SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS. Contents

Transcription

1 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Contents Preface 2 Jan Aug Jan 2 A 3 Jan 2 B 8 Aug 2 23 Aug 2 28 Jan Jan May Aug Jan 24 55

2 2 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Preface The following pages contain solutions to problems from qualifying exams in Analysis given at Rice University dating back to 999. The solutions are due to the combined efforts of Amanda Knecht, Eric Samansky, Steve Wallace, and Charlie Bingham with great aid from Professors Gao, Hardt, and Jones. These solutions have been compiled, typeset, and are currently being maintained by Charlie Bingham These solutions were last updated on March, 24. A good number of mistakes were remedied. Most were only minor, but a few solutions were complete nonsense. In addition January 24 exam solutions were added. Corrections are still welcome. Onward to 999, Year of the Rabbit...

3 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 3 Jan 999 Problem : Evaluate (e 2πz + ) 2 dz z = where the integral is taken in the counterclockwise direction. Solution : The zeros of e 2πz + are i and i. The possible poles of the integrand 2 2 function are thus i and i. So let s determine the residues incurred at the two points 2 2 respectively. Expanding e 2πz in its power series about i 2 (e 2πz +) = [ 2π(z i 2 we obtain: 4π2 )+ (z i 2 2 ) ] = (z i 2 )( 2π 2π2 (z i 2 )+...) (e 2πz + ) 2 = (z i 2 ) 2 ( 2π 2π 2 (z i 2 ) +...) 2 = (z i 2 ) 2 P (z) Note: here we may observe that i is a pole of order 2 since P ( i ) =. Now recall 2 2 4π 2 the binomial formula: ( ) α ( + w) α = w k k k= So setting w = (π(z i ) +...) we have the following: 2 P (z) = ( 2π) 2 ( + π(z i 2 ) +...) 2 = ( 2π) 2 ( + π(z i 2 ) + ) 2 = = 4π ( 2π(z i ) + higher order terms) 2 2 Recall from above that (e 2πz + ) 2 = (z i 2 ) 2 P (z) Hence the residue at i is 2 ( )( 2π) = 4π 2 residue at z = i 2 z = 2π. Since all of the above is unaltered for the calculation of the, we may conclude: (e 2πz + ) 2 dz = 2πi (residues of (e 2πz + ) 2 with z < ) = = 2πi( 2π 2π ) = 2i

4 4 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT (Another) Solution : Problem 2: Give an example of a subset of R having uncountably many connected components. Can such a subset be open? Closed? Solution 2: An example of such a subset is the Ternary Cantor Set (the Middle Thirds Cantor Set). This set is the compliment of a union of open intervals, hence it is closed. So we need only address the second question: Can such a subset be open? An open subset of R must contain an open interval (a, b). By density of Q in R, each open interval (a, b) must contain a rational number. Since Q is countable, we can have at most a countable number of open connected components. Hence a subset with uncountable connected components can t be open. Problem 3: Let f be a meromorphic function on C which is analytic in a neighborhood of. Suppose its Taylor Series at is: a k z k k= With a k. Let r = min{ z : f has a pole at z } <. Prove f has a pole at z = r where r R. Solution 3: By elementary arguments and the hypothesis a k following: we obtain the ( ) a k z k a k z k k= k= ak z k = a k z k k= k=

5 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 5 We know we have a minimum modulus pole on z = r, say z. Since power series representations diverge at poles, utilizing ( ) we have = a k z k a k z k = a k r k k= k= k= This implies the power series diverges at z = r as well. Therefore z = r is a pole. Problem 4: Let f(x) be R-valued, defined x, satisfying f() = and: f (x) = /(x 2 + f(x) 2 ) Prove lim x f(x) exists and lim x f(x) + π/4. Solution 4: By the Fundamental Theorem of Calculus, we have: f(x) = f() + x f (t)dt = + x f (t)dt Now, since f (t) > t, the function x f (t)dt is increasing in x and hence has a limit x. So taking a limit on both sides yields: lim f(x) = + lim x x Furthermore, f (t) t implies: lim x f(x) + lim x x x x /(t 2 + f(t) 2 )dt /(t 2 + f(t) 2 )dt x /(t 2 + )dt /(t 2 + )dt = + (arctan ( ) arctan ()) = + π/4 Problem 5: Suppose f : [, ] R is continuous. a. Prove that lim n xn f(x)dx = b. Prove that lim n n xn f(x)dx = f()

6 6 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Solution 5: (a) Note that f is continuous on [, ], and x n is continuous on [, ] n. Hence their product is continuous on [, ], a compact subset of R. Therefore x n f is integrable on [, ]. Furthermore, since x n on [, ], x n f is dominated by f. Thus by Lebesgue s Dominated Convergence Theorem, we have: lim n x n f(x)dx = lim n xn f(x)dx = dx = (b) Let u = x n. Then x = n u and du = nx n. Substitution into the integral yields (bringing the constant n inside): ( n u)f( n u)du Note that the domain of integration is not altered by the change of variables. Since composition of continuous functions preserves continuity, f( n u) is continuous. So by the same argument as in (a), we can interchange the limit and the integral s.t. lim n ( n u)f( n u)du = lim ( n u)f( n u)du = n f()du = [f()u] = f() Problem 6: Let α be a complex number and ɛ a positive number. Prove that the function f(z) = sin z + z α has infinitely many zeros in the strip Imz < ɛ. Solution 6: This problem stinks of smelly cheese. So our first instinct should be to apply Rouche s Theorem. Since we know where the zeros of sin x lie, we should apply Rouche to f(z) = sin z + z α and g(z) = sin z with appropriate contour.

7 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 7 Let Z = {z i } be the set of zeros of f(z). Let {x i } be the subset of Z on the the R interval [, 2π]. Since f is non-constant, there are only finitely many, say N zeros on the interval. So let U = {U i } N i= be an open cover of the interval [, 2π] such that each U i does not contain any other zero (on or off the R line). Furthermore, there exists a contour γ contained in U such that γ is the oriented boundary of a rectangle with vertices δi, 2π + δi, δi, 2π δi shifted slightly to the right or left to avoid zeros on the vertical edges of both f and g. We can also take δ < ɛ. γ is a compact subset of C hence it achieves its infimum. By construction, f is nonzero on γ. So let m = min γ f(z). Now recall that sin z has period 2π. Therefore we may shift γ sufficiently far to the right to some contour γ (or left for that matter) so that we have z α m on γ. We can do this since α is fixed and m is unchanged by 2π shifting. We may now apply Rouche s Theorem to obtain that f and g have the same number of roots inside γ. Note that there is at least one zero of sin z inside γ since sin x has period 2π. We may obtain the result be repeating this last step on {γ k } the set of 2kπ shifts of γ.

8 8 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Aug 999 Problem : Prove that, for any family F of upper semi-continuous functions on R, the function g(x) = inf{f(x) f F} is upper semi-continuous. Solution : f is upper semicontinuous if and only if for any t > f(x) there exists a neighborhood of x such that t > f(x) in that neighborhood. Let t > g(x). Then t > f(x) for at least one function f F. But then that function has a neighborhood as mentioned above. There we have t > f(y) g(y). Hence g is upper semicontinuous by the above equivalence. Problem 2: Suppose that f is a holomorphic function on {z : z < 3R}, f() =, M R = sup z R f(z), and N R = sup z R f (z). a. Estimate M R (from above) in terms of N R. b. Estimate N R (from above) in terms of M 2R. Solution 2: (a) Take any contour γ along a radii of the circle of radius R. z z = R. Then: f(z ) = f() + γ f (z)dz f(z ) + RN R Let Since this holds for arbitrary z on z = R, we have M R RN R. (b) Since f is holomorphic on z < 3R, Cauchy s Integral Formula gives us: f(z ) = f(ζ) dζ 2πi ζ z γ

9 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 9 f(ζ) dζ. Differ- ζ z where γ is the boundary of z 2R positively oriented, and z on z = R. So restricting our attention to z = we have f(z) = 2πi entiating on both sides we have: f (z) = d dz 2πi γ f(ζ) ζ z dζ = 2πi γ d dz f(ζ) ζ z dζ = 2πi γ γ f(ζ) (ζ z) 2 dζ Note that the second equality above is justified by the fact that the integral is on a compact subset of the domain of analyticity of an analytic function. Now taking absolute values on both sides of the equality, we obtain: f (z) = 2πi γ f(ζ) (ζ z) dζ 2 2π γ f(ζ) (ζ z) dζ 2 = 2π 2π2RM 2R 2π(2R R) = 2M 2R 2 R γ f(ζ) (ζ z) 2 dζ Thus we have N R 2M 2R /R. Problem 3: Suppose that f(x) is defined on [, ], and that f (x) is continuous. Show that the series converges. (n(f( n ) f( n )) 2f ()) n= Solution 3: Utilizing the Taylor series expansion about, we have: f(x) = f() + f ()x + f () 2 x 2 + f (t) x 3 6 f( x) = f() f ()x + f () x 2 f (s) x for some s, t [, ]. Thus we have the following: f(x) f( x) = 2f ()x + x3 6 [f (s) + f (t)]

10 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Now let x = /n to obtain: f(/n) f( /n) = 2f () n + 6n 3 [f (s) + f (t)] A n := [n(f(/n)) f( /n)) 2f ()] = 6n 2 [f (s) + f (t)] Since f (x) is continuous on [, ], f (x) is bounded by some M so [f (s) + f (t)] 2M. Thus A n M 3n 2 A n converges. Problem 4: Let f be an analytic function such that f(z) = +z +z for z <. Define a sequence of real numbers a, a, a 2,... by f(z) = a n (z + 2) n What is the radius of convergence of this new series n= a n(z + 2) n? n= Solution 4: + z + z = /( z) z = as n. So by uniqueness n of power series representations of analytic functions, f(z) = /( z) on D. In fact, f(z) = /( z) on the latter s domain of analyticity, which is C\{}. Hence the radius of convergence for the power series representation of f(z) about 2 is d( 2, ) = 3. Problem 5: Suppose that f, f 2,... are nonnegative continuous functions on [, ] with f n(x)dx M. a. Show that there exists a point a [, ] with f (a) 2M and f 2 (a) 2M. b. Does there exist a better estimate? That is, a number N < M so that inf max{f (a), f 2 (a)} 2N a for all such f, f 2. If so, find the smallest such N. If not, give a counterexample.

11 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS c. Show that there always exists an a [, ] so that f n (a) M for infinitely many n. Solution 5 (a) Suppose not. Then f (a) + f 2 (a) > 2M for all a. Hence we have: [f (x) + f 2 (x)]dx > 2M either f > M or f 2 > M, a contradiction to our hypothesis. (b) Draw a picture, or: Fix N < M, say M N = ɛ. So 2M 2N = 2ɛ. Now let: (2M ɛ) ifx [, ] 2 f, ɛ = (2M ɛ)x + (2M ɛ)( 2 +δ) for x [, + δ]. δ δ 2 2 otherwise f 2, ɛ = (2M ɛ)x δ (2M ɛ)( 2 δ) δ for x [ 2 δ, 2 ] (2M ɛ) forx [ 2, ] otherwise Note that f, ɛ and f 2, ɛ are are reflections about x = /2 of each other. From this observation we note: f, ɛ(x)dx = f 2, ɛ(x)dx. Since the area under these curves is the area of a rectangle and a right triangle with base; height ; (2M ɛ) and δ; 2 2M ɛ respectively, we have the following: f, ɛ (x)dx = f 2, ɛ (x)dx = 2 (2M ɛ) + δ 2 (2M ɛ) = (2M ɛ)( + δ) 2 We want the above functional restricted to our two functions to be bounded above by M. i.e. we want (2M ɛ)( + δ) M. So we simplify the inequality: 2 2 (2M ɛ)( + δ) = (2M + 2Mδ ɛ ɛδ) M 2 2Mδ ɛ ɛδ δ(2m ɛ) ɛ Then let δ = ɛ/(2m ɛ). We thus have equality above which is sufficient to satisfy the hypothesis of concern. So substituting the δ value computed above into our definition of f, ɛ and f 2, ɛ yields a pair of counterexample functions for any N < M..

12 2 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT (c) The statement is false. Consider the following counterexample: M( nx) if x [, /n] For n 2 define f n (x) = M( + n x) if x [/n, 2/n] 2 M if x [2/n, ] Now let g n (x) = f n(x) n Note that n g n(x)dx = M and for each a [, ], g n (a) M for only finitely many n. So g n serves as a counterexample to the statement in (c).. Problem 6: Prove that there is no one-to-one conformal map of the punctured disc G = {z C : < z < } onto the annulus A = {z C : < z < 2}. Solution 6: Suppose there exists such a map, g : D\{} A. Then g has an analytic continuation G on D. G() must lie in A (which is the open annulus) since intd G() intā by the Open Mapping Theorem. But g onto implies that there exists some z D\{} such that g(z ) = G(). Now take open disjoint neighborhoods about z and, U z and U respectively. Push them forward to the range space. Since G() = g(z ), G(U ) g(u z ). Therefore, points in our two domain neighborhoods share image points. Since G D\{} = g, we have a contradiction to confomality (conformal maps are injective).

13 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 3 Jan 2 A Problem : Suppose g(x) = lim n g n (x) for x [, ] where {g n } are positive and continuous on [, ]. Also suppose that: a. Is it always true that g(x)dx? b. Is it always true that g(x)dx? g n dx = Solution : a. Yes. This is a direct application of Fatou s Lemma: lim inf g n(x)dx lim inf g n (x)dx n n g(x)dx b. No. Consider the following family of functions on [, ]: n 2 x ( n2 n) if x [, ] 2 2 n 2 g n (x) = n 2 x + ( n2 + n) if x [, + ] n otherwise. The graphs of these functions look like triangles with height n peaks at x = 2 and with bases on [, + ]. The ratio of the base to the height is 2 n but 2 n 2 n lim n f n = δ = almost everywhere. Hence f(x) = δ 2 =. 2 Problem 2: Let f be C-valued in D(, ) such that g = f 2 and h = f 3 are both analytic. Prove f is analytic in D(, ). Solution 2: Define k = h/g = f 3 /f 2 = f. Now we must be careful to show that this function k is in fact analytic. The only possible singularities of a quotient of two

14 4 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT analytic functions is at the zeros of the denominator function. So suppose f 2 (z ) =, then f 3 (z ) =. Let m be the multiplicity of the zero of f 3 at z, and n be the multiplicity of the zero of f 2 at z. That is, f 3 (z) = α m (z z ) m +... and f 2 (z) = β n (z z ) n +... Now (f 3 ) 2 = f 6 = (f 2 ) 3 2m = 3n m > n. Therefore the limit of k(z) as z z exists (in fact it is zero). Thus z is a removable singularity of k, and so k is indeed analytic there by the Riemann Removable Singularity Theorem. Since z was arbitrary, we are done. Problem 3: Construct an open set U [, ] such that U is dense in [, ], the Lebesgue measure µ(u) <, and that µ(u (a, b)) > for any interval (a, b) [, ]. Solution 3: Order Q [, ], say {q, q 2,...}. Let A i be the interval (q i, 2 i+2 q i + ). Let B 2 i+2 i = A i [, ]. Note that B i is open i and µ(b i ). So we have: 2 i+ µ( B i) = /( ) =. Hence µ( B i). Q B 2 i, so for any open interval (a, b) [, ], q Q such that q (a, b). And by construction, we have: µ((a, b) ( B i )) >. Problem 4: Let f, g, g 2,... be entire functions. Assume that the kth derivatives at satisfy: a. g (k) n f (k) for all n and k; b. lim n g (k) n () exists for all k. Prove that the sequence {g n } converges uniformly on compact sets and thats its limit is an entire function.

15 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 5 Solution 4: The {g n } are entire, so they have power series representations at : g = α + α z + α 2 z g 2 = α 2 + α 2 z + α 22 z g n = α n + α n z + α n2 z Also let f(z) = β + β z + β 2 z be the power series representation of f at. and f (k) () = (k!)(β k ). Observe that g (k) n () = (k!)(α nk ) By hypothesis (ii), g (k) n () converge for each k. Therefore lim n α nk exists for each k since the sequence is merely a constant multiple (k!) times the given sequence. Let α i = lim n α ni. From our observation above we have α i β i. Now we claim that g n (z) converges uniformly on compacta to g(z) where g(z) = i= α iz i. That is, given ɛ > and z R, N such that n N g(z) g n (z) < ɛ for all z such that z R. More precisely, restricting to our compact domain we desire the following: n N α i z i α in z i < ɛ And since α i z i ı= ı= ı= ı= α in z i = (α i α in )z i it is sufficient to bound the following by ɛ for n N α i α in z i ( ) ı= ı= (αi α in )z i So let ɛ > and R R be given. Let z R. Now, the power series expansions of analytic functions converge absolutely in the interior of the function s domain of ı=

16 6 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT analyticity. Since f is entire, on z R we have K N such that the second inequality holds below. βi z i β i R i < ɛ 4 i=k Substituting the estimate α i α in 2 β i into ( ) we obtain αi z i α ni z i 2 βi z i ɛ < 2 i=k i=k We now have an ɛ/2-bound for the infinite tail of the sum which is independent of n and z, so all we need do is bound the first K terms by ɛ/2 independent of z. Since the α ni converge to the α i, N i such that n N i α i α ni < ɛ 2KR i Put N = max K i= {N i}. Then n N ensures: K K K α i α ni z i ɛ ɛ < 2KR i Ri = 2K = ɛ 2 i= i= i= Combining our two bounds, we may indeed conclude that proved that the limit function g is entire as well. i=k Problem 5: Let f be a continuous, strictly increasing function from [, ) onto [, ) and let g = f be the inverse of f. Prove that: a for all positive numbers a and b. f(x)dx + b g(x)dx ab Solution 5: Since f is monotone increasing, f is monotone increasing. We have one of the following f(a) = b, f(a) > b or f(a) < b. However, since (f ) = f we may assume without loss of generality that f(a) b. Thus we have

17 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 7 a f(x)dx + b ( a = f(x)dx + ( a = f(x)dx + ( a g(x)dx = f(x)dx + f(a) f(a) f(a) ) ( b g(x)dx + a dx + f(a) ) g(x)dx + b f(a) a dx g(x)dx + b f(a) b f(a) ) + b f(a) g(x)dx = ) (g(x) a)dx (g(x) a)dx = = = (ab) + b f(a) (g(x) a)dx ab Problem 6: Suppose that f(z) is analytic and satisfies f( ) = f(z) for all z C\{}. z a. Write down the general Laurent Expansion for f. b. Show that the coefficients of this expansion are all real if this f has real values on the unit circle z = Solution 6: (a) The condition f(z) = f( ) forces the z ith and i th coefficients to be equal: f(z) = α + α i [(z z ) i + (z z ) i ] (b) For all k we have the following identity: ( ) α k = 2πi ζ = f(ζ) dζ ζk+ Since f(z) R on z =, trivially f(z) = f(z) there. Also on z = we have z = z z z = z z 2 = z

18 8 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Using this fact we obtain the following: dz = z 2 dz Now conjugating both sides of ( ) and incorporating the above facts yields: α k = f(ζ) f(ζ) dζ = dζ = f(ζ) 2πi ζ = ζk+ 2πi ζ = ζ k+ 2πi ζ = ζ k = 2πi ζ = f(ζ) ζ k+ dζ = α k = α k ζ 2 dζ = From the above string of equalities, we obtain α k = α k. Hence α k R for all k Z. Jan 2 B Problem : Does there exist a function f(z) that is holomorphic near the origin and that satisfies: Why or why not? f( n ) = f( n ) =, n =, 2, 3,...? n3 Solution : No. By hypothesis, f(z) agrees with z 3 on the set { n } which has as a limit point. By uniqueness of power series expansions, f(z) = z 3 near zero. But n 3 n 3. Problem 2: Let f : R R be continuous, with f(x) dx <. Show that there is a sequence x n R such that x n, x n f(x n ) and x n f( x n ) as n. Solution 2: Assume to the contrary that there does not exist such a sequence {x n }. Then either lim inf x x f(x) > or lim inf x x f( x) >. So either M + N

19 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 9 or M N such that or x M + x f(x) k > f(x) k/x x M x f( x) c > But then we would have or M + f(x) dx f( x) dx M contradicting integrability. f( x) c/x M + k x dx = M c x dx = Problem 3: Suppose that f(z) = a + a z + a 2 z and g(z) = b 2 z 2 + b z + b + b z + b 2 z where the two series n= a nz n and n= b nz n converge for z < 2. Find Γ f(z)g(z)dz in terms of the a n, b n where Γ is the positively-oriented unit circle. Solution 3: f(z)g(z) = a b 2 z 2 + a b z + a b 2 z +... Hence by the Residue Formula: f(z)g(z)dz = 2πi(a b + a b 2 ) Γ Problem 4: Suppose f n is a sequence of nonnegative measurable functions on R, and lim n f 2 n(x)dx =.

20 2 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT a. Show that the Lebesgue measure λ{x : f n (x) > δ} as n for every δ >. b. Show that lim n f n(x)g(x)dx = for every g L 2 (R). c. Find a sequence g n of nonnegative measurable functions on R such that lim n λ{x : g n (x) > δ} = for every δ > AND lim n g n(x)g(x)dx = for every g L 2 (R) BUT g2 n(x)dx =. Solution 4: (a) Recall the Chebyshev Inequality for L 2 : λ{x : f n (x) > δ} f n 2 2 δ 2 Fix δ > then the RHS as n since f n 2 2 by hypothesis. (b) Via a basic inequality and Hölder s Inequality we obtain: Taking a limit as n yields: lim n f n (x)g(x)dx f n g f n 2 g 2 f n (x)g(x)dx lim n f n 2 g 2 = The last equality follows from the fact that g 2 < and the hypothesis on f n. (c) Define g n : R R by Note that g n (x) = χ [n,2n] n for n =, 2, 3,... g 2 n(x)dx = n n = Also note that g n (x)g(x) is dominated by the restriction of g to a compact domain [n, 2n]. Since L 2 L on compact subsets of R, we obtain the following via Lebesgue s Dominated Convergence Theorem (since g n (x)g(x) is dominated by.g(x)) lim n g n (x)g(x)dx = lim g n(x)g(x)dx = n g(x)dx = a.e.

21 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 2 Problem 5: a. Prove that if f is a holomorphic map from the unit disk D = {z : z < } to itself with f() =, then f(z) z for all z D. b. Which of these f admit a point a in D with f(a) = a? c. Let h be a holomorphic map of the unit disk D into itself which is not the identity map of D. Show that h can have at most one fixed point. Solution 5: Note that parts (a)-(c) are known as the Schwarz Lemma. (a) Define g : D D by g(z) = { f(z)/z x f () x = Note that this function is analytic in < z < Observe that lim z g(z) = f () = g(). Since f is analytic at, f () is analytic at. Hence our g(z) is analytic in D. By the maximum principle for analytic functions g(z) f(z) z z So g(z) is bounded by on every closed disk of radius r. Letting r and applying r the Maximum Principle for analytic functions we obtain the inequality g(z) on D. Therefore f(z) z there.. (b) Suppose there exists such a point. Then g(a) =. The Maximum Principle then implies that g(z) is constant in D. Hence f(z) must be a rotation about the origin. That is, f(z) = kz with k =. (c) Suppose h : D D is not the identity, and suppose h(a) = a and h(b) = b, a, b D. Let g(z) = a z ā z

22 22 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Note that g(a) = and g() = a. Furthermore, this mapping is an analytic automorphism of D such that g = g since a <. Now consider the composition f = g h g : D D. After observing that f() =, we apply the Schwarz Lemma to obtain f(z) z. Furthermore f = g h g g h = f g. Therefore f ( g(b) ) = g ( h(b) ) ( ) ( ) a b a b f = āb āb Therefore f satisfies the Schwarz Lemma and has a fixed point other than inside the disk. Hence f(z) = z. But this in turn implies that h = g id g = g g = id, a contradiction. Problem 6: Suppose h(x) = lim n h n (x) where h h 2 h 3... is a decreasing sequence of nonnegative continuous functions on R. a. Give a specific example of such h and h n where h is not continuous at. b. Prove that, in general, such an h is always upper semi-continuous; that is, for all a R. h(a) lim sup h(x) x a Solution 6: (a) Consider the following sequence of functions: { e nx x h n (x) = e nx x > Note that lim n h n (x) = for x and lim n h n (x) = for x =.. (b) Lemma: Let F be any family of upper semicontinuous R R functions. Then g(x) = inf{f(x) : f F} is upper semicontinuous. Proof of Lemma: f is upper semicontinuous if and only if for any t > f(x) there exists a neighborhood of x such that t > f(x) in that neighborhood. Let t > g(x).

23 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 23 Then t > f(x) for at least one function in f F. But then that function has a neighborhood as mentioned above. There we have t > f(y) g(y). Hence g is upper semicontinuous by the above equivalence. Now note that h n (x) is a monotone decreasing sequence of bounded continuous functions. Hence inf n h n (x) = lim n h n (x) = h(x). From our Lemma we have that h is upper semicontinuous. Aug 2 Problem : Let f be a real valued function defined and integrable over a measurable set E R. Prove that given ɛ >, there is a δ >, such that f(x)dx < ɛ for every measurable subset A E with λ(a) < δ. A Solution : Fix ɛ. Since f L (E), there exists a simple function s such that s f and: s(t)dt > f(t)dt ɛ/2 E E Now s assumes a finite number of values, so let M = max{s(x) : x E}. Then f(t)dt = s(t)dt + (f s)(t)dt A A A A f(t)dt < M[λ(A)] + ɛ/2 So let δ = ɛ/2m, and we have our result.

24 24 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Problem 2: For each of the following conditions, decide if there is a non-constant, holomorphic function defined on the whole complex plane with the given property. a. realf(z) > z b. f(z) < + z log( + z ) z c. A function f which has and as its only asymptotic values. (α C is an asymptotic value of f if there is an unbound path γ C s.t. lim z, z γ f(z) = α) Solution 2: (a) Since f(z) is entire, e f(z) is entire. Now e f(z) = e ref(z) > by hypothesis. Since this function does not have a zero, /e f(z) is entire as well. But /e f(z) < thus Liouville s Theorem implies /e f(z) is constant which in turn implies that f(z) is constant. So there does not exist such a function. (b) Consider the following: f(z) < + z log ( + z ) z f(z) < log ( + z ) + z z log ( + z ) f(z) < + z z such that z (e ). So by the Extended Liouville Theorem, f(z) is a polynomial of degree at most. But we can do better. I.e. our first argument gave up a lot of ground on estimating the growth of f(z). Suppose f(z) = αz + β, for some α, β C. αz + β α z + β. If we consider z with large enough modulus, then eventually /(log ( + z ) will be smaller than both α and β. Hence for some M, f(z) > M f(z) > + z log ( + z )

25 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 25 Therefore, if an entire function satisfies such an inequality, it must be constant. (c) Claim: e z satisfies the criterion. First, if we take the paths from the origin to infinity along the positive x-axis and negative x-axis we obtain the asymptotic values of and respectively. Now, suppose α C is some other asymptotic value. Then α = k, k (, ). This implies that the asymptotic path must tend towards x = e k and hence the path approaches in the imaginary direction. But if so e z argument never approaches a single value. Roughly speaking, along this path e z asymptotically cycles the circle of modulus e k. Therefore α is not really an asymptotic value. (This argument could be made more precise by a delta-epsilon argument) Problem 3: Let g(z) be analytic in the right half-plane {z : real(z) > }, with g(z) < for all such z. If g() =, how large can g(2) be? Solution 3: There exists an analytic isomorphism of the right half-plane with the open unit disk, h : RHP D given by h(z) = ( z +z ) with inverse h (z) = ( +z z ). So consider g(h (z)) : D D. g(h ()) = g() = so we may apply Schwarz s Lemma to our composition of analytic functions. But first note: h : 3 2. Hence: g(h (z)) z g(2) 3 Problem 4: Let f be a C 2 function on the real line. Assume f is bounded with bounded second derivative. Let Prove that A = sup x R f(x) and B = sup f (x) x R sup f (x) 2 AB x R

26 26 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Solution 4: The Mean Value Theorem states that (a, b) R, c (a, b) such that f (c) = f(b) f(a). A simple estimate provides f (c) 2A. Applying the MVT in an b a b a identical manner, ( ) picking an x (a, b) we also have f (y) = f (x) f (c) for some x c y (x, c) (Assume x < c WLOG). Thus we have the following: f (y) = f (x) f (c) f (x) = f (c) + (x c)f (y) x c f (x) f (c) + x c f (y) f (x) 2A + (b a)b Now at ( ) if we select x = b a 2 b a, we obtain a slightly better estimate: f (x) 2A b a + b a 2 B Setting l = (b a) and minimizing the RHS of the inequality, we may obtain a best estimate. I.e. let F (l) = 2A + Bl Then l 2 F (l) = 2A + B l 2 2 = 2A l = B 2 2 l2 = 4 A A B l = 2 B This critical value of F (l) is indeed a global minimum for R + for F (l) = 4A l 3 Plugging this value of l into our inequality yields: f (x) 2A B 2 A + 2B A 2 B = 2 AB l. Problem 5: Let f(x, y) be defined on the unit square S = {(x, y) R 2 : x and y } and suppose that f has the following properties: i. For each fixed x, the function f(x, y) is an integrable function of y on the unit interval. exists at every point (x, y) S, and is a bounded func- ii. The partial derivative f x tion on S.

27 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 27 Show that: a. The partial derivative f x b. d dx is a measurable function of y for each x. f(x, y)dy = f (x, y)dy x Solution 5: (a) Consider the following function in the variable y (fix x, h): F h (y) = f(x + h, y) + f(x, y) h Since h is a constant, F h is a measurable function by application of (i). Since a limit of measurable functions is also measurable, we obtain f x x by observing: f f(x + h, y) + f(x, y) (x, y) = lim x h h is measurable in y for each = lim h F h (y) (b) All of the following equalities are elementary or definitional except the second, which is an application of Lebesgue s Dominated Convergence Theorem justified by hypothesis (ii). f (x, y)dy x = lim h lim h (f(x + h, y) + f(x, y))dy h (f(x + h, y) + f(x, y))dy h (f(x + h, y) + f(x, y))dy = lim h h d dx f(x, y)dy = Problem 6: Suppose that f(z) is a non-constant holomorphic function on a connected open set U C. Suppose that V is an open set such that its closure V is a

28 28 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT compact subset of U, and suppose that f(z) is constant, say k, on the boundary of V. Show that f has at least one zero in V. Solution 6: By the maximum principle, we have f(z) < z, for all z V, z on V. So f(z) < k on V. Hence f maps V into the open disk of radius k. Now suppose there is no point z of V such that f : z. Then /f is holomorphic on V as well. By the maximum principle and compactness of domain, it obtains a maximum on V. When /f is maximized, f is minimized, hence f obtains a minimum on V. Therefore, f has constant modulus V f is constant there f is constant on U, a contradiction. Whence f indeed has a zero in V. Aug 2 Problem : Suppose f is a positive bounded measurable function on [, ] and F (x) = x f(t)dt for x. a. Show that F is continuous on [, ]. b. Show that F is differentiable at almost every point in [, ]. Solution : (a) F (x) F (y) = x f(t)dt y f(t)dt Let ɛ be given. By hypothesis, f M. Without loss of generality, let y x. Then we have x So let δ = ɛ/2m. f(t)dt y y y f(t)dt = f(t)dt f(t)dt + x = f(t)dt M x y y x y f(t)dt =

29 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 29 (b) This is an application of Lebesgue s Differentiation Theorem. Since f is bounded, f L loc (R). Then we have F (x) = f(x) almost everywhere.

30 3 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Problem 2: Suppose f : C C is a holomorphic function, m and n are integers, and is bounded on C. a. Prove that f is a polynomial (2 + z m ) dn f dz n b. Estimate the degree of f in terms of m and n. Solution 2: (a) f is entire so d n f/dz n is entire. Furthermore, by the Extended Louiville Theorem, i.e. since d n f dz n 2M + M z m d n f/dz n is a polynomial of degree at most m. Integrating backwards, we have f must be a polynomial. (b) In (a) we integrate backwards n times. Therefore deg(f) m + n. Problem 3: Suppose f and g are integrable functions on R. Show that the convolution h(x) = f(x y)g(y)dy is integrable with h(x)dx f(x)dx g(x)dx Solution 3: Consider the absolute value of the given product function. By Fubini s Theorem for non-negative measurable functions, we have: dx f(y)g(x y) dy = dy Let ( f g )(x) denote the convolution LHS = f(y)g(x y) dx f(y)g(x y) dy. Now observe ( f g )(x)dx

31 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 3 RHS = f(y) dy g(x y) dx = f g Since f, g L, ( f g )(x) exists a.e. and is integrable. Now note the following inequalities f g f g = f(y)g(x y)dy f(y)g(x y)dy = f g So incorporating all of the above and using the fact that h L h L, we obtain f g f g Problem 4: Find the following integral: dx x 4 + Solution 4: Let f(z) = z 4 + γ R z 4 + dz =. Consider the following contour integral R x 4 + dx + R S R z 4 + dz where γ R is the contour that runs from R to R on the x-axis then over the half circle in the upper half-plane of radius R, and S R is the half-circle component of the contour. By the residue formula, we have ( ) γ R z 4 + dz = 2πi (residues of f(z) in the UHP) Now since z 4 + the following is true: z 4 R 4 lim R S R z 4 + dz = Since ( ) holds for all R, we obtain via the limit R that x 4 + dx = 2πi (residues of f(z) in the UHP) So now all we need to do is calculate the residues in the UHP. The resides of f occur at the singularities of f, which are the zeros of z 4 +. Those zeros are

32 32 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT e πi/4, e 3πi/4, e 5πi/4, e 7πi/4. Only the first two lie in the UHP. The residues are given by the following function (evaluated at each singularity) Therefore f (z) = 4z 3 x 4 + dx = 2πi( 4 e 3πi/4 + 4 e πi/4 ) = ( πi 2 )(e πi/4 )( + e πi/2 ) = = ( πi 2 )( i 2 )( i) = π 2 Problem 5: a. Show that any sequence f n of non-negative integrable functions on [, ] with f 2 ndx n 3 must converge to zero almost everywhere. b. Is there a sequence g n of non-negative integrable functions on [, ] satisfying g2 ndx which does not converge to zero almost everywhere? Explain. Solution 5: (a) The first equality in the following line is a corollary of Lebesgue s Monotone Convergence Theorem (using non-negativity). ( fn(x))dx 2 = f 2 n(x)dx n 3 < ( fn(x))dx 2 < fn(x) 2 < a.e. fn(x) 2 a.e. f n (x) a.e.

33 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 33 (b) Define g n : R [, ) by g n (x) = χ An where A = [, ], A 2 = [, 2 ], A 3 = [, ], A 2 4 = [, ], A 3 5 = [, 2], A = [ 2, ], A 3 7 = [, ],... 4 lim n g n (x)dx = lim n λ(a n ) = Yet for each point, no matter how much time passes, there will be infinitely many A n that cover it. Hence no point has a limit. Notice that the integral tends very slowly to. This is caused by requiring a covering of [, ] for each interval width. Problem 6: Suppose A = {z C : < z < 2}. a. Is the function the uniform limit of a sequence of polynomials on A. Explain. z b. Is the function the uniform limit of a sequence of polynomials on A. Explain. z 3 Solution 6: (a) No. If polynomials converge uniformly for z = R then they converge uniformly for all r R. But blows up at and hence cannot be such a limit. z (b) Yes. At z =, has a radius of convergence r = 3, the distance to its singularity z = 3. Since the annulus is inside its radius of convergence, the power series z 3 for converges absolutely and uniformly on A. Thus let p z 3 n(z) = n a nz n, the partial sums of the power seires.

34 34 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Jan 22 Problem : Let Ω = C\{R } c. Is there a non-constant bounded holomorphic function on Ω? Justify your answer. Solution : Ω is star-convex (take as the star center) hence simply connected. Thus the Riemann Mapping Theorem applies to Ω. I.e. Ω is analytically isomorphic to the unit disk. Therefore there exists an analytic map h : Ω D. Problem 2: A function f : R R is convex if f(( t)a + tb) ( t)f(a) + tf(b) a, b R and t. a. Prove that a convex function is continuous b. Prove that a twice differentiable function f is convex if f. Solution 2: (a) The convexity condition above is equivalent to a < b < c f(b) f(a) b a f(c) f(b) c b So let x R be given, and choose s < l < x < r < t. Then our equivalent convexity condition implies and Therefore f(l) f(s) l s f(x) f(s) x s f(x) f(l) x l f(r) f(x) r x f(t) f(x) t x f(t) f(r) t r either f(x) f(l) x l f(t) f(x) t x ( ) or f(x) f(l) x l f(l) f(s) l s ( )

35 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 35 and either f(r) f(x) r x f(t) f(r) t r ( ) or f(r) f(x) r x f(x) f(s) x s ( ) (*) and (****) are independent of l and r respectively thus left and right continuity holds in those cases (let l x and r x). Using the following estimates we similarly have left and right continuity for (**) and (***) respectively { } f(l) max f(s), f(x) f(x) + sf(t) t x { } f(r) max f(t), f(x) f(x) + tf(s) s x (b) DO THIS PART Problem 3: If f is a nowhere zero, entire holomorphic function, is there necessarily an entire holomorphic function g such that e g = f. Prove your answer. Solution 3: Define z f (t) h(z) z f(t) dt Hence h (z) = f (z)/f(z). Thus we have ( ) f = ( fe h) = f e h + ( h )fe h = e h (f f ) = e h And so f/e h is constant, which implies e h = αf. So let g h + ln α. Problem 4: a. Prove that if f n is a sequence of Lebesgue measurable functions on [, ] with

36 36 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT lim n f 2 ndx =, then lim n f n dx =. b. Does it also follow that lim n f n (x) = for a.e. x [, ]? c. Is the implication in (a) still true if [, ] is replaced by the whole real line R? Solution 4: (a) lim n f 2 ndx = lim n ( f 2 ndx) 2 =. I.e. lim n f 2 =. Since f n f n 2 on the unit interval, we have lim n f =. (b) No. Consider the following family of sets, {A i } i= where A = [, ], A 2 = [, ], A 2 3 = [, ], A 2 4 = [, ], A 3 5 = [, 2], A = [ 2, ],... 3 Let f n = χ An. Then we have lim n f n(x) dx = but lim n f n (x) does not exist for any x [, ]. (c) The conclusion does not extend to R. Consider the following counterexample. ( ) f n (x) = χ [,n] n Note that while lim n ( ) fn(x)dx 2 = lim (n) = lim n n 2 n n = lim n f n (x) dx = Problem 5: Does there exist an analytic function mapping the annulus A = {z : z 4} onto B = {z : z 2} and taking C C, C 4 C 2, where C r is the circle of radius r? Why or why not?

37 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 37 Solution 5: Define g(z) = z 2. Consider g : B A. g : C C and g : C 2 C 4. Then: (g f)(z) z = = z (g f)(z) But then the maximum modulus principle implies that g f = c, a constant with c =. In turn this implies that f(z) = c z. But z is not a surjective mapping from A to B for z : A B UHP. Since c = c z is some rotation of the half annulus, we cannot possibily find a surjective map from A to B. Problem 6: Suppose that f L (R). Find a necessary and sufficient condition such that exists for all g L (R). f + tg f lim t t Solution 6: (Missing) Jan 23 Problem : How many roots of the equation f(z) = z 4 + 8z 3 + 3z 2 + 8z + 3 = lie in the right half plane? Hint: Apply the argument principle to a large half disk. Solution : Let γ R be the contour which travels along the circle of radius R in the right half-plane and then from ir to ir on the imaginary axis. On the half-circle, i.e. z = Re it for π t π, we have: 2 2 f(z) = f(re it ) = R 4 e 4it ( + 8 Re it + 3 R 2 e 2it + 8 R 3 e 3it + 3 R 4 e 4it ) = R4 e 4it ( + ζ)

38 38 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT where ζ 22/R < ɛ for any ɛ for R large enough. So the argument of f(re it ) is approximately the argument of e 4it. As we move along the half-circle, t goes from π 2 to π, thus the argument of f(z) changes by approximately 4π along this part of the 2 contour. Now lets consider the argument of f(z) along the imaginary axis section of γ. I.e. z = iy where y [ R, R]. So f(iy) = y 4 8iy 3 3y 2 + 8iy + 3. Hence: ref(iy) = y 4 3y and imf(iy) = 8y 3 + 8y So for R large enough, f(ir) s real part (quartic) dominates its imaginary part (cubic). Hence for R large enough the change in argument on the vertical segment is approximately zero. Therefore by the Argument Principle, the number of roots in the RHP is (4π + ) = 2. 2π Problem 2: Suppose that f and g are entire functions, and f(z) g(z) for all z with z. Show that f/g is a rational function. Solution 2: g is entire but g may have some zeros. But there can only be finitely many. For if there are infinitely many in Ω = {z : z }, the set of zeros has a limit point there. So then g = on Ω. Hence g = on C which implies f = on C. So let {α} N be the set of roots of g in Ω. Let q(z) = (z α )(z α 2 )...(z α N ). Then qf/g is entire. Furthermore, by hypothesis qf/g q for all z such that z. The Extended Liouville Theorem implies qf/g is a polynomial, p of degree at most N. Therefore f/g = p/q is a rational function. Problem 3: Suppose that f is a nonvanishing holomorphic function on the punctured disk Ω {z : < z < 2}, and lim z f(z) = +. a. Prove that lim z z N f(z) = α for some positive integer N and nonzero complex number α.

39 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 39 b. Find a formula for α in terms of an integral over the unit circle of some expression in terms of f and N. c. Find a formula for N in terms of an integral over the unit circle of some expression in terms of f and f. Solution 3: (a) Consider g(z) /f(z). Since f is non-vanishing on Ω, g : Ω C is holomorphic. Furthermore by hypothesis lim z =. Hence is a removable singularity for g. So g may be extended to a holomorphic function G : Ω {} C given by: { g(z) on Ω G(z) =. at Hence for some unique positive n we have that G(z) = z N Q(z) where Q is holomorphic on Ω {} and Q(). G(z) is non-vanishing on Ω implies that Q(z) is non-vanishing on Ω {}. It thus follows that /Q(z) is a holomorphic function on Ω {}. Therefore on Ω {} For < z < 2 /Q(z) = β j z j j= f(z) = G(z) = z N Q(z) ( ) = z N β j z j = j= N j= β j+n z j By uniqueness of Laurent expansions, this series agrees with f on Ω so it must indeed be it s Laurent expansion. Our desired result follows by observing that ( lim zn f(z) ) ( ) = lim z N β j+n z j = β z z j= N

40 4 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Note: This proof was tailored from the development in Function Theory of One Complex Variable by Greene & Krantz. Their development of meromorphic functions is excellent. (b) Let γ be the curve that traverses the unit circle once in the positive orientation. Cauchy s Integral Formula states f(z)dz = 2πi Res (f) γ Recall that Res (f) is defined as the i = coefficient in the Laurent expansion at for f. So to find α in the statement (β in the proof of part (a)), we need only take the following integral z N f(z)dz 2πi γ (c) The following is a consequence of the Reside Theorem. Assume f is meromorphic in U, with finite number of zeros and poles there. Let a, a 2,..., a r be the zeros in U and b, b 2,..., b n be the poles in U. Also assume that γ is a closed -homologous curve in U, then γ ( r f f = 2πi m j ord aj f j= ) n m k mul bk f where m i is the winding number relative γ at each point a i (likewise for the poles). In this case, we have that inside D(, 2) f has no zeros (non-vanishing), has precisely one singularity at, and γ winds about once. Hence the formula above becomes: f f = 2πi( mul f) Solving for mul f we obtain γ k= N = mul f = f 2πi γ f

41 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 4 Problem 4: Calculate Show all your work and justify all steps. /n n lim n + n 2 x 2 + n 6 x dx 8 Solution 4: Let y = nx, then dy = ndx so /n n A = lim dx = lim n + n 2 x 2 + n 6 x8 n + y 2 + n 2 y 8 dy + y 2 + y is dominated by n 8 + y 2 L ([, ]) 2 So Lebesgue s Dominated Convergence Theorem implies A = lim n + y 2 dy = π 4 = π 4 Problem 5: Suppose f is a nonnegative measurable function defined on R k. Let a(n) = f n, for n =, 2, 3,... R k a. Suppose that the infinite series a(n) converges. Show that f < almost everywhere and that f/( f) is in L. b. Prove the converse. Solution 5: (a) First suppose that f on a set B R k with λ(b) = ɛ. Then a(n) = n= n= f n f n R k B n= ɛ = n=

42 42 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Which is a contradiction to the hypothesis. Hence f < a.e.. Using this result we obtain that f f = Since f is non-negative, as a consequence of Lebesgue s Monotone Convergence Theorem we have R k f f = R k Therefore f/( f) L. f n = n= n= n= f n R k f n = a(n) < n= (b) The above arguments are reversible, i.e. f > f = f n = R k R k n= n= R k f n = a(n) n=

43 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 43 Problem 6: a. Suppose f : R R is a smooth function with f() =, f( x) = f(x), and f (x) 3 for all x R. Find the best bound M so that f(x) M for all x [ 2, 2]. b. Suppose that f n is a sequence of functions satisfying all the conditions on f from (a). Prove that f n contains at least one subsequence that converges uniformly on every bounded subset of R to a continuous function. Solution 6: (a) First note that the hypothesis f( x) = f(x) and f() = implies that f () =. Now by the Fundamental Theorem of Calculus we have x x x ( f(x) = f() + f (t)dt = f (t)dt = f () + = x ( t ) f (s)ds dt Taking absolute value on both sides we obtain: x ( t ) x ( t f(x) f (s)ds dt = ) 3ds dt = t x ) f (s)ds dt 3tdt = 3x2 2 Hence any function satisfying the hypothesis is absolutely bounded by 6 on [ 2, 2]. This bound is shown to be a best estimate by the function 3x 2. (b) Let F be a family of functions that satisfy the hypothesis, and suppose f F. On any interval [ R, R] we have the following y x y f(y) f(x) = f (t)dt f (t)dt = y f (t)dt f (t) dt y x x 3Rdt = 3R y x The above inequlality implies that F is equicontinuous. For given an ɛ > we may set δ = ɛ/6r to guarantee that y x < δ f(y) f(x) < ɛ for all f F. Hence the Arzela-Ascoli Theorem implies that on every interval [ R, R], F has a uniformly convergent subsequence (which converges to a continous function). We may find a x

44 44 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT convergent subsequence on R via the following diagonal argument. Let F n be the convergent subsequence on [, ], F 2 n be the convergent subsequence on [ 2, 2],..., F k n be the convergent subsequence on [ k, k],... Now define a new subsequence F n by setting F i = F i i. From the construction it is clear that F n converges uniformly on all bounded subsets of R. May 23 Problem : Suppose that x > and x n+ = (2 + x n ) for n =, 2, 3,.... Prove that the sequence x n converges and find its limit. Solution : First we may find the fixed points for the mapping by setting the following equality x n = x n+ x n = (2 + x n ) x 2 n + 2x n = So we obtain that the fixed points for the map are + 2. Since x > and the map preserves positivity, the possible limit point is the fixed point + 2. We need only prove that the map indeed converges. Let M(x n ) = (2 + x n ). Now observe: M(x n+ ) M(x n ) = 2 + x n+ 2 + x n = = x n x n x n + 2x n+ + x n x n+ 4 x n+ x n The inequality follows from the fact that x n, x n+ are positive n (since x > ). So now we invoke Banach s Fixed Point Theorem to obtain that iteration of the mapping does converge to our fixed point.

45 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 45 Problem 2: Evaluate where a >. log x x 2 + a 2 dx Solution 2: C r as usual denotes the circle of radius R. Let γ be the contour that traverses C R from z = R to z = R in the positive orientation (counter-clockwise). Let γ 2 be the contour that traverses C δ from z = δ to z = δ in the negative orientation. Suppose at all stages to come that δ < and R > 2. Let γ 2 3 be the straight-line contour on the x-axis from δ to R, and γ 4 be the straight-line contour on the x-axis from R to δ. Now consider the contour integral = γ γ + γ 4 + γ 2 + By the Residue Theorem, we know that f(z)dz = 2πi (residues of f in the UHP ) γ Now put f(z) = log z z 2 + a 2 First we show that γ f(z)dz as R γ f(z)dz = π log Re it R 2 e 2it + a 2 ireit dt = Note that the RHS tends to as R approaches. γ 3 π log R + it idt Re it + a2 Re it Next we will show that γ 2 f(z)dz as δ. π log (δe it ) π (log δ) ( δe it + δe it (it) ) f(z)dz = γ 2 δ 2 e 2it + a 2 iδeit dt = idt δ 2 e 2it + a 2 π (δ log δ) ( e it + ite it) π ( e it + ite it) = idt = (δ log δ) δ 2 e 2it + a 2 δ 2 e 2it + a idt 2

46 46 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Therefore lim δ (δ log δ) = γ 2 f(z)dz =. Since the Residue Theorem result holds for all δ, R, taking the respective limits as we did above, we obtain that γ f(z)dz = lim ( δ δ = 2 log ( x) + iπ dx + x 2 + a 2 log (x) x 2 + a 2 dx + δ iπ x 2 + a 2 dx ) log (x) x 2 + a dx 2 Substituting 2πi(Res f (ai)) for f(z)dz we obtain that γ log (x) ( ) x 2 + a dx = ( ) iπ 2πi(Res 2 f (ai)) 2 x 2 + a dx 2 We have iπ iπ dx = x 2 + a2 a 2 ( x a ) 2 dx = iπ ( ( )) arctan x/a = iπ + a a π 2 = iπ2 a And Res f (ai) = ( ) log (z) = 2z z=ai log a + iπ 2ai Substituting these two values into ( ) we have log (x) x 2 + a dx = ( π log a + iπ a ) iπ2 = π log a a 2a Problem 3: Suppose f : R R is differentiable and f(x) dx <. a. Show that the Lebesgue measure of {x : f(x) > t} approaches as t. b. Show that the additional assumption f (x) dx < implies that f(x) as x.

47 SOLUTIONS TO QUALIFYING EXAM PROBLEMS IN ANALYSIS 47 Solution 3: (a) Recall the Chebyshev Inequality for L (R): λ{x : f(x) > t} f t Taking the limit as t we obtain the desired inequality. (b) The additional hypothesis implies that f (x) as x. Since f is differentiable, the Fundamental Theorem of Calculus gives us: f(x) = f(x ) + x x f (t)dt. Thus we obtain: f(x) f(x ) + x x Now taking a limit yields the result: x f (t)dt f(x ) + x f (t) dt x lim f(x) lim ( f(x ) + f (t) dt) = + = x x x Note: the last equality follows from the application of Lebesgue s Dominated Convergence Theorem. We can apply it here because f (t) is a continuous function integrated over a compact domain. One could alternatively view the limit as x on the integral as a limit of a family of continuous functions defined on [, ]. I.e. g n : [, ] R given by g n (x) = f (x + (n )). The application of LDCT from this viewpoint is more easily justified. Problem 4: Find all entire functions f which satisfy Re(f(z)) 2 z for z >. Prove your answer. Solution 4: Since f is entire, e f is entire. Furthermore, e f(z) = e Ref(z) e 2 for z. The maximum modulus principle implies that e f(z) e 2 on D(, ). So e f is entire and bounded on C. By Liouville s Theorem e f is constant which implies f is constant.

48 48 AS GIVEN BY THE RICE UNIVERSITY MATHEMATICS DEPARTMENT Problem 5: Suppose that f is a bounded measurable function on R and g(x) dx <. Prove that lim f(x)[g(x + t) g(x)]dx = t Solution 5: Suppose f is bounded on R by M, we may write: f(x)[g(x+t) g(x)]dx f(x)[g(x + t) g(x)]dx f(x)[g(x+t) g(x)] dx M [g(x + t) g(x)] dx Now since g L by hypothesis, we have continuity of translation. I.e. if we take the limits on both sides of the string of inequalities above, we obtain: lim f(x)[g(x + t) g(x)]dx lim M [g(x + t) g(x)] dx = t t = M lim [g(x + t) g(x)] dx = M = t Problem 6: a. State Rouche s Theorem. b. State Schwarz s Lemma. c. Suppose f is holomorphic in the unit disk z < with f(z) and f() =. Prove that for any integer n, f(z) 2 n z n has precisely n zeros (counting multiplicity) in the disk z <. 2 Solution 6: (a) Rouche s Theorem: Let γ be a closed path homologous to in U and assume that γ has an interior. Let h, g be analytic on U, and h(z) g(z) < h(z) for z on γ.