Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be one-to-one, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true if X is not compct? (b) Suppose tht f is uniformly continuous nd f is continuous. If Y is complete, prove tht X is complete. If X is complete, does it follow tht Y is complete? () A continuous function mps compct sets to compct sets, nd subset of compct metric spce is compct if nd only if it is closed, so f mps closed sets to closed sets, which implies tht f is continuous. Hence, f is homeomorphism. (b) A uniformly continuous function mps Cuchy sequences to Cuchy sequences: Suppose ( n ) is Cuchy nd y n = f( n ). Then, given ɛ > 0, there eists δ > 0 such tht d(, b) < δ implies tht d(f(), f(b)) < ɛ, nd there eists N N such tht m, n > N implies tht d( m, n ) < δ. Hence m, n > N implies tht d(y m, y n ) < ɛ, so (y n ) is Cuchy. Suppose tht f is uniformly continuous nd Y is complete. Let ( n ) be Cuchy sequence in X nd y n = f( n ). Then (y n ) is Cuchy in Y, so there eists subsequence (y nk ) such tht y nk y for some y Y. Then nk = f (y nk ) f (y) =, since f is continuous, which proves tht X is compct. Completeness of X does not imply completeness of Y. For emple, define f : R ( π/2, π/2) by f() = tn, where R nd ( π/2, π/2) re equipped with their stndrd metric. Then f is uniformly continuous, since 0 < f (), nd f is continuous, but R is complete while ( π/2, π/2) is not. Remrk. The lst emple shows tht completeness is not topologicl property, since it need not be preserved by homeomorphisms. To mke sense of completeness in topologicl spces, one hs to introduce n dditionl uniform structure on the spce.
Problem 2. Consider the spce of continuously differentible functions, with the C -norm, C ([0, ]) = {f : [0, ] R f, f re continuous}, f C = sup f() + sup f (). 0 0 Prove tht C ([0, ]) is Bnch spce. It is immedite to verify tht C is normed liner spce, so the only nontrivil property to check is its completeness. Suppose tht (f n ) is Cuchy sequence in C. Then (f n ) nd (f n) re Cuchy sequences in C, so by the completeness of C, there eist f, g C such tht f n f nd f n g uniformly. We clim tht f C nd g = f, in which cse f n f in C, so C is complete. By the fundmentl theorem of clculus, f n () = f n (0) + 0 f n(t) dt for every n N. Tking the limit n, nd using the fct tht we cn echnge uniform limits with the Riemnn integrl, we get tht f() = f(0) + 0 g(t) dt. The fundmentl theorem of clculus then implies tht f C nd f = g, which proves the result. 2
Problem 3. () Suppose tht f, g C([, b]). Prove the Cuchy-Schwrz inequlity: b ( b f()g() d ) /2 ( b /2 f() 2 d g() d) 2. (b) Let F = { f C ([0, ]) : f(0) = f() = 0, 0 } f () 2 d. Prove tht F is precompct in (C([0, ]), ). () If r, s 0, then r 2 2rs + s 2 = (r s) 2 0, so rs ( r 2 + s 2). 2 Let ( b /2 ( b /2 f = f() d) 2, g = g() d) 2. Then, writing r = f()/ f nd s = g()/ g, we get tht f()g() f g ( ) f() 2 2 f + g()2. 2 g 2 Integrtion of this inequlity with respect to yields which proves the result. f g b (b) For, y [0, ], we hve f() f(y) = 3 f()g() d, y f (t) dt.
If f F, then by using the Cuchy-Schwrz inequlity in (), we get tht y ( y ) /2 ( y /2 f (t) dt d f () d) 2 y /2, so f() f(y) y /2 for ll, y [0, ] nd ll f F. It follows tht F is uniformly equicontinuous (given ɛ > 0, tke δ = ɛ 2 ). In prticulr, if y = 0 in the previous inequlity, then f(y) = 0, nd we get tht f() /2 for ll [0, ] nd ll f F, which shows tht F is uniformly bounded. Hence, the Arzelà-Ascoli theorem implies tht F is precompct in C([0, ]). Remrk. Let 0 < α. If function f : [, b] R stisfies f() f(y) C y α for ll, y [, b] for some constnt C 0, then f is sid to be Hölder continuous on [, b] with eponent α. If α =, then f is Lipschitz continuous. The rgument in (b) shows tht if function f : [, b] R hs squre-integrble (wek) derivtive, then f is Hölder continuous with eponent /2. This result is simple emple of Morrey s inequlity, which is bsic Sobolev embedding theorem. 4
Problem 4. () If f : R R is continuous, show tht h +h f(t) dt f() s h 0. (b) Given f C([0, ]), etend f to function f : R R by f() if > f() = f() if 0 f(0) if < 0 Define f h : [0, ] R for h R by f h () = h +h f(t) dt. Show tht f h C ([0, ]). Wht is f h? (c) Show tht f h f uniformly s h 0, mening tht C ([0, ]) is dense in (C([0, ]), ). +h () Using the fct tht c = h we hve f h () f() = h h c dt to bring f() inside the integrl, +h +h (f(t) f()) dt f(t) f() dt. Since f is continuous t, there eists δ > 0 such tht f(t) f() < ɛ for t < δ, nd then h < δ implies tht f h () f() < ɛ, which shows tht f h () f() s h 0. 5
Using the result in (), we hve ( ) f h() fh ( + ɛ) f h () = lim ɛ 0 ɛ = lim ɛ 0 hɛ = h lim ɛ 0 = h ( +h+ɛ +ɛ +h+ɛ f(t) dt ( ɛ +h ( ) f( + h) f(). f(t) dt ɛ +h +ɛ ) f(t) dt ) f(t) dt Alterntively, you cn use the fundmentl theorem of clculus. (c) Since f is continuous, it is uniformly continuous on [, 2]. Given ɛ > 0, choose 0 < δ such tht f() f(y) < ɛ for ll, y [, 2] such tht y < δ. Then the sme rgument s in () shows tht f h () f() < ɛ for ll [0, ] when h < δ, which proves tht f h f uniformly s h 0. 6
Problem 5. Let X be metric spce nd let B(X) denote the normed liner spce of ll bounded (but not necessrily continuous) functions f : X R with the sup-norm f = sup f(). X Fi 0 X, nd for ech X define φ : X R by φ () = d(, ) d(, 0 ). Finlly, define Φ : X B(X) by Φ() = φ. () Show tht B(X) is complete. (b) Show tht Φ is n isometric embedding of X into B(X). (c) Use this result to show tht every metric spce X hs completion. () See Theorem 7 in the notes on continuous functions. (b) Suppose tht, b X. Then d (Φ(), Φ(b)) = sup φ () φ b () X = sup d(, ) d(, b). X By the reverse tringle inequlity, d(, ) d(, b) d(, b), with equlity when =, b, so d (Φ(), Φ(b)) = d(, b), mening tht Φ is n isometric embedding. (c) Let X = Φ(X) be the closure of the imge of X under Φ in B(X). Then Φ(X) is dense in X, nd X is complete becuse it is closed subspce of complete spces. Thus, X is the completion of X, which is unique up to n isometric isomorphism. 7
Problem 6. Suppose tht A X is dense subset of metric spce X nd f : A Y is uniformly continuous function into complete metric spce Y. Prove tht there is unique continuous etension f : X Y of f, such tht f = f. A For ny X, there eists sequence ( n ) in A such tht n. Let y n = f( n ). Since ( n ) is convergent, it is Cuchy, nd since f is uniformly continuous, (y n ) is Cuchy. Hence, y n y for some y Y, since Y is complete. We then define f() = y. The vlue f() is well-defined. Suppose tht ( n ) nd ( n) re sequences in A tht converge to X. Let y n = f( n ) nd y n = f( n). Then d( n, n) 0 s n, so the uniform continuity of f implies tht d(y n, y n) 0, nd therefore lim y n = lim y n. If A, then we cn use the sequence () with n = for every n N, so f() = f(), nd f is n etension of f. Finlly, we show tht f is uniformly continuous. Let ɛ > 0. By the uniform continuity of f, there eists δ > 0 such tht d (f(), f( )) < ɛ 3 for ll, A with d(, ) < δ. If, X, then there eist, A with d(, ) < δ ( ) 3, d f(), f() < ɛ 3, d(, ) < δ ( ) 3, d f( ), f( ) < ɛ 3. It follows tht if d(, ) < δ/3, then nd ( ) d f(), f( ) d(, ) d(, ) + d(, ) + d(, ) < δ, ( ) ( d f(), f() + d (f(), f( )) + d f( ), f( ) ) < ɛ which shows tht f : X Y is uniformly continuous. 8