity holds for all u (even those for which T u = 0, where it is trivial) we find that kT k 6 c, as required. Exercise 32.1. Use the closed graph theorem to show that if T is a linear map from a Hilbert space to itself, satisfying hT u, vi = hu, T vi for all u, v ∈ H, then T is an operator (i.e., it is bounded). Definition 32.3. An operator U is unitary if U U ∗ = U ∗ U = I. Recall that the condition U ∗ U = I is equivalent to kU xk2 = kxk2 , that is the property of being an isometry. The example of the unilateral shift shows that an isometry need not be invertible; a unitary is simply an invertible isometry. Example 32.4. A multiplication operator Mf is a unitary if and only if the complex number f (x) has absolute value 1 for almost all x. Remark 32.5. Self-adjoint and unitary operators are both examples of the more general class of normal operators: an operator T is normal if T T ∗ = T ∗T . When doing analysis in infinite dimensional Hilbert spaces one has two notions of ‘smallness’ available: something can be small in norm, or small in the sense of being restricted to a finite dimensional subspace. The tradeoff between these yields the following idea. Definition 32.6. An operator T on H is compact if, for every bounded subset B of H, the image T (B) is precompact. Recall that a subset of a metric space is precompact if, for every , it has a finite cover by -balls. In a complete metric space such as H, a subset is precompact if and only if its closure is compact. We proved this last semester. It is easy to see that T is compact iff T (B) is precompact where B is the unit ball. For every bounded set is contained in some multiple of this B. Proposition 32.7. The compact operators form a closed subset of B(H). Proof. Let Tn be a sequence of compact operators converging to T , and let B be the unit ball. Let > 0 and choose n so large that kTn − T k < /2. The set Tn (B) has a finite cover by /2-balls; but then the -balls with the same centers must cover T (B). Since is arbitrary, T (B) is precompact. 103

Proposition 32.8. The adjoint of a compact operator is compact. Proof. Let T be a compact operator on H. Let B be the unit ball in H and let X be the compact metric space T (B). For every y ∈ B we obtain a continuous function φy : X → C by φy (x) = hx, yi. Moreover we have |φy (x) − φy (x0 )| 6 kx − x0 k for all y ∈ B, which implies that the collection Φ = {φy : y ∈ B} is an equicontinuous subset of C(X). By the Arzela-Ascoli theorem, then, Φ is precompact in C(X). Now consider T ∗ (B). Since kT ∗ yk = suphz, T ∗ yi = suphT z, yi = sup hx, yi = kφy k, z∈B

z∈B

x∈X

the map T ∗ y 7→ φy is a (well-defined) isometry of T ∗ (B) into Φ. Since Φ is precompact, T ∗ (B) is also. Proposition 32.9. The compact operators form an ideal in the ring B(H). Proof. We must show that if T, T 0 are compact so is T + T 0 , and that if T is compact and S is bounded, then ST and T S are compact. It is easy to check that the sum of two precompact sets (in a normed vector space) is again precompact; this proves that T + T 0 is compact. Since S maps bounded sets to bounded sets, it is also clear that T S is compact. That ST is compact follows from this, since ST = (T ∗ S ∗ )∗ and the adjoint of a compact operator is compact.

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Lecture 33 Calculations with compact operators An operator T is of finite rank if its range is a finite-dimensional space. Recall (from last semester) that any finite-dimensional normed space is linearly homeomorphic to Rn or Cn (for some n). These sets have the HeineBorel property that bounded sets are precompact. If T is a finite rank operator, then, and B is the unit ball, T (B) is a bounded set in a finitedimensional normed vector space, hence is precompact. Thus Proposition 33.1. Every finite rank operator is compact. From Proposition 32.7 it follows that if Tn → T is a convergent sequence of operators, and each Tn has finite rank, then T is compact. This is the most common way to check that a particular operator is compact. Example 33.2. Let {an } ∈ `∞ be a bounded sequence of integers. It defines a multiplication operator M on `2 which sends the sequence {cn } to the sequence {an cn }. The norm of this multiplication operator is (of course) equal to sup |an |. When is M compact? If and only if the sequence an tends to zero. To see this, notice that if the sequence {an } tends to zero, then the multiplication operators Mk by the sequence a1 , a2 , . . . , ak , 0, 0, . . . are of finite rank and Mk → M in norm. Thus M is the limit of a sequence of finite rank operators, hence compact. In the other direction, if the sequence {an } does not tend to 0 then for some > 0 there are infinitely many n’s, say n1 , n2 , . . ., such that |ani | > . It follows that M (B), the image under M of the unit ball, contains an infinite set of mutually orthogonal vectors of length /2. No /2-ball can contain more than one of these vectors, so M (B) cannot be precompact. Therefore, M is not a compact operator. Exercise 33.1. Show that an operator T is compact if and only if it transforms weakly convergent sequences to norm convergent ones. Use this to get another proof that if the multiplication operator M in the previous example is compact, then the sequence {an } tends to zero. Example 33.3. Let (X, µ) be a σ-finite measure space, and consider a Hilbert-Schmidt integral operator on H = L2 (X, µ) (Example 31.6) of the form Z T f (x) = k(x, y)f (y)dµ(y), where k belongs to L2 (X × X, µ × µ). This operator is always compact. To see this, first consider the special case where k is the characteristic function 105

of a finite measurable rectangle (a set of the form A × B, where A, B are finite-measure subsets of X). In this case T f = χA hχB , f i and so T has finite rank—its range is the one-dimensional subspace spanned by χA . Therefore T has finite rank whenever k belongs to the space L of (finite) linear combination of characteristic functions of finite measurable rectangles. Such linear combinations are dense in L2 (X × X), so for any k we can find a sequence kn ∈ L with kn → k in L2 norm. Let Tn be the Hilbert-Schmidt operators with kernels kn . In Example 31.6 we showed that the operator norm of a Hilbert-Schmidt operator is dominated by the L2 norm of its kernel. Thus Tn → T in norm, and each Tn has finite rank. It follows that T is compact. We shall be particularly interested in compact operators that are selfadjoint. Proposition 33.4. Let T be a compact, self-adjoint operator on a Hilbert space H. Then either +kT k or −kT k (or both) is an eigenvalue for T . Unless T is the zero operator, the corresponding eigenspace is finite-dimensional. The terms eigenvalue, eigenvector, eigenspace are defined just as in ordinary linear algebra: if the equation T u = λu has nonzero solutions u ∈ H, then λ is called an eigenvalue for T , u is an eigenvector, and the space of all solutions is the eigenspace corresponding to λ. Proof. Assume T 6= 0. By Proposition 32.2, there exists a sequence {xn } of unit vectors in H such that hT xn , xn i converges to λ = ±kT k. Since T is a compact operator, the sequence {T xn } has a convergent subsequence. We may therefore assume (by passing to a subsequence if necessary) that the sequence {T xn } itself converges. Consider now the expansion kT xn − λxn k2 = kT xn k2 − 2λhT xn , xn i + λ2 kxn k2 . The first term is bounded above by kT k2 = λ2 , the second term tends to −2λ2 by hypothesis, and the third term equals λ2 . Thus lim sup kT xn − λxn k2 6 λ2 − 2λ2 + λ2 = 0. n→∞

It follows that T xn − λxn → 0. Since λ 6= 0 and T xn converges, this implies that xn converges, say to x, and that T x − λx = 0. For the final statement (the finite-dimensionality of the eigenspace) notice that the identity operator on the eigenspace is compact (because it is 106

equal to λ−1 T ) and thus the unit ball of the eigenspace is compact. But a Hilbert space (or a normed vector space) whose unit ball is compact must be finite-dimensional.

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Lecture 34 The spectral theorem for compact operators Lemma 34.1. Let T be a self-adjoint operator on a Hilbert space H, and let E 6 H be an eigenspace for T , that is the set {y : T y = λy} for some λ. Then the orthogonal complement E ⊥ is also T -invariant. Proof. Let x ∈ E ⊥ . Then, for all y ∈ E, hT x, yi = hx, T yi = λhx, yi = 0 because x ∈ E ⊥ and T y = λy. Lemma 34.2. The eigenspaces belonging to distinct eigenvalues of a selfadjoint operator are orthogonal. Proof. Let x1 , x2 be eigenvectors of T , having distinct eigenvalues λ1 , λ2 . Then λ1 hx1 , x2 i = hT x1 , x2 i = hx1 , T x2 i = λ2 hx1 , x2 i and so hx1 , x2 i = 0. Exercise 34.1. Extend the two lemmas to normal operators T (those for which T commutes with T ∗ ). Theorem 34.3. Let T be a compact, self-adjoint operator on a (separable) Hilbert space H. Then there is an orthonormal basis {en } of H made up of eigenvectors for T . The corresponding eigenvalues λn form a sequence of real numbers converging to zero. In other words, every compact self-adjoint operator can be reduced to the form of Example 33.2 by a suitable choice of basis. Proof. Let E1 be the eigenspace for T corresponding to eigenvalue +kT k, and let E−1 be the eigenspace corresponding to the eigenvalue −kT k. These spaces are finite-dimensional and at least one of them is nonzero, by Proposition 33.4 from last time. Let H1 be the orthogonal complement (E1 ⊕ E−1 )⊥ . Then T maps H1 to itself (by lemma 34.1); let T1 be the restriction of T to H1 . It is a compact self-adjoint operator, of norm 6 kT k. In fact, its norm is strictly less than kT k, because by construction neither ±kT k is an eigenvalue of T1 . Repeat the construction with T1 , H1 in place of T, H and continue inductively. We obtain in this way a sequence of orthogonal finite-dimensional eigenspaces E1 , E−1 , E2 , E−2 , corresponding to eigenvalues λ1 , −λ1 , λ2 , −λ2 , . . . 108

where λk is a decreasing sequence of positive real numbers. The process may stop at some finite stage k = n, if the restriction of T to Hn = (E1 ⊕ · · · ⊕ E−n )⊥ is the zero operator. Otherwise, the process continues indefinitely. If the process continues indefinitely, the eigenvalues λk must converge to zero. If not, we could find an infinite sequence ej of mutually orthogonal unit eigenvectors whose eigenvalues all had absolute value greater than some > 0. The T ej would then also be orthogonal and have length greater than , so they would have no convergent subsequence, contradicting the compactness of T . (This is the same argument as in Example 33.2.) StillTconsidering the case where the process continues indefinitely, let H∞ = Hn be the orthogonal complement of the sum of all the eigenspaces En and E−n . Then T restricts to a map T∞ from H∞ to itself. I claim T∞ is actually zero. Indeed, by construction, any eigenvalue of T on Hn must have absolute value less than λn . Since λn → 0, any eigenvalue of T on H∞ must be zero. By Proposition 33.4, the norm kT∞ k must be zero, so T∞ = 0. Now we have shown that H is the orthogonal direct sum of a (finite or infinite) sequence of finite-dimensional eigenspaces for T , together with an orthogonal complement on which T is zero. Putting together orthonormal bases for the eigenspaces and the complement, we get the result. Remark 34.4. The same argument applies to a non-separable Hilbert space, except that the orthogonal complement H∞ will now be non-separable and thus won’t have a complete orthonormal sequence. Thus a compact operator on a non-separable Hilbert space is always the sum of a compact operator on a separable subspace and the zero operator on the complement. Informally, the spectral theorem says that every compact, self-adjoint operator can be ‘diagonalized’ by a suitable choice of orthonormal basis. Proposition 34.5. Every compact operator on a Hilbert space is a limit of finite rank operators. Proof. For a compact self-adjoint operator T this is an immediate consequence of the spectral theorem. If {en } is an orthonormal basis of eigenvectors for T with corresponding eigenvalues λn tending to 0, let Pk be the orthogonal projection on the span of {e1 , . . . , ek }. Then Tk = T Pk = Pk T has finite rank and kTk − T k = sup{|λn | : n > k} → 0 as k → ∞. Now let T be a general compact operator. Then R = (T + T ∗ ) and S = i(T − T ∗ ) are compact and self-adjoint. Therefore they are limits of

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finite rank operators, and thus T = (R − iS)/2 is a limit of finite rank operators too. Example 34.6. As an important special case we are going to consider the so-called Sturm-Liouville theory of ordinary differential equations. Consider the operator D defined by d du Du = − p(x) + q(x)u dx dx acting on functions on the interval [a, b]. Here p and q are given, smooth, real-valued functions. Our objective is to sketch the proof that L2 [a, b] possesses a complete orthonormal system made up of eigenfunctions for D, that is solutions of Du = λu, which in addition satisfy the boundary conditions u(a) = u(b) = 0. Such eigenfunction decompositions are important in many areas of applied mathematics. The proof begins with the construction of the Green’s function G(x, y) on [a, b] × [a, b]. This is a continuous real-valued symmetric (G(x, y) = G(y, x)) function with the characteristic property that for any continuous v on [a, b], the function Z b

u : x 7→

G(x, y)v(y)dy a

is twice differentiable, satisfies the boundary conditions and Du = v. The construction of this function is a standard topic in differential equations. Now let T be the Hilbert-Schmidt integral operator with kernel G. By Examples 31.6 and 33.3, T is compact and self-adjoint. Thus, the spectral theorem tells us that there is an orthonormal basis for L2 [a, b] made up of eigenvectors for T . Consider such an eigenvector, f , with a nonzero eigenvalue λ. A priori, f is merely an L2 function. However, since G is continuous, the dominated convergence theorem shows that the range of T consists of continuous functions. Thus f = λ−1 T f is actually continuous. Now by the characteristic property of the Green’s function, f is twice differentiable, satisfies the boundary conditions and D(λf ) = f , which is to say that f is an eigenfunction of D with eigenvalue λ−1 . An additional argument shows that T has no zero eigenvalues and thus completes the proof.

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Lecture 35 General spectral theory Definition 35.1. The spectrum of a bounded operator T on a Banach space is the set of complex numbers λ for which T − λI fails to have a (bounded, two-sided) inverse. The spectrum of T is denoted σ(T ). Example 35.2. The number 0 always belongs to the spectrum of a compact operator on an infinite-dimensional Hilbert space. For if a compact operator T were invertible, then I = T T −1 would be compact and therefore the closed unit ball would be compact, which implies that our space is finitedimensional. Remark 35.3. Every eigenvalue is of course in the spectrum of T , since if T −λI has non-trivial kernel then it cannot be invertible. The example above shows that the spectrum need not consist entirely of eigenvalues. However, by the spectral theorem for compact operators, every non-zero point in the spectrum of a compact self-adjoint operator is an eigenvalue. For suppose that T has an orthonormal basis {en } of eigenvectors, with corresponding eigenvalue sequence {λn }. Since λn → 0, if λ is not an eigenvalue, then there P is some P > 0 such that |λ − λn | > for all n. The map M that sends αn en to (λn − λ)−1 αn en is therefore well-defined, linear and bounded (by −1 ) and it is a two-sided inverse for T − λI. Thus λ ∈ / σ(T ). For general operators T the spectrum need not contain any eigenvalues at all. Example 35.4. Let H = L2 [0, 1] and consider the multiplication operator Mf , where f is a continuous function on [0, 1]. Then Mf − λI is the multiplication operator by the function f − λ, and this will have a (bounded) inverse if and only if λ is not in the range of the function f . Thus in this case the spectrum of Mf is the range of f . However, the number λ will be an eigenvalue of Mf if and only if the set {x ∈ [0, 1] : f (x) = λ} has positive Lebesgue measure. Thus ‘in general’ (for a specific example, take f (x) = x) the operator Mf will have no eigenvalues. Lemma 35.5. The set of invertible operators on a Hilbert space H is an open subset of B(H). (The proof, as well as most of the rest of this lecture, works just the same for a Banach space but we’ll stick to the Hilbert space case.)

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Proof. Begin by noting that if kT k < 1 then the series S=

∞ X

Tn

n=0

converges in the Banach space B(H) to an operator S that satisfies ST = T S =

∞ X

T n = S − I.

n=1

Thus S is a two-sided inverse of I − T . We conclude that if X ∈ B(H) and kX − 1k < 1 then X is invertible. Now let Y ∈ B(H) be invertible and suppose that kX − Y k < kY −1 k−1 . Then kXY −1 − Ik < 1 so XY −1 is invertible and thus X is invertible. Proposition 35.6. The spectrum of an operator T is a closed subset of C, and it is contained in the closed disk D(0; kT k). Proof. The spectrum of T is the inverse image of the closed set of noninvertible operators under the continuous map λ 7→ T −λI. Thus it is closed. If λ > kT k then T − λI = −λ(I − λ−1 T ) is invertible since kλ−1 T k < 1. Proposition 35.7. The spectrum of an operator is always non-empty. Proof. Let T be an operator and let Ω = C \ σ(T ) (this is sometimes called the resolvent set of T ). Consider the continuous function f : z 7→ (T − zI)−1 from Ω to B(H). The proof of Lemma 35.5 shows that this function is analytic in the following sense: around each point z0 ∈ Ω there is a disk of some radius r0 , such that for z ∈ D(z0 ; r0 ) the function f is represented by an absolutely convergent power series f (z) =

∞ X

An (z − z0 )n

n=0

with coefficients An ∈ B(H). Moreover, when |z| > 2kT k we have f (z) = −z −1 (I − z −1 T )−1 and therefore kf (z)k 6 12 kT k−1 · 2 = kT k−1 . We conclude that if Ω = C, then f (z) is a bounded, analytic, B(H)valued function on C. For every φ ∈ B(H)∗ , then, φ◦f is a bounded analytic function C → C: that is a bounded entire function, which (by Liouville’s theorem) must be constant. Since the linear functionals φ separate points on B(H) by the Hahn-Banach theorem, f itself must be constant. But this contradicts the definition of f . 112

How are the spectra of T and T n , or more generally T and some polynomial p(T ), related? The answer is given by the spectral mapping theorem: Proposition 35.8. Let T be an operator and p a polynomial (with complex coefficients). Then σ(p(T )) = p(σ(T )). Proof. We will consider the special case p(λ) = λn . The factorization T n − λn I =

n−1 Y

T − λe2kπi/n I

k=0

shows that T n −λn I is invertible if and only if all the (commuting) operators T − λe2kπi/n I are invertible. Thus µ ∈ σ(T n ) if and only if there is some λ ∈ σ(T ) with λn = µ. Exercise 35.1. Prove the spectral mapping theorem for general polynomials p. (The method is the same.)

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Lecture 36 The spectral radius Example 36.1. Let H = L2 [0, 1] and consider the Volterra integral operator Z x f (t)dt. V f (x) = 0

It is easy to see that V defines a bounded operator on H. Indeed, by Cauchy-Schwarz, |V f (x)| 6 kf k for all x and thus kV f k 6 kf k, so kV k 6 1. Define f0 (x) = f (x) and for n > 1, Z x 1 fn (x) = (x − t)n−1 f (t)dt. (n − 1)! 0 Then f1 (x) = V f (x) and dfn (x) 1 = dx (n − 2)!

Z

x

(x − t)n−2 f (t)dt = fn−1 (x)

0

by the fundamental theorem of calculus. We conclude that fn (x) = V n f (x). The same estimate as above now shows that kV n k 6 1/(n − 1)!. The spectrum of V n is therefore contained in the disk of radius 1/(n−1)!. It follows from the spectral mapping theorem that the spectrum of V is contained in the disk of radius (1/(n−1)!)1/n . But, as n → ∞, (1/(n−1)!)1/n tends to zero8 and thus the spectrum of V contains only the single point 0. An operator of this sort is called quasinilpotent. This example suggests the following definition. Definition 36.2. Let T be an operator. The spectral radius of T is ρ(T ) := sup{|λ| : λ ∈ σ(T )}. In other words, ρ(T ) is the radius of the smallest disk (centered at the origin) that contains the spectrum of T . Proposition 35.6 gives us the inequality ρ(T ) 6 kT k. Theorem 36.3. For any operator T , ρ(T ) = lim kT n k1/n . n→∞

The result of this theorem is called the spectral radius formula. It is part of the assertion of the theorem that the limit exists. 8

This follows from Stirling’s formula 12.6, though it is easy to give a direct argument also.

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Proof. By the spectral mapping theorem ρ(T )n = ρ(T n ) 6 kT n k. Thus ρ(T ) 6 kT n k1/n for all n and so ρ(T ) 6 lim inf kT n k1/n . (This is the argument of Example 36.1.) To get an inequality in the opposite direction we use techniques based on holomorphic functions (compare the proof of Proposition 35.7). Let φ ∈ B(H)∗ be a continuous linear functional of norm 1 and consider the function gφ : z 7→ φ((I − zT )−1 ) which we know is locally representable by power series, and hence holomorphic, inside the disk D(0; 1/ρ(T )). It has a power series expansion gφ (z) =

∞ X

φ(T n )z n .

n=0

For any r < 1/ρ, let M = sup{k(I − zT )−1 k : |z| = r}, which is finite. Since φ has norm 1, |gφ (z)| 6 M for |z| = r. The Cauchy estimates give us a bound for the Taylor coefficients of gφ , namely |φ(T n )| 6 M r−n . This holds for all φ, and hence kT n k 6 M r−n by the Hahn-Banach theorem. Consequently lim sup kT n k1/n 6 1/r. But r < 1/ρ is arbitrary, so lim sup kT n k1/n 6 ρ. This completes the proof. Proposition 36.4. The norm of a self-adjoint operator is exactly equal to its spectral radius. Proof. Recall the C ∗ -identity kT ∗ T k = kT k2 . When T is self-adjoint this n n takes the form kT 2 k = kT k2 , and by induction kT 2 k = kT k2 . The result now follows from the spectral radius formula. Proposition 36.5. The spectrum of a self-adjoint operator is a subset of R.

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Proof. Let T be self-adjoint. For each y ∈ R let Sy = T + iyI. Then Sy∗ Sy = (T − iyI)(T + iyI) = T 2 + y 2 I, so kSy k2 6 kT k2 + y 2 by the triangle inequality and the C ∗ -identity. Thus the spectrum of Sy is contained within the disk D(0; (kT k2 + y 2 )1/2 ), and so the spectrum of T is contained within the disk D(−iy; (kT k2 + y 2 )1/2 ). But the intersection of all these disks is the interval [−kT k, kT k] on the real axis. Let T be a self-adjoint operator on H. Denote by A = A(T ) the closure of the set of polynomials p(T ) with complex coefficients. A is a unital, commutative ring. Example 36.6. Suppose T is compact and self-adjoint. For each unit eigenvector v of T with eigenvalue λ we have hv, p(T )vi = p(λ). We see therefore that for every eigenvalue λ we can recover the complex number p(λ) from the operator p(T ), and moreover that |p(λ)| 6 kp(T )k. In fact, the spectral theorem tells us that kp(T )k = sup{|p(λ)|}, where the supremum is taken over all the eigenvalues of T . Let σ(T ) be the spectrum of T (the eigenvalues, together possibly with zero). We have shown that the assignment p(T ) 7→ p|σ(T ) is a well-defined, isometric homomorphism from the ring of polynomials in T to the ring C(σ(T )) of continuous functions on σ(T ). It therefore extends by continuity to an isometric homomorphism of rings from A to C(σ(T )). The range of this homomorphism is closed (because it’s isometric) and dense (by the Stone-Weierstrass theorem). So it is the whole of C(σ(T )). Our conclusion is that A(T ) is isometrically isomorphic to C(σ(T )). The statement that A(T ) is isomorphic to C(σ(T )) may be regarded as a form of the spectral theorem, because it represents T as a ‘multiplication’ (namely, an element of an algebra of functions). In this form it makes no reference to compactness, eigenvalues, or eigenvectors; and in fact this is a form of the spectral theorem that is valid for all self-adjoint operators, compact or not. Theorem 36.7 (Simplified Gelfand-Naimark). Let T be any self-adjoint operator. Then the algebra A(T ) defined above is isometrically isomorphic to C(σ(T )), by an isomorphism which takes the ∗-operation on A to complex conjugation on C(σ(T )). The isomorphism takes T itself to multiplication by the identity function f (x) = x. 116

We don’t have time to prove this (though it is not very difficult), but let’s see how it can be used as a version of the spectral theorem. Example 36.8. Let T be any self-adjoint operator such that hT v, vi > 0 for all v. We’ll show that T = S 2 where S is another self-adjoint operator. The first step is to show that the spectrum σ(T ) consists od positive real numbers. Assuming that this is done, T corresponds under the GelfandNaimark isomorphism to the positive real-valued function x on σ(T ). The function x1/2 on σ(T ) is then real-valued and continuous, so it corresponds to a self-adjoint operator S ∈ A(T ) whose square is T . To check that the condition indeed implies that T has positive spectrum, suppose that kT k 6 1. By Proposition 32.2, kI − T k = sup{|h(I − T )v, vi| : kvk = 1}. But 0 6 hT v, vi 6 1 for unit vectors v, so 1 > h(I − T )v, vi > 0. It follows that kI − T k 6 1, so σ(T ) ⊆ [0, 2] and thus is positive.

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Lecture 37 Harmonic functions The final topic of the course is an item of real analysis which is closely related to complex (and functional) analysis. Definition 37.1. Let Ω be an open subset of Rn . A smooth function h : Ω → R is harmonic if it satisfies the equation ∂2h ∂2h + · · · + =0 ∂x2n ∂x21 called Laplace’s equation. We shall consider the case n = 2 and we shall identify R2 with C. We have already seen that the real part of a holomorphic function on Ω must be harmonic (this is a consequence of the Cauchy-Riemann equations, see Remark 2.4). Conversely we have Proposition 37.2. Any harmonic function on a simply connected Ω ⊆ C is the real part of a holomorphic function. Proof. Suppose that h is harmonic on Ω and define g = u + iv = hx − ihy (we use subscripts to denote partial derivatives). Then ux − vy = hxx + hyy = 0,

uy + vx = hxy − hyx = 0

so g satisfies the Cauchy-Riemann equations. Since g is smooth, it is holomorphic. Because Ω is simply connected, the holomorphic function g has an antiderivative on Ω, a holomorphic f such that f 0 = g. Then (Re f )x = Re g = hx ,

(Re f )y = −(Im f )x = − Im g = hy

and it follows that h − Re f is a constant c. Thus h is the real part of the holomorphic function f + c. Remark 37.3. By definition, an antiholomorphic function of z is a holomorphic function of z¯. Since the Laplace equation is invariant under complex conjugation (x 7→ x, y 7→ −y), the real part of an antiholomorphic function is also harmonic. We’ll use this fact in a moment. Exercise 37.1. Show that the assumption of simple connectedness is necessary in the above theorem. (Consider the function g(x + iy) = log(x2 + y 2 ) on C \ {0}. )

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Corollary 37.4 (Maximum principle for harmonic functions). Let h be a non-constant harmonic function on the connected open set Ω. Then h cannot attain a local maximum (or minimum) at any point of Ω. Proof. If h has a local maximum then eh does so too. Fix a disk in Ω containing the supposed local maximum. On this disk we can write h = Re f with f holomorphic. But now eh = eRe f = |ef |, so the result follows from the maximum principle for holomorphic functions (Theorem 7.4) applied to the function ef . It follows that “a harmonic function is determined by its boundary values”. More precisely, suppose that Ω is an open set with compact closure Ω, and let ∂Ω = Ω \ Ω. Then for each continuous function g on ∂Ω there is at most one harmonic function h on Ω which extends g continuously. If there were two such functions, we could take their difference to get a harmonic function h that is zero on ∂Ω. By the maximum principle, h must attain its maximum and minimum values on the boundary ∂Ω; so it must be identically zero. This is a uniqueness statement. The corresponding existence statement is called the Dirichlet Problem: given a continuous function g on ∂Ω, find a harmonic function h having g as its boundary values. We will begin our investigation of the Dirichlet problem by looking at the simplest case — the unit disk. Proposition 37.5. The Dirichlet problem for the disk has a (unique) solution for any continuous boundary data g. In fact, the solution is given explicitly by the Poisson integral formula Z 2π 1 1 − r2 iθ h(re ) = g(eiφ )dφ. 2π 0 1 − 2r cos(θ − φ) + r2 Proof. First we want to show that the expression 1 − r2 , 1 − 2r cos(θ − φ) + r2 called the Poisson kernel, is a harmonic function of w = reiθ . Put z = eiφ ,

119

w = reiθ , and consider 1 1 − z − w z − 1/w ¯ =

w − 1/w ¯ 1 − |w|2 = (z − w)(z − 1/w) ¯ z(1 − w¯ z )(1 − wz) ¯ 1 1 − r2 = · z 1 − 2r cos(θ − φ) + r2

This shows that the Poisson kernel is (the real part of) the sum of a holomorphic and an antiholomorphic function of w, so it is harmonic. By differentiation under the integral sign, it follows that the function h defined by the Poisson integral formula is harmonic. It remains to show that h has g as its boundary values, which is to say that h(reiθ ) approaches g(eiθ ), uniformly in θ, as r → 1. It is convenient to allow ourselves to consider complex-valued g and h. Let us check it first for the functions g(z) = z n . The calculations above show that we can rewrite the Poisson integral formula as I I 1 g(z)dz 1 g(z)dz h(w) = − . 2πi z−w 2πi z − 1/w ¯ If g(z) = z n for n = 0, 1, 2 . . . then the first integral is wn and the second is 0, by the Cauchy integral formula and the Cauchy theorem. Thus h(w) = wn and h has g as its boundary values. Because the Poisson kernel is real-valued, the same statement applies when g(z) = z −n = z¯n when h(w) = w ¯n. Let g be a trigonometric polynomial (a finite sum of positive and negative powers of z), and denote by P g the function h produced by the Poisson integral formula. From the above it follows that P g is continuous on the closed disk, harmonic on the open disk, and restricts to g on the unit circle. By the maximum principle, kP gk 6 kgk where the norms are supremum norms. Now for a general continuous function g on the circle, the Stone-Weierstrass theorem produces a sequence gn of trigonometric polynomials converging uniformly to g. Then P gn is a Cauchy sequence of continuous functions on the closed disk, hence converges uniformly to a continuous function h. On the other hand, P gn converges pointwise on the open disk to P g, by continuity properties of integration. Hence P g = h is continuous on the closed disk and has boundary values g.

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Lecture 38 The mean value property In the previous lecture we have investigated the Poisson integral formula, which produces a harmonic function h on the unit disk having given continuous boundary values. A simple rescaling produces a version of the formula valid for any disk. On a disk center a radius R, this takes the form Z 2π 1 R2 − r 2 h(a + reiθ ) = g(a + Reiφ )dφ 2π 0 R2 − 2Rr cos(θ − φ) + r2 with r < R. If h is harmonic on an open set Ω and D(a; R) ⊆ Ω, then the Poisson integral formula can be applied to h to produce a harmonic function on D(a; R) having the same boundary values as h. By the maximum principle, this harmonic function actually is h. Thus we get the formula Z 2π 1 R2 − r 2 iθ h(a + re ) = h(a + Reiφ )dφ 2π 0 R2 − 2Rr cos(θ − φ) + r2 which is a harmonic counterpart of the Cauchy integral formula. In particular a harmonic function has the mean value property Z 2π 1 h(a + Reiφ )dφ. h(a) = 2π 0 Definition 38.1. A continuous function f on Ω is subharmonic if it has the sub mean value property: for every a ∈ Ω we have Z 2π 1 f (a) 6 f (a + Reiφ )dφ 2π 0 for all sufficiently small R > 0. Thus, harmonic functions are subharmonic. Subharmonic functions also satisfy the maximum principle: Proposition 38.2. Let Ω ⊆ C be open and connected and have compact closure. Let f be nonconstant, continuous on Ω and subharmonic on Ω. Then the maximum value of f is attained at a point of ∂Ω = Ω \ Ω. Proof. Suppose that f attains its maximum at a ∈ Ω. Then for all sufficiently small R, Z 2π Z 2π 1 1 iφ f (a + Re )dφ 6 f (a)dφ = f (a) f (a) 6 2π 0 2π 0 121

and equality must hold everywhere, so f (a + Reiφ ) = f (a) for all φ and f is constant on a disk around a. It follows that the set of points a ∈ Ω where f attains its maximum value is open. Clearly it is also closed, so it is either Ω or ∅ because Ω is connected. In the first case, f would be constant. Proposition 38.3. A continuous function on an open subset of C that has the mean value property is harmonic. Proof. Let f be continuous on Ω and have the MVP. Let D be any disk whose closure is contained in Ω. Let h be the harmonic function on D (provided by the Poisson integral formula) that agrees with f on ∂D. Then f − h has the MVP and is zero on ∂D. By the maximum principle, f − h is zero on D, so f is harmonic on D. Since D is arbitrary, f is harmonic. Lemma 38.4. Lat h be a nonnegative harmonic function defined on a domain containing the closed disk D(a; R). Then for r < R, R−r R+r h(a) 6 h(a + reiθ ) 6 h(a). R+r R−r Proof. The Poisson integral formula is Z 2π R2 − r 2 1 iθ h(a + re ) = h(a + Reiφ )dφ. 2π 0 R2 − 2Rr cos(θ − φ) + r2 Observe that (R − r)2 6 R2 − 2Rr cos(θ − φ) + r2 6 (R + r)2 and thus R−r R2 − r 2 R+r 6 2 6 . 2 R+r R − 2Rr cos(θ − φ) + r R−r If C denotes the L1 norm of the function φ 7→ (2π)−1 h(a + Reiφ ), simple estimates give R−r R+r C 6 h(a + reiθ ) 6 C. R+r R−r But since h is nonnegative, the mean value property tells us that C is exactly h(a). Corollary 38.5. A harmonic function on the plane that is bounded below is constant. Proof. We may assume the function is nonnegative. Apply the lemma and let R → ∞. 122

This corollary is a version of Liouville’s theorem for harmonic functions. Proposition 38.6 (Harnack’s Principle). Let {hn } be a monotone increasing sequence of harmonic functions on a connected open set Ω ⊆ C. Then either hn converges uniformly on compact subsets of Ω to a harmonic function, or hn → ∞ everywhere. Proof. Replacing {hn } by {hn − h1 } we may assume that all the hn are nonnegative. Let a ∈ Ω and suppose that D(a; R) ⊆ Ω. If hn (a) converges to a finite limit, then by the right-hand inequality in lemma 38.4, (hm − hn )(a + reiθ ) 6 3(hm − hn )(a) for any m > n and any r 6 R/2. Thus hm converges uniformly on the disk of radius R/2. If hn (a) increases to infinity, a similar argument using the left-hand inequality in lemma 38.4 shows that hm increases to infinity on the disk of radius R/2. These observations show that the set of points a where hn (a) converges to a finite limit and the set of points a where hn (a) increases to infinity are open (and disjoint). By connectedness, one of them is empty. Finally, in the case that the hn converge to a finite limit h, we have shown that each a ∈ Ω has a neighborhood on which the converge is uniform. Therefore the convergence is uniform on compact sets. The limit function h is therefore continuous and satisfies the mean value property. Thus, it is harmonic.

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Lecture 39 The Perron construction Let Ω be a connected open subset of C with compact closure. Recall that the Dirichlet problem is this: given a continuous function g on ∂Ω, find a function h that is continuous on Ω, harmonic on Ω, and extends g. Example 39.1. Let Ω be the punctured disk D(0; 1) \ {0}. If we take g to be 0 on the unit circle and 1 at the origin, then g is continuous on ∂Ω. But there is no harmonic function h on Ω that continuously extends g, as this would contradict the maximum principle. This shows that some condition on ∂Ω is necessary for the solvability of the Dirichlet problem. In this lecture we shall solve the Dirichlet problem subject to a ‘regularity’ condition on ∂Ω. The method is due to Perron. Definition 39.2. Let Ω be as above and let g be a continuous function on ∂Ω. The Perron family associated to g is the collection Fg of all continuous functions f : Ω → R that are subharmonic on Ω and have boundary values 6 g. Every function f ∈ Fg is bounded above by sup g, by the maximum principle. Moreover, Fg is nonempty since it contains each constant function 6 inf g. Thus the supremum in the next definition exists: Definition 39.3. The Perron function P g associated to g is the function (defined on Ω) P g(z) = sup{f (z) : f ∈ Fg }. Proposition 39.4. The Perron function P g is harmonic on Ω. Proof. Let us begin by making two observations. (i) If f1 , f2 belong to the Perron family then so does max{f1 , f2 }. (Indeed, we only need to check that the maximum of two subharmonic functions is subharmonic.) (ii) Let D be a closed disk in Ω and let f belong to the Perron family. Define a new function fD to be equal to f on Ω \ D, and to be equal on D itself to the unique harmonic function (provided by the Poisson integral formula) that agrees with f on ∂D. We must check that this construction produces a subharmonic function. Notice first that fD > f (since, on D, f − fD is a subharmonic function with zero boundary values, hence it is 6 0 by the maximum principle). Now to check that fD has the sub mean value property, the only points a where

124

this is not obvious are those on ∂D. But for such points fD (a) = f (a) and Z 2π Z 2π 1 1 iφ f (a) 6 f (a + Re )dφ 6 fD (a + Reiφ )dφ 2π 0 2π 0 which gives the result. Now let a ∈ Ω and choose R > 0 so that D(a; R) ⊆ Ω. There is a sequence of functions fn ∈ F with fn (a) → P g(a), and because of (i) and (ii) above we can choose each fn to be harmonic on D(a; R) and fn+1 > fn . By Harnack’s principle (Proposition 38.6), the sequence {fn } converges locally uniformly on D(a; R) to a harmonic function h, where by construction h(a) = P g(a). Let b ∈ D(a; R) be another point and choose a sequence of functions fn0 ∈ F such that fn0 (b) → f (b). Using (i) and (ii) in the same way as before, we can assume that fn0 > fn , that the sequence {fn0 } is increasing and that it consists of functions harmonic on D(a; R). The limit function h0 is therefore harmonic on D(a; R) and has h0 > h and h0 (b) = P g(b). By the definition of P g(a) as a supremum we have fn0 (a) 6 h(a) = P g(a) for all n, so h0 (a) 6 h(a) and therefore h0 (a) = h(a). The harmonic function h0 − h then has a minimum at a, so it is identically 0 on D(a; R). We have therefore proved that P g(b) = h0 (b) = h(b). Thus the functions P g and h agree on D(a; R), so P g is harmonic there. Since a was arbitrary, P g is harmonic everywhere. We should like to say that the Perron function P g solves the Dirichlet problem. There remain two issues in doing this: show that the Perron function is continuous on Ω, and show that its boundary values are exactly g (obviously they are 6 g). Definition 39.5. A boundary point a ∈ ∂Ω is regular if there exists a continuous function φa on Ω, subharmonic on Ω, with φa (a) = 0 and φa (z) < 0 for all z ∈ ∂Ω \ {a}. (The function φa is called a barrier at a.) Example 39.6. If there is a straight line segment [a, b] in C that meets Ω only at the boundary point a, then a is regular. For, without loss of generality, take a = 0 and b ∈ R+ . Then define w to be the branch of p z/(z − b) (defined on C \ [a, b]) which has positive real part everywhere. Then φa (z) = − Re w(z) is a barrier at a. Proposition 39.7. The Dirichlet problem is solvable for Ω if and only if every a ∈ ∂Ω is regular. 125

Proof. If the Dirichlet problem is solvable, its solution with boundary data g(z) = −|z − a| is a barrier. Conversely, suppose that a is a regular boundary point. Let φa be a barrier at a, let g ∈ C(∂Ω) be given and let h = P g be the associated Perron function. Let > 0, and choose a neighborhood U of a such that |g(z) − g(a)| < for all z ∈ U ∩ ∂Ω. Consider the function u(z) = g(a) − + Cφa (z) where C is a positive constant. Then u is subharmonic and u 6 g on U ∩∂Ω. By taking C sufficiently large we can arrange that u 6 g on ∂Ω \ U also (because φa is bounded above on this set by a strictly negative constant). If we choose C in this way then u belongs to the Perron family, so we have h > u. Since φa is continuous, we have lim inf h(z) > lim inf u(z) = g(a) − . z→a

z→a

Letting → 0 we find that lim inf h(z) > g(a). z→a

Now consider the function v(z) = g(a) + − Cφa (z) which is super harmonic (i.e. −u is subharmonic); we choose C so that v > g on ∂Ω. If f belongs to the Perron family, then f − v is subharmonic and 6 0 on ∂Ω, hence 6 0 everywhere on Ω by the maximum principle. It follows that h 6 v on Ω, so lim sup h(z) 6 lim inf v(z) = g(a) + . z→a

z→a

Letting → 0 we get lim sup h(z) 6 g(a). z→a

Together with the earlier inequality this implies that h is continuous at a with h(a) = g(a).

126

Math 502 Homework 1 Due Friday, January 20th (1) Let f = u + iv be a holomorphic function of z = x + iy. Show that the function g = log |f |2 = log(u2 + v 2 ) satisfies Laplace’s equation ∂2g ∂2g + = 0. ∂x2 ∂y 2 (You may assume that u and v are twice continuously differentiable.) (2) Let z be a complex 5th root of 1, z 6= 1. Show that 1 + z + z 2 + z 3 + z 4 = 0. By taking real parts deduce that 1 + 2 cos(2π/5) + 2 cos(4π/5) = 0, and hence show that

√ cos(2π/5) =

5−1 . 4

(3) Let z be a complex number with positive real part. By induction on n, show that Z 1 n! . tz−1 (1 − t)n dt = z(z + 1) · · · (z + n) 0 Substitute t = u/n and let n → ∞ to obtain n!nz−1 , n→∞ z(z + 1) · · · (z + n − 1)

Γ(z) = lim

where Γ is the Gamma function defined in class. (Do your best to justify the limit processes that are used in this argument.)

1

Math 502 Homework 2 Due Friday, January 27th (1) Let a and b be complex numbers with strictly negative real part. Prove the inequality |ea − eb | 6 |a − b|.

(2) True or false: There exists a sequence of complex polynomials pn (z) such that pn (z) → 1/z uniformly on the unit circle {z : |z| = 1}? Give careful reasons. (3) By considering the contour integral I dz z taken around an elliptical contour whose Cartesian equation is x2 /a2 + y 2 /b2 = 1, show that Z 0

2π

a2

cos2

dt 2π . 2 = 2 ab t + b sin t

1

Math 502 Homework 3 Due Friday, February 3rd (1) Evaluate the integral I cos z dz z around the unit circle. Deduce that Z 2π cos(cos θ) cosh(sin θ) dθ = 2π. 0

(2) Suppose that the function f has a Taylor expansion f (z) =

∞ X

cn z n

n=0

valid in C. Prove that for each R > 0, ∞ X

|cn |Rn 6 2M (2R),

n=0

where M (r) = sup{|f (z)| : |z| = r}. (3) Find all the possible values of the integral Z dz 2 Γ z(z + 1) over a cycle Γ in C \ {0, i, −i}.

1

Math 502 Homework 4 Due Friday, February 10th (1) Prove that the equation z 5 + 15z + 1 = 0 has precisely four solutions in the annulus {z : 23 < |z| < 2}. (2) Find the residues of the following meromorphic functions at the indicated poles: • z sec2 z at z = π/2. • 1/(z 4 sin z) at z = 0. • tanh(z/2) at z = iπ. (3) Show that the series f (z) =

∞ X 1 (−1)n − 2z z π 2 n2 − z 2 n=1

converges to a meromorphic function f on C, which has the property that f (z) = −f (z + π) for all z ∈ C. Show further that the poles of f are simple ones occurring precisely at the points kπ, and find their residues. Deduce that f (z) = 1/ sin(z) (use Liouville’s theorem).

1

Math 502 Homework 5 Due Friday, February 17th (1) Evaluate the integral Z ∞ 2x2 dx 4 −∞ 4 + x by means of complex integration. Be sure to justify carefully all the steps in your argument. (2) Prove that ∞ Y Γ(a + 1)Γ(b + 1) k(a + b + k) = (a + k)(b + k) Γ(a + b + 1) k=1

for any complex numbers a, b not equal to negative integers. (You should include a proof that the infinite product is convergent.) (3) A theorem of Carlson states that if f (z) is holomorphic and bounded in Ω = {z : Im(z) > −}, > 0, and if moreover f (z) = 0 whenever z = ki, k = 1, 2, . . ., then f is identically zero. Complete the following outline to prove this theorem: (a) Show that for any a ∈ R+ and any n = 1, 2, . . . we have Z in (a − 1) · · · (a − n) ∞ f (x) f (ia) = dx. 2πi −∞ (x − ia)(x − i)(x − 2i) · · · (x − ni) (b) Show that for a > 1 the absolute value of the integral appearing above is no more than M π/n!, where M is an upper bound for |f | in Ω. (c) By letting n → ∞ deduce from (a) above that f (ia) = 0 for all a > 1. Carlson’s theorem now follows by analytic continuation.

1

Math 502 Homework 6 Due Friday, February 24th (1) By contour integration find the value of √ Z ∞ x dx. 2 x + 5x + 6 0 (Begin by making a substitution to get rid of the square root term.) (2) The function f (z) is entire, and it is known that |f (z)| 6 1 whenever |z| = 1 and that |f (z)| 6 10 whenever |z| = 10. Show that |f (z)| 6 5 whenever |z| = 5. What can be said about |f (0)|? (3) Find a conformal equivalence from the region √ √ {z ∈ C : |z − 1| < 2 and |z + 1| < 2} to the unit disk U.

1

Math 502 Homework 7 Due Friday. March 17th (1) Let E be a normed vector space and let F be a closed subspace. For x ∈ E, let [x] denote the coset of x in the quotient space E/F . Show that the formula k[x]k = inf{kx + yk : y ∈ F } defines a norm on the quotient space E/F , and that if E is a Banach space, then E/F is a Banach space too. (2) Let 1 < p < q < ∞. Let (X, µ) be a σ-finite measure space. Let E be the space of functions Lp (X, µ) + Lq (X, µ). Show that the expression kf k = inf{kgkp + khkq : f = g + h, g ∈ Lp , h ∈ Lq } defines a norm on E that makes it into a Banach space. (3) Give a concrete description of the dual of the Banach space E appearing in Question 2. (Your description should identify E ∗ with a space of functions and say what the norm is.)

1

Math 502 Homework 8 Due Friday, March 24th REVISED: March 27th (1) Let K be a closed subset of the Riemann sphere S, whose complement is connected and nonempty. Let p ∈ S \ K. Show that any function f holomorphic on a neighborhood of K can be approximated, uniformly on K, by rational functions with poles only at p. (Hint: Use a suitable conformal transformation to reduce this to the version of Runge’s theorem that we proved in class.) (2) Let Ω be the set {x +iy ∈ C : x > −1, |y| < e−x }. Using the previous question, show that there is a sequence of rational functions fn on C having the following properties: (a) f1 (z) = 1/z, (b) for n > 1, fn (z) is a rational function with pole only at 2n − 1, and (c) |fn+1 (z) − fn (z)| < 2−n for all z ∈ D(0; 2n − 2) ∪ C \ Ω. Deduce that the functions fn converge uniformly on compact subsets of C to an entire function which is bounded on the complement of Ω. (The function grows extremely rapidly along Ω itself. This is a standard way to use Runge’s theorem to construct (counter)examples.) (2) Let E be a Banach space and E ∗ its dual space. Carry out the details of the following argument to show that if E ∗ is separable, then E is separable. (a) Let {φn } be a countable dense subset of the unit sphere of E ∗ . For each n, show that there is a unit vector xn ∈ E such that |φn (xn )| > 21 . (b) Use the Hahn-Banach theorem to show that the closed linear span of the {xn } is all of E. (c) Hence prove that E is separable.

1

Math 502 Homework 9 Due Friday, March 31st (1) Let E = `∞ be the space of all bounded sequences a = {an } of complex numbers. Show that there is a continuous linear functional φ : E → C which has the property that if lim an exists, then φ(a) = lim an . Let B : E → E take the sequence {an } to the sequence {bn } where bn =

a1 + · · · + an . n

Show that B : E → E is a bounded linear operator. Show that ψ = φ ◦ B is a linear functional on E which has the property that if lim an exists, then ψ(a) = lim an . In addition, show that ψ(a) = ψ(a0 ), where a0n = an+1 . (ψ is called a Banach limit.) (2) Let E, F be Banach spaces and let T : E → F be a bounded linear map. Must T be continuous relative to the weak topologies on E and F ? Give a proof or counterexample as appropriate. (3) Let V be a normed vector space. Suppose that the closed unit ball of V can be covered by finitely many balls B(xk ; 12 ), k = 1, . . . , n, of radius 1 . Show that the {xk } span V . (First show that the span of the {xk } is a 2 closed subspace, then show that it contains the unit ball of V .) Deduce that every locally compact normed vector space is finite dimensional. (N.B. This question does not have anything to do with the Hahn-Banach theorem. You do need to know the theorem that a finite dimensional normed vector space is automatically complete. We proved this last semester.)

1

Math 502 Homework 10 Due Friday, April 7th (1) Let E be a Banach space. Show that every weakly convergent sequence in E is bounded (with respect to the norm). The Banach space `p is the space of sequences a = {an } with norm P kakp = ( |an |p )1/p . In other words, it is the Lp space of the integers with counting measure. Show that if 1 < p < ∞ then there exist weakly convergent sequences in `p that do not converge in norm. (2) Show that a sequence in `1 is weakly convergent if and only if it is convergent in the norm. Show that, nevertheless, the weak topology and the norm topology on `1 are not the same. Explain the apparent incompatibility here. (3) Let {an }∞ n=−∞ be a two-way infinite sequence of complex numbers. It is a Fourier multiplier if it has the following property: whenever {cn } is the sequence of Fourier coefficients of a continuous 2π-periodic function, {an cn } is the sequence of Fourier coefficients of a continuous 2π-periodic function also. Let C(T) denote the Banach space of continuous 2π-periodic functions (with the supremum norm). The Fourier multiplier {an } defines a linear map C(T) → C(T). Using the closed graph theorem, show that this linear map is continuous. Deduce that there exists a complex Borel measure µ on [0, 2π] such that Z 2π an = e−itn dµ(t). 0

1

Math 502 Homework 11 Due Friday, April 14th (1) Let f : [−π, π] → C be a continuously differentiable function which has f (−π) = f (π). How do the Fourier coefficients of the derivative f 0 relate to the Fourier coefficients of f ? Prove Poincar´e’s inequality: if Z π f (x) dx = 0, −π

then

Z

π

Z

2

π

|f (x)| dx 6

|f 0 (x)|2 dx.

−π

−π

When does equality hold? (2) An operator P on a Hilbert space is called Fredholm if there is another operator Q such that P Q − I and QP − I are compact. Show that a Fredholm operator has finite-dimensional kernel and closed range. (3) Let A be a compact self-adjoint operator on a Hilbert space H. Suppose that for all x ∈ H, hAx, xi > 0. Prove that there is a unique compact self-adjoint operator B such that hBx, xi > 0 for all x and B 2 = A. Now let A be the operator on `2 defined by A(x1 , x2 , x3 , · · · ) = (x2 /2, x3 /3, x4 /4, · · · ). Prove that A is compact, but that there is no operator B such that B 2 = A.

1

Math 502 Homework 12 Due Friday, April 21st (1) Let T be a compact, self-adjoint operator on an infinite-dimensional Hilbert space. Suppose that there exists some polynomial p such that p(T ) = 0. Prove that the range of T is finite-dimensional. Prove also that the polynomial p has no constant term. (2) Consider the linear mapping T from L2 [0, ∞) to itself defined by ( e−x f (x − 1) (x > 1) T f (x) = 0 (x < 1) Show that the spectrum of T is the set {0}. (3) Show that the function " h(z) = Im

1+z 1−z

2 #

is harmonic and nonconstant on the unit disk. Show furthermore that for every θ, the radial limit limr→1 h(reiθ ) exists and equals 0. How is this consistent with the maximum principle?

1