n 1 < n 2 < n 3 <..., x n1 > x n2 > x n3 >... x nj is a decreasing subsequence. If there are only finitely many peaks,

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1 Question 1 (a Prove that every sequence of real numbers either has a non-decreasing subsequence or a non-increasing subsequence. (b Deduce that every bounded sequence of real numbers has a convergent subsequence. Solution Let us denote our sequence by (x n n=. (a We say that the sequence has a peak at n if x n > x n for all n > n. If there are infinitely many peaks, say at (n k k=1, where then n 1 < n 2 < n 3 <..., x n1 > x n2 > x n3 >... so ( x nj is a decreasing subsequence. If there are only finitely many peaks, j=1 let N be the last peak. Then for all n > N, there exists m > n with x m x n. (If not, there would be a peak at n, imposible. Then we can construct an increasing subsequence, using induction. To see this, let n 1 = N. Since there is not a peak at n 1, there exists n 2 > n 1 such that x n2 x n1. Next, there is not a peak at n 2, so there exists n 3 > n 2 such that x n3 x n2. Assuming that we have chosen n 1 < n 2 <... < n k, we can choose n k+1 > n k such that x nk+1 > x nk, as there is not a peak at n k. Then (x nk k=1 is increasing. (b As (x n n= is bounded, there exists A > such that x n A, n. By (a, (x n n= has either an increasing or decreasing subsequence ( x nj j=1. If this subsequence is increasing, then it is increasing and bounded above by A, so converges. If this subsequence is decreasing, then it is decreasing and bounded below by A, and so converges. 1

2 Question 2 Let (f n be a sequence of nonnegative Lebesgue measurable functions on [, 1], and let (E m be a sequence of Lebesgue measurable subsets of [, 1]. (a Suppose that there is an integrable function f such that for n 1 and almost every x [, 1], f n (x f (x. (1 Prove that f n Is the hypothesis (1 for some integrable f necessary? (b Let E n [, 1] for n 1 and let (i Prove that E = n=1 m=n f n. (2 E m. meas (E n meas (E. (ii Either prove that equality holds in part (i or give an example in which strict inequality holds. (iii Prove that meas (E k < meas (E =. k=1 Solutions (a We apply Fatou s lemma to the non-negative measurable functions f f n (They are non-negative a.e. and we can ignore the set of measure. We have lim inf Since f does not depend on n, so Also lim inf (f f n lim inf (f f n = f + lim inf ( f n lim inf (f f n = ( f f n lim inf (f f n. (1 = f = = 2 f f n, f + lim inf f f n. ( f n. f n

3 Plugging these into (1 gives and hence the result. f f n f f n The condition (1 for some integrable f is necessary. We can use the same type of counterexamples that are used to show we need a dominating function in Lebesgue s Dominated Convergence Theorem. For example, let { n f n (x = 2 x, x [ ], 1 n, x [ 1 n, 1]. Then we see that if x (, 1], we have f n (x = for n > 1 x, so Also f n ( = for n 1. So lim f n (x =. and then But so lim f n (x =, x [, 1] f n = =. /n f n = n 2 x dx = n 2 1 2n 2 = 1 2, f n = 1 2 > = f n. (b (i Note that x E iff x E n for infinitely many n, that is χ En (x = 1 for infinitely many n. So Hence x E χ En (x = 1. χ E (x = χ En (x. Since characteristic functions are bounded above by 1, we can apply (b to deduce that χ En = 3 χ En χ E.

4 That is, meas (E n meas (E. (ii No, we don t have equality always. For example, let E n = [, 1 2 if n is odd and E n = [ 1 2, 1] if n is even. Then E = E n = [, 1] as every point of [, 1] belongs to infinitely many of the {E n }, and conversely, each E n [, 1]. So χ E = But for each n, E n has linear measure 1 2, so (iii We have for each n, so E meas (E m=n 1 = 1. χ En = 1 2. E m meas (E m. m=n As n, the right-hand side approaches (because of convergence and hence meas (E =. 4

5 Question 3 (a State a necessary and sufficient criterion for a function f on [, 1] to be Riemann integrable. Your criterion must involve Lebesgue measure. (b Let f : [, 1] R be Lebesgue integrable. Set f (x = f ( for x < and f (x = f (1 for x > 1. Prove that lim h f (x + h f (x dx =. (You may assume results about approximation of Lebesgue integrable functions by continuous functions. (c Let g : [, 1] [, 1] be measurable. Prove that if f is (bounded and Riemann integrable in [, 1], then lim h f (x + hg (x f (x dx =. Solution (a For f to be Riemann integrable, it is necessary and sufficient that f be continuous a.e. (b If first f is continuous in [, 1], then because of the way we extended it, it will be continuous in [ 1, 2]. Then f is uniformly continuous there (a continuous function on a compact interval is unformly continuous. Then given ε >, we can find δ (, 1 such that So x [, 1] and h < δ f (x + h f (x < ε. h < δ f (x + h f (x dx ε dx = ε. Then the result follows for continuous f. Now suppose that we only know f is Lebesgue integrable. Then we can find continuous g : [, 1] R such that f g < ε/3. We may also assume that g ( = f ( and g (1 = f (1 (just change g in small neighborhoods of, 1 if necessary. Extend g to the real line in the same way we did for f. Then < ε/3 + f (x + h f (x dx f (x + h g (x + h dx + g (x + h g (x dx + ε/3. g (x + h g (x dx + g (x f (x dx 5

6 For h small enough, as g is continuous, we have g (x + h g (x dx < ε/3 and then f (x + h f (x dx < ε. (c Suppose M is such that We have f (x M for all x. f (x + hg (x f (x 2M for all x [, 1]. Moreover, as g is bounded above by 1 and below by, we have at each point of continuity of f, lim f (x + hg (x = f (x. h Hence a.e. (recall f is Riemann integrable, lim f (x + hg (x f (x =. h By Lebesgue s Dominated Convergence Theorem, lim h f (x + hg (x f (x dx =. 6

7 Question 4 Let 1 < p < and {f n } be a sequence of functions in L p [, 1]. Give a proof of, or counterexample to, the following assertions: (a If f n f weakly in L p [, 1] as n, then there is a subsequence {f nk } that converges a.e. to f. (b If f n f in norm in L p [, 1] as n, then there is a subsequence {f nk } that converges a.e. to f. Solutions (a This is false. Let ϕ be the periodic function { 1, t < 1 ϕ (t = 2, 1 1, 2 t < 1. For n 1, let Then f n (t = ϕ (2 n t, t [, 1]. lim f n χ A = for every measurable subset A of [, 1]. Here χ A denotes the characteristic function of A. (We can first see this for dyadic intervals [ ] j, k 2 l 2, and then for general l intervals, then for finite unions of intervals, and by dominated convergence for any measurable set. It follows that f n weakly as n, but no subsequence converges a.e., since f n (t = 1 for all n and t. (b This is true. Let ε > and meas denote linear Lebesgue measure. For n 1, It follows that for each ε >, ε meas ({t : f n f (t ε} f n f p {t: f n f (t ε} f n f p, n. meas ({t : f n f (t ε} as n. That is, f n f in measure. By a standard argument, there is a subsequence of {f n } that converges a.e. to f. Outline of this argument: Choose n 1 < n 2 < n 3 <... such that for k 1, E k = {t : f nk f (t 12 } k has meas (E k 1 2 k. 7

8 Let E = E k = k m=1 k=m E k. Then it is easy to check that meas (E = and for t / E, lim f n k (t = f (t. k 8

9 Question 5 Let f be a function defined on R 2 with continuous second partial derivatives. Use Fubini s theorem to give an easy proof that x y = 2 f y x. (Hint: Assume it is not true, and integrate 2 f y x over a suitable square. Solution Let us suppose the result is false. Then at some point x y 2 f x y 2 f y x. Let us suppose that it is positive at that point. By continuity, we may find a square S = [a, a + h] [b, b + h] containing that point, in which Then we know Here I = S x y 2 f y x >. [ 2 ] f x y 2 f dx dy >. (1 y x 2 f I = dx dy dx dy S x y S y x = I 1 I 2. (2 Because we are dealing with a continuous integrand, we may write these integrals as iterated integrals, and may change the order of integration, and may also use the fundamental theorem of calculus: I 1 = = b+h b b+h b ( a+h a x y dx dy ( f f (a + h, y (a, y y y dy = f (a + h, b + h f (a + h, b f (a, b + h + f (a, b. Next, I 2 = = a+h a a+h a ( b+h b y x dy dx ( f f (x, b + h (x, b x x dx = f (a + h, b + h f (a, b + h f (a + h, b + f (a, b. Thus I 1 = I 2 and (2 gives I =, contradicting (1. 9

10 Question 6 (a Show that there is a bounded linear functional ϕ on l such that ϕ (x = lim x i for every convergent sequence x = (x i in l. (b Is this linear functional unique? Solution (a Let c denote the set of all convergent sequences in l. It is a closed subspace of l. Note that ϕ is a linear functional on c. This follows from the linearity of limits. Moreover, ϕ is a bounded/continuous linear functional. Indeed if x = (x i c, then ϕ (x = lim x i = lim x i sup i x i = x. So ϕ has norm at most one. By the Hahn-Banach theorem, ϕ has a bounded/continuous extension to all of l. (b It is not unique. Let c e denote the set of all sequences in l whose even index components converge. Thus x = (x i c e iff lim x 2i exists. Similarly, let c o denote the set of all sequences in l whose odd index components converge. Thus x = (x i c e iff lim x 2i+1 exists. Suppose we first extend ϕ above to c e by ϕ (x = lim x 2i. It clearly is a bounded linear extension. Now we extend via Hahn-Banach to all of l. Call the resulting extension ϕ 1. Next, extend our original ϕ from c to c o by ϕ (x = lim x 2i+1. Now we extend via Hahn-Banach to all of l. Call the resulting extension ϕ 2. Both ϕ 1 and ϕ 2 are bounded linear functional extending ϕ but they are not equal. To see this, let x denote the sequence with for all i. We have x c e c and x 2i = ; x 2i+1 = 1 ϕ 1 (x = lim x 2i = ; ϕ 2 (x = lim x 2i+1 = 1. 1

11 Question 7 Prove that there is no norm under which the vector space P of polynomials with real coefficients is a Banach space. Solution There is no norm under which P is complete. Let us assume there is, and derive a contradiction. We claim that P n = {polynomials P of degree n} is a closed subspace of P, and is also a nowhere dense subset of P. Indeed, suppose (p k is a sequence of polynomials of degree n with p k p, k, for some function p. Now P n is a finite dimensional subspace of a Banach space, so K = {q P n : p q 1}, which is closed and bounded,is also compact. As p k K for large enough k, we deduce that p P n also. Thus P n is closed. To see that it is nowhere dense, we must show it has empty interior. that is obvious: if l > n and ε is small enough, while p P n, then But q (x = εx l + p (x / P n but q p = ε x l may be made as small as we please. So P n must have empty interior. Then P = n=1 is a countable union of nowhere dense sets, so is of the first category. This contradicts Baire s theorem for closed metric spaces (and hence Banach spaces. P n 11