Lecture 1 The complex plane Exercise 1.1. Show that the modulus obeys the triangle inequality z ± w 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. Solution. See Proposition 1.8, last semester. Exercise 1.2. Show that the dot product of two complex numbers z and w (considered as vectors in R 2 ) 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. Solution. If z = x + yi and w = p + qi then z w = (x + yi)(p qi) = (xp + yq) + (yp xq)i. The real part of this is the dot product of z and w, thought of as vectors in R 2. Now let w and w be complex numbers. By what we just remarked, the angle between them is ( ) Re cos 1 w w w w. The angle between zw and zw is cos 1 ( Re w z z w zw zw ). Since z z = z 2 the factors of z cancel and the angles are equal. 1
Lecture 2 Holomorphic Maps 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 Möbius transformations map circlines to circlines. Solution. If you take the equation z p 2 = λ 2 z q 2 and expand in terms of (x, y) with z = x + iy, you get (1 λ 2 )(x 2 + y 2 ) + linear terms in x, y + constant = 0. This represents a straight line if λ = 1, and a circle otherwise. If λ = 1, z p = z q represents the perpendicular bisector of the segment pq; any line can be so represented. For circles, consider wlog the circle center 0 and radius r. If 0 < p < r define q = r 2 /p and λ = p/r. The equation z p = λ z q then becomes x 2 + y 2 2px + p 2 = p2 r 2 After some algebra this gives (x 2 + y 2 2r2 p x + r4 p 2 x 2 + y 2 = r 2, the equation of the original circle. Notice that for a general Mobius transformation M(z) = (az + b)/(cz + d) az + b cz + d = a z p c z q, where p = b/a and q = c/d. Thus the equation M(z) = constant represents a circline, and every circline can be so represented. Since the composite of Mobius transformations is Mobius, it follows that Mobius transformations map circlines to circlines. ). 2
Lecture 3 Some examples of holomorphic functions Exercise 3.1. Show that the zeta function can also be defined by the integral ζ(s) = 1 t s 1 Γ(s) 0 e t 1 dt, provided that Re s > 1. Solution. For t > 0 we can write 1 e t 1 = e t 1 e t = From the dominated convergence theorem, then, 0 t s 1 e t 1 dt = But the substitution u = nt shows that and so as required. 0 n=1 0 n=1 e nt. t s 1 e nt dt. t s 1 e nt dt = n s u s 1 e u du = n s Γ(s) t s 1 0 e t 1 dt = n=1 0 n s Γ(s) = Γ(s)ζ(s), 3
Lecture 4 Plane topology and complex analysis Exercise 4.1. Prove the facts corresponding to those above for the logarithm function. That is, show that every nonzero w 0 has a neighborhood on which there is defined a holomorphic function g with e g(w) = w, but that there is no function defined on all of C \ {0} that has this property. Solution. Let w 0 = r 0 e iθ0 be nonzero. Define Ω to be the strip {x + iy : x R, θ 0 π < yθ 0 + π} in C. Then the exponential function f(z) = e z maps Ω bijectively onto U = {re iθ : r > 0, θ 0 π < θ < θ 0 + π}, which is an open set (a cut plane) containing w 0. Lemma?? now shows that there is a holomorphic logarithm function defined on U. If a holomorphic logarithm function existed on the whole of C \ {0}, it would in particular give a continuous map from T to ir (the imaginary axis) whose composite with the exponential map ir T would be the identity. But since ir is contractible, the induced map on fundamental groups π 1 (T) π 1 (ir) = 0 π 1 (T) would be the zero map, and this is impossible. 4
Lecture 5 Path integrals and Cauchy s theorem 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.) Solution. First of all, consider the case where the bad point is the vertex A of a triangle ABC. Since f is continuous at A, given ɛ > 0 we can find points B (on AB) and C (on AC) sufficiently close to A that f(z)dz < ɛ. AB C Now ABC f(z)dz = f(z)dz + AB C BB C f(z)dz + f(z)dz CB C and the last two integrals vanish by Cauchy s theorem (original version). Thus f(z)dz < ɛ ABC and letting ɛ 0 we get the result. If the bad point is on an edge or inside the triangle, subdivide into two or three smaller triangles with the bad point at the vertex and apply the previous result. 5
Lecture 6 Consequences of Cauchy s theorem Exercise 6.1. (The reflection principle) Let f be continuous on the closed upper halfplane, 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. Solution. We shall use Morera s theorem. If T is a triangle in the open upper halfplane, then f(z)dz = 0 by cauchy s theorem. This result extends to triangles in T the closed upper half-plane, since f is continuous and any triangle in the closed upper half-plane can be regarded as the limit of a sequence of triangles in the open upper half-plane. If T is a triangle in the open lower half-plane then T f(z)dz = T f( z)dz = T f(w)dw = 0 using the substitution w = z. Again, this extends by continuity to triangles in the closed lower half-plane. Any triangle can be subdivided into a sum of triangles in the upper and lower half planes. Thus the integral of f around any triangle is zero. Morera s theorem now implies that f is holomorphic. 6
Lecture 7 The rigidity of holomorphic functions Exercise 7.1. Show that if f is an entire function and f(z) C( z n + 1) for some constant C, then f is in fact a polynomial (of degree at most n). Solution. Since f is entire, it has a Taylor series expansion f(z) = convergent in C. Let p(z) be the polynomial p(z) = a k z k k=0 n a k z k. k=0 Then (f(z) p(z))/z n is a bounded holomorphic function that vanishes at 0. By Liouville s theorem, it is identically zero, and so f p. 7
Lecture 8 The Global Cauchy Theorem Exercise 8.1. Show that every 1-cycle is equivalent to a formal linear combination of closed paths. Solution. Let Γ be a 1-cycle, and let it be represented as j k j[γ j ]. We can replace it by an equivalent cycle where all the numbers k j are positive; to do this, notice that if γ : [a, b] C is a path, then the cycle [γ] is equivalent to [γ ], where γ is defined by γ (t) = γ(a + b t). Assume then that all the k j are positive, and define the height Γ to be the sum of the k j. We ll prove by induction of Γ that every 1-cycle is equivalent to a combination of closed paths. This is apparent if Γ = 1 since then Γ = [γ] and the cycle condition implies that the beginning and ending points of γ are the same. Now given a general cycle Γ = k j [γ j ] (with all k s positive), the cycle condition implies that each end point of one γ j is the start point of another γ j. Thus, starting at γ 1, we can find a chain of γ s each of which begins where the previous one ends. Because there are only finitely many γ s this chain must close at some point, so there is a finite sequence of γ s by renumbering we may assume that it is γ 1,..., γ p such that the end of γ j is the start of γ j+1 for 1 j p 1, and the end of γ p is the start of γ 1. But then Γ = [γ 1 ] + + [γ p ] is a cycle, and it is equivalent to the closed path obtained by concatenating γ 1,..., γ p. Moreover, Γ Γ is a cycle, and its height Γ Γ = Γ p < γ. By induction, Γ Γ is equivalent to a sum of closed paths; so Γ = (Γ Γ ) + Γ is equivalent to a sum of closed paths as well. 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). 8
Lecture 9 Laurent series and the residue theorem Exercise 9.1. Suppose that f 1 and f 2 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 f 1 + f 2 at a, in terms of the orders of f 1 and f 2? Solution. Let f 1 and f 2 have poles of orders N 1 and N 2 respectively. If N 1 N 2 then the worst singularity wins : f 1 + f 2 has a pole of order max{n 1, N 2 }. However, if N 1 = N 2, then the sum f 1 + f 2 may have a pole of order N 1 or of any lesser order (including no pole at all). These statements are easily verified by adding Laurent zeries. 9
Lecture 10 Counting zeroes and poles 10
Lecture 11 Calculations with the residue theorem 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 ( f ) (a) 2 g (a) f(a)g (a) 3g (a) 2. Verify that this is consistent with our solution to Example??. Solution. Write g(z) = (z a) 2 h(z) where h is holomorphic and h(a) 0. Then (either by direct calculation using the product rule for differentiation, or more simply by comparing Taylor series on the left and right hand sides of the above identity) we find h(a) = 1 2 g (a), h (a) = 1 6 g (a). Now write f(z) g(z) = f(z)/h(z) (z a) 2. The numerator is holomorphic near a, so the residue is [ ] d f(z) = f (a) dz h(z) h(a) f(a)h (a) h(a) 2 = 2f (a) g (a) 2f(a)g (a) 3g (a) 2 as required. z=a 11
Lecture 12 The Gamma Function 12
Lecture 13 More examples of contour integration 13
Lecture 14 More about the Riemann Sphere Exercise 14.1. Let S be a compact connected Riemann surface. Show that the only holomorphic functions S C are the constant functions. Solution. Let f : S C be holomorphic. Since S is compact, f attains a maximum at some point a S. By the maximum principle (applied to f φ 1, where φ is a chart near a), f must be constant (say equal to c) on some neighborhood of a. Consider now the set of all points x S such that f = c on a neighborhood of x. By definition this set is open. By the principle of isolated zeroes, it is closed. It is nonempty as we have seen, and thus (by connectedness) it is the whole of S. 14
Lecture 15 Automorphisms Exercise 15.1. Show that the automorphism groups of S and C are transitive. Solution. Translations z z + c are automorphisms of C, and it is obvious that the group of translations acts transitively on C. Since translations also give automorphisms of S, we see that any two finite points of S are equivalent under the automorphism group. The transformation z 1/z is also an automorphism of S and maps to a finite point, so in fact all points of S are equivalent. 15
Lecture 16 Hyperbolic Geometry 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. Solution. Consider first the special case of a triangle with a right angle at C. Write d(a, B) = c, d(b, C) = a, d(c, A) = b. We have cosh c = cosh a cosh b Rewrite the second equation to give cosh b = cosh a cosh c sinh a sinh c cos β sin 2 β = 1 cos 2 β = sinh2 a sinh 2 c (cosh b cosh a cosh c) 2 sinh 2 a sinh 2. c Substitute cosh a = cosh c/ cosh b from the first equation and sinh 2 a = cosh 2 a 1 (standard identity). After canceling a common factor of cosh 2 c cosh 2 b we get cosh 2 b cosh 2 c + sinh 2 c sinh 2 c = sinh2 b sinh 2 c so sin β = sinh b/ sinh c. This is the sine rule for a right triangle. The general case follows by dropping a perpendicular from a vertex of the triangle to the opposite side, and thus dividing it into two right triangles (exactly as in Euclidean geometry). 16
Lecture 17 The Riemann mapping theorem 17
Lecture 18 Multi-valued functions Exercise 18.1. Find a region on which a branch of the inverse cosine function is defined. (Start by proving that ( cos 1 (z) = i log z + ) z 2 1 and then look for a branch of the function on the right.) Solution. If z = cos w = 1 2 (eiw + e iw ), then e iw is a root of the quadratic equation t 2 2zt + 1 = 0. The roots of this equation are t = z ± z 2 1, which gives the formula w = i log ( z + z 2 1 ) from the question. There are branch points of z 2 1 at ±1. Since (z + z 2 1)(z z 2 1) = 1, the expression under the logarithm sign is never zero, so no further branch points are introduced by the logarithm term. The most natural way to cut the plane here is to cut from to 1 and 1 to. The complement of these two cuts is a simply connected region, on which branches of z 2 1 and then i log ( z + z 2 1 ) can be defined. Note that it does not suffice to cut from 1 to 1; the square root is well-defined on this cut region, but the logarithm is not. 18
Lecture 19 Constructing conformal maps 19
Lecture 20 Contour integrals with multi-valued functions Exercise 20.1. Evaluate the integral of homework 6, problem 1, x x 2 + 5x + 6 dx, 0 by means of a contour integral involving a suitably defined branch of the square root function. Solution. Let I denote the desired integral. Cut the plane along the positive real axis and let f(z) be the branch of the function z/(z 2 + 5z + 6) that is positive real-valued just above the real axis. The function f has poles at 2 and 3, with residues i 2 and i 3 respectively. Integrate around the keyhole contour described above. The integrals along the top and bottom sides of the keyhole approach 2I, and the integrals along the inner and outer loops tend to 0. Thus, we obtain so I = π( 3 2). 2I = 2πi.(i 2 i 3), 20
Lecture 21 Analytic Continuation Exercise 21.1. Complete the proof that w is continuous. Solution. With notation as in the previous proof, let Ω C be open and let p w 1 (Ω). Let the germ p be represented by (U, f, a), with f(a) Ω. Then f(u f 1 (Ω)) is a basic neighborhood of p and is contained in w 1 (Ω). Thus w 1 (Ω) is open, so w is continuous. 21
Lecture 22 The monodromy theorem 22
Lecture 23 Basics of Banach Spaces Exercise 23.1. Show that L is a normed vector space. Solution. First let us prove that if f L then there is a null set such that f = sup{ f(x) : x X \ N}. Indeed, by definition, for every k there is a null set N k such that f = sup{ f(x) : x X \ N k } + 1/k. Take N = N k, which is a null set. In particular, this shows that if f = 0, then f = 0 almost everywhere. Now to prove the triangle inequality. Let f 1 and f 2 belong to L. By the above there are null sets N 1 and N 2 such that f i = sup{ f i (x) : x X \ N i }, i = 1, 2. Let N = N 1 N 2 and f = f 1 + f 2. Then f sup{ f 1 (x) + f 2 (x) : x X \ N} sup{ f 1 (x) : x X \ N} + sup{ f 2 (x) : x X \ N} f 1 + f 2 23
Lecture 24 The Spaces L p (X, µ) 24
Lecture 25 The Hahn-Banach Theorem Exercise 25.1. For any Banach space E construct an isometric injection E E. Solution. Any x E defines a linear map e x : E C via the equation e x (φ) = φ(x). Since e x (φ) = φ(x) φ x, we see that e x is a bounded linear functional on E, with norm e x x. Thus e x E, and x e x is a linear map from E to E. Let us show that this linear map is an isometry (and so, in particular, that it is injective.) Let x E. By the Hahn-Banach theorem there is a linear functional φ E, of norm 1, such that φ(x) = x (define φ first on the one dimensional subspace spanned by x, and extend by Hahn-Banach). Then e x (φ) = x so e x x. We proved the opposite inequality above, so e x = x and x e x is an isometric injection. 25
Lecture 26 Applications of the Hahn-Banach Theorem 26
Lecture 27 Convexity and the Hahn-Banach Theorem Exercise 27.1. Give an example of an absorbing subset in a normed vector space which does not contain any neighborhood of the origin. Solution. There are many possible solutions depending on how sophisticated you want to get. Here are a couple. Take C as a 2-dimensional vector space over R, and let A be the set {re iθ : 0 < θ 2π, 0 r θ}. This is easily seen to be absorbing, but it contains no disk around 0. Let E be the vector space of differentiable functions [ 1, 1] R, with the supremum norm, and let A = {f : f (0) < 1}. Then A is absorbing, but it contains no neighborhood of 0 (because one can have very small functions with very large derivatives). Both of these examples feel a bit like cheating: in (a), the set A is not convex and in (b), the space E is not complete. Can one have a convex absorbing set in a complete space which is not a 0-neighborhood? The answer is yes. Indeed, if φ: E R is a discontinuous linear functional, then {x E : φ(x) < 1} is an absorbing set that cannot contain a neighborhood of the origin. To produce such a linear functional on a complete space needs some transfinite machinery again. For instance, take E = C[ 1, 1], F the subspace of functions differentiable at 0, and φ(f) = f (0) for f F. Extend φ to E using the technique of the Hahn-Banach theorem (but without bothering about the norms at all). 27
Lecture 28 Weak topologies Exercise 28.1. The argument above, as stated, works for real vector spaces only. Work out how to generalize it to the complex case. Solution. If E is a complex vector space, and A and b are as above, the argument will produce a bounded real-linear functional φ such that φ(x) < c < φ(b) for all x A. This φ is the real part of the bounded complex-linear functional ψ(x) = φ(x) iφ(ix) (this is the same complexification trick that we used in the proof of the complex form of the Hahn-Banach theorem). Now define F b to be the inverse image under ψ of the closed half-plane {z C : Re(z) c} and complete the proof as before. 28
Lecture 29 Applications of the Baire category theorem Exercise 29.1. Give an example of a separately continuous map that is not jointly continuous. Solution. The function (from R 2 to R) (x, y) f(x, y) = { xy x 2 +y 2 (x, y) (0, 0) 0 (x, y) = (0, 0) is separately continuous. Indeed, if y 0, f(x, y) is clearly a continuous function of x, and if y = 0, f(x, y) = 0 identically. Similarly for x fixed as a function of y. However f is not jointly continuous, because any neighborhood of 0 contains points (ɛ, ɛ), for which f(x, y) = 1 2, whereas f(0, 0) = 0. 29
Lecture 30 The open mapping and closed graph theorems Exercise 30.1. Prove the above statements. Solution. If g L 2 [0, 1] is supported in the interval [ɛ, 1] for some ɛ > 0, then the function f(x) = g(x)/x is well-defined and belongs to L 2 with f ɛ 1 g. Thus g belongs to the range of T. But now for any g L 2 the functions g n = gχ [1/n,1] belong to the range of T and g n g in L 2 by the dominated convergence theorem. Thus the range of T is dense. However, the constant function 1 is not in the range of T, since if T f = 1 then f(x) = 1/x almost everywhere, and the function 1/x is not in L 2 [0, 1]. Thus the range of T, being dense, cannot be closed (otherwise it would be all of L 2 [0, 1]). 30
Lecture 31 Operators on Hilbert Space Exercise 31.1. Show that in the above situation the norm of M f is exactly equal to the L norm of f. Solution. We need to show that for any ɛ > 0 there is g L 2 such that M f g 2 (C ɛ) f g 2. (The subscripts denote the norms in L 2 and L.) By definition of the L norm, there is a subset E X of positive measure such that f(x) C ɛ for all x E. Since we are working with a σ-finite measure space, E contains a subset of positive finite measure; replace E by this subset and we may assume that E itself has finite measure. Let g = χ E. Then g belongs to L 2 with g 2 = µ(e) 1/2 and as required. ( M f g 2 = E ) 1/2 f 2 dµ (C ɛ)µ(e) 1/2 31
Lecture 32 More about operators Exercise 32.1. Use the closed graph theorem to show that if T is a linear map from a Hilbert space to itself, satisfying T u, v = u, T v for all u, v H, then T is an operator (i.e., it is bounded). Solution. Suppose that u n u and T u n w. Then T u, v = u, T v = lim u n, T v = lim T u n, v = w, v for every v H. Thus T u w, v = 0 for every v, whence T u = w. This proves that the graph of T is closed, so T is continuous by the closed graph theorem. This is a very early theorem (Hellinger-Toeplitz, around 1903). 32
Lecture 33 Calculations with compact operators 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 {a n } tends to zero. 33
Lecture 34 The spectral theorem for compact operators Exercise 34.1. Extend the two lemmas to normal operators T (those for which T commutes with T ). Solution. Let T be a normal operator. Then T x 2 = T T x, x = T T x, x = T x 2 and thus ker(t ) = ker(t ). Applying this to the normal operator T λi instead of T gives ker(t λi) = ker(t λi). In other words, every eigenvector T (with eigenvalue λ) is also an eigenvector for T (with eigenvalue λ). Now let E be the eigenspace ker(t λi). If x E and y E then T x, y = x, T y = x, λy = 0. Thus T x E also. Let x 1, x 2 be eigenvectors for T with eigenvalues λ 1, λ 2. Then λ 2 x 1, x 2 = x 1, T x 2 = T x 1, x 2 = λ 1, x 2 = λ 1 x 1, x 2 so either λ 1 = λ 2 or x 1, x 2 = 0. 34
Lecture 35 General spectral theory Exercise 35.1. Prove the spectral mapping theorem for general polynomials p. (The method is the same.) Solution. Let p be a polynomial of degree n. By the fundamental theorem of algebra, for each µ there exist λ 1,..., λ n such that p(z) µ = c(z λ 1 ) (z λ n ) where c is a constant. The set {λ 1,..., λ n } is exactly p 1 {µ}. Arguing as in the special case we see that p(t ) µi is invertible if and only if all the T λ k I are invertible. That is, µ belongs to the spectrum of p(t ) if and only if one of the λ k belongs to the spectrum of T. But this is exactly to say that µ belongs to p(σ(t )). 35
Lecture 36 The spectral radius 36
Lecture 37 Harmonic functions Exercise 37.1. Show that the assumption of simple connectedness is necessary in the above theorem. (Consider the function g(x + iy) = log(x 2 + y 2 ) on C \ {0}. ) Solution. One can show that g is harmonic by direct calculation, or by observing that it is locally the real part of a suitably defined branch of log z 2. But if g were globally the real part of a holomorphic function f we would have df dz = g x i g y by the Cauchy-Riemann equations. Now evaluate f (z)dz around the unit circle z = cos θ + i sin θ. We obtain 2π f (z)dz = 2(cos θ i sin θ) ( sin θ + i cos θ)dθ = 4πi. 0 This contradicts the fundamental theorem of calculus, so no such function f can exist. 37
Lecture 38 The mean value property 38
Lecture 39 The Perron construction 39