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ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY Contents Spring 2011 Fall 2010 Spring 2010 Fall 2009 Spring 2009 Fall 2008 Spring 2008 Fall 2007 Winter 2007 1 7 13 19 23 28 33 38 43 I make no claims about the infallibility of these solutions. In addition to my own solutions, I also received considerable aid from Ben Hayes and Quinn Maurmann, as well as the previous work by William Meyerson, Dave Gaebler, and Jon Handy. Spring 2011 Problem 1: (a) Define what it means to say that fn → f weakly in L2 ([0, 1]). (b) Suppose fn ∈ L2 ([0, 1]) converge weakly to f ∈ L2 ([0, 1]) and define ‘primitive’ functions: Z Fn (x) := x Z fn (t)dt and F (x) = 0 x f (t)dt. 0 Show that Fn , F ∈ C([0, 1]) and that Fn → F uniformly on [0, 1]. Proof. (a) We say that fn → f weakly in L2 ([0, 1]) if for all g ∈ L2 ([0, 1]), we have Z 1 Z fn (x)g(x)dx → 0 1 f (x)g(x)dx as n → ∞. 0 1 2 ANALYSIS QUALS (b) WLOG, assume 0 ≤ y ≤ x ≤ 1. Then Z x Z y fn (t)dt fn (t)dt − |Fn (x) − Fn (y)| = 0 Z0 x fn (t)dt = y Z 1 fn (t)χ[y,x] (t)dt = Z 0 1 |fn (t)| χ[y,x] (t)dt ≤ 0 Z 1 ≤ 1/2 Z |fn (t)| dt 2 1 1/2 χ[y,x] (t) dt 2 0 0 = kfn kL2 (x − y) 1/2 → 0 as x − y → 0. Hence, Fn ∈ C([0, 1]). Similarly, |F (x) − F (y)| ≤ kf kL2 (x − y)1/2 , so F ∈ C([0, 1]) as well. It is left to show that Fn → F uniformly on [0, 1]. Note that x 7→ |Fn (x) − F (x)| is a continuous function on [0, 1], so as [0, 1] is compact, it attains a maximum. Thus, there exists y ∈ [0, 1] such that |Fn (x) − F (x)| ≤ |Fn (y) − F (y)| for every x ∈ [0, 1]. Let ε > 0. Note that χ[0,y] ∈ L2 ([0, 1]). As fn → f weakly, we may choose N such that Z 1 Z 1 fn (t)χ[0,y] (t)dt − f (t)χ[0,y] (t)dt < ε when n ≥ N. 0 0 Let x ∈ [0, 1] and suppose n ≥ N . Then we have that Z 1 Z 1 |Fn (x) − F (x)| ≤ |Fn (y) − F (y)| = fn (t)χ[0,y] (t)dt − f (t)χ[0,y] (t)dt < ε. 0 0 Thus, Fn → F uniformly. Problem 5: (a) Show that l∞ (Z) contains continuum many functions xα : Z → R obeying kxα kl∞ = 1 and kxα − xβ kl∞ ≥ 1 whenever α 6= β. (b) Deduce (assuming the axiom of choice) that the Banach space dual of l∞ (Z) cannot contain a countable dense subset. (c) Deduce that l1 (Z) is not reflexive. ANALYSIS QUALS 3 Proof. (a) First, we note that P(Z), the power set of Z, contains continuum many elements. For α ∈ P(Z), α 6= ∅, let xα = χα , the characteristic function of α. Then |xα kL∞ = 1. But for α 6= β, there exists n ∈ α∆β. Then |xα (n) − xβ (n)| = 1, so kxα − xβ kL∞ ≥ 1. (b) Assume for sake of contradiction that l∞ (Z)∗ has a countable dense subset {fn }. That is, l∞ (Z)∗ is separable. For each n, choose yn ∈ l∞ (Z) such that |fn (yn )| ≥ 21 kfn k. Let S be the set of linear combinations of {yn } with rational coefficients. Then S is countable. We claim S = l∞ (Z). If not, S is a closed proper subspace of l∞ (Z). By the Hahn-Banach theorem, we may choose f ∈ l∞ (Z)∗ such that f (y) = 0 for all y ∈ S, but kf k = 1. Choose {fnj } for our countable dense set such that fnj − f → 0 as j → ∞. But fnj − f ≥ fnj (ynj ) − f (ynj ) = fnj (ynj ) ≥ 1 fnj . 2 Thus, we must have fnj → 0, so f = 0, which contradictions the assumption that S was a proper subspace. Hence, l∞ (Z) is separable. Thus, we let {yn } be a countable dense set in l∞ (Z). We have that for every ε > 0 and for every non-empty α ∈ P(Z), there exists n such that kyn − xα kL∞ < ε. But there are uncountably many xα with kxα − xβ k ≥ 1 if α 6= β, so by the Pigeonhole Principle (and Axiom of Choice), the xα ’s cannot all be approximated by the yn ’s, so we have a contradiction to the assumption that l∞ (Z)∗ has a countable dense subset. (c) By definition, l1 (Z) reflexive means that l1 (Z) ∼ = (l1 (Z))∗∗ . But l1 (Z) 1 ∗∗ ∼ ∞ ∗ is separable, and (l (Z)) = (l (Z)) is not separable by part (b). Hence, l1 (Z) is not reflexive. Note: I lost a point on the qual with this solution for not elaborating on the fact that l1 (Z) is separable. I don’t know whether they wanted me to prove this, or just mention a countable dense set. For the record, the set of linear combinations with rational coefficients of the sequences ek consisting of all zeroes with 1 in the kth spot is a countable dense set. Problem 6: Suppose µ and ν are finite positive (regular) Borel measures on Rn . Prove the existence and uniqueness of the Lebesgue decomposition: There are a unique pair of positive Borel measures µa and µs so that µ = µa + µs , µa << ν, and µs ⊥ ν That is, µa is absolutely continuous to ν, while µs is mutually singular to ν. Solution. Let Z F = f : Rn → [0, ∞] : f dν ≤ µ(E) for all E ∈ M E Note that if f Rand g are in F, then h = max(f, g) ∈ F. Let a = sup{ X f dν : f ∈ F }. Let {fn } ⊂ F be a sequence such that 4 ANALYSIS QUALS R f dν → a. Let gn = max{f1 , . . . , fn }. Then gn ∈ RF for all n. FurtherX n more, gn increases pointwise to f = sup{f R n }. Then X gn dν → a, and by the monotone convergence theorem, a = X f dν and f ∈ F. RDefine dµs = dµ − f dν, and dµa = f dν. Then if ν(E) = 0, we have that f dν = 0, so µa << ν. Therefore, it suffices to show that µs ⊥ ν. E Suppose not. Then we may choose ε > 0 and E ∈ M such that ν(E) = 0 and E is a positive set for µs −εν. That is, µs ≥ εν on E. But then εχE dν ≤ χE dµs , so εχE dν ≤ χE (dµ − f dν), which implies that (εχE + f )dν ≤ µ(E). But this contradicts our selection of a as the supremum of all such integrals. Therefore, we conclude that µs ⊥ ν. This completes the proof. Problem 7: Prove Gorsat’s theorem: If f : C → C is complex differentiable (and so continuous), then for every triangle ∆ ⊂ C Z f (z)dz = 0 ∂∆ where line integral is over the three sides of the triangle. Solution. Let ∆ be a closed triangle in C and let a, b, c be the vertices of ∆. Let a0 , b0 , c0 be the midpoints of [b, c], [a, c], [a, b], respectively. Consider the four triangles ∆j , j = 1, 2, 3, 4, with vertices {a, c0 , b0 }, {b, a0 , c0 }, {c, b0 , a0 }, and {a0 , b0 , c0 }. Let Z J= f (z)dz. ∂∆ Then J= 4 Z X j=1 f (z)dz. ∂∆j Hence, Z ∂∆k f (z)dz ≥ |J/4| for some k. Define ∆1 to be this ∆k . Replace ∆ with ∆1 and repeat. This yields a sequence of triangles {∆n } such that ∆ ⊃ ∆1 ⊃ ∆2 ⊃ . . . ⊃ ∆n ⊃ ∆n+1 ⊃ . . . and the length of ∂∆n is 2−n L, where L is the length of ∂∆. Thus, we have that Z n |J| ≤ 4 f (z)dz for all n. ∂∆n T∞ Choose z0 ∈ n=1 ∆n , and we note that in fact this z0 is unique. Let ε > 0. As f is complex differentiable, we may choose r > 0 such that if |z − z0 | < r, then |f (z) − f (z0 ) − f 0 (z0 )(z − z0 )| ≤ ε |z − z0 | for z ∈ ∆. Choose n such that |z − z0 | < r for all z ∈ ∆n . Note also that |z − z0 | ≤ 2−n L for all z ∈ ∆n . We use the fact that R γ z n dz = 0 for any closed path γ ⊂ C and n = ANALYSIS QUALS 5 0, 1, 2, . . ., which can be seen considering antiderivatives, parametrization, and FTC. Then Z Z f (z)dz = f (z) − f (z0 ) − f 0 (z0 )(z − z0 )dz, ∂∆n ∂∆n so Z ∂∆n Z f (z)dz ≤ |f (z) − f (z0 ) − f 0 (z0 )(z − z0 )| dz ≤ ∂∆n Z ε |z − z0| ≤ ε(2−n L)2 . ∂∆n Then |J| ≤ εL2 . Hence, J = 0. Problem 8: (a) Define upper-semicontinuous for functions f : C → [−∞, ∞). (b) Define what it means for such an upper-semicontinuous function to be subharmonic. (c) Prove for refute (with a counterexample) each of the following: The pointwise supremum of a bounded family of subharmonic functions is subharmonic. The pointwise infimum of a family of subharmonic functions is subharmonic. (d) Let A(z) be a 2×2 matrix-valued holomorphic function (i.e., the entries are holomorphic). Show that z 7→ log (kA(z)k) is subharmonic where kA(z)k is the norm as an operator on the Hilbert space C2 . Proof. (a) f is defined to be upper-semicontinuous if for every α ∈ R, the set {z ∈ C : f (z) < α} is an open set. (b) f is defined to be subharmonic if for any closed ball B and every continuous h : B → R that is harmonic on the interior of B such that f (z) ≤ h(z) on the boundary of B, we have that f (z) ≤ h(z) for all z ∈ B. (c) First, let {fi }i∈I be a bounded family of subharmonic functions and define f (z) = supI fi (z). Let B be a closed ball. Suppose that h is continuous on B, harmonic on the interior, and f (z) ≤ h(z) on the boundary. Then for all i ∈ I, fi (z) ≤ h(z) on the boundary as well, so since fi is subharmonic, we conclude that fi (z) ≤ h(z) on all of B for all i ∈ I. Thus, we have that f (z) ≤ h(z) on B as well. The second claim is false. Consider fn (z) = Re(z n ), with f (z) = inf n fn (z). Let B be the closed unit ball. For almost every z ∈ ∂B, f (z) = −1, but f (0) = 0, so the sub-mean-value-property is violated: Z 1 f (0) 6≤ f (z)dz. 2π ∂B (d) We write kA(z)k = sup{Re hA(z)v, wi : v, w ∈ C2 , kvk = kwk = 1}. Thus, we have the family of functions fv,w defined by fv,w (z) = log Re hA(z)v, wi for v, w unit vectors of C2 . For each fixed v, w, this is the real part of a holomorphic function, so it is harmonic, and thus subharmonic. By part c, this means that z 7→ log (kA(z)k) is subharmonic, since it is the supremum of subharmonic functions. 6 ANALYSIS QUALS Problem 10: Let D = {z : |z| < 1} and let Ω = {z ∈ D : Imz > 0}. Evaluate i sup{Ref 0 ( )|f : Ω → D is holomorphic}. 2 Proof. Note: When I did this on the qual, I misread the problem and thought that Ω was the upper half-plane instead of merely the upper halfdisc, which caused the already awful calculations to get worse. Here is my attempt at the correct calculations, but it seems too ugly to be correct. Follow at your own peril. First, note that by rotation, it suffices to find the sup of f 0 ( 2i ) instead of Re f 0 ( 2i ). The idea is to use Schwarz Lemma. That is, we want g : D → Ω in order to have f ◦ g map D to D, and then tweak by automorphisms of the disc to get the map to send 0 to 0. √ 1+z We have that z 7→ i 1−z takes D to the upper half-plane, z 7→ z maps the upper half-plane to the first quadrant, and z 7→ z−1 z+1 maps the first quadrant to Ω. Thus, let r 1+z i 1−z −1 g(z) = r . 1+z i 1−z + 1 Let f1 be the automorphism of the disc taking g(0) to 2i , and f2 be the automorphism of the disc taking f ( 2i ) to 0. These guys can be written out explicitly; for example we recall that f2 (z) = z−f ( 2i ) 1−f ( 2i )z . Let h = f2 ◦ f ◦ f1 ◦ g. Then h : D → D with h(0) = 0. By Schwarz Lemma, |h0 (0)k ≤ 1. But we note that i i h0 (0) = g 0 (0)f10 (g(0))f 0 ( )f20 (f ( )). 2 2 By calculation, we have that i 1 i , f20 (f ( )) = 2 1 − f ( 2 ) and 1 g 0 (0) = √ √ , i( i + 1)2 and f10 (g(0)) = where we note that 1+ i 2 1 − |g(0)| √ i − 1 . |g(0)| = √ i + 1 2, ANALYSIS QUALS 7 Thus, as |h0 (0)| ≤ 1, we have that √ 2 0 i (1 − f ( 2i )) i + 1 (1 − |g(0)|2 ) f ( ) ≤ , 2 1 + i 2 and so √ 2 √ 2 0 i (1 − f ( 2i ))( i + 1 − i − 1 ) f ( ) ≤ . 2 1 + i 2 We have that equality holds if and only if h is a rotation. To maximize the right-hand-side of the above, we may take f ( 2i ) = 0 (which of course eliminates the need for f2 ). Then we conclude that 2 √ 2 √ 0 i i + 1 − i − 1 . sup f ( ) = 1 + i 2 2 This completes the proof. Fall 2010 Problem 1: For this problem, consider just Lebesgue measurable functions f : [0, 1] → R together with the Lebesgue measure. (a) State Fatou’s lemma (no proof required). (b) State and prove the Dominated ConvergenceR Theorem. (c) Give an example where fn (x) → 0 a.e., but fn (x)dx → 1. Solution. (a) Fatou: Let {fn } be a sequence of nonegative R R functions such that f = lim fn exists a.e. Then f (x)dx ≤ lim inf fn (x)dx. (b) DCT: Let {fn } be a sequence of functions such that fn → f pointwise a.e., and suppose there exists Lebesgue integrable g with kfn } ≤ g a.e. R R for every n. Then fn (x)dx → f (x)dx. Proof: Rg + fn → g + fR a.e., and g + fn is nonnegative a.e., so by Fatou, g + f ≤ lim inf (g + fn ). Then Z Z Z Z g + f ≤ g + lim inf fn so we have that Z Z f ≤ lim inf fn Similarly, R g − fn → g −Rf a.e. and g − fn is nonnegative a.e., so by Fatou, g − f ≤ lim inf (g − fn ). Then Z Z Z Z g − f ≤ g − lim sup fn so we have that Z lim sup R R Z fn ≤ f Hence, f = lim fn . (c) Consider the pointwise-linear bump functions fn defined to be 0 at 0, n at 1/n, 0 at 2/n, and linear between. 8 ANALYSIS QUALS Problem 2: Prove the following form of Jensen’s inequality: If f : [0, 1] → R is continuous, then Z 1 Z 1 f (x)dx . ef (x) dx ≥ exp 0 0 Moreover, if equality occurs then f is a constant function. R1 Solution. Let a = min[0,1] f , b = max[0,1] f , and t = 0 f (x)dx. Let et − es . a<s<t t − s β = sup Then for a < s < t, we have that β(t − s) ≥ et − es ⇒ β(s − t) ≤ es − et ⇒ β(s − t) + et ≤ es u t −e Then since ex is a convex function, β ≤ eu−t for all u ∈ (t, b). Thus, for t s t ≤ s < b, β(s − t) + e ≤ e . Hence this inequality holds for all s ∈ (a, b). Thus, for all x ∈ [0, 1], β(f (x) − t) + et ≤ ef (x) . Then Z 1 Z f (x) e dx ≥ 0 0 1 t 1 Z β(f (x) − t)dx + et β(f (x) − t) + e dx = 0 Z 1 f (x)dx − βt + et Z 1 = βt − βt + exp f (x)dx . =β 0 0 Clearly if f is constant, equality holds. If equality holds, then a = b, so f constant. Problem 5: Let R/Z denote the torus (whose elements we will write as cosets) and fix an irrational number α > 0. (a) Show that Z 1 N −1 1 X lim f (nα + Z) = f (x + Z)dx N →∞ N 0 n=0 for all continuous functions f : R/Z → R. (Hint: consider first f (x) = e2πikx ). (b) Show that the conclusion is also true when f is the characteristic function of a closed interval. Solution. (a) First, let f (x) = e2πikx . If k = 0, the statement is clearly true. Suppose k 6= 0. Then the statement is that Z 1 N −1 1 X 2πinkα e → e2πikx dx. N n=0 0 ANALYSIS QUALS 9 But by the geometric series, the LHS is equal to 1 1 − e2πiN αk N 1 − e2πiαk and since α ∈ / Q, theR denominator is never 0. Hence, this approaches 1 0 as N → ∞, and 0 e2πikx dx = 0, so the statement is true for all trigonometric polynomials. Now let ε > 0, f a continuous function from R/Z to R. As trigonometric polynomials are uniformly dense in the space of continuous periodic functions, we may choose P (x) trigonometric such that ε sup |f (x) − P (x)| < . 3 x∈R/Z Then for N sufficiently large, we have that N −1 Z 1 1 X ε P (nα) − P (x)dx < . N 3 0 n=0 Then Z 1 −1 1 NX f (nα) − f (x)dx N 0 n=0 Z Z N −1 N −1 1 X 1 1 1 X ≤ |f (nα) − P (nα)| + P (nα) − P (x)dx + |P (x) − f (x)| dx N N n=0 0 0 n=0 < ε. Hence, the statement holds for general f . ( (b) Choose fε+ , fε− continuous on the torus so that fε x) ≤ χ[a,b] (x) ≤ fε+ (x) and Z 1 Z 1 − b − a − 2ε ≤ fε (x)dx, fε+ (x)dx < b − a + 2ε. 0 Let SN = 1 N 1 N PN −1 n=0 N −1 X 0 χ[a,b] (nα). Then fε− (nα) ≤ SN ≤ n=0 But the left side approaches R1 + f (x)dx. Thus 0 ε R1 0 N −1 1 X + f (nα). N n=0 ε fε− (x)dx and the right side approaches b − a − 2ε ≤ lim SN ≤ b − a + 2ε. N →∞ As this holds for all ε > 0, we have that Z 1 lim SN = b − a = χ[a,b] (x)dx. n→∞ 0 Problem 6: Let D := {z ∈ C : |z| ≤ 1} and consider the (complex) Hilbert space ( ) ∞ ∞ X X 2 H := f : D → C|f (z) = fˆ(k)z k with kf k := (1 + |k|2 )|fˆ(k)|2 < ∞ k=0 k=0 10 ANALYSIS QUALS (a) Prove that the linear functional L : f 7→ f (1) is bounded. (b) Find the element g ∈ H representing L. (c) Show that f 7→ ReL(f ) achieves its maximal value on the set B := {f ∈ H : kf k ≤ 1 and f (0) = 0} , that this maximum occurs at a unique point, and determine this maximal value. Solution. (a) Using Cauchy-Schwarz, ∞ ∞ X X ˆ ˆ |L(f )| = f (k) ≤ f (k) k=0 k=0 !1/2 ∞ ! ∞ 2 1/2 X X 1 + |k|2 ˆ 1 2 ˆ p = (1 + |k| ) f (k) f (k) ≤ 1 + |k|2 1 + |k|2 k=0 k=0 k=0 !1/2 ∞ X 1 kf k . = 1 + |k|2 k=0 P∞ 1 k (b) Let ĝ(k) = 1+|k| 2 , g(z) = k=0 ĝ(k)z . Then g(z) converges for all ∞ X p z ∈ D, and ∞ X (1 + |k|2 )(1 + |k|2 )−2 < ∞, k=0 so g ∈ H. Then hf, gi = = ∞ X (1 + |k|2 )fˆ(k)ĝ(k) k=0 ∞ X k=0 1 + |k|2 ˆ f (k) 1 + |k|2 = f (1) = L(f ) (c) B is a compact, convex set in a Hilbert space, so a bounded linear functional on it must achieve a maximum at a unique point. We claim the maximizer is f (z) = √z2 . This is in B, and the handwavey argument of why this is the maximizer is that we want all possible weight on the smallest possible k, since the computation of the norm gives increasing penalty to larger k. Since f (0) = 0, fˆ(0) = 0, so we cannot put any weight on k = 0, and thus we put all possible weight on k = 1. Problem 7: Suppose that f : C → C is continuous on C and holomorphic on C − R. Prove that f is entire. Solution. Let Γ be a triangle in C. If Γ does not cross the real axis, R then as f is holomorphic off R, Cauchy integral theorem guarantees that Γ f (z)dz = 0. Now suppose Γ crosses the real axis. For k > 0, we cut a 1/k neighborhood (1) (2) out about the real axis and let Γk and Γk be the resulting polygons above and below the real axis, respectively, as shown in the figure. ANALYSIS QUALS Then 11 (j) R f (z)dx = 0 for all k, j = 1, 2. But Γk converges uniformly R R R R (j) to Γ with Γ(1) f (z)dz + Γ(2) f (z)dz = Γ f (z)dz. Hence, Γ f (z)dz = 0. Thus, by Morera’s theorem, f is entire. (j) Γk Problem 9: Let f (z) = ∞ X an z n n=0 be a holomorphic function in D. Show that if ∞ X n|an | ≤ |a1 | n=2 with a1 6= 0 then f is injective. P∞ Solution. Note that f 0 (z) = n=1 nan z n−1 . Then ∞ ∞ ∞ X X X 0 n−1 |f (z) − a1 | = nan z n|an ||z|n−1 < n|an | ≤ |a1 |. ≤ n=2 n=2 n=2 0 Hence, |f (z) − a1 | < |a1 |. Thus, we may choose θ so that Re(eiθ f 0 (z)) > 0 for all z ∈ D. Then if z1 6= z2 , Z z1 Z 1 0 |f (z1 ) − f (z2 )| = f (z)dz > |z2 − z1 | eiθ Re(f 0 (z1 + t(z2 − z1 )))dt > 0. z2 Thus, f (z2 ) 6= f (z1 ), so f is injective. 0 Problem 10: Prove that the punctured disc {z ∈ C : 0 < |z| < 1} and the annulus given by {z ∈ C : 1 < |z| < 2} are not conformally equivalent. 12 ANALYSIS QUALS Problem 11: Let Ω ⊂ C be a non-empty open connected set. If f : Ω → C is harmonic and f 2 is also harmonic, show that either f or f is holomorphic on Ω. Proof. Write f (x + iy) = u(x, y) + iv(x, y). By the Cauchy-Riemann equations, it suffices to show that either ∂v ∂v ∂u ∂u ∂v ∂v ∂u ∂u = and = − , or =− and = . ∂x ∂y ∂x ∂y ∂x ∂y ∂x ∂y Then as ∆f = 0, we have that ∆u = ∆v = 0. Note that f 2 = u2 (x, y) − v 2 (x, y) + 2iu(x, y)v(x, y). Thus, # " 2 2 ∂v ∂u ∂2u ∂2v 2 u(x, y) − v(x, y) ∆(f ) = 2 + − ∂x ∂x2 ∂x ∂x2 " # 2 2 ∂u ∂v ∂2u ∂2v +2 + 2 u(x, y) − − 2 v(x, y) ∂y ∂y ∂y ∂y ∂u ∂v ∂2u ∂2v ∂u ∂v ∂2u ∂2v +2i 2 + v(x, y) + u(x, y) + 2 + 2 v(x, y) + 2 u(x, y) ∂x ∂x ∂x2 ∂x2 ∂y ∂y ∂y ∂y But using ∆u = ∆v = 0, we get that " 2 2 2 # 2 ∂u ∂v ∂v ∂u ∂v ∂u ∂v ∂u 2 + − − + 4i + ∆(f ) = 2 ∂x ∂y ∂x ∂y ∂x ∂x ∂y ∂y We set this equal to 0 and separate real and imaginary parts to find that 2 2 2 2 ∂u ∂v ∂u ∂v ∂u ∂u ∂v ∂v =− and + = + . ∂x ∂x ∂y ∂y ∂x ∂y ∂x ∂y 2 ∂u ∂v 2 ∂v ∂u ∂v ∂v ∂u , so Thus, ∂u − i = + i − i = ± + i ∂x ∂x ∂y ∂y ∂x ∂x ∂y ∂y . This yields the result. Problem 12: Let F be the family of functions f holomorphic on D with Z Z 2 |f (x + iy)| dxdy < 1. x2 +y 2 <1 Prove that for each compact subset K ⊂ D there is a constant A so that |f (z)| < A for all z ∈ K and all f ∈ F. Solution. Let K ⊂ D be compact. Let B = B(x, r) ⊂ D. Let f ∈ F. As f is Principle, we have that Z 1 f (z) = f (w)dw = Area(B) B C ) r = d(K,D . Let x ∈ K. Then 2 holomorphic, by the Mean Value 1 πr2 Z f (w)dw. B Hence we have that Z Z 1 1 |f (z)| ≤ 2 |f (w)| dw = 2 |f (w)| |χB (w)| dw. πr B πr D Then using Cauchy-Schwarz, we have that Z 1/2 Z 1/2 1 2 2 |f (w)| ≤ 2 |f (w)| dw |χB (w)| dw πr D D ANALYSIS QUALS 13 Z 1/2 1 1 2 =√ <√ , |f (w)| πr πr D since f ∈ F. Thus, let A = √1πr , which depends only on K and does not depend on x or f . Spring 2010 Problem 1: (a) Let 1 ≤ p < ∞. Show that if a sequence of real-valued functions {fn }n≥1 converges in Lp (R), then it contains a subsequence that converges almost everywhere. (b) Give an example of a sequence of functions converging to zero in L2 (R) that does not converge almost everywhere. Solution. (a) By Chebyshev’s inequality, p µ ({x : |f (x)| > α}) ≤ kf kp αp . Then we have that p µ ({x : |fn (x) − f (x)| > ε}) ≤ kfn − f kp εp → 0 as n → ∞, so fn → f in measure. Then we may choose a subsequence {fnj } such that µ x : fnj (x) − fnj+1 (x) > 2−j ≤ 2−j . S∞ Let Ej = {x : |fnj (x) − fnj+1 (x)| > 2−j }. Let Fk = j=k Ej . Then µ(Fk ) ≤ ∞ X µ(Ej ) ≤ 21−k . j=k For x ∈ / Fk and k ≤ j ≤ i, we have that i−1 i−1 X X fn (x) − fn (x) ≤ fn (x) − fn (x) ≤ 2−l ≤ 21−j , j i l l+1 l=j l=j T∞ so {fnj } is pointwise Cauchy on Let F = k=1 Fk . Then µ(F ) = 0 and fnj converges on F C , so fnj converges almost everywhere. (b) We use the example of the moving dyadic intervals. Let f1 = χ[0,1] , f2 = χ[0,1/2] , f3 = χ[1/2,1] , and in general, for n = 2k + j < 2k+1 , let fn = χ[j/2k ,(j+1)/2k ] . Then fn → 0 in L2 , but for every x ∈ [0, 1], fn (x) 6→ 0 as n → ∞. FkC . Problem 3: For an f : R → R belonging to L1 (R), we define the HardyLittlewood maximal function as follows: Z x+h 1 |f (y)|dy. (M f )(x) := sup h>0 2h x−h Prove that it has the following property: There is a constant A such that for any λ > 0, A |{x ∈ R (M f )(x) > λ}| ≤ kf kL1 λ 14 ANALYSIS QUALS where |E| denotes the Lebesgue measure of E. If you use a covering lemma, you should prove it. Solution. We do employ the following covering lemma, due to Weiner in the style of Vitali. S Lemma: Let C be a collection of open balls in Rn , and let U = B∈C B. k X If c < m(U ), there exist disjoint B1 , B2 , . . . , Bk ∈ C with m(Bj ) > 3−n c. j=1 Pf: Choose compact K ⊂ U with m(K) > c. Then finitely many balls A, A2 , . . . , Am ∈ C cover K. Let B1 be the largest of the Ai , let B2 be the largest of the Ai disjoint from B1 , and inductively, let Bk be the largest of the Ai disjoint from the previously chosen Bj s. Then for any i, if Ai is not one of the Bj , it has nonempty intersection with a Bj . Choosing the smallest such j, we have diam(Bj ) ≥diam(Ai ). ∗ Let Bj∗ be S the∗ball concentric with Bj with thrice the radius. Then Ai ⊂ Bj , so K ⊂ j Bj . Hence [ X X c < m(K) ≤ m( Bj∗ ) = m(Bj∗ ) = 3n m(Bj ). This proves the lemma. From this lemma, the problem is solved easily: Let Eλ = {x : (M f )(x) > λ}. For each x ∈ Eλ , choose rx > 0 such that Z 1 |f (y)|dy > λ. m(B(x, rx )) B(x,rx ) The balls B(x, rx ) cover Eλ , so by the lemma, if C <Pm(Eλ ), some finite collection of disjoint balls {Bj = B(xj , rxj )} satisfies m(Bj ) > 31 c. But then Z X 3X 3 c<3 m(Bj ) ≤ |f (y)|dy ≤ kf kL1 . λ λ Bj j j Letting c → m(Eλ ) yields the result. Problem 4: Let f be continuous on D and analytic on D with f (0) 6= 0. (a) Prove that if 0 < r < 1 and inf |z|=r |f (z)| > 0, then Z 2π 1 log |f (reiθ )|dθ ≥ log |f (0)|. 2π 0 (b) Use (a) to prove that |{θ ∈ [0, 2π] : f (eiθ ) = 0}| = 0. Solution. (a) This is a version of Jensen’s Theorem, a result of which it is said that the proof is easier to remember than the theorem. Let zj enumerate the zeroes of f in |z| ≤ r with multiplicity. Consider the Blashke factor Y r(z − zj ) B(z) = . r2 − zj z j ANALYSIS QUALS 15 Then B has the same zeroes as f , and |B(z)| = 1 when |z| = r. Then f (z) log B(z) is harmonic, so by the mean value principle, we have that Z 2π f (0) = 1 log log f (reiθ ) dθ. B(0) 2π 0 But z X f (0) j = log |f (0)| − log log , B(0) r j and thus Z 2π z X j log f (reiθ ) dθ + log . r 0 j zj zj P But r < 1, so j log r ≤ 0, so we have that Z 2π 1 log |f (reiθ )|dθ ≥ log |f (0)|. 2π 0 R 2π (b) If |{θ ∈ [0, 2π] : f (eiθ ) = 0}| > 0, then 0 log f (reiθ ) dθ → −∞ as r → 1, which contradicts the inequality in (a). log |f (0)| = 1 2π Problem 5: (a) For f ∈ L2 (R) and {xn } converging to zero, define fn (x) = f (x + xn ). Show that {fn } converges to f in L2 . (b) Let W ⊂ R be Lebesgue measurable of positive measure. Show that W − W = {x − y : x, y ∈ W } contains an open neighborhood of the origin. Solution. (a) We want to show that kfn − f k2 → 0. Since Cc (R) is dense in L2 , choose g ∈ Cc (R) such that kf − gk2 < ε. Then kfn − f k2 ≤ kfn − gn k2 + kgn − gk2 + kg − f k2 < 2ε + kgn − gk2 Chose δ so that |g(x) − g(y)| < ε when |x − y| < δ, and choose N so that if n > N , |xn | < δ. Then for n > N , we have that Z 2 2 kgn − gk2 = |g(x + xn ) − g(x)| dx ≤ ε2 2supp(g). R Thus, kfn − f k2 → 0 as n → ∞. (b) Here are two solutions, neither of which uses part (a). (1) We use the fact that Z f (x) = χW (y)χW (y − x)dy R is continuous, which is Q3 in Fall 2004. Then since W has positive measure, f (0) > 0. As f is continuous, there exists δ > 0 such that f (t) >R 0 for t ∈ (−δ, δ). Then for each t in this interval, we have that R χW (y)χW (y − t)dy > 0, so there exists yt such that χW (yt ) = χW (yt − t) = 1. Thus yt ∈ W and yt − t ∈ W . Hence t = yt − (yt − t) ∈ W − W . Hence (−δ, δ) ⊂ W − W . 16 ANALYSIS QUALS (2) As W has positive measure, then by the definition of outer meaS sure, there exists a collection of closed intervals In such that W ⊂ In and ∞ X |In | ≤ (21/20)m(W ). n=1 There exists n such that m(W ∩ I) ≥ (9/10)m(In ), for if not, then we have X X 9 9 m(In ) ≤ 2120m(W ) < m(W ), m(W ) ≤ m(W ∩ In ) < 10 10 which is a contradiction. Thus, let In = I = [a, b], so m(W ∩ I) ≥ (9/10)m(I). Let W0 = W ∩ I. Then it suffices to show that W0 − W0 contains an open neighborhood. Suppose not. Then for all n, choose an ∈ (−1/n, 1/n) \ (W0 − W0 ). Consider W0 ∩ (W0 + an ). If x ∈ W0 ∩ (W0 + an ), then x = y + an for y ∈ W0 , so an ∈ W0 − W0 . Hence, W0 ∩ (W0 + an ) = ∅ for all n. But W0 ∪ (W0 + an ) ⊂ I ∪ (I + an ) ⊂ [a − |an |, b + |an |], and thus m(W0 ∪ (W0 + an )) ≤ b − a + 2|an | → m(I) as n → ∞. However, since an 6∈ W0 − W0 , we have that 18 m(I) m(W0 ∪ (W0 + an )) = 2m(W0 ) ≥ 10 so we must have m(I) = 0, which is a contradiction. This completes the proof. Problem 7: Let H be a Hilbert space and let E be a closed convex subset of H. Prove that there exists a unique element x ∈ E such that kxk = inf kyk . y∈E Solution. Let d = inf y∈E kyk. Choose {yn } ∈ E such that kyn k → d. Then yn + ym 2 2 2 2 . kyn − ym k = 2 kyn k + 2 kym k − 4 2 2 m m ∈ E, so yn +y As E is convex, we have that yn +y ≥ d2 . Thus, 2 2 2 2 2 kyn − ym k ≤ 2 kyn k + 2 kym k − 4d2 → 0 as n, m → ∞. Thus, {yn } is Cauchy, and since H is complete, the sequence converges to a point x. Then x ∈ E since E is closed, and kxk = d. This proves existence. If y also satisfies kyk = d, then x + y 2 x + y 2 2 2 2 2 ≤ 4d2 − 4d2 = 0, 0 ≤ kx − yk = 2 kxk + 2 kyk − 4 = 4d − 4 2 2 so x = y. This proves uniqueness. ANALYSIS QUALS 17 Problem 8: Let F (z) be a non-constant meromorphic function on the complex plane C such that for all z ∈ C, F (z + 1) = F (z) and F (z + i) = F (z). Let Q be a square with vertices z, z + 1, z + i and z + (1 + i) such that F has no zeros and no poles on ∂Q. Prove that inside Q the functions F has the same number of zeros as poles (counting multiplicities). Solution. As F has no zeroes or poles on ∂Q, by the argument principle, we have that Z F 0 (w) dw = 2πi(#zeroes − #poles). ∂Q F (w) By periodicity, F (w) = F (w + 1) = F (w + i) and F 0 (w) = F 0 (w + 1) = F 0 (w + i). Thus, we have that Z z+i+1 0 Z z+1 0 F (w) F (w) dw = dw, F (w) F (w) z+i z and Z z+i z Z ∂Q F 0 (w) dw = F (w) Z z+1+i z+1 F 0 (w) dw. F (w) Thus, we have that Z z+1 0 Z z+1+i 0 Z z+i+1 0 Z z+i 0 0 F (w) F (w) F (w) F (w) F (w) dw = dw+ dw− dw− dw F (w) F (w) F (w) F (w) F (w) z z+1 z+i z so we conclude that Z ∂Q F 0 (w) dw = 0, F (w) and hence the number of zeroes is equal to the number of poles. Problem 10: Let Ω ⊂ C be a connected open set, let z0 ∈ Ω, and let U be the set of positive harmonic functions u on Ω such that u(z0 ) = 1. Prove for every compact set K ⊂ Ω there is a finite constant M (depending on Ω, z0 , and K) such that sup sup u(z) ≤ M. u∈U z∈K You may use Harnack’s inequality for the disk without proving it, provided you state it correctly. Solution. We will use Harnack’s inequality for the disk without proving, so let’s hope we state it correctly. Here I am using the version from Conway. Harnack’s inequality states that if u is a non-negative harmonic function on B(a, R), then for every 0 ≤ r < R and every θ, we have R+r R−r u(a) ≤ u(a + reiθ ) ≤ u(a). R+r R−r First, we note that it suffices to prove that for every z ∈ Ω, there exists an open neighborhood Uz of z and a constant Cz such that sup sup u(w) ≤ Cz . u∈U w∈Uz 18 ANALYSIS QUALS If this holds, then since any compact K can be covered by finitely many Uz , we can let M be the maximum of the resulting finitely many Cz , and we will be done. Let E = {z ∈ Ω : there exists Uz , Cz as above}. We claim that E is nonempty, open, and closed. Since Ω is connected, this will show that E = Ω. First, E is obviously open: for z ∈ E, every w ∈ Uz is also in E with, for example, Cw = Ck and appropriately chosen Uw ⊂ Uz . E is nonempty: Choose ε such that B(z0 , ε) ⊂ Ω. We apply Harnack with a = z0 , R = ε, r = |w − z0 |, and see that for all u ∈ U, w ∈ B(z0 , ε/2), we have that ε + |w − z0 | u(z0 ) ≤ Cz0 , u(w) ≤ ε − |w − z0 | for some constant Cz0 . Thus, z0 ∈ E. E is closed: let z ∈ Ω, zn ∈ E such that zn → z. Choose ε so that B(z, ε) ⊂ Ω. Choose n large enough so that |zn − z| < 4ε . Let Uz = B(z, 4ε ). Let u ∈ U and w ∈ Uz . Then |w − zn | < 2e . We apply Harnack’s inequality with a = zn , R = 3ε 4 , and r = |w − zn |, and we see that u(w) ≤ 3ε 4 3ε 4 + |w − zn | u(zn ) ≤ KCzn , − |w − zn | where K is some constant. We let Cz = KCzn . Thus, z ∈ E, so E is closed. This completes the proof. Problem 11: Let φ : R → R be a continuous function with compact support. (a) Prove there is a constant A such that kf ∗ φkLq ≤ A kf kLp for all 1 ≤ p ≤ q ≤ ∞ and all f ∈ Lp . If you use Young’s (convolution) inequality, you should prove it. (b) Show by example that such a general inequality cannot hold for p > q. Problem 13: Let X and Y be two Banach spaces. We say that a bounded linear transformation A : X → Y is compact if for every bounded sequence {xn }n≥1 ⊂ X, the sequence {Axn }n≥1 has a convergent subsequence in Y . Suppose X is reflexive (that is, (X ∗ )∗ = X) and X ∗ is separable. Show that a linear transformation A : X → Y is compact if and only if for every bounded sequence {xn }n≥1 ⊂ X, there exists a subsequence {xnj } and a vector φ ∈ X such that xnj = φ + rnj and Arnj → 0 in Y . Solution. The reverse direction is easy: Let {xn } be bounded. Choose {xnj } such that xnj = φ + rnj and Arnj → 0 in Y . Then Axnj = Aφ + Arnj → Aφ as j → ∞, which proves this direction. For the forward direction, suppose A is compact and let {xn } be bounded. WLOG, pass to a subsequence and assume Axn → y ∈ Y . Suffices to show y ∈ A(X). ANALYSIS QUALS 19 Consider x̂n ∈ X ∗∗ . This is a bounded sequence, so by Banach-Alaoglu, there is a weak* convergent subsequence x̂nj → x̂. That is, for every f ∈ X ∗ , x̂nj (f ) → x̂(f ). But then f (xnj ) → f (φ), where φ is the canonical preimage of x̂ in X, using reflexivity. Hence, xnj → φ weakly in X. We want A(xnj ) → A(φ); that is, A(φ) = y. Assume for sake of contradiction that this does not hold. Then by the Hahn-Banach theorem, we may choose g ∈ Y ∗ separating A(φ) from y. Then g ◦ A ∈ X ∗ , and g ◦ A(xnj ) → g(y) 6= g(A(φ)). But xnj → φ weakly, so (g ◦ A)(xnj ) → (g ◦ A)(φ), a contradiction. Thus, A(xnj ) → A(φ), and we may let rnj = xnj − φ, and we have that A(rnj ) → 0. This completes the proof. Fall 2009 Problem 1: Find a non-empty closed set in the Hilbert space L2 ([0, 1]) that does not contain an element of smallest norm. Prove your assertion. Solution. Let en (x) = (1 + 2−|n| )e2πinx . Then ken k → 1 as |n| → ∞, but ken k = 6 1 for all n. 2 Also, ken − em k2 = (1 + 2−|n| )2 + (1 + 2−|m| )2 ≥ 2, so the set is discrete, and therefore any L2 convergent sequence must be constant. Hence, the set is closed. Problem 4: Prove the following variant of the Lebesgue Differentiation theorem: Let µ be a finite Borel measure on R, singular with respect to Lebesgue measure. Then for Lebesgue-almost every x ∈ R, µ([x − ε, x + ε]) = 0. lim ε→0 2ε Proof. Let m be Lebesgue measure. Since µ ⊥ m, we may choose A Borel such that µ(A) = m(Ac ) = 0. Note that µ([x − ε, x + ε]) |µ| ([x − ε, x + ε]) ≤ , 2ε 2ε so it suffices to assume µ is positive. Let 1 µ([x − ε, x + ε]) Fk = {x ∈ A : lim sup ≥ }. 2ε k ε→0 S Then F = Fk is the set of x in which the conclusion fails. Thus, it suffices to show that m(Fk ) = 0 for all k. For δ > 0, choose Uδ ⊃ A open such that λ(Uδ ) < δ. For every x ∈ Fk , there S exists a ball Bx = B(x, rx ) ⊂ Uδ such that kµ(Bx ) > m(Bx ). Let Vδ = x∈Fk Bx . By the Weiner covering lemma, for c < m(Vδ ), there exist P disjoint Bx1 , . . . , Bxl such that m(Bxj ) > 3−n c. Thus, we have that c < 3n l X m(Bxj ) ≤ 3n k j=1 l X µ(Bxj ) ≤ 3n kµ(Uδ ) ≤ 3n kδ. j=1 n Thus, m(Vδ ) ≤ 3 kδ, so we must have that m(Fk ) = 0. 20 ANALYSIS QUALS Problem 5: Construct a Borel subset E of the real line R such that for all intervals [a, b] we have 0 < m(E ∩ [a, b]) < |b − a| where m denotes Lebesgue measure. Proof. We will use the Baire Category Theorem on the complete metric space X = {χE ∈ L1 : E Borel . First, to see that X is complete, it suffices to show that X is closed in L1 . Suppose fn → f in L1 , where fn ∈ X. Then there is a subsequence fnk that converges to f pointwise a.e. (cf. another common qual problem). But as characteristic functions converge pointwise a.e. to f , we must have that f is almost everywhere a characteristic function of some set F . As every set differs by a measure zero set from a Borel set, we can assume that F is Borel. Thus, f ∈ X, so X is complete. Now for a non-trivial interval I = [a, b], we define the set UI = {χE ∈ X : 0 < m(E ∩ I) < m(I)}. We claim that UI is open and dense in X. First, UI is dense in X: if χE 6∈ UI , then either m(E ∩I) = 0 or m(E ∩I) = m(I). In the former case, we can add a small amount of measure to E, and in the latter case, we can remove a small amount of measure from E to obtain E 0 with kE − E 0 kL1 < ε, and E 0 ∈ X. This gives density. To see that UI is open in X, we note that χE 7→ m(E ∩ I) is a continuous function on L1 restricted to X. Thus, if 0 < m(E ∩ I) < m(I), then 0 < m(E 0 ∩ I) < m(I) for all E 0 such that χE 0 is L1 -close to χE , so UI is open. Thus, by the Baire Category Theorem, \ UI is dense in X, I=[a,b],a,b∈Q so in particular there exists a Borel F such that χF is in the intersection. Since every non-trivial interval contains an interval with rational endpoints, this F does the job, which completes the proof. Problem 7: a) Define unitary operator on a complex Hilbert space. b) Let S be a unitary operator on a complex Hilbert space. Using your definition, prove that for every complex number |λ| < 1 the operator S − λI is invertible. Here I denotes the identity operator. c) For a fixed vector v in the Hilbert space and all {λ ∈ C : |λ < 1}, we define h(λ) = (S + λI)(S − λI)−1 v, v . Show that Re h is a positive harmonic function. [You may not invoke the spectral theorem - this is part of a proof of that theorem.] Solution. a) An operator U : H → H is a unitary operator if U is bounded, linear, and U ∗ U = U U ∗ = I. ANALYSIS QUALS 21 b) Suppose S − λI is not invertible. Choose v ∈ H such that |v| = 1, and (S − λI)v = 0. Then Sv = λv. Hence 2 2 1 = hv, vi = h(S ∗ S)v, vi = hSv, Svi = hλv, λvi = |λ| hv, vi = |λ| . Hence, |λ| = 1. Thus, conversely, |λ| < 1 implies that S − λI is invertible. c) We will prove that h is holomorphic, which means that Re h is harmonic, and then we will prove that Re h > 0. First, if v = 0, the statement is false. Let v 6= 0. Then (S + λI)(S − λI)−1 = (S + λI)(I − λS −1 )−1 S −1 = (S + λI)(I + λS −1 + λ2 S −2 + . . .)S −1 = (S + λI)(S −1 + λS −2 + λ2 S −3 + . . .) = I + 2λS −1 + 2λ2 S −2 + . . . Hence, (S + λI)(S − λI)−1 v, v = hv, vi + 2λ S −1 v, v + 2λ2 S −2 v, v + . . . This is a power series expansion of h about 0, so h is analytic in the disc. Hence, h is holomorphic on |λ| < 1, and thus Re h is harmonic. Now let w = (S − λI)−1 v, and as v 6= 0, we have that w 6= 0 by part b. Then h(λ) = h(S + λI)w, (S − λI)wi = hSw, Sw − λwi + hλw, Sw − λwi = hSw, Swi − hSw, λwi + hλw, Swi − hλw, λwi 2 2 2 2 2 2 = kSwk − |λ| kwk + hSw, λwi − hSw, λwi = kSwk − |λ| kwk + 2iIm hSw, λwi 2 2 2 = kwk − |λ| kwk + 2iIm hSw, λwi 2 2 2 Hence, Re h(λ) = kwk (1 − |λ| ) > 0 as |λ| < 1. Problem 8: Let Ω be an open convex region in the complex plane. Assume f is a holomorphic function on Ω and the real part of its derivative is positive: Re(f 0 (z)) > 0 for all z ∈ Ω. a) Prove that f is one-to-one. b) Show by example that the word “convex”’ cannot be replaced by “connected and simply connected”’. Solution. a) Suppose z1 6= z2 ∈ Ω. Then as Ω is convex, the line between z1 and z2 is contained in Ω. Thus, Z z 1 Z z1 Z z1 |f (z1 ) − f (z2 )| = f 0 (z)dz = Re(f 0 (z))dz + i Im(f 0 (z))dz z2 z2 z2 Z z1 ≥ Re(f 0 (z))dz > 0. z2 22 ANALYSIS QUALS Thus, f (z1 ) 6= f (z2 ), so f is injective. 3π b) Let Ω = {z : − 3π 4 < arg(z) < 4 }. This is clearly connected and 5/3 simply connected. Let f (z) = z , with an appropriate branch cut so that f is holomorphic on Ω. Then f 0 (z) = But π2 < 23 θ < 5 2/3 . 3z π , and 2 Let reiθ ∈ Ω. Then f 0 (reiθ ) = so Re(f 0 (z)) > 0 for all z ∈ Ω. 5 2/3 2iθ/3 e . 3r However, f (e7πi/10 ) = e7πi/6 = e−5πi/6 = f (e−π/2 ). Hence f is not injective on Ω. Problem 9: Let f be a non-constant meromorphic function on the complex plane C that obeys √ √ f (z) = f (z + 2) = f (z + i 2). (In particular, the poles of these three functions coincide.) Assume f has at most one pole in the closed unit disc D. a) Prove that f has exactly one pole in D. b) Prove that this is not a simple pole. Solution. a) Let z = e5πi/4 . Then by the periodicity, √ f takes√ all its possible values on the square with corners at z, z + 2, z + i 2, and √ √ z + 2 + 2. This square is then contained in D, so it suffices to show that there is a pole in this square, P . If not, then f is entire. But since ¶ is compact, f is bounded, so by Liousville, f would be constant, a contradiction. Thus, f has a pole in P . b) Using the periodicity, we have that Z z+√2 Z z+√2+i√2 f (w)dw = f (w)dw √ z and Z z+i 2 √ z+i 2 Z f (w)dw = z √ √ z+ 2+i 2 √ z+ 2 f (w)dw. Thus, we have that Z f (w)dw = 0. ∂P But by the residue theorem, we have that Z X ∂P f (w)dw = 2πi Res(f, z0 ). poles z0 P Thus, Res(f ) = 0. Thus, counting multiplicities, f has at least two poles. But as f has only one pole, it cannot be simple. ANALYSIS QUALS 23 Problem 12: Let f be a non-constant meromorphic function in the complex plane. Assume that if f has a pole at the point z ∈ C, then z is of the form nπ with an integer n ∈ Z. Assume that for all non-real z we have that estimate −1 |f (z)| ≤ (1 + |Im(z)| )e−|Im(z)| Prove that for every integer n ∈ Z, f has a pole at the point πn. Proof. By the estimate, we have that f (πn + ir) = O(r−1 ) for r real, so all poles are simple poles with residue of norm at most one. Note that |sin(z)| ≤ e|Im(z)| Then letting g(z) = f (z) sin(z), it suffices to show that g is constant, which would imply that f is a constant multiple of cosecant, which has poles at nπ for all n. First, since sin(z) has simple zeroes for all nπ, we have that g is entire. Secondly, using our estimates for f and sin, we have that |g(z)| ≤ 1 + |Im(z)| −1 for z ∈ C\R. Thus, |g| ≤ 2 off the strip |Im(z)| ≤ 1. We consider g on this strip, and in particular, on each rectangle with vertices ±nπ ± i. On the horizontal segments of the rectangles, we have the same estimate as before, |g| ≤ 2. For the vertical segments, we use the fact that sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y). Then we have that |sin(nπ + iy)| = |sinh(y)| ≤ C|y| = C|Im(z)| where the constant C does not depend on n. Thus, −1 g ≤ C |Im(z)| (1 + |Im(z)| )e−|Im(z)| < K on the vertical segments. Therefore, g is bounded on the edges of the rectangle, and the bound depends not on n. Thus, by the maximum modulus principle, g is bounded on the interior of this rectangle, and as this bound does not depend on n, we may conclude that g is bounded on the strip. Thus, g is bounded on C, so by Liousville, g is constant. This completes the proof. Spring 2009 Problem 1: Let f and g be real-valued integrable functions on a measure space (X, B, µ), and define Ft = {x ∈ X : f (x) > t}, Gt = {x ∈ X : g(x) > t}. Prove Z Z ∞ µ((Ft \Gt ) ∪ (Gt \Ft ))dt. |f − g|dµ = −∞ 24 ANALYSIS QUALS Solution. Let E1 = {(x, t) : x ∈ X, t ∈ R, f (x) > t and g(x) < t}. Let E2 = {(x, t) : x ∈ X, t ∈ R, f (x) < t and g(x) > t}. Then Z Z Z |f − g|dµ = |f − g|dµ + |f − g|dµ {f >g} {f <g} Z Z = f − gdµ + g − f dµ {f >g} {g>f } Z Z ∞ = {f >g} ∞ Z χE1 (x, t)dt dµ(x) + −∞ Z {g>f } ∞ χE2 (x, t)dt dµ(x) −∞ Z Z = (χE1 + χE2 )(x, t)dtdµ(x), −∞ since E1 and E2 are disjoint, and χE1 vanishes off {f > g} in X, and χE2 vanishes off {g > f } in X. Then by Fubini, we have that Z Z ∞ Z ∞Z (χE1 + χE2 )(x, t)dt, dµ(x) = (χE1 + χE2 )(x, t)dµ(x)dt. −∞ −∞ But Z Z χE1 dµ = µ(Ft \Gt ) and χE2 dµ = µ(Gt \Ft ), and as these sets are disjoint, µ((Ft \Gt )∪(Gt \Ft )) = µ(Ft \Gt )+µ(Gt \Ft ). Thus, Z ∞Z Z ∞ (χE1 + χE2 )(x, t)dµ(x)dt = µ((Ft \Gt ) ∪ (Gt \Ft ))dt. −∞ −∞ Problem 2: Let H be an infinite dimensional real Hilbert space. a) Prove the unit sphere S of H is weakly dense in the unit ball B of H. b) Prove there is a sequence Tn of bounded linear operators from H to H such that kTn k = 1 for all n but lim Tn (x) = 0 for all x ∈ H. Proof. a) Let x0 ∈ B, ε > 0. Let Uε (x0 ) be an ε-neighborhood of x0 in the weak topology. That is, ( ) Uε (x0 ) = x ∈ H : sup |hx − x0 , yi i| < ε for k ∈ N, yi ∈ H . i=1,...,k Let fi (x) = hx, yi i. Then fi is a linear functional. As H is infinite dimensional, the intersection of the kernels of a finite number of linear functionals is a nontrivial linear subspace. Thus, x0 + k \ ker(fi ) i=1 is contained in Uε (x0 ), so there is a line L through x0 contained in Uε (x0 ). This line intersects S, so choose xε ∈ L ∩ S. Then hx0 , yi = lim hxε , yi for all y ∈ H. ε→0 Thus, S is weakly dense in B. ANALYSIS QUALS 25 b) I present the proof in the (simple) case in which H is a separable Hilbert space (which is how Stein-Shakarchi defines their Hilbert spaces). In this case, let {en } be an orthonormal basis for H. Define Tn (y) = 1 kyk hen , yi y. Then kTn (y)k ≤ kyk , and kTn (en )k = 1, so kTn k = 1. But by Bessel’s inequality, X 2 2 |hen , yi| ≤ kyk , so hen , yi → 0 for all y. n Thus, lim Tn (y) = 0 for all y ∈ H. Problem 3: Let X be a Banach space and let X ∗ be its dual Banach space. Prove that if X ∗ is separable the X is separable. Solution. Let {fn } be a countable dense set of X ∗ . For each n, choose xn ∈ X such that kxn k = 1 and |fn (xn )| ≥ 12 kfn k. Let S be the set of all linear combinations of the xn with rational coefficients (rational real and imaginary parts if X is a Banach space over C). Thus, S is countable. We aim to show that S is dense in X. Let Y = S and assume for sake of contradiction that Y 6= X. Then by Hahn-Banach theorem, we may choose f ∈ X ∗ such that f (y) = 0 for all y ∈ Y and f 6= 0. Then by density of {fn }, choose a sequence fin → f . Then 1 kf − fin k ≥ |(f − fin )(xin )| = |fin (xin )| ≥ kfin k 2 for all n ∈ N. But then as fin → f , we must have fin → 0, which means that f = 0, a contradiction. Thus Y = X. Problem 4: Let f (x) be a non-decreasing function on [0, 1]. You may assume the theorem thatR f is differentiable almost everywhere. 1 a) Prove that 0 f 0 (x)dx ≤ f (1) − f (0). b) Let {fn } be a sequence of non-decreasing P∞ functions on the unit interval [0, 1], such that the seriesP F (x) = n=1 fn (x) converges for all x ∈ ∞ [a, b]. Prove that F 0 (x) = n=1 fn0 (x) almost everywhere on [0, 1]. Proof. a) Define the difference quotient gn (x) = f (x + 1/n) − f (x) , 1/n and then by the theorem we are permitted to assume, gn (x) → f 0 (x) as n → ∞ for a.e. x. As f is non-decreasing, we have that gn is non-negative, so by Fatou, we have that Z 1 Z 1 0 f (x)dx ≤ lim inf gn (x)dx. 0 0 26 ANALYSIS QUALS Z 0 1 But then Z 1 Z 1 1 1 gn (x)dx = f (x + 1/n)dx − f (x)dx 1/n 0 1/n 0 Z 1+1/n Z 1/n 1 1 = f (x)dx − f (x)dx → f (1) − f (0) as n → ∞. 1/n 1 1/n 0 P b) We let rn (x) = k≥n fk (x). Hence, for a.e. x, we can write F (x) = n−1 X fj (x) + rn (x), j=1 and so F 0 (x) = n−1 X fj0 (x) + rn0 (x). j=1 Thus, it suffices to show that rn0 (x) → 0 as n → ∞. Note that as fn is non-decreasing for all n, we have that rn is also non-decreasing for all n, and thus rn0 is non-negative. Note further that since F is a convergent series, we have that rn (x) → 0 for all x ∈ [0, 1]. Thus, rn (1) − rn (0) → 0. But by the first part, we have that Z 1 rn0 (x)dx ≤ rn (1) − rn (0). 0 Thus, R1 r0 (x)dx 0 n → 0 as n → ∞, so rn0 (x) → 0 for a.e. x ∈ [0, 1]. Problem 5: Let I = I0,0 = [0, 1] be the unit interval, and for n = 0, 1, 2, . . . , and 0 ≤ j ≤ 2n − 1 let In,j = [j2−n , (j + 1)2−n ]. P2n −1 R For f ∈ L1 (I, dx) define En f (x) = j=0 2n In,j f dt χIn,j . Prove that if f ∈ L1 (I, dx) then limn→∞ En f (x) = f (x) almost everywhere on I. Solution. Let S = {f ∈ L1 (I, dx) : En f → f a.e. on I}. Then S contains the characteristic functions of dyadic intervals, as Em χIn,j = χIn,j for m > n. But we can approximate a positive L1 function f by a sequence of positive linear combinations of characteristic functions of dyadic intervals. Thus, such an f is in S by monotone convergence. S is a vector space, so writing f as the difference of its positive part and negative part yields the result. Problem 8: Let f be an entire non-constant function that satisfies the functional equation f (1 − z) = 1 − f (z) for all z ∈ C. Show that f (C) = C. ANALYSIS QUALS 27 Solution. Picard’s Big Theorem states that if f is entire and non-constant, its range omits at most one value. Suppose w 6∈ f (C). Then 1 − w 6∈ f (C). Note that t = 1 − t implies t = 1/2, and f (1/2) = 1 − f (1/2), so f (1/2) = 1/2 ∈ f (C). Thus, t 6= 1 − t, and neither is in the range, so the range of f omits at least two values, which gives a contradiction of Picard. Hence, f (C) = C. Problem 10: Let D be the open unit disc and µ be Lebesgue measure on D. Let H be the subspace of L2 (D, µ) consisting of holomorphic functions. Show that H is complete. Solution. First, since L2 (D, µ) is complete, it suffices to show that H is a closed subspace of L2 (D, µ). Let f ∈ H. Let K be a compact subset of D. Let r = dist(K, U c ). Let z ∈ K, B = B(z, r/2) ⊂ U . By mean value principle and Cauchy-Schwarz, we have that Z 1 f (w)dw |f (z)| = Area(B) B Z 4 ≤ 2 |f (w)| dw πr B Z 4 |f (w)| χB dw = 2 πr U Z 1/2 Z 1/2 4 2 2 ≤ 2 |f (w)| dw χB dw πr U U 1/2 Z 1/2 4 = |f | πr2 U 1/2 , and note this depends only on K. Thus, we let CK = πr4 2 Now suppose {fn } ∈ H and fn → f in L2 (D, µ). Then for each compact K ⊂ D and each z ∈ K, we have that |fn (z) − f (z)| ≤ CK kfn − f kL2 (D,µ) → 0. Hence, fn → f uniformly on compact subsets of D, and thus, f is holomorphic on D. Therefore, H is complete. Problem 11: Suppose that f : D → C is holomorphic and injective in some annulus {z : r < |z| < 1}. Show that f is injective in D. Solution. Let Ar be the annulus, and let r < R < 1 be arbitrary. Let γ be the closed curve γ = {Reiθ : −π ≤ θ < π} so that γ ⊂ Ar . Using the Jordan curve theorem and the injectivity of f , we have that f (γ) is a simple, closed curve in C, and in particular it is non-intersecting, and f (int(γ)) = int(f (γ)). 28 ANALYSIS QUALS Since the winding number is constant on connected components, the winding number of f (z) must be 1 for all z ∈ int(γ), again using injectivity of f in Ar . Let b ∈ f (int(γ)). Then by the Argument Principle, the number of points in int(γ) where f takes the value b is the winding number, 1. Hence, f is injective on int(γ). As this holds for every R < 1, f is injective on D. Fall 2008 Problem 1: Fix 1 ≤ p < ∞ and let {fn } be a sequence of Lebesgue measurable functions fn : [0, 1] → C. Suppose there exists f in Lp ([0, 1]) so that fn → f in the Lp sense, that is, Z 1 p |fn (x) − f (x)| dx → 0. 0 (a) Show that fn → f in measure, that is, lim µ ({x : |fn (x) − f (x)| ≥ ε}) = 0 n→∞ for all ε > 0. (b) Show that there is a subsequence fnk such that fnk (x) → f (x) almost everywhere. Solution. (a) By Chebyshev’s inequality, p µ ({x : |f (x)| > α}) ≤ kf kp αp . Then we have that p µ ({x : |fn (x) − f (x)| > ε}) ≤ kfn − f kp εp → 0 as n → ∞, so fn → f in measure. (b) As fn → f in measure, we may choose a subsequence {fnj } such that µ x : fnj (x) − fnj+1 (x) > 2−j ≤ 2−j . S∞ Let Ej = {x : |fnj (x) − fnj+1 (x)| > 2−j }. Let Fk = j=k Ej . Then µ(Fk ) ≤ ∞ X µ(Ej ) ≤ 21−k . j=k For x ∈ / Fk and k ≤ j ≤ i, we have that i−1 i−1 X X fnj (x) − fni (x) ≤ fn (x) − fn (x) ≤ 2−l ≤ 21−j , l l+1 l=j l=j T∞ so {fnj } is pointwise Cauchy on Let F = k=1 Fk . Then µ(F ) = 0 and fnj converges on F C , so fnj converges almost everywhere. FkC . ANALYSIS QUALS 29 Problem 2: Is every vector space isomorphic as a vector space to some Banach space? Prove your answer. (Banach space = complete normed vector space, as usual). Solution. No. Let V be the vector space consisting of all infinite sequences (over R, for example) with only finitely many non-zero terms. Assume for sake of contradiction that V can be given a Banach space structure. Let Vn be the subspace of V consisting of the sequences whose components are all zero after the nth. Then ∞ [ V = Vn . n=1 Note that Vn is finite dimensional, so it is isomorphic as a Banach space to Rn , and thus is a closed subspace. However, Vn is nowhere dense, as V Sn contains no open balls in V . Thus, by the Baire Category Theorem, Vn 6= V , a contradiction. Problem 3: Prove: If f : [0, 1] → R is an arbitrary function, not necessarily measurable, then the set of points at which f is continuous is a Lebesguemeasurable set. Proof. Let Sx (δ) = sup{|f (x1 ) − f (x2 )| : x1 , x2 ∈ (x − δ, x + delta)}, φ(x) = inf{Sx (δ) : δ > 0}. Note that 1 1 ≥ Sx k k+1 for every k ∈ N. We claim that f is continuous at x if and only if φ(x) = 0. First suppose that φ(x) = 0. Let ε > 0 and choose δ > 0 such that Sx (δ) < ε. Thus, for any y ∈ (x − δ, x + δ), we have that |f (y) − f (x)| < ε, so f is continuous at x. Now assuming f is continuous at x, we have that for all ε > 0 we may choose δ such that |x − y| < δ implies |f (x) − f (y)| < ε, and hence Sx (δ) < 2ε. Thus, φ(x) = 0. Next, we claim that {x : φ(x) < α} is an open set for every α (that is, φ is upper-semicontinuous). Suppose φ(x) < α and choose N such that 1 Sx < α whenever k > N. k 1 1 1 1 Suppose k > N . Let y ∈ x − 2k , x + 2k . Note that then y − 2k , y + 2k ⊂ 1 1 1 (x−1/k, x+1/k), so φ(y) ≤ Sy 2k ≤ Sx k1 < α. Thus, x − 2k , x + 2k ⊂ {x : φ(x) < α}, so the set is open. But then we have that ∞ \ 1 {x : φ(x) = 0} = {x : φ(x) < }, k Sx k=1 and thus the set of points of continuity of f is a Gδ set, and hence is measurable. 30 ANALYSIS QUALS Problem 6: Define for each n = 1, 2, 3, . . . , the Cantor-like set Cn as [0, 1] with its central open interval of length 21n · 13 removed, then with the two central open intervals of length 21n · 312 removed from the remaining two closed intervals and so on (at the jth stage, 2j−1 intervals of length 21n · 31j are removed), continuing with j = 1, 2, 3, . . .. S∞ (a) With µ=Lebesgue measure, show that µ([0, 1]\ n=1 Cn ) = 0. (b) Show that if E is a subset of [0, 1] which is not Lebesgue measurable (you may assume such an E exists without proof), then for some n ≥ 1, E ∪ Cn fails to be Lebesgue measurable. (c) Use part (b) to show that there is a continuous, strictly increasing function f : R → R with f (R) = R and a Lebesgue measurable set A ⊂ R such that f (A) is not Lebesgue measurable. Solution. (a) Cn is constructed by removing intervals of total length 2−n 1 3 at step j. But ∞ j X 2 j=0 3 = 1 1− 2 3 2 j−1 · 3 = 3, −n so C total measure 1−2−n , Sn has a total of 2 removed. Thus, Cn has S∞ so n Cn has measure 1, and hence, µ([0, 1]\ n=1 Cn ) = 0. (b) E not measurable implies that there exists ε > 0 and a set A such that µ(E) < µ(E ∩A)+µ(E ∩Ac )−ε. Choose n such that µ(Cn ) > 1−ε/10. Then µ(E ∩ Cn ) < µ(E) < µ(E ∩ Cn ∩ A) + µ(E ∩ Cn ∩ Ac ) − 4ε/5, so E ∩ Cn is not measurable. (c) Let f be a strictly increasing homeomorphism from C0 to Cn , where n as in part (b). Such a function exists: both sets are homeomorphic to {0, 1}N by considering at each point in a Cantor set, whether it is in the left or right side at each stage of removing central intervals. Thus, just as the strictly monotonic Cantor-Lebesgue function can be constructed for C0 onto [0, 1], so too can such a function be created for Cn , and taking an appropriate composition yields f . Extend f linearly to all of R. Then f −1 (E∩Cn ) is a subset of C0 , which has measure zero. Therefore f −1 (E ∩ Cn ) is measurable. However, by part (b), E ∩ Cn is not measurable. Problem 8: Suppose f : U → C is a holomorphic function with intU |f |2 < ∞ where U = {z ∈ C : 0 < |z| < 1} and the integral is the usual R2 -area integral. Prove that f has a removable singularity at z = 0. Solution. Let w be a point in the annulus around z = 0 of radii ε and 1/2, for 0 < ε < 1/2. Then by Cauchy integral formula, we have that Z Z f (ξ) f (ξ) dξ − dξ. f (w) = ξ − w ξ |ξ|=ε − w |ξ|=1/2 ANALYSIS QUALS 31 Then Z δf (w) = δ |ξ|=1/2 f (ξ) dξ − ξ−w Z 0 δ Z f (ξ) dξdε ξ−w |ξ|=ε and thus Z Z Z f (ξ) f (ξ) 1 δ dξ ≤ dξdε . f (w) − ξ − w δ ξ − w 0 |ξ|=ε |ξ|=1/2 We claim that this tends to 0 as δ → 0, which would prove that f (w) = R f (ξ) dξ for |w| < 12 , and this is holomorphic, so any singularity at 0 |ξ|=1/2 ξ−w would be removable. But using Fubini, Cauchy-Schwarz, and basic estimations, we have that Z Z Z Z δ f (ξ) δ 2π f (εeiθ ) iθ dξ = iεe dθdε 0 |ξ|=ε ξ − w 0 0 εeiθ − w Z δ Z 2π |f (εeiθ )| ≤ εdθdε |εeiθ − w| 0 0 Z 2π Z δ 2 ≤ |f (εeiθ )|εdεdθ |w| 0 0 !1/2 Z !1/2 Z 2π Z δ 2π Z δ 2 ≤ |f (εeiθ )|2 εdεdθ εdεdθ |w| 0 0 0 0 !1/2 Z √ 2 πδ. |f |2 ≤ |w| {0<|z|<δ} Thus, we have that √ Z f (ξ) 2 π dξ ≤ f (w) − |w| |ξ|=1/2 ξ − w Z !1/2 2 |f | → 0 as δ → 0 {0<|z|<δ} for every fixed w ∈ {0 < |z| < 1/2}, and so f is holomorphic on D. (Note: 1 it appears that I pretty uniformly forgot the 2πi in Cauchy’s formula. Obviously, this constant doesn’t change things.) Problem 9: Let D denote the open unit disk in the complex plane and let H denote the upper half plane. (a) Explicitly describe all conformal mappings g from H onto D that obey g(i) = 0. (b) Suppose f : D → H has f (0) = i, f holomorphic. Show that Im(f (x)) ≥ 1−x 1+x for all x ∈ (0, 1). Solution. (a) Let g : H → D be conformal such that g(i) = 0. Define f : H → D be defined by z−i f (z) = . z+i Then f is a conformal map from H → D with f (i) = 0. Let h = g◦f −1 . Then h is a conformal map from D to D with h(0) = 0, so h is an automorphism of the disk, and thus by Schwarz lemma, h is of the 32 ANALYSIS QUALS form eiθ z for some θ. Thus, we have that g(z) = h ◦ f (z) = eiθ z−i z+i for some θ. (b) Since f is holomorphic, its imaginary part is a harmonic function on the disk. Then by Harnack’s inequality, we have that 1 − |z| 1 + |z| Im(f (0)) ≤ Im(f (z)) ≤ Im(f (0)). 1 + |z| 1 − |z| Hence, as Im(f (0)) = 1, we have that Im(f (x)) ≥ 1−x 1+x . Problem 10: Suppose U is a bounded connected open set in C and z0 ∈ U . Let F = {f : U → D, f holomorphic, f (z0 ) = 0}. (a) Show that if K is a compact subset of U , then there is a constant MK > 0 such that |f 0 (z)| ≤ MK for all z ∈ K, f ∈ F . (b) Use part (a) to show that if {fn } is a sequence in F , then there is a subsequence {fnj } which converges uniformly on every compact subset of U to a function f0 ∈ F . (Note: Part of this is to show f0 (U ) ⊂ D.) Solution. (a) Let r = dist(K, U c ). Let γ be a closed curve in U around K with dist(K, γ) ≥ 2r . Then by Cauchy Integral formula, we have that Z 1 f (z) 0 f (w) = dz. 2πi γ (z − w)2 Hence for w ∈ K, we have that Z 1 1 1 4 4M 0 |f (w)| ≤ dz ≤ length(γ) ≤ 2 2π γ |z − w|2 2π r2 r where M is the bound on |z| in U . Thus, let MK = 4M r2 . (b) As f : U → D, |f (z)| < 1 for all z ∈ U and f ∈ F . Hence by Montel’s theorem, there exists a subsequence {fnj } converging uniformly on compact subsets of U to a function f0 . But then fnj (z0 ) → f0 (z0 ), so f0 (z0 ) = 0. Since |fnj (z)| < 1 for all z ∈ U , we must have that |f0 (z)| ≤ 1 for all z ∈ U . But as f0 is holomorphic, it is an open mapping by the Open Mapping theorem, so as U is open, f0 (U ) is open. Thus, f0 (U ) ⊂ D, so f0 ∈ F . Problem 11: Let D denote the open unit disk in the complex plane let D denote its closure. Suppose f : D → C is continuous on D and analytic in its interior. Show that if f takes only real values on ∂D, then f must be constant. Solution. Let g(z) be defined to be f (z) for z ∈ D, and f z1 for z1 ∈ D. Note that this is well-defined: if both conditions are satisfied, then z must be of the form eiθ , in which case f z1 = f (eiθ ) = f (eiθ ) since f takes real values on the circle. Thus, by Schwarz Reflection Principle, g is entire. ANALYSIS QUALS 33 But as D is compact, g is bounded on D, and hence is bounded on C by definition of g. Thus g is entire and bounded, and hence constant by Liousville. As g is constant, we conclude that f is constant. Problem 12: Evaluate Z 0 π a2 dθ + sin2 θ for all real numbers a > 0. Solution. First, note that sin2 θ = − 14 (e2iθ − 2 + e−2iθ ). Let z = e2iθ , so as θ ranges from 0 to 2π, z will range around the unit circle T. Then dz = 2izdθ, and the integral becomes Z Z 1 dz 4dz = 2 − (z − 2 + z −1 )/4) 2 + 2)z − z 2 − 1 2iz(a 2i (4a T T =π X residues of poles of poles −4 inside T. z 2 − (4a2 + 2)z + 1 The poles are at 4a2 + 2 ± √ p 16a4 + 16a2 = 2a2 + 1 ± 2a a2 + 1. 2 The + root is outside the unit circle, so it is of no interest to us. We √ calculate the residue of the other root, z0 = 2a2 + 1 − 2a a2 + 1. z 2 − (4a2 + 2)z + 1 = 2z − 2 − 4a2 , z − z0 so Res(f, z0 ) = −4 1 −4 √ √ = = √ . 2(2a2 + 1 − 2a a2 + 1) − 2 − 4a2 −4a a2 + 1 a a2 + 1 Thus, Z π π dθ = √ . 2 + sin2 θ 2+1 a a a 0 Warning: I have little faith in my calculations on this one. Spring 2008 Problem 2: Let R/Z be the unit circle with the usual Lebesgue measure. + For each n = 1, 2, 3, R . . . , let Kn : R/Z → R be aR non-negative integrable function such that R/Z Kn (t)dt = 1 and limn→∞ ε<|t|<1/2 Kn (t)dt = 0 for every 0 < ε < 1/2, where we identify R/Z with (−1/2, 1/2] in the usual manner. Let f : R/Z → R be continuous, and define the convolutions f ∗ Kn : R/Z → R by Z f ∗ Kn (x) := f (x − t)Kn (t)dt. R/Z Show that f ∗ Kn converges uniformly to f . 34 ANALYSIS QUALS Proof. First, since f is continuous on a compact set, it is uniformly continuous, and the image is compact. Let M be the maximum value of f . Let ε > 0. Choose δ > 0 such that |f (x − t) − f (x)| < ε whenever |t| < δ for every x ∈ R/Z. Then for any x ∈ R/Z, we have that Z |f ∗ Kn (x) − f (x)| = [f (x − t) − f (x)]Kn (t)dt R/Z Z Z [f (x − t) − f (x)]Kn (t)dt + [f (x − t) − f (x)]Kn (t)dt = |t|<δ δ≤|t|<1/2 Z Z |f (x − t) − f (x)| Kn (t)dt + 2M Kn (t)dt ≤ |t|<δ δ≤|t|<1/2 Z Z ≤ε Kn (t)dt + 2M Kn (t)dt |t|<δ δ≤|t|<1/2 Z ≤ ε + 2M Kn (t)dt → ε + 0 as n → ∞ δ≤|t|<1/2 This completes the proof. Problem 3: Let X be a compact metric space. (a) Show that X is separable (i.e. it has a countable dense subset). (b) Show that X is second countable (i.e. there exists a countable base for the topology). (c) Show that C(X) (the space of continuous functions f : X → R with the uniform topology) is separable. (Hint: use part (b), Urysohn’s lemma and the Stone-Weierstrass theorem.) S Solution. (a) Note that X ⊂ x∈X B(x, n1 ) for every n ∈ N. Then as (n) (n) X is compact, there is a finite set {x1 , . . . , xk } such that X ⊂ Sk (n) (n) 1 i=1 B(xi , n ). Then {xi }n∈N,i is countable and dense, so X is separable. (n) 1 (n) 1 (b) Using notation S∞ from the previous part, we let Bn = {B(x1 , n ), . . . , B(xk , n )}. Let B = n=1 Bn . Then B is countable. We claim that this is a base for the topology. Clearly, the elements of B cover X. (n1 ) Let x ∈ B(xi (n2 ) , n11 ) ∪ B(xj , n12 ), elements of B. The intersection (n1 ) must be open, so there exists ε > 0 such that B(x, ε) ⊂ B(xi (n2 ) B(xj , n12 ). Choose N so that that x ∈ (2N ) 1 B(xk , 2N ), (2N ) Let y ∈ B(xk 1 N (2N ) < ε, and we may choose xk since B2N is an open cover. 1 , 2N ). Then (2N ) d(x, y) ≤ d(x, xk (2N ) ) + d(xk , n11 ) ∪ , y) < 1 < ε, N so ANALYSIS QUALS (n1 ) so y ∈ B(x, ε), so y ∈ B(xi (2N ) 35 (n2 ) , n11 ) ∪ B(xj , n12 ). Thus, we have that (n ) (n ) 1 B(xk , 2N ) ⊂ B(xi 1 , n11 ) ∪ B(xj 2 , n12 ). Hence, B is a base for the topology, so X is second countable. (c) Let {x1 , x2 , . . .} be the dense subset of X found in (a). By Urysohn’s Lemma, for n, m ≥ 1, there exists ψn,m ∈ C(X) such that ψn,m = 1 1 2 on B(xn , m ) and ψn,m is supported on B(xn , m ). Then {ψn,m } separates points, as the balls are a base for the topology by (b). By the Stone-Weierstrass theorem, the algebra of rational polynomial combinations of ψn,m is dense, so C(X) is separable. Problem 4: Let f, g ∈ L2 (R) be two square-integrable functions on R (with the usual Lebesgue measure). Show that the convolution Z f ∗ g(x) := f (y)g(x − y)dy R of f and g is a bounded continuous function on R. Solution. First, we show it is bounded: Z |f ∗ g(x)| = f (y)g(x − y)dy R Z ≤ 2 1/2 Z |f (y)| dy R 2 1/2 |g(x − y)| dy R = kf k2 kgk2 . Hence, f ∗ g is bounded. Now we show that f ∗ g is continuous. Z Z f (y)g(x2 − y)dy |f ∗ g(x1 ) − f ∗ g(x2 )| = f (y)g(x1 − y)dy − R ZR = f (y) [g(x1 − y) − g(x2 − y)] dy ZR = f (x1 + y) [g(y) − g(x2 − x1 − y)] dy R ≤ kf k2 g − g(x2 −x1 ) 2 , where g(x2 −x1 ) is the translation. But g − g(x2 −x1 ) 2 → 0 as |x2 − x1 | → 0. Hence f ∗ g is continuous. Problem 5: Let H be a Hilbert space, and let T : H → H be a bounded linear operator on H. (a) Show that if the operator norm kT k of T is strictly less than 1, then the operator 1 − T is invertible. (b) Let σ(T ) denote the set of all complex numbers z such that T − zI is not invertible. (This set is known as the spectrum of T .) Show that σ(T ) is a compact subset of C. 36 ANALYSIS QUALS Solution. (a) Let (1 − T )−1 = ∞ X T n . This converges, as n=0 ∞ ∞ X X n n kT k < ∞. T ≤ n=0 n=0 −1 But then (1 − T )(1 − T ) = I, so 1 − T is invertible. (b) We claim σ(T ) is closed and bounded. First, we prove bounded. Let |α| > kT k. Then αI − T = α(I − Tα ), and Tα < 1, so by part (a), we have that (I − Tα ) is invertible, so α 6∈ σ(T ). Thus, σ(T ) ⊂ {α ∈ C : |α| ≤ kT k}, so σ(T ) is bounded. Next, we aim to show that σ(T ) is closed. Let G be the set of invertible operators on H. We 1 claim that G is open. Let S ∈ G and suppose that kT − Sk < S −1 . Then S −1 T − I = S −1 (T − S) < 1, so by part (a), S −1 T is invertible, and therefore T −1 = (S −1 T )−1 S −1 , so T is invertible. Therefore, G is open. Define f : C → B(H) by α 7→ (αI − T ). Then f is continuous, so f −1 (G) is open. But f −1 (G) = {z ∈ C : T − zI invertible} = C\σ(T ), so σ(T ) is closed. Thus, σ(T ) is compact. Problem 8: Let H be a real Hilbert space, let K be a closed non-empty subset of H, and let v be a point in H. Show that there exists a unique w ∈ K which minimizes the distance to v in the sense that kv − wk < kv − w0 k for all w0 ∈ K\{w}. (Hint: you may find the parallelogram law to be useful.) Solution. Note that the problem is incorrect as stated, as can be seen by Problem 1 on Fall 2009. We employ the additional assumption that K is convex. Let d = inf w∈K kv − wk. Choose {wn } ∈ K such that kv − wn k → d. Then wn + wm 2 2 2 2 . kwn − wm k = 2 kwn k + 2 kwm k − 4 2 2 m m As K is convex, we have that wn +w ∈ K, so wn +w ≥ d2 . Thus, 2 2 2 2 2 kwn − wm k ≤ 2 kwn k + 2 kwm k − 4d2 → 0 as n, m → ∞. Thus, {wn } is Cauchy, and since H is complete, the sequence converges to a point w. Then w ∈ K since K is closed, and kv − wk = d. This proves existence. If y also satisfies kv − yk = d, then 2 w + y 2 2 2 2 = 4d2 −4 w + y ≤ 4d2 −4d2 = 0, 0 ≤ kw − yk = 2 kw − vk +2 ky − vk −4 2 2 so w = y. This proves uniqueness. ANALYSIS QUALS 37 Problem 10: Let the power series f (z) = ∞ X an z n have radius of conver- n=0 gence r > 0. For each ρ with 0 < ρ < r let Mf (ρ) := sup{|f (z)|; |z| = ρ}. Show that the following holds for each such ρ: ∞ X |an |2 ρ2n ≤ Mf (ρ)2 . n=0 Solution. Let φ be a function on the unit circle T defined by φ(z) = f (ρz). P P 2 Then the Fourier expansion of φ is an ρn z n , so kφk2 = 2π |an |2 ρ2n , by Parseval. But we have that 2 2 kφk2 ≤ 2π kφk∞ and since kφk∞ = Mf (ρ), we have that ∞ X |an |2 ρ2n ≤ Mf (ρ)2 . n=0 Problem 11: Let f : D → D be a holomorphic map having two unequal fixed points a, b ∈ D. Show that f (z) = z for all z ∈ D. (Hint: use Schwarz’s lemma.) z−a . Then φ is an automorphism of the disc. Let Solution. Let φ(z) = 1−az −1 g = φ ◦ f ◦ φ . Then g : D → D, g(0) = φ ◦ f (a) = φ(a) = 0, and g(φ(b)) = φ ◦ f ◦ φ−1 (φ(b)) = φ(b). Furthermore, g is holomorphic. Thus, we may apply Schwarz Lemma, and as g(φ(b)) = φ(b), we may conclude that g(z) = z for all z ∈ D. Then f = φ−1 ◦ g ◦ φ is also the identity, f (z) = z. Problem 12: Consider the annulus A := {z ∈ C : r < |z| < R}, where 0 < r < R. Show that the function f (z) = 1/z cannot be uniformly approximated in A by complex polynomials. Solution. Let P (z) = n X ck z k be a complex polynomial. Let r < η < R be k=0 arbitrary. Then Z 2π Z 1 2π it it it f (ηe )P (ηe )dt = e (c0 + ηc1 eit + c2 η 2 e2it + . . .)dt η 0 0 Z Z 2π c0 2π it = e dt + c1 e2it dt + . . . η 0 0 =0 38 ANALYSIS QUALS But as R 2π 0 f (ηeit )f (ηeit )dt = 2π η 2 , we have that Z 2π 2π it it it = f (ηe ) f (ηe ) − P (ηe ) dt η2 0 Z 2π f (ηeit ) kf − P k dt ≤ ∞ 0 2π = kf − P k∞ η Thus, kf − P k∞ ≥ η1 for every such choice of η, so kf − P k ≥ R1 for any polynomial P , and hence there can be no sequence of polynomials that converges uniformly to f . Problem 13: Let Ω ⊂ C be an open set containing the closed unit disk D, and let fn : Ω → C be a sequence of holomorphic functions on Ω which converge uniformly on compact subsets of Ω to a limit f : Ω → C. Suppose that |f (z)| = 6 0 whenever |z| = 1. Show that there is a positive integer N such that for n ≥ N , the functions fn and f have the same number of zeros in the unit disk D. Solution. As |f (z)| = 6 0 and {|z| = 1} is compact, we can let m = min|z|=1 |f (z)|. As fn → f uniformly, we may choose N so that n ≥ N implies that |fn (z) − f (z)| < m 2 for all z ∈ ∂D. Then |fn − f | < |f | on ∂D, so by Rouche’s Theorem, fn and f have the same number of zeros inside D. Fall 2007 Problem 1: Let f : R → R be a function of class C 1 , periodic of period 2π, and satisfying Z π f (t)dt = 0. −π Prove that Z π 2 Z π −π 2 |f 0 (t)| dt. |f (t)| dt ≤ −π Find all such functions for which equality holds. Proof. Write fˆ(n) for the nth Fourier coefficient of f . Then we have that fˆ0 (n) = infˆ(n). By Parseval, we know that Z π ∞ X ˆ 2 2 |f (t)| dt = 2π f (n) , −π n=−∞ and Z ∞ ∞ X X ˆ 2 ˆ 2 2π f (n) ≤ 2π nf (n) = n=−∞ n=−∞ π −π 2 |f 0 (t)| dt, ANALYSIS QUALS 39 where the inequality follows since fˆ(0) = 0 by assumption, and the equality also comes from Parseval. If equality holds, then fˆ(n) = 0 for all n, so f is identically zero, as seen below in problem 8 of this exam. Problem 2: Is there a closed uncountable subset of R which contains no rational numbers? Prove your answer. Solution. Let {qn } be an enumeration of the rational numbers. Then we have that ∞ [ Q= B(qn , 2−n ) n=1 is a countable union of open sets, and hence is open. Thus, R\Q is closed and contains no rational numbers. Furthermore, µ(Q) ≤ ∞ X 21−n < ∞, n=1 so R\Q has strictly positive Lebesgue measure, and hence is uncountable. Problem 3: (a) Prove that a complete normed vector space (a Banach space) is either finite-dimensional or has uncountable dimension in the vector space sense, i.e., it is not generated as finite linear combinations of elements of some countable subset. (b) Use part (a) to give an example of a vector space that cannot be given the structure of a Banach space, that is, it is not complete in any norm. Solution. (a) Assume X is a an infinite-dimensional Banach space, and assume for sake of contradiction that (xi )is a countable Hamel basis S∞ for X. Let Vn = span{x1 , . . . , xn }. Then X = i=1 Vn . Note that Vn is finite dimensional, so it is isomorphic as a Banach space to Rn , and thus is a closed subspace. However, Vn is nowhere dense, as Vn contains no open balls in V . Thus, by the Baire Category Theorem, S Vn 6= X, a contradiction. (b) Let V be the vector space consisting of all infinite sequences with finitely many non-zero terms. Then V has a countable Hamel basis: let xi be the sequence of all zeros with a 1 in the ith coordinate. Every element of V can be written uniquely as a finite linear combination of the xi . Thus, by part (a), V cannot be a Banach space. Problem 4: Prove that non every subset of [0, 1] is Lebesgue measurable. Solution. Consider the relation x ∼ y is x − y ∈ Q. This is an equivalence relation. For α ∈ [0, 1], let Eα be the equivalence class of α. Then [0, 1] ⊂ S E . By the axiom of choice, we may choose exactly one element xα ∈ α α [0, 1] from each equivalence class, and let N be the set of these elements. We claim that N is not Lebesgue measurable. Assume for sake of contradiction that N is measurable. Let {rk } be an 40 ANALYSIS QUALS enumeration of Q ∩ [0, 1]. Let Nk = N + rk for each k. Then by definition of the equivalence relation, Nk ∩ Nj = if k 6= j. Then [0, 1] ⊂ ∞ [ Nk ≤ [−1, 2]. k=1 If N is measurable, then Nk is measurable, and µ(N ) = µ(Nk ). Thus, we have ∞ X 1≤ µ(Nk ) ≤ 3, k=1 and so 1≤ ∞ X µ(N ) ≤ 3, k=1 which yields a contradiction, as µ(N ) can be neither 0 nor strictly positive. Problem 7: Suppose H is a separable Hilbert space. (a) Prove: If T : H → H is a linear mapping such that kI − T k < 1, where I is the identity map of H to itself, then T is invertible. (b) Suppose {en } is a complete, orthonormal set in H (a Hilbert space basis). Suppose also that {fn } is an orthonormal set in H such that P 2 ken − fn k < 1. Prove that fn is a complete orthonormal set in H. Solution. (a) Since kI − T k < 1, we have that ∞ X (I − T )n is convergent n−0 in the norm topology to a map which is the inverse of I − (I − T ) = T . Thus, T is invertible. (b) An invertible bounded linear map T sends orthonormal bases to orthonormal bases, so it suffices to find such a map T with T (ej ) = fj . Or rather, we define T by T (ej ) = fj , and we show that T is invertible. T can be considered as an infinite matrix where the i, j entry is hfj , ei i. Then let D = I − T , considered as an infinite matrix. The jth column represents the vector ej − fj , so the sum of the squares of the norms of the entries is strictly less than 1, by assumption. Let αi be the sum of squares of norms of elements in the ith row. Write x = and 2 P kbn k ≤ an en . (I − T )x = X 2 2 |Dn,j | Hence, k(I − T )xk ≤ kxk X 2 P |aj | 2 bn en . Then bn = P j Dn,j aj ej 2 = αn kxk . P i αi , so we have that sX αi < 1. kI − T k ≤ i Thus by part (a), T is invertible. ANALYSIS QUALS 41 Problem 8: Let f : R → R be a continuous function that is periodic with period 2π. Show that if all the Fourier coefficients of f are 0, then f is identically 0. Proof. By Parseval, we know that Z π ∞ X ˆ 2 2 |f (t)| dt = 2π f (n) . −π n=−∞ By assumption, all Fourier coefficients are 0, so the right-hand side is 0. Thus, Z π 2 |f (t)| dt = 0. −π But this implies that f = 0 a.e., so since f is continuous, we have that f is identically 0. Problem 10: Let U be a bounded connected open set in C. Prove that if K is a compact subset of U , then there is a constant CK such that for every point R 1/2 2 z ∈ K and every holomorphic L2 function f on U , |f (z)| ≤ CK U |f | . Solution. Let r = dist(K, U c ). Let z ∈ K, B = B(z, r/2) ⊂ U . By mean value principle and Cauchy-Schwarz, we have that Z 1 |f (z)| = f (w)dw Area(B) B Z 4 ≤ 2 |f (w)| dw πr B Z 4 |f (w)| χB dw = 2 πr U Z 1/2 Z 1/2 4 2 2 ≤ 2 |f (w)| dw χB dw πr U U 1/2 Z 1/2 4 |f | = πr2 U 1/2 Thus, we let CK = πr4 2 , and note this depends only on K. This completes the proof. Problem 11: Let U be a bounded connected open set in C. Suppose that 0 belongs to U and that F : U → U is a holomorphic function with F (0) = 0 and with F 0 (0) = 1. Prove that F is the identity map. [Suggestion: Consider the power series of F composed with itself many times]. Solution. In a neighborhood of 0, we can write F (z) = z +a2 z 2 +a3 z 3 +. . .. Let F (n) (z) = F ◦ F ◦ · · · ◦ F (z), where there are n compositions. Then F (n) : U → U . We have that X F (2) (z) = F (z) + ak F (z)k = z + 2a2 z 2 + 2a3 z 3 + . . . . k≥2 Similarly, we have that F (n) (z) = z + na2 z 2 + na3 z 3 + . . . . 42 ANALYSIS QUALS As U is bounded, there exists M with |z| ≤ M for all z ∈ U . Then by the Cauchy estimates, for B(0, r) ⊂ U , we have that k d M (n) Cr,n M (n) ≤ (z) (F (z)) sup . = F dz k rk B(0,r) rk C M Thus, |nak | ≤ r,n . But since Cr,n ≤ M , if we have ak 6= 0, we get a rk contradiction by fixing r and taking n → ∞. Thus, ak = 0 for all k ≥ 2, which means that F (z) = z. Problem 12: Find a conformal map to the unit disc of the half disc {z : |z| < 1, Rez > 0}. You may write your answer as a composition of simpler conformal maps. Solution. Let f1 (z) = z−1 z+1 . This maps the half disc to the upper left quad2 rant. Let f2 (z) = z . This maps the upper left quadrant to the lower half z+i plane. Let f3 (z) = z−i . This maps the lower half plane to the unit disc. Thus, f = f3 ◦ f2 ◦ f1 is the desired conformal map. Problem 13: Suppose R is a positive number and U = {z : |z| < R}. (a) Show that for each holomorphic function f on U , there is a power ∞ X an z n which converges at each point z ∈ U to f (z). series n=0 (b) Show that there is only one such power series. Proof. (a) We have that for some fixed ε, z <1−ε ξ for all ξ ∈ ∂D. Thus, the geometric series yields ∞ n X z ξ 1 = = . ξ ξ−z 1 − zξ n=0 But then by the Cauchy integral formula, we have that Z Z ∞ n 1 1 f (ξ) 1X z dξ = , f (z) = f (ξ) 2πi ∂D ξ − z 2πi ∂D ξ n=0 ξ and since we have uniform convergence in the geometric series, we may switch the sum and integral and get Z ∞ X 1 f (ξ) f (z) = dξ z n n+1 2πi (ξ) ∂D n=0 ∞ X f (n) (0) n z . n! n=0 P P (b) If we had an z n = bn z n for all z ∈ D, then by matching coefficients, we have an = bn for all n. = ANALYSIS QUALS 43 Winter 2007 Problem 1: Let fn : R → R be non-negative integrable functions with kfn k1 = 1. Suppose fn → f pointwise a.e. with kf k = 1. Show that Z Z fn (x)dx → f (x)dx A A uniformly in the choice of Borel set A ⊂ R. Solution. By Fatou’s lemma, we have that Z Z lim inf (−|fn − f | + |fn | + |f |) ≤ lim inf −|fn − f | + |fn | + |f |. Hence, Z 2 kf k1 = 2 ≤ 2 − lim sup |fn − f | and thus kfn − f k1 → 0 as n → ∞. Then Z Z Z fn (x)dx − |fn (x) − f (x)| dx ≤ kfn − f k1 . f (x)dx ≤ A A A R R Thus, A fn (x)dx → A f (x)dx uniformly. Problem 2: Consider the function f : (0, ∞) × (0, ∞) → R defined by ∞ X x . f (x, y) = 2 x + yn2 n=0 Show that the limit g(y) := limn→∞ f (x, y) exists for all y > 0 and compute g(y). Solution. We have that f (x, y) = ∞ X x 1/y · 2 + yn2 1/y x n=0 = ∞ xX y n=0 = 1 x + x y n=1 1 x2 y + ∞ X n2 x2 y 1 . + n2 But we also have that √ n=∞ Z ∞ √ √ ∞ X y n y y π 1 1 ≤ dn = arctan · . = x2 x2 2 2 x x x 2 +n 0 n=0 y +n n=1 y Thus, f (x, y) ≤ 1 x + π √ 2 y. Similarly, f (x, y) ≥ π √ 2 y. Thus, π lim f (x, y) = √ . 2 y x→∞ R Problem 3: Let f ∗ g(x) = R f (x − y)g(y)dy denote the convolution of f and g. Fix g ∈ L1 (R). Do the following: (1) Show that Ag (f ) := f ∗ g is a bounded operator L1 (R) → L1 (R). (2) Suppose in addition g ≥ 0. Find the corresponding norm kAg k. 44 ANALYSIS QUALS (3) Show that the only f ∈ L1 (R) for which f ∗ f = f is f = 0. Solution. (1) We have that Z Z ≤ kf k kgk , f (x − y)g(y)dydx kAg (f )k = 1 1 R R so as g ∈ L1 , we have that Ag is a bounded operator. (2) By part (1), we have that kAg k ≤ kgk. We have that Z Z Z Z Z 2 kAg (g)k = g(x−y)g(y)dydx = g(y) g(x−y)dxdy = g(y) kgk dy = kgk , R R R R R and so kAg k ≥ kgk, and hence kAg k = kgk. (3) Suppose that f ∗ f = f . Then by taking Fourier transforms, we have that f ˆ ∗ f = fˆ. But using properties of Fourier transforms, we have ˆ f ∗ f = fˆ · fˆ. Thus, fˆ2 − fˆ = 0 a.e. Hence fˆ = 0 or 1 a.e. But since f ∈ L1 , we have that fˆ is continuous and vanishes at ∞ by Riemann-Lebesgue theorem, so we conclude that fˆ = 0 everywhere. Then by Fourier inversion, we conclude that f = 0. Problem 4: Let α ∈ R\Q and let T : L2 ([0, 1]) → L2 ([0, 1]) be defined by (T f )(x) = f (x + α mod 1). Denote Sn f = f + T f + T 2 f + · · · + T n−1 f . Do the following: (1) For any f ∈ L2 ([0, 1]), prove that n1 Sn f converges in L2 . Identify the limit. (2) Suppose f : [0, 1] → R is continuous with f (1) = f (0). Show that the convergence in (1) is uniform. R1 Solution. We prove both at once. We claim that n1 Sn f (x) → 0 f (t)dt for every x ∈ [0, 1], and that the convergence is uniform for continuous periodic functions. First, let f (x) = e2πikx . If k = 0, the statement is clearly true. Suppose k 6= 0. Then the statement is that Z 1 N −1 1 X 2πik(x+nα) e → e2πikt dt. N n=0 0 But by the geometric series, the LHS is equal to 1 1 − e2πiN αk N 1 − e2πiαk and since α ∈ /RQ, the denominator is never 0. Hence, this approaches 0 as 1 N → ∞, and 0 e2πikx dx = 0, so the statement is true for all trigonometric polynomials. e2πikx Now let ε > 0, f a continuous function from R/Z to R. As trigonometric polynomials are uniformly dense in the space of continuous periodic functions, we may choose P (x) trigonometric such that ε sup |f (x) − P (x)| < . 3 x∈R/Z ANALYSIS QUALS 45 Then for N sufficiently large, we have that N −1 Z 1 1 X ε P (t)dt < . P (x + nα) − N 3 0 n=0 Then N −1 Z 1 1 X f (t)dt f (x + nα) − N 0 n=0 Z 1 N −1 −1 1 NX 1 X ≤ P (t)dt |f (x + nα) − P (x + nα)| + P (x + nα) − N N n=0 0 n=0 Z 1 |P (t) − f (t)| dt < ε. + 0 Thus, the convergence is uniform, which proves part (2). Part (1) follows by noting that continuous periodic functions are dense in L2 ([0, 1]). R n Problem 5: Let An (f ) = n1 0 f (x)dx. Show that there exists a continuous linear functional A : L∞ (R+ ) → R such that A(f ) = lim An (f ) n→∞ whenever the limit exists. Here, R+ = (0, ∞). Solution. We define A as above, and prove that it is a continuous linear functional on L∞ (R+ ). Clearly, A is a linear functional, as for c, d ∈ R, A(cf + dg) = lim An (cf + dg) = lim cAn (f ) + dAn (g) = cA(f ) + dA(g). n→ n→∞ Thus, it suffices to show that A is bounded. But we have that Z 1 n |A(f )| = lim f (x)dx n→∞ n 0 Z n 1 = lim f (x)dx n→∞ n 0 Z 1 n ≤ lim |f (x)| dx n→∞ n 0 Z n 1 kf kL∞ dx ≤ lim n→∞ n 0 = lim kf kL∞ n→∞ = kf kL∞ . Thus, A is a bounded linear functional, with kAk ≤ 1. Problem 6: Let X be a Banach space and let A : X → X be a linear map. Define ρ(A) = {λ ∈ C : (λ − A) maps X onto X} Show that ρ(A) is an open subset of C. 46 ANALYSIS QUALS Solution. The problem is false as stated, as some assumption of boundedness must be made. A counterexample is to let X be l∞ , the space of bounded sequences (x1 , . . . , xn , . . .), and let A : (x1 , . . . , xn , . . .) 7→ (x1 , x2 /2, . . . , xn /n, . . .). Then A is linear and surjective, so 0 ∈ ρ(A), but 1/n 6∈ ρ(A) for any n ∈ N, so ρ(A) is not open. That said, we alter the problem and change the definition of ρ(A). Let ρ(A) = {λ ∈ C : (λ − A)−1 exists and is bounded} Let G be the set of operators on X with bounded inverse. 1 We claim that G is open. Let S ∈ G and suppose that kT − Sk < S −1 . Then S −1 T − I = S −1 (T − S) < 1, so we have that X (S −1 T )−1 = (I − (I − S −1 T ))−1 = (I − S −1 T )i exists. Therefore T −1 = (S −1 T )−1 S −1 , so T is invertible. Therefore, G is open. Define f : C → B(X) by α 7→ (αI − T ). Then f is continuous, so f −1 (G) is open. But f −1 (G) = ρ(A), so ρ(A) is open. Problem 8: Determine the number of zeros of the polynomial p(z) = z 4 + z 3 + 4z 2 + 2z + 3 in the right half-plane {z : Re z > 0}. Solution. Let ΓR be the contour that is the right half of the circle of radius R centered at the origin. Let γR be the semi-circle part of the contour, and let γˆR be the segment from Ri to −Ri. Note that on {Re z = 0}, we have that Re p(it) = t4 − 4t2 + 3 = (t2 − 3)(t2 − 1), and Im p(it) = −t3 + 2t = −t(t2 − 2). Thus, there are no zeros on {Re z = 0}. Thus, we can use the argument principle in this contour, which says that R p0 (z) dz = 2πi(number of zeros in the interior of ΓR ). By taking R → ∞, ΓR p(z) we can calculate the total number of zeros in {z : Re z > 0}. 0 Z γR (z) On γR , p(z) = z 4 + O(|z|3 ), and p0 (z) = 4z 3 + O(|z|2 ), so pp(z) = 4z −1 + O(|z|−2 ). Thus, Z π p0 (z) e−θ dz = Ri · 4 · dθ + O(|R|−1 ) = 4πi + O(|R|−1 ) → 4πi as R → ∞. p(z) R 0 On γˆR , note that critical points (points where p crosses either the real√or √ √ imaginary √ axes) are at t = 0, ±1, ± 2, ± 3 for z = it. Suppose R > 3. Then on ( 3i, Ri), p takes values in the 4th quadrant (by noticing that the real part is positive and the imaginary part is negative). Similarly, we ANALYSIS QUALS 47 √ √ √ see that on ( 2i, 3i), p is in the 3rd quadrant; on (i, 2i), p is in 2nd quadrant; on (0, i), p is in 1st quadrant; on √ √ (−i,√0), p is in 4th quadrant; on (− 2i, −i), p √ is in 3rd quadrant; on (− 3i, − 2i), p is in 2nd quadrant; and on (−Ri, 3i), p is in the 1st quadrant. Thus, moving from Ri to −Ri gives two clockwise revolutions around the R 0 (z) origin, meaning the winding number is 2, which means that γˆR pp(z) dz → −4πi as R → ∞. Hence, R p0 (z) dz ΓR p(z) → 4πi − 4πi = 0, so there are no zeros in {Re z > 0}. Problem 9: Let f (z) be analytic for 0 < |z| < 1. Suppose there are C > 0 and m ≥ 1 such that C (m) f (z) ≤ m , 0 < |z| < 1. |z| Show that f (z) has a removable singularity at z = 0. Solution. Suppose that f (z) has Laurent series ∞ X cn z n = n=−∞ −1 X n=−∞ cn z n + ∞ X cn z n . n=0 Then f (m) (z) = −1 X n=−∞ cn ∞ X n! n! cn z n−m + z n−m . (n − m)! (n − m)! n=0 Then as f (m) (z) ≤ |z|Cm for 0 < |z| < 1, we have that cn = 0 for n < 0. ∞ X cn z n , so f is analytic on D. Thus, f (z) = n=0 Problem 10: Let J = {iy : 1 ≤ y < ∞} and H be the open upper half plane. Consider the domain D = H\J. Find a bounded harmonic function u : D → R such that u(x + iy) → 0 as y ↓ 0 and u(z) → 1 as z → J. It is fine to represent the solution in terms of a composition of conformal maps. Proof. Let’s map this slit half plane to something better, keeping track of where we’re sending J. Let f1 (z) = z−i z+i . Then f1 takes D to the slit disc, and takes J √ to the slit. Let f2 (z) = z, under the branch cut that makes sense (positive real axis). Then f2 takes the slit disc to the upper half disc, taking the image of J to R ∩ D. Let f3 (z) = z−1 z+1 . Then f3 takes the upper half disc to the second quadrant of the plane and takes the image of J to the negative real axis. Let f4 (z) = iz 2 . This takes the second quadrant to the right half plane, taking the image of J to the positive imaginary axis. Let f5 (z) = Log(z), with the negative real axis branch cut. Then f5 takes the right half plane to the horizontal strip between iπ/2 and −iπ/2, which takes J to the line Im(z) = π/2. 48 ANALYSIS QUALS Let f = f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 . Then f is a conformal map taking D to a strip, which takes J to the line Im(z) = π/2, and takes R to the line Im(z) = −π/2. Let u(z) = 21 + π1 Im(f (z)). Then since the imaginary part of a holomorphic function is harmonic, we have that u is harmonic on D, and u(x + iy) → 0 as y ↓ 0 and u(z) → 1 as z → J. Problem 11: Prove that a meromorphic function f (z) in the extended complex plane C∗ = C ∪ {∞} is the sum of the principal parts at its poles. Solution. In a neighborhood of each pole a of f , we have c−m+1 c−1 c−m + + ··· + + c0 + c1 (z − a) + · · · , f (z) = (z − a)m (z − a)m−1 z−a c−1 c−m where the principal part is (z−a) m + · · · + z−a . Either f vanishes at infinity or not. That is, either f (1/z) has a pole at 0 or a removable singularity at 0. If f (1/z) has a pole at 0, then f has a pole at ∞. Including possibly ∞, let z1 , . . . , zm be the poles of f . Near each non-infinite zk , as above, we may write f (z) = fk (z) + gk (z), where fk is the principal part and gk is holomorphic If zk = ∞, we can write f (1/z) = f˜∞ (z)+g∞ (z), where f˜∞ (z) is the principal part at 0 of f (1/z), and g∞ is holomorphic in a neighborhood of 0. Define fk (z) = f˜∞ (1/z). Let H = f − m X fk . Near each pole, we have subtracted the principal k=1 part of f , so we see that H is entire (the singularities at the points zk are all removable). But since we also subtracted f˜∞ (1/z), we have that H is bounded. Hence by Liousville, H is constant. But by the definition of H, we must have that H = 0 everywhere. Thus, f is the sum of the principal parts of its poles. Problem 12: Let D be a domain (connected open set) in C and let (un ) be a sequence of harmonic functions un : D → (0, ∞). Show that if un (z0 ) → 0 for some z0 ∈ D, then un → 0 uniformly on compact subsets of D. Solution. As un (z0 ) → 0, we have that {un (z0 )} is Cauchy. Thus, for every ε > 0 there exists N such that 0 ≤ un (z0 ) − um (z0 ) < ε for all N ≤ n ≤ m. Let K ⊂ D be compact. Harnack’s inequality states that there exists a constant CK > 0 so that maxx∈K u(x) ≤ CK minx∈K u(x) for u : D → (0, ∞) harmonic. Applying this to un − um , we have that max(un (x) − um (x)) ≤ CK min(un (x) − um (x)). x∈K x∈K If z0 ∈ K, then minx∈K (un (x) − um (x)) < ε, and thus un → 0 uniformly on K. ANALYSIS QUALS 49 If z0 6∈ K, choose K 0 such that K 0 ∩ K 6= ∅ and z0 ∈ K 0 , and choose z1 ∈ K 0 ∩ K. Since un (z1 ) → 0, the argument can be repeated to show that un → 0 uniformly on K. Thus, un → 0 uniformly on compact subsets of D.