Analysis Comprehensive Exam Questions Fall 2. Let f L 2 (, ) be given. (a) Prove that ( x 2 f(t) dt) 2 x x t f(t) 2 dt. (b) Given part (a), prove that F L 2 (, ) 2 f L 2 (, ), where F(x) = x (a) Using Hölder s inequality we have ( x 2 ( x ) 2 f(t)dt) = t 4 f(t)dt Set (b) By (a) t 4 x F 2 (x) 2 x 3 D = {(t, x)r 2 < t x < } x t 2 x f(t)dt. x dt t 2 f 2 (t)dt = 2 x x tf(t) 2 dt. tf 2 (t)dt. and g(t, x) = 2 t x 3 χ D(t, x)f 2 (t). We conclude that F 2 L 2 (, ) ( x 2 t 3 f2 (t)dt ) dx = g(t, x)dtdx. x R 2 As g is nonnegative we may apply Tonelli s theorem to obtain that F 2 L 2 (, ) R 2 g(t, x)dxdt = = 2 tf 2 (t) ( 2 tf 2 (t)dt 2 t dx = 4 f 2 L 2 (, ). t x 3 dx) dt
2 2. Let X be a compact metric space, and let X n X n X 2 X be a nested sequence of closed nonempty subsets of X. (a) Show that S := n=x n is nonempty. (b) Suppose that none of the X n is contained in the disjoint union of two nonempty open sets. Show that S cannot be contained in the disjoint union of two nonempty open sets (hence S is connected). (a) Denote by X c n the complement of X n. I if Y was empty, {X c n} n= would be an open cover of X. As X is compact, we may a finite subcover {X ni } k i= where n < n 2, n k are positive integers. This would imply that such that = k i= X n i = X nk which yields a contradiction. (b) Suppose on the contrary that S U V where U and V are open, non empty and disjoint. Then the sets F n = X n \ (U V ) are closed and {F n } n= is a nested family whose intersection is empty. Thus, by (a) there exists n such that F n is empty. This yields a contradiction.
3. Let U : H H be a unitary mapping on a Hilbert space H. Let I be the identity map on H, set M = ker(u I), let P be the orthogonal projection of H onto M, and for each integer N > define S N = N where we take U = I. Show that S N P strongly, i.e., for each f H we have S N f Pf as N. N Hint: Show that M = range(u I). Consider f M and f range(u I) first. We have so Therefore k= U k, Uf f = f U f =, ker(u I) = ker(u I) = ker(u I). M = ker(u I) = ker(u I) = range((u I) ) = range(u I). If f M = ker(u I), then Uf = f and therefore S N f = f for every N. Hence S N f Pf in this case. If f range(u I), then f = Ug g for some g H. Therefore Hence S N f = N N k= U k f = N (U k+ g U k g) = UN g N N g N. k= S N f UN g N + g N = 2 g N. Suppose f M = range(u I), and fix ε >. Then there exists some g range(u I) such that f g < ε. Hence S N f S N g + S N (f g) S N g + N N k= = S N g + f g S N g + ε. U k (f g) Consequently lim sup S N f lim sup S N g + ε = ε. N N Since ε is arbitrary, it follows that S N f as N. Finally, given an arbitrary vector f H, we have f = g + h where g = Pf M and h M. Therefore, by combining the cases above we see that S N f = S N g + S N h g + = Pf. 3
4 4. Let X be a complete metric space. Prove that if U, U 2,... are open, dense subsets of X, then U n is dense in X. The sets E n = Un C = X \ U n are closed and have empty interiors. If U n is not dense in X, then we can find some f X and r > such that ( ) C B r (f) U n = E n. n= The closed ball Y = B r/2 (f) is a closed subset of X and hence is a complete metric space (using the metric on X). We have Y = (E n Y ). n= n= Each set E n Y is closed in Y, and its complement in Y is Y \ (E n Y ) = Y U n. Since U n is dense in X, the set Y U n is dense in Y. Hence the interior of E n Y must be empty. Therefore we have written Y as a countable union of closed sets that each have empty interiors. The Baire Category Theorem says that a complete metric space is nonmeager in itself, so this is a contradiction.
5. Given f : R R and y R, define f y (x) = f(x y). (a) Show that if f is continuous and has compact support, then lim y f y f L (R) =. (b) Show that if f L p (R) and p <, then lim y f y f L p (R) =. (c) Either prove or give a counterexample: if f L (R), then lim y f y f L (R) =. (a) Suppose f is continuous its support is contained in [ r, r] and r >. Note that if y < then f y (x) f(x) = f(x y) f(x) = unless x 2r. Hence, if ǫ (, ) l(ǫ) := sup f y f L (R) = sup{ f(x y) f(x) x [ 2r, 2r], y ǫ}. y ǫ x,y But, f is uniformly continuous on [ 4r, 4r] and so, lim sup { f(x y) f(x) x [ 2r, 2r], y ǫ} =. ǫ + x,y This proves that lim ǫ + sup y ǫ f y f L (R) which proves (a). (b) Suppose f L p (R). For each integer n we may find g n : R R continuous with compact support such that f g n L p (R) < /n. We use the triangle inequality to obtain that f y f L p (R) f y g y n L p (R) + g y n g n L p (R) + g n f L p (R). A simple change of variables reveal that f y g y n L p (R) = g n f L p (R) < /n and so, f y f L p (R) g y n g n L p (R) + 2 n. () Assume that g n is supported by [ r n +, r n ] and y. Then g y n g n is supported by [ r n, r n ] and so, g y n g n L p (R) = g y n g n L p ([ r n,r n]) (2r n ) p g y n g n L (R). This, together with (a) implies lim sup y We combine () and (2) to conclude that gn y g n L p (R) lim sup(2r n ) p g y n g n L (R) =. (2) y lim sup f y f L p (R) 2/n. y Since n is arbitrary we have that lim sup y f y f L p (R) and so, lim y f y f L p (R) =. (c) Let f = χ (,). In other words, f(x) = if x (, ) and f(x) = if x (, ). For < y < and x (, y), f y (x) f(x) =. Hence, f y f L (R) f y f L (,y) =. This proves that we don t have lim y f y f L (R) =. 5
6 6. Assume f is absolutely continuous on an interval [a, b], and there is a continuous function g such that f = g a.e. Show that f is differentiable at all points of [a, b], and f (x) = g(x) for all x [a, b]. First proof. Since g is continuous, every point is a Lebesgue point of g. Suppose that x (a, b). If h is small enough, then we have x+h (a, b) as well, so we can compute that g(x) = lim h h = lim h h x+h x x+h x g(t) dt f (t) dt Fund. Thm. Calculus since f = g a.e. f(x + h) f(x) = lim since f is absolutely continuous. h h Therefore f is differentiable at all points in (a, b), and f (x) = g(x) for x (a, b). a similar proof works at the endpoints x = a and x = b if we take appropriate limits from the left or right. Second proof. Since g is continuous, its antiderivative F(x) = x g(t) dt is absolutely a continuous, differentiable at all points, and satisfies F (x) = g(x) for every x [a, b]. Hence (F f) = F f = g g = a.e. An absolutely continuous function whose derivative is zero almost everywhere must be a constant. Therefore F f is constant, so f = F +c and f is differentiable at all points. Also f (x) = F (x) = g(x) for all x [a, b].
7. Let Y be a dense subspace of a normed linear space X, and let Z be a Banach space. Let L: Y Z be a bounded linear operator. (a) Prove that there exists a unique bounded linear operator L: X Z whose restriction to Y is L. Prove that L = L. (b) Prove that if L: Y range(l) is a topological isomorphism (L is a bijection and L, L are both continuous) then L: X range(l) is also a topological isomorphism. (a) Fix any f X. Since Y is dense in X, there exist g n Y such that g n f. Since L is bounded, we have Lg m Lg n L g m g n. But {g n } n N is Cauchy in X, so this implies that {Lg n } n N is Cauchy in Z. Since Z is a Banach space, we conclude that there exists an h Z such that Lg n h. Define Lf = h. To see that L is well-defined, suppose that we also had g n f for some g n Y. Then Lg n Lg n L g n g n. Since Lg n h, it follows that Lg n = Lg n+(lg n Lg n) h + = h. Thus L is well-defined, and similarly it is linear. To see that L is an extension of L, suppose that g Y is fixed. If we set g n = g, then g n g and Lg n Lg, so by definition we have Lg = Lg. Hence the restriction of L to Y is L. Consequently, L = sup Lf f X, f = sup Lf = f Y, f = sup Lf = L. f Y, f = Now suppose that f X. Then there exist g n Y such that g n f and Lg n Lf, so Lf = lim n Lg n lim n L g n = L f. Hence L L. Combining this with the opposite inequality derived above, we conclude that L = L. Finally, we must show that L is unique. Suppose that A B(X, Y ) also satisfied A Y = L. Then Af = Lf for all f Y. Since Y is dense, this extends by continuity to all f X, which implies that A = L. (b) Suppose that L: Y range(y ) is a topological isomorphism. We already know that L: X range( L) is bounded. We need to show that L is injective, that L : range( L) X is bounded, and that range( L) = range(l). Fix any f X. Then there exist g n Y such that g n f and Lg n Lf. Since L is a topological isomorphism, g n = L Lg n L Lg n. Hence g n Lf = lim Lg n lim n n L = f L. Consequently, L is injective and for any h range( L) we have Therefore L : range( L) X is bounded. L h L L( L h) = L h. 7
8 It remains only to show that the range of L is the closure of the range of L. If f X, then by definition there exist g n Y such that g n f and Lg n Lf. Hence Lf range(l), so range( L) range(l). On the other hand, suppose that h range(l). Then there exist g n Y such that Lg n h. Since L is bounded and L extends L, we conclude that g n = L (Lg n ) L (h). Hence f = L (h), so f range( L).
8. Assume that E R d is Lebesgue measurable and m(e) >, where m denotes Lebesgue measure. Show that there exists a point x E such that for every δ > we have m(e B δ (x)) >. Here, B δ (x) denotes the open ball with center x and radius δ. Suppose on the contrary that for every x E there exists δ x > such that m(e B δx (x)) =. Set F := { B δ (x) x E, < δ < min{δ x, }}. By Vitali s Covering Lemma there exists G, a countable family of disjoint balls in F, such that B F B B G ˆB where ˆ Bδ (x) = B 5δ (x). As E B F B we have E B G ˆB and so, This contradicts the fact that m(e) >. m(e) B G m( ˆB) =. 9