# SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

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1 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such tht µ ν nd ν µ. (Recll tht µ ν mens tht µ is bsolutely continuous with respect to ν, tht is, for ny E M, ν(e implies µ(e. Solution. If µ is finite, tke ν µ. Hence we ssume tht µ(x. By the σ-finiteness ssumption, we cn write X n F n for mesurble sets F n such tht µ(f n < for ech n N. For ny mesurble set A M, define µ(a F n ν(a 2 n ( + µ(f n. n One checks tht ν is mesure by writing ν(a f dµ with A f(x 2 n ( + µ(f n whenever x F n. (The function f cn lso be written s ( f 2 n χ Fn. ( + µ(f n n Alterntively, show directly tht ν is mesure: clerly ν(, nd countble dditivity of ν follows from countble dditivity of µ. Clerly ν(x <. If µ(a, then µ(a F n for ech n N, whence ν(a. On the other hnd, if ν(a, then µ(a F n for ech n N. This implies tht µ(a by countble dditivity. 2. Let (X, µ be mesure spce with µ(x <. For f L 2 (X, µ, prove tht f log( f is in L (X, µ. (Tke y log(y when y. If (X, µ (R, m, is it still the cse tht f L 2 implies f log( f L (X, µ? Solution. First consider the cse µ(x <. Since y y log(y is continuous on (, ] nd lim y + y log(y exists (it is zero, there is constnt M such tht y log(y M for ll y [, ]. (In fct, methods of elementry clculus show tht for ll y (, ], we hve y log(y e, so we cn tke M e. But we don t need the exct bound. Set E { x X : f(x }. For x E, log ( f(x < f(x. Hence f(x log ( f(x < f(x 2. Therefore, using f L 2 (X, µ t the lst step, f(x log ( f(x dµ f(x log ( f(x dµ + f(x log ( f(x dµ X E E X X\E f 2 dµ + M dµ X\E f 2 dµ + Mµ(X \ E X f 2 dµ + Mµ(X <. If (X, µ (R, m, the sttement fils. Here is counterexmple. (A counterexmple, with proof, must be present in correct solution. Set { x 3/4 x f(x x <. Then R f 2 dm x 3/2 dx 2,

2 2 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 but f log( f dm x 3/4 log(x 3/4 dx 3 4 R x 3/4 log(x dx > 3 4 e x 3/4 dx. Alternte solution to the second prt (sketch. For ny α ( 2, ], the function { x α x f(x x < is counterexmple. There re mny others. 3. Let f : [, ] R be continuous function. ( Prove tht lim (b Prove tht lim x n f(x dx. (3 points. (n + x n f(x dx f(. (7 points Solution to 3.(. For n N, set g n (x x n f(x for ll x [, ]. Then lim g n (x for lmost ll x [, ] (in fct, for ll x [, ] except possibly x, f bounded nd hence is integrble on [, ], nd g n (x f(x for ll x [, ] nd ll n N. The Lebesgue Dominted Convergence Theorem therefore implies tht lim x n f(x dx ( lim g n(x dx. Alternte solution to 3.(. Let >. We find N N such tht n N implies xn f(x dx <. Set M sup f(x nd α x [,] 2M +. Without loss of generlity, is smll enough tht α >. Choose N N so lrge tht α N < 2M +. Let n N stisfy n N. Then α x n f(x dx x n f(x dx x n f(x dx + x n f(x dx α ( α Mα n + M( α < Mα N + M( α < M α 2M + α n M dx + ( + M α 2M + M dx <. Solution to 3.(b. Using the chnge of vribles y x n+, so x y /(n+ nd dy (n + x n dx, we get (n + x n f(x dx f ( y /(n+ dy. Set M sup x [,] f(x. For n N, set g n (x f ( y /(n+ for ll y [, ]. Then lim g n (y f( for ll y [, ] except y, the constnt function M is integrble on [, ], nd g n (y M for ll y [, ] nd ll n N. The Lebesgue Dominted Convergence Theorem therefore implies tht lim g n (x dx We cn lso prove this by direct estimtes. f( dx f(.

3 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 3 Alternte solution to 3.(b. Let >. Set M sup x [,] f(x. Choose δ (, sufficiently smll tht f(x f( < 3 for x [ δ, ]. Choose N N so lrge tht ( δ N+ < 3M +. Let n N stisfy n N. Then Now nd (n + x n f(x dx δ δ δ (n + x n f(x dx + (n + x n f( dx + δ δ (n + x n f(x dx (n + x n [f(x f(] dx + δ (n + x n f(x dx. (n + x n f( dx f( f([ ( δ n+ ] f( ( δ n+ f( < 3, δ (n + x n [f(x f(] dx 3 δ (n + x n dx 3, (n + x n f(x dx < M( δn+ < 3. Putting the lst three inequlities together gives (n + x n f(x dx f( < Let C([, ] be the spce of continuous functions on [, ]. Prove tht ( /2 f f 2 dm [,] defines norm on C([, ]. Is C([, ] Bnch spce with respect to this norm? Justify your nswer. We give direct proof first. Solution. We check tht the formul given defines norm. It is obvious tht f for ll f C([, ] nd tht. It is cler tht αf α f for ll f C([, ] nd ll α C. Next, suppose tht f C([, ] nd f. Then f lmost everywhere with respect to m. For every nonempty open set U [, ], we hve m(u >. So W {x [, : f(x } contins no nonempty open set. Since f is continuous, W. Thus f is the zero element of C([, ]. Finlly, let f, g C([, ]; we clim tht f + g f + g. We my ssume f + g. We hve, using Hölder s inequlity t the fourth step, f + g 2 f + g 2 dm f + g ( f + g dm f + g f dm + f + g g dm [,] [,] ( /2 ( /2 ( /2 ( f + g 2 dm f dm 2 + f + g 2 dm g 2 dm [,] [,] [,] [,] f + g ( f + g. The clim follows by dividing by f + g. However, C([, ] with this norm is not Bnch spce, since C([, ] is not complete with respect this norm. There re mny exmples one cn construct. For exmple, for n N define f n C([, to be the piecewise liner function x [, 2 ] f n (x 2 n ( x 2 [,] 2n x ( 2 2n, 2 + 2n 2n, ]. x [ 2 [,] /2

4 4 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 We clim tht (f n n N is Cuchy sequence. Let >. Choose N N with N > 4/ 2. Let m, n N stisfy m, n N. Then f m (x f n (x for x [, ]\ [ 2 2N, 2 + ] 2N nd fm (x f n (x f m (x+f n (x 2 for x [ 2 2N, 2 + ] 2N. Therefore ([ f m f n 2 4m 2 2N, 2 + ] 4 2N N < 2. The clim is proved. We finish the proof by showing tht lim f n does not exist. Suppose tht f C([, ] nd lim f n f. Since f n is rel for ll n N, it is esy to see tht f n Re(f f n f. Since limits re unique if they exist, f must be rel. Choose δ > such tht ( x 2 < δ implies f(x f 2 < 4. First suppose tht f( 2 2. For every n > 2/δ, if x ( 2 + δ 4, 2 + ] δ 2, then f(x > 4 while f n(x. So ( [ /2+δ/2 /2 ( ( ] 2 /2 δ δ f n f f n f dm 2 /2+δ/ This contrdicts lim f n f. Otherwise, f ( 2 < 2. Then similr rgument on [ 2 δ 2, 2 ] δ 4 shows f n f > δ/8 for n > 2/δ. Agin, we hve contrdicted lim f n f. This completes the solution. It is not enough to just sy f n χ [, 2 ] in the norm given, nd tht χ [, 2 ] C([, ]. It isn t even enough to prove tht there is no function in C([, ] which is equl lmost everywhere to χ [, 2 ]. One must show tht (f n n N does not converge in C([, ]. (Note, though, tht this pproch does work in the lternte solution, becuse the lternte solution explicitly uses the embedding of C([, ] in L 2 ([, ]. An lternte proof of the tringle inequlity cn be derived from the fct tht the norm defined on C([, ] comes from sclr product. Alternte solution. There is n obvious liner mp I : C([, ] L 2 ([, ]. We clim tht I is injective. Suppose I(f. Then f lmost everywhere with respect to m. For every nonempty open set U [, ], we hve m(u >. So W {x [, : f(x } contins no nonempty open set. Since f is continuous, W. Thus f is the zero element of C([, ]. Since I is injective nd liner, the formul f I(f L2 ([,] is norm on C([, ]. We know tht the rnge of I is dense in L 2 ([, ]. We clim tht it is not ll of L 2 ([, ]. Given this, C([, ] is nonclosed subspce of L 2 ([, ], nd therefore not complete; this will finish the solution. Set f χ [, 2 ]. If f were in the rnge of I, then there would be g C([, ] such tht f g lmost everywhere. But this is clerly impossible. 5. Let X, Y, nd Z be Bnch spces. Let S : Y Z be bounded injective liner opertor nd let T : X Y be liner opertor. Suppose S T : X Z is bounded. Prove tht T is bounded. (Hint: use the Closed Grph Theorem. Solution. We prove tht T hs closed grph nd then pply the Closed Grph Theorem to conclude tht T is bounded. To show tht T hs closed grph, it is sufficient to prove tht whenever (ξ n n N is sequence in X such tht lim ξ n nd lim T ξ n η, then η. Since S is bounded, we hve lim S(T ξ n Sη. Since S T is bounded, we hve lim (S T ξ n. Hence Sη, nd injectivity of S implies tht η. 6. Let m be Lebesgue mesure on [, 28]. For f L 2 ([, 28], let T f : L 2 ([, 28] C be the liner functionl T f (g fg dm [, 28] for g L 2 ([, 28]. Suppose (f n n N is sequence of functions in L 2 ([, 28] such tht f n L2 ([, 28] for ech n N. Prove tht if (f n n N converges to zero lmost everywhere, then for ech g L 2 ([, 28], lim T f n (g. (In other words, prove tht (f n n N converges wekly to zero in L 2 ([, 28].

5 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 5 Solution. To simplify the nottion, set I [, 28] nd bbrevite L2 ([, 28] to. Let >. Since g 2 L (I, there exists δ > such tht for ny mesurble set A I with m(a < δ, g 2 dm < 2 4. Set For n N define A E n { x I : f n (x > } 2 28( g +. nd T n E k. Define Z n T n. Then f n lmost everywhere implies m(z. Since m(i <, there is N N such tht for ll n N, m(t n < δ. Suppose tht n N. Then, using f n nd m(t n < δ t the second step, ( /2 ( /2 f n g dm f n 2 dm g 2 dm < T n T n T n 2. Also, using f n on I \ T n t the second step, ( /2 ( f n g dm f n 2 dm g 2 dm I\T n I\T n I\T n Therefore m(i \ T n /2 ( I I kn /2 /2 g 2 dm m(i /2 g < 2. T fn (g f n g dm < We cn lso use Egoroff s Theorem. The rgument involving the sets E n in the first solution is prt of the proof of Egoroff s Theorem, but one sees from tht solution tht we don t need the full strength of Egoroff s Theorem. Alternte solution. To simplify the nottion, set I [, 28] nd bbrevite L2 ([, 28] to. Let >. Choose δ > s in the first solution. Use Egoroff s Theorem to find subset B I such tht m(i \ B < δ nd f n uniformly on B. Choose N N such tht for ll n N, sup f n (x < x B 2 28( g +. For n N, we hve T fn (g f n g dm I I\B f n g dm + f n g dm. B The first integrl on the right is less thn 2 by the rgument used in the first solution for T n f n g dm, nd the second integrl on the right is less thn 2 by the rgument used in the first solution for I\T n f n g dm. So T fn (g <. 7. Find ll entire functions f such tht f(z z 5/2 for ll z C. Solution. We show tht the only such function is the constnt function f(z for ll z C. First, clerly f(. Suppose f is not identiclly zero. Then there re n entire function h with h( nd k N such tht f(z z k h(z for ll z C. We clim tht k 3. To prove the clim, choose δ > such tht h(z h( < h( /2 for ll z C with z < δ. Then for z < δ, we hve h(z > h( /2, so z k 2 zk h(z h( 2 z 5/2 h(. For this to be true for ll z C with z < δ, we must hve k 5/2. Since k N, it follows tht k 3.

6 6 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Given the clim, whenever z, h(z f(z z k f(z. z 5/2 Hence h is bounded entire function, nd h must be constnt by Liouville s theorem. So f(z h(z k for ll z C. Then h( z 5/2 k for ll z C \ {}. Since k 3, this contrdicts h(. Hence f is identiclly zero. Alternte solution. We show tht the only such function is the constnt function f(z for ll z C. Set g(z z 2 f(z for z C \ {}. Then g(z z /2 for z C \ {}, so lim z g(z, Therefore g hs removble singulrity t, tht is, there is n entire function h such tht h(z g(z for ll z C \ {}. Moreover, h( lim z g(z. Therefore there is n entire function k such tht h(z zk(z for ll z C. Tht is, f(z z 3 k(z for ll z C. Now z 3 k(z z 5/2 for ll z C, so k(z z /2 for ll z C \ {}. Therefore lim z k(z. Since k is entire, Liouville s theorem implies tht k(z for ll z C. Therefore f(z for ll z C. 8. Let n, nd let f(z z n + n z n + + z + be monic polynomil with complex coefficients. Prove tht mx z f(z. Solution. Set g(z + n z + n 2 z z n for z C. Then g( nd g(z z n f(/z for z. Suppose mx z f(z <. Then mx z g(z <. Hence g hs locl mximum inside the open unit disk D. This forces g to be constnt. Hence g. This contrdicts mx z g(z <. Alternte solution. Suppose mx z f(z <. Then for z we hve z n (z n f(z < z n. So Rouché s Theorem sys tht z z n nd z z n f(z hve the sme number of zeros in D {z C: z < }, counting multiplicity. But z z n hs n zeros in D, nd z z n f(z, being polynomil of degree t most n, hs t most n zeros in the whole complex plne. This contrdiction shows tht mx z f(z. 9. Suppose f L ((, (using Lebesgue mesure. Prove tht F (z defines holomorphic function on {z C: Im(z > }. f(te itz dt Solution. Set Ω {z C: Im(z > }. We prove tht if Ω then exists, by showing tht it is equl to F (z F ( lim z z f(tite it dt. It suffices to let (z n n N be ny sequence in Ω \ {} such tht lim z n, nd prove tht F (z n F ( ( lim z n z n We will use the Lebesgue Dominted Convergence Theorem. For n N define g n : (, C by for t (,. Then g n (t eitzn e it z n F (z n F ( z n f(tite it dt. f(tg n (t dt for ll n N. Since for ech t (, the function z e itz is holomorphic function with derivtive z ite itz, we hve lim g n (t ite it for ll t (,. So lim f(tg n (t f(tite it for ll t (,.

7 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 7 (2 For z C \ {} nd t (,, we estimte: e itz z (it k z k k! t t k z k t k! k k k t k z k (k! te t z. Set b Im(. Then b >. Choose N N so lrge tht n N implies z n < b/2. For such n, (2 implies (3 e it(zn z n tebt/2, so (4 g n (t e it e it(zn z n te bt/2. The function t te bt/2 is bounded on (,, so t f(t te bt/2 is integrble on (,. By (4, we hve f(tg n (t f(t te bt/2 for ll n N nd ll t (,, nd we lredy sw tht lim f(tg n (t f(tite it for ll t (,, so ( follows from the Lebesgue Dominted Convergence Theorem. This completes the solution. Alternte solution. This solution differs only in the method used to get the estimte (3. As in the first solution, set b Im( >, nd choose N N so lrge tht n N implies z n < b/2. For such n, we hve Im(z n > b/2, so Im(z n Im(z n Im( > b 2. Further let t (,. Then, if α [, ], we hve exp(itα(zn exp ( Re(itα(zn exp ( tαim(z n < exp ( tαb/2 e bt/2. Setting h t (z e itz for z C, we then hve e it(z n ht (z h t ( z n sup The estimte (3 follows. α [,] h t (α(z n z n sup it exp(itα(zn t zn e bt/2. α [,] One cn lso solve this problem by combining Fubini s Theorem nd Morer s Theorem. This is in principle net wy to do it. Unfortuntely, one must prove tht F is continuous in order to use Morer s Theorem. Second lternte solution. Set Ω {z C: Im(z > }. We first clim tht F is continuous on Ω. To prove the clim, let z Ω, nd let (z n n N be ny sequence in Ω such tht lim z n z. For n N nd t (,, set g n (t f(te itzn. Then lim g n (t f(te itz for ll t (,. For ll n N nd t (,, using Im(z n > t the lst step, we hve e itzn exp(re(itz n exp( tim(z n. Therefore g n (t f(t. Since f is integrble on (,, the Lebesgue Dominted Convergence Theorem implies tht F (z n f(te itzn dt f(te itz dt F (z s n. The clim is proved. Now let be ny tringle in Ω. We show tht F (z dz, by using Fubini s Theorem nd the fct tht the integrnd in the definition of F is holomorphic s function of z. Assume tht is defined on [, b]. Then, by definition, b F (z dz F ((s (s ds. Define function g : [, b] (, C by g(s, t f(t exp(it(s (s. The function (s, t exp(it(s (s is continuous except on the product of finite subset of [, b] with (,. Therefore it is mesurble. The function (s, t f(t is clerly mesurble. Therefore g is the product of

8 8 SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 two mesurble functions, hence mesurble. So g is mesurble. (Checking tht g is mesurble is n essentil prt of the solution. We cn now use Fubini s Theorem for nonnegtive functions to clculte b ( g d(m m f(t exp(it(s (s b b dt ds f (s ds f (s ds, [,b] (, which is finite. Therefore Fubini s Theorem for integrble functions cn be pplied, to get b ( F (z dz g(s, t dt ds ( b f(t exp(it(s (s ds dt ( f(t exp(itz dz dt. Since Ω is convex, Cuchy s Theorem implies tht the inner integrl on the right is zero for ll t (,. Therefore F (z dz. Since is n rbitrry tringle in Ω, Morer s Theorem now implies tht F is holomorphic.

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