Phil Wertheimer UMD Math Qualifying Exam Solutions Analysis  January, 2015


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1 Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function g in L m [, b] such tht f(x) = c + g dm for ll x [, b] Suppose tht f is Lipschitz on [, b] with Lipschitz constnt K. Recll tht this mens for ll x, y [, b] we hve f(x) f(y) K x y. First note tht f is bsolutely continuous on [, b] since if ɛ > nd n {( i, b i )} n re disjoint in [, b] with b i i < ɛ i then n f(b i ) f( i ) n K b i i = K n ɛ b i i < K K = ɛ Now, since f is bsolutely continuous on [, b], it is differentible.e. on [, b] nd we hve for ll x [, b], f(x) f() = f (t) dm It remins to show tht f L m [, b]. This follows directly from the Lipschitz condition: f f(x + h) f(x) f(x + h) f(x) (x) = lim = lim K h h h (x + h) x Conversely, suppose there is constct c nd function g in L m [, b] such tht f(x) = c + g dm for ll x [, b] Then for ll x, y [, b] (without loss of generlity tke x < y): ( y ) ( f(y) f(x) = c + g dm c + g dm) = Therefore f is Lipschitz with constnt g. y x g dm g y x (b) Find function f on [, 1] tht is bsolutely continuous but is not Lipschitz nd verify tht it is bsolutely continuous but not Lipschitz. Tke f(x) = x. Then since f (x) = 1 2 is unbounded on [, 1], f is not Lipschitz on [, 1]. And f is x bsolutely continuous becuse we my write f(x) = where the integrnd, f is Lebesgueintegrble (even though it s n improper integrl). dt 2 t Pge 1 of 6
2 Problem 2 Let f be n entire function stisfying Prove or disprove: f is constnt. f(z) 1 + Im z. f is entire so it hs Tylor series round, sy f(z) = n= nz n. By Cuchy s Integrl formul, for ny R >, we hve Since n = f (n), we hve n! f (n) () = n! 2πi z=re iθ f(z) n! 2π dz = zn+1 2πi n 1 2π 2π = 1 2π 1 2π 2π 2π 1 + I(Re iθ ) R n 1 + R sin θ R n 1 + R R n = 1 + R R n f(re iθ ) (Re iθ ) n+1 Rieiθ dθ for n 2. Therefore n = for n 2, which implies f(z) = z + b is liner. If, then for z R the reverse tringle inequlity would give dθ 1 = 1 + I(z) f(z) Then choosing z so tht z > b + 1, this becomes = z + b z b 1 z b > 1 This contrdiction shows tht must be zero. Tht is, f is constnt. dθ dθ Pge 2 of 6
3 Problem 3 Let m denote the Lebesgue mesure on [, 1], nd let {f n } n 1 be sequence of Lebesgue mesurble functions on [, 1]. Suppose tht for ech x [, 1], sup f n (x) <. Show tht for ech ɛ >, there is n mesurble set A ɛ [, 1] nd rel number B ɛ > so tht () m([, 1] \ A ɛ ) < ɛ, (b) f n (x) B ɛ for ll x A ɛ nd ll n 1. For k N define E k := {x : f n (x) > k}, the set of ll x for which t lest one of the f n hs f n (x) > k. Observe tht {E k } k=1 sequence of mesurble sets. Next define n=1 is descending E := E k = k=1 k=1 n=1 {x : f n (x) > k}, the set of ll x for which f n (x) = for some n. By hypothesis, for ech x [, 1], there is no n with f n (x) =. In other words, m(e) =. Since {E k } k=1 is decresing sequence of mesurble sets, by continuity of mesure we hve ( ) = m(e) = m E k = lim m(e k) k Therefore for ny ɛ >, k ɛ such tht m(e kɛ ) < ɛ. Define A ɛ := E c k ɛ. Then k=1 m([, 1] \ A ɛ ) = m(a c ɛ) = m(e kɛ ) < ɛ so condition () is stisfied. To prove (b), choose x A ɛ = E c k ɛ. Since ( c Ek c ɛ = {x : f n (x) > k ɛ }) = ({x : f n (x) > k ɛ }) c, n=1 n=1 we hve for ll n tht f n (x) k ɛ. Therefore (b) is stisfied. Pge 3 of 6
4 Problem 4 Let D C denote the unit disc nd let S denote the strip {z C : Im z < π 2 }. () Find conforml mp f from D onto S stisfying f() = nd f () = 2. The mp h : z e z tkes S into the right hlf plne {z : Rez > }, nd the mp k : z z 1 z + 1 tkes the right hlf plne into D. Therefore their composition k h = ez 1 e z is conforml mp S D, so + 1 (k h) 1 is conforml mp from D to S. Define ( ) 1 + z f = (k h) 1 = h 1 k 1 = log 1 z Then by the quotient nd chin rules we hve f (z) = 1 z 1 + z ( ) (1 z) + (1 + z) (1 z) 2 We now compute so f stisfies the desired conditions. f() = log 1 = f () = = 2 (b) Show tht if g : D S is nlytic with g() = then g () 2. Let h(z), k(z) be s in prt (). Then F := k h g : D D is n nlytic mp of the unit disk. By the Schwrz Lemm, F () 1. By construction, F is given by so by the quotient rule nd chin rule we hve Using the fct tht g() = this simplifies to F (z) = eg(z) 1 e g(z) + 1 F (z) = (eg(z) + 1) e g(z) g (z) (e g(z) 1) g (z) e g(z) (e g(z) + 1) 2 Finlly, F () 1 gives F (z) = 2g (z) 4 F () = g () 2 = g (z) 2 1 g () 2 Pge 4 of 6
5 Problem 5 Let {u n } n 1 be sequence of Lebesgue mesurble functions on [, 1] nd ssume lim n u n (x) =.e. on [, 1], nd lso u n L2 [,1] 1 for ll n. Prove lim n u n L1 [,1] =. For ny mesurble set E [, 1], define g = χ E L 2 [, 1]. Then Hölder s inequlity gives u n L1 (E) u n L2 (E) g L2 (E) 1 m(e) = m(e) Let ɛ >. [, 1] hs finite mesure so by Egoroff s Theorem, there is mesurble set A [, 1] such tht u n on A nd m([, 1] \ A) < ɛ2. By uniform convergence, there exists N such tht n N implies 4 u n (x) < ɛ 2 for ll x A. Then for n N, u n 1 = A u n + u n [,1]\A < ɛ 2 m(a) + u n L1 ([,1]\A) ɛ 2 + m([, 1] \ A) Therefore u n 1. < ɛ 2 + ɛ 2 = ɛ Pge 5 of 6
6 Problem 6 Suppose O is region in C nd B is n open disc with B O. () Assume tht f is nonconstnt nlytic function on O with f constnt on B. Prove tht f hs zero in B. Denote the constnt by M, so f = M on B. By the mximum modulus principle, f M in B. If f hs no zeroes in B, then 1 f is lso nlytic on O, nd 1 f 1 M on B. Applying MMP gin gives 1 f 1 M in B, which implies f M in B. Since B is compct, f ttins its mx. nd min. vlues on B. Therefore f is constnt nd equl to M on ll of B. By the open mpping theorem, either f is constnt on B or f(b) must be n open set. But since f is constnt on B, f(b) { z = M} cnnot be open, so f is constnt on B. Since it is nlytic on O, it must lso be constnt on O. This contrdiction proves f must hve zero in B. (b) Prove tht if g is nlytic in O nd Re(g) is constnt on B, then g is constnt. Since g is nlytic, so is f(z) = e g(z). Note tht f(z) = e g(z) = e Re(g). Thus f is constnt on B. Since f on O, we cn conclude by prt () tht f is constnt in O, which implies g is constnt. Pge 6 of 6
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