Analysis Comprehensive Exam Questions Fall 2008

Transcription

1 Analysis Comprehensive xam Questions Fall 28. (a) Let R be measurable with finite Lebesgue measure. Suppose that {f n } n N is a bounded sequence in L 2 () and there exists a function f such that f n (x) f(x) for a.e. x. Show that f f n as n. (b) Show that the conclusion of part (a) can fail if. (a) Choose ε >, and let C sup n f n 2 <. By Fatou s Lemma, we have f 2 2 lim f n 2 lim inf f n 2 lim inf f n 2 2 C. n n n Hence f L 2 (). By gorov s Theorem, there exists A such that ( ) 2 ε \A < 4C and f n f uniformly on A. Therefore, we can find an N such that (f f n ) χ A < ε, all n > N. 2 Then for n > N we have by Cauchy Schwarz that f f n f f n + f f n A \A ( ) /2 ( A (f f n ) χ A + f f n 2 \A \A ε < A 2 + f f n 2 \A /2 < ε 2 + 2C ε 4C ε. (b) Let f n χ [n,n+]. Then f n 2 for every n, and f n (x) for every x. However, f n does not converge to the zero function in L -norm, since f n. ) /2

2 2 2. Let X be a Banach space and let T, S be bounded linear operators on X. Prove that: (a) I TS has a bounded inverse if and only if I ST has a bounded inverse. (b) σ(ts)\{} σ(st)\{}. Remark: σ(a) denotes the spectrum of A. (a) Suppose that I TS has a bounded inverse. In particular, I TS is injective. Suppose that (I ST)v for some v X. Then we have T(I ST)v (I TS)Tv, so Tv. But this implies that v (I ST)v. Hence also I ST is injective. On the other hand since I TS is surjective we have that for every z X there exists an x X such that (I TS)x Tz. Observe that this implies that x T(X) since x T(Sx + z) Ty. We thus have that T(I ST)y Tz, or (I ST)y z + v with v Ker(T). But then, setting w y v, we have that (I ST)w z and I ST is surjective. Thus I ST is a bounded bijection of X onto itself, and therefore has a bounded inverse by the Open Mapping Theorem. (b) Suppose λ / σ(ts) and λ. Then TS λi has a bounded inverse, so I T S has λ a bounded inverse. By part (a) it follows that I S T and thus ST λi has a bounded λ inverse, so λ / σ(st).

3 3. Let f, g be absolutely continuous functions on [, ]. Show that for x [, ] we have x f(t) g (t) dt f(x)g(x) f()g() x f (t) g(t) dt. Since f, g are absolutely continuous, we know that they are differentiable almost everywhere and that f, g L [, ]. Consequently, f (s) g (t) L ( [, ] 2). Letting {(s, t) [, x] 2 : s t}, we compute that x ( t ) f (s) g (t) ds dt f (s) ds g (t) dt On the other hand, f (s) g (t) dt ds x x x x x g(x) ( f(t) f() ) g (t) dt f(t) g (t) dt f() x g (t) dt f(t) g (t) dt f() ( g(x) g() ). ( x ) f (s) g (t) dt ds s f (s) ( g(x) g(s) ) dt x f (s) ds x g(x) ( f(x) f() ) x f (s) g(s) ds f (s) g(s) ds. Finally, Fubini s Theorem implies that these two integrals are equal, so the result follows. 3

4 4 4. Let f : [, ] R a bounded function whose set of discontinuities D is closed and nowhere dense. (a) Is it true that every such f is Riemann integrable? (b) Prove that for every such f there exists an homeomorphism h : [, ] [, ] such that f h is Riemann integrable. Remark: A homeomorphism is a continuous bijection that has a continuous inverse. (a) Clearly no. Let {q, q 2,...,q n,...} be an ordering of the rational numbers in (, ) and set I n B(q n, ε2 n ), where B(x, r) (x r/2, x + r/2) (, ). Thus I ε but I is open and dense. Thus J [, ]\I is closed and nowhere dense but with large positive measure. Observe that f χ J is continuous for every x I since I is open, but it is discontinuous for every x J since I is dense. Hence f is discontinuous on a closed nowhere dense set of positive measure and thus it is not Riemann integrable. (b) Let D be the set of discontinuities of f and D c [, ]\D. We can define g(x) D x χ D c(t) dt. Observe that g(), g(), and g is continuous and strictly increasing. Indeed, if x < y, there exists an open interval I (x, y) such that I D c since D is closed and nowhere dense. From this we have that y g(y) g(x) χ D c(t) dt I >. D x Thus g is an invertible function and its inverse is continuous. Finally since D is closed we have that D c is the union of countably many open interval I i. Observe that g(i i ) D I i χ D c(t) dt D I i, so g(d c ) and g(d). Hence we can choose h g, for then f h is discontinuous on the set g(d), which has measure zero, and therefore f g is Riemann integrable.

5 5. Let X be a Banach space with norm X. Assume that Y is proper subspace of X that is dense in X with respect to X, and that there is another norm Y on Y with respect to which Y is a Banach space. Show that if there exists a constant C such that x X C x Y for all x Y, then there exists a continuous linear functional on (Y, Y ) that has no extension to a continuous linear functional on (X, X ). The hypotheses imply that Y is continuously embedded into X, i.e., if i: Y X is given by i(x) x for x Y then i is continuous and i C. The adjoint of i is the restriction map R: X Y given by R(µ) µ Y. Hence R is bounded, with R C. That is, µ Y Y C µ X for each µ X. This can also be proved without recourse to adjoints by observing that if x Y and µ X then x, µ Y x, µ µ X x X C µ X x Y, so µ Y Y C µ X (we are using the linear functional notation x, µ µ(x)). Suppose now that every continuous linear functional on (Y, Y ) had an extension to a continuous linear functional on (X, X ). Then R is onto. Further, if µ X and R(µ) µ Y, then µ since µ is continuous and Y is dense in X. Therefore R is injective. Thus R: Y X is a bounded bijection, so the Inverse Mapping Theorem implies that R is bounded. Combining this with the above facts, there exist c, C > such that µ X, c µ X µ Y Y C µ X. Now fix any x Y. Then by Hahn Banach, there exists a ν Y such that ν Y and x, ν x Y. By hypothesis, there exists an extension of ν to a continuous linear functional on (X, X ). Call this extension µ, so we have µ Y ν. Then x Y x, ν x, µ x X µ X x X c µ Y Y x X c ν Y c x X. Since we also have x X C x Y, we conclude that X and Y are equivalent norms on Y. But Y is complete with respect to Y, and therefore it is complete with respect to X. Consequently, Y is closed with respect to X. However, Y is dense in X with respect to X, which implies that Y X, a contradiction. 5

6 6 6. Let G be an unbounded open subset of R. Prove that is dense in R. H {x R : kx G for infinitely many k Z} If kx belongs G for infinitely many k then, for every n >, x belongs to G/k k >n where G/k {y R : ky G}. Vice versa, if x k >n G/k for every n >, then kx G for infinitely many k. Thus H G/k. n k >n Clearly k >n G/k is an open set. By the Baire Category Theorem, it is therefore enough to prove that k>n G/k is dense, for then H must be dense. Let D (z, z + ) be any open interval. If D G/k, then k >n k >n kd G. Without loss of generality, assume that z >. Then for k large enough we have that (k + )z > kz +, and hence k>n kd contains a subset of the form (d, ). By considering negative k we likewise conclude that k>n kd contains (, d). Consequently, G cannot contain (, d) (d, ), which contradicts the fact that G is unbounded.

7 7. Let µ, µ 2 be bounded signed Borel measures on R. Show that there exists a unique bounded signed Borel measure µ such that ( ) f dµ f(x + y) dµ (x) dµ 2 (y), f C c (R). 7 Show further that µ µ µ 2. Note: Scalars in this problem are real. If is any Borel set in R, then χ (x + y) d µ (x) d µ 2 (y) d µ (x) d µ 2 (y) µ µ 2 <. Hence, by Fubini s Theorem, we can define µ() χ (x + y) dµ (x) dµ 2 (y), and we have µ() µ µ 2. We claim that µ defined in this way is a signed Borel measure. The above work shows that µ() is a finite real number for every Borel set, and we clearly have that µ( ). Hence we need only show that µ is countably additive. Suppose that, 2,... are disjoint Borel sets, and let j. For each x and y, we have that N χ j (x + y) χ (x + y) L (µ µ 2 ). j Therefore, by the Dominated Convergence Theorem, µ() χ (x + y) dµ (x) dµ 2 (y) lim j lim j lim j N j N j N µ( j ) j µ( j ). j Therefore µ is a signed Borel measure. χ j (x + y) dµ (x) dµ 2 (y) χ j (x + y) dµ (x) dµ 2 (y)

8 8 If we let R P N be a Hahn decomposition of R for µ, then µ µ (R) µ(p) µ(n) χ P (x + y) dµ (x) dµ 2 (y) χ P (x + y) d µ (x) d µ 2 (y) + d µ (x) d µ 2 (y) µ µ 2. If φ n k a k χ k is any simple function, then n n φ dµ a k χ k dµ a k k k χ N (x + y) dµ (x) dµ 2 (y) χ N (x + y) d µ (x) dµ 2 (y) χ k (x + y) dµ (x) dµ 2 (y) φ(x + y) dµ (x) dµ 2 (y). If we fix f C c (R), then there exist simple functions φ k such that φ k f and φ k f pointwise. Since f L (µ) and f(x+y) L (µ µ 2 ), we therefore have by the Dominated Convergence Theorem that f(x + y) dµ (x) dµ 2 (y) lim φ k (x + y) dµ (x) dµ 2 (y) k lim φ k dµ f dµ. k It remains only to show that µ is unique. If ν is another signed Borel measure that satisfies ( ) f dν f(x + y) dµ (x) dµ 2 (y), f C c (R), () then we have f d(µ ν) for every f C c (R). By the Riesz Representation Theorem, C c (R) M b (R), the space of finite signed Borel measures on R. Therefore we must have µ ν. As the Riesz Representation Theorem for C c (X) is not part of the Comprehensive xam syllabus, we give an alternative direct proof. As above, suppose that ν is another signed Borel measure that satisfies equation (). Fix any open interval (a, b). Let f n C c (R) be such that f n and f n χ (a,b) pointwise. Then by the Dominated Convergence Theorem, we have µ(a, b) lim n f n dµ lim n f n dν ν(a, b). This extends from open intervals to all Borel sets, so we conclude that µ ν.

9 8. Given p < and f n L p (R), prove that {f n } n N is a Cauchy sequence in L p (R) if and only if the following three conditions hold ( denotes Lebesgue measure). (a) {f n } n N is Cauchy in measure. (b) For every ε > there exists a δ > such that if < δ then f n p < ε for every n. (c) For every ε > there exists a set with < such that C f n p < ε for every n.. Assume that {f n } n N is Cauchy in L p (R). Since L p (R) is complete, there exists a function f L p (R) such that f n f in L p -norm. (a) By Tchebyshev s inequality, so {f n } n N is Cauchy in measure. { f m f n ε} ε p f m f n p p, (b) Given ε >, we have by standard arguments that for each n there exists a δ n > such that if < δ n then f n p < ε. Since f n f, there exists an N such that f n f p < ε for all n N. Set δ min{δ, δ,..., δ N }, and suppose that < δ. Then we have f n p ε for n N, and if n > N then ( ) /p ( ) /p ( ) /p f n p f n f p + f p f n f p + ε < 2ε. Hence statement (b) holds. (c) Choose ε >. Since for each f L p (R) we have x >m f p as m, for each n we can find a set n with n < such that f n p < ε p, all n. C n Let N, where N is such that f n f p < ε for all n N. Then <, and if n > N then ( ) /p ( ) /p ( ) /p f n p f f n p + f p f f n p + ε 2ε. C C C Since,..., N, we also have the required inequality for n N, so statement (c) holds.. Assume statements (a) (c) hold and choose ε >. Let the set be given as in statement (c). Set { ( ) /p } ε A mn f m f n. 9

10 Let δ be as given in statement (b). By statement (a), there exists an N such that A mn < δ for all m, n N. Hence f m f n p p f m f n p + f m f n p + f m f n p A mn \A mn C 2 p ( f m p + f n p ε ) + A mn + 2 p ( f m p + f n p ) C 2 p+ ε + ε + 2 p+ ε. Hence {f n } n N is Cauchy in L p (R). \A mn