ANALYSIS QUALIFYING EXAM FALL 2016: SOLUTIONS. = lim. F n

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1 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS Problem. Let m be Lebesgue measure on R. For a subset E R and r (0, ), define E r = { x R: dist(x, E) < r}. Let E R be compact. Prove that m(e) = lim m(e /n). Solution. Recall that if (X, µ) is a measure space, and (F n ) n Z>0 is a sequence of measurable subsets of X, and then F F 2 F 3 and µ(f ) <, µ ( F n ) = lim m(f n). Choose M [0, ) such that E [ M, M]. Then E E /2 E /3 E /4 and E [ M, M + ]. Since E is closed, we have E = E /n. Since m(e ) 2M + 2 <, it follows that ( ) m E /n = lim m(e /n). Therefore m(e) = lim m(e /n). Alternate solution. Substitute the Dominated Convergence Theorem, used on the characteristic functions χ E/n, for the measure theoretic theorem stated at the beginning of the first solution. Problem 2. Show that, for any n 2, the function ( + x/n) n x /n is integrable on [, ), and compute, with justification, lim Solution. For n Z >0 and x [, ), define f n (x) = dx. ( + x/n) n x/n ( + x/n) n x /n.

2 2 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS For n 2 we have n n 2, so ( + x ) n ( ( ) x n(n ) ( x ) 2 + n + n n) 2 n ( ) n 2 ( x ) 2 x 2 + x + = + x + 4 n 4 > x2 4. Therefore 0 f n (x) ( + x/n) n 4 x 2. The function g(x) = 4/x 2 is integrable on [, ), so f n is integrable on [0, ). Since lim ( + x/n) n = e x and lim x /n = for all x [, ), we can apply the Dominated Convergence Theorem with the dominating function g to get the first step in the following calculation: lim This completes the solution. dx = ( + x/n) n x/n e x dx = e. Problem 3. Let (X, µ) be a finite measure space. Let (f n ) n Z>0 be a sequence of integrable functions on X. Suppose that there is an integrable function f on X such that f n (x) f(x) for almost every x X. Prove that, for every ε > 0, there are M [0, ) and a measurable subset E X such that µ(e) < ε and such that for all n Z >0 and x X \ E we have f n (x) M. Solution. We first observe that for any integrable function g : X C and any r > 0, we have () µ ({ x X : g(x) r }) r g. Indeed, if we set then g rχ E, so as desired. Now let ε > 0. Set E = { x X : g(x) r }, g dµ rχ E dµ = µ(e), r X r R R = 3 f + and G = { x X : f(x) R }. ε Apply () with g = f and r = R, getting µ(g) ε 3. Next, use Egoroff s Theorem to find a measurable set F X such that µ(f ) < ε 3 and f n X\F f X\F uniformly. In particular, there is N Z >0 such that for all x X \ F and all n N, we have f n (x) f(x) <. It follows that for all x X \ (F G) and all n N, we have f n (x) < R +. Now, for n =, 2,..., N set M n = 3N f n + ε and E n = { x X : f n (x) M n }.

3 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS 3 Apply () with g = f n and r = M n, getting µ(e n ) Then ε 3N. Define N E = F G E n and M = max(r +, M, M 2,..., M N ). N µ(e) µ(f ) + µ(g) + µ(e n ) < ε 3 + ε ( ε ) 3 + (N ) < ε. 3N Moreover, if x X \ E, then for n N we have f n (x) < R + because x F G, and f (x) M, f 2 (x) M 2,..., f N (x) M N, since x N E n. So f n (x) M for all n Z >0. Problem 4. For any compact interval [a, b] R and any real valued function f on [a, b], let V [a,b] (f) denote the total variation of f on [a, b]. Let (f n ) n Z>0 be a sequence of real valued functions on [0, ]. Assume that f n (0) < and V [0,] (f n ) <. Show that f n (t) < for all t [0, ]. Further, define f(t) = f n(t), and prove that V [0,] (f) V [0,] (f n ). Solution. For any t [0, ], we have f n (t) f n (0) + f n (t) f n (0) f n (0) + V [0,] (f n ) <. Now, for any partition P = {t 0, t,..., t n } with 0 = t 0 < t < < t m =, we have m m f(t j ) f(t j ) = f n (t j ) f n (t j ) j= j= m ( = fn (t j ) f n (t j ) ) j= j= m f n (t j ) f n (t j ) V [0,] (f n ). By taking the supremum over all such partitions P, we get V [0,] (f) V [0,] (f n ) This completes the proof.

4 4 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS Problem 5. The space l 2 is defined by { (x n ) n Z>0 : x n C for all n Z >0 and } x n 2 <. This space is a complex normed vector space with the norm ( ) /2. (x n ) n Z>0 = x n 2 (You may use this fact without proof.) Prove that the closed unit ball of l 2 is not compact. Solution. For n Z >0, define e n l 2 by { k = n (e n ) k = 0 k n. Clearly if m, n Z >0 are distinct, then e m e n 2 = 2, so e m e n = 2. Therefore the sequence (e n ) n Z>0 has no convergent subsequence. Problem 6. Let E be a complex Banach space. Suppose that ξ, ξ 2,..., ξ n E are linearly independent, and that η, η 2,..., η n E are n elements. Prove that there is a bounded linear map T : E E such that T ξ k = η k for k =, 2,..., n. Solution. For k =, 2,..., n, define the subspace Z k E by Z k = span ({ ξ j : j {, 2,..., n} \ {k} }). Then ξ k Z k because ξ, ξ 2,..., ξ n are linearly independent. Since Z k is finite dimensional, it is closed, so the Hahn-Banach Theorem provides a bounded linear functional ω k : E C such that ω k (ξ k ) = and ω Zk = 0. Now define T L(E) by n T ξ = ω k (ξ)η k k= for ξ E. Clearly T ξ k = η k for k =, 2,..., n. Also, T is bounded because for ξ E we have n T ξ ω k η k ξ. This completes the solution. k= Problem 7. Let ε > 0, and let f : B +ε (0) C be a holomorphic function such that f(z) = for all z C with z =, and such that f(z) 0 for all z B (0). Prove that f is constant. Solution. The Maximum Modulus Theorem implies that f(z) for all z B (0). Also, f(z) 0 for all z B (0), so there is an open set U B +ε (0) such that f(z) 0 for all z U. The function g(z) = f(z) satisfies g(z) = for all z C such that z =, so the Maximum Modulus Theorem implies that g(z) for all z B (0). Therefore f(z) for all z B (0). So f has a local maximum at every point of B (0). Thus f is constant by the condition for equality in the Maximum Modulus Theorem.

5 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS 5 Problem 8. Let Ω C be a nonempty open set, and let f : [0, ] Ω C be a continuous function. For t [0, ] define f t : Ω C by f t (z) = f(t, z) for z Ω. Suppose that f t is holomorphic for every t (0, ]. Prove that f 0 is holomorphic. Solution. We use Morera s Theorem. So let be a triangle in Ω. Let B be the boundary of as a subset of C (rather than as a path). Then [0, ] B is compact, and f is continuous on this set. Therefore there is M [0, ) such that f(t, z) M for all t [0, ] and z B. Let γ : [a, b] C be a piecewise linear parametrization of the boundary path of. Then L = sup s [a,b] γ (s) is finite. For n Z >0 define h n : [a, b] C by h n (s) = f(/n, γ(s))γ (s). Also define h: [a, b] C by h(s) = f(0, γ(s))γ (s). Then h n (s) LM for all n Z >0 and s [a, b]. Also, since f is continuous, we have lim h n (s) = h(s) for all s [a, b] except for the finite set at which γ (s) does not exist. Using the Dominated Convergence Theorem at the second step, with the dominating function being the constant LM, we get f 0 (z) dz = b a h(s) ds = lim b a h n (s) ds = lim f /n (z) dz = lim 0 = 0. Since is arbitrary and f 0 is continuous, Morera s Theorem implies that f 0 is holomorphic. Alternate solution. For z 0 C and r > 0, we let B r (z 0 ) be the open ball B r (z 0 ) = { z C: z z 0 < r }. We claim that f /n f 0 uniformly on compact subsets of Ω. To prove this, let K Ω be compact and let ε > 0. For z K, use continuity of f at (0, z) to choose δ(z) > 0 such that whenever w Ω and t [0, ] satisfy w z < δ(z) and t < δ(z), then w Ω and f(t, w) f(z, 0) < ε 2. Since K is compact, there are z, z 2,..., z l K such that Choose N Z >0 such that K l B δ(zk )(z k ). k= N < min ( δ(z ), δ(z 2 ),..., δ(z l ) ). Now let n Z >0 satisfy n N. Let z K. Choose k {, 2,..., l} such that z B δ(zk )(z k ). Then n < δ(z k), so f /n (z) f 0 (z) = f(/n, z) f(0, z) f(/n, z) f(0, z k ) + f(0, z k ) f(0, z) < ε 2 + ε 2 = ε. This completes the proof of the claim. Given the claim, we know that f /n is holomorphic for all n Z >0, so f is holomorphic by Theorem 0.28 of Rudin s book. Problem 9. Prove that there is no entire function f such that f ( ) n = 2 n n Z >0. for all

6 6 ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS Solution. We argue by contradiction. Suppose there is such a function f. Then f(0) = 0 by continuity. Therefore there is an entire function g such that f(z) = zg(z) for all z C. The function g is continuous at 0. The formula f(z) = zg(z) implies that g ( ) n = n for n 2 Z>0, which implies that lim z 0 g(z) does not exist. This contradiction shows that there is no such function f.