Functional Analysis: Assignment Set # 2 Spring 2009 Professor: Fengbo Hang February 18, 2009

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1 Eduardo Coroa Fuctioal Aalysis: Assigmet Set # Sprig 9 Professor: Fegbo Hag February 8, 9 C4. Show that if i Sectio 4. we take S N, Y C the space of coverget sequeces, l de ed by (4), the fuctio p give by (4) is the same as de ed by (). That is: p(x) iffl(y) : x y; y Y g if fx k g Solutio () Let s k k fx k g. By de itio of i mum, 8" > 9N(") s.t. if s s N if s + ". Now, sice s is odecreasig, we ca pick a subsequece fs (j) g such that if Now, de ig () ad s () k fx k g, let y s s (j) if s + j X s (i) [(i);(i+)) () i That is, y is costat for f(i); (i + ) g; y k(i) fx k g x 8 f(i); (i + ) g (hece y x), ad l(y) lim j! s (j) if s. So, iffl(y) : x y; y Y g if k fx k g. () Now, let y : y x; y Y. The, sice y is coverget, lim y if fy k g if fx k g! k This is true for all such y s; so takig i mum, we get: iffl(y) : x y; y Y g if k k fx k g k C4. Show that a Baach Limit ca be chose so that for ay bouded sequece (c ; c ; :::) that is Cesaro summable; amely such that the arithmetic meas of the partial sums coverge to c; LIM! c c Solutio We have to show that the limit geeralized o the set C of Cesaro summable sequeces is a valid extesio i the sese of Theorem. If this is the case, we ca the exted this to a Baach limit o B l so that its restrictio to C has the desired properties: (i) Lets call l P A : C! R the liear fuctioal that assigs ay sequece i C the limit of the meas of partial averages. We eed to show Y C; ad l P A j Y l. Let fy g Y; y! y ad for " >, N(") s.t. N ) jy yj < ". The: y X y k X jy k yj i < " + N(") X i jy l P A (y)j < " 8" >. jy k yj! "

2 So, l P A (fy g) l(fy g) y. This shows that Y C; ad that the limit of Cesaro meas coicides with the stadard limit for coverget sequeces. (ii) c; d C ) l P A (c + d) l P A (c) + l P A (d): This just follows from the liearity of averages: P i c k + d k P i c k + P i d k 8. (iii) l P A (fc +t g) l P A (fc g) (traslatio ivariace): For >> t, X X+t c k c k tx jc k j + X+t jc k j t kck! (iv) Now, we eed to show if k fc k g l P A (c) if k fc k g. For all t N, > t we have: Fially, by traslatio ivariace, this meas: if fc kg X+t c k fc k g k if fc kg l P A (fc +t g) fc k g if fc kg l P A (fc g) fc k g 8t N Ad takig remum o oe side ad i mum o the other gives us the desired result. Thus, the limit de ed by the Cesaro meas is a valid extesio of the usual limit, ad it ca i tur be exteded usig the equivariat Hah-Baach theorem to obtai a Baach limit. C4. Show that a geeralized limit as t! ca be assiged to all bouded fuctios x(t) de ed o t that has properties (i) to (iv) i Theorem. Solutio To do this, we wat to emulate the procedure i Theorem. Here B L ; l(x) lim t! x(t) ad Y fy B : l(y) existsg. Fially, we take p as: p(x) if t fx(s)g st The proof is the idetical: all we eed to show is that p(y) l(y) 8y Y; ad that p is traslatio ivariat. The, properties (i) to (iv) result from applyig the equivariat Hah-Baach to obtai this geeralized limit. These two results are almost immediate: Let y Y, 8" > 9N(") s.t. x N ) jy(x) l(y)j < ". ) st fy(s)g y(s) < " 8t N ) p(y) l(y). Also, lim t st fx(s + p)g lim t st+p fx(s)g lim t st fx(s)g. Applyig Hah- Baach, properties (i) through (iv) state: (i) LIM is a extesio (ii) LIM is liear (iii) LIM is traslatio ivariat ad (iv) LIM(x) p(x) 8x B. (which agai, turs ito the double iequality with lim ad lim if whe we write p( x) l(x) p(x) 8x B). C5. Show that (6) are orms. Show that they are equivalet. Solutio 4 We rst observe that the fuctios i (6) are othig else tha a compositio of the orms i Z ad U with a particular orm of R (i this exercise, the p orms for p ; ; ). Hece, we ca prove a much more geeral statemet: Let jj be a orm for R such that: j(a; b)j j(; )j 8(; ) (a; b) (; )

3 Not all orms i R ll this coditio: for this to be true, the itersectio of the uit ball ad the rst orthat is cotaied i the rectagle geerated by its itersectio with the coordiate axis. I particular, the p-orms do posess this property: (a p + b p ) p ( p + p ) p 8(; ) (a; b) (; ) fa; bg f; g 8(; ) (a; b) (; ) Uder this hypothesis, we check the coditios for a orm o Z U: (i) Positivity: j(; )j is already positive, ad j(kzk Z )j () (kzk Z ) (; ) () (z; u) (sice jj ; kk Z ad kk U are orms). (ii) Homogeeity: 8 R, j(kzk Z )j j(jj kzk Z ; jj kuk U )j jj j(kzk Z )j This is, agai, by homogeeity of these three orms. We ote that oly positive homogeeity is required i the case of jj. This meas we ca choose a Mikowski gauge istead of jj ; sice the uit ball is ot required to be balaced. (iii) Subadditive: Here, we use our hypothesis, as well as subadditivity of all orms ivolved: Let z ; z Z ad u ; u U: j(kz + z k Z ; ku + u k U )j j(kz k Z + kz k Z ; ku k U + ku k U )j j(kz k Z ; ku k U ) + (kz k Z ; ku k U )j j(kz k Z ; ku k U )j + j(kz k Z ; ku k U )j We eeded this extra hypothesis to obtai the rst iequality by subadditivity of the orms i Z ad U. C5. Show that if X is a Baach space, Y a closed subspace of X, the quotiet space XY is complete. Solutio 5 Let f[q ]g m; M ) be a Cauchy sequece i the quotiet space. That is, 8" > ; 9M(") > such that k[q ] [q m ]k XY if yy fk(q q m ) yk X g < " From this, we pick the subsequece q (k) q M(k ): that is, fq (k) g is such that: q (k) q(k+) XY < k (this by de itio of the M( k ); which we pick carefully so that M( k ) < M( (k+) ) 8k). The, X X q (k) q(k+) XY if f (q(k) q (k+) ) y yy X g 6 Now, by de ito of i mum, for each k 9y k Y s.t. (q(k) q (k+) ) y X k if yy f (q(k) q (k+) ) y X g+ k : Thus: X X (q(k) q (k+) ) y X k if f (q(k) q (k+) ) y yy X g + k But X is a Baach space, ad so, every absolutely summable sequece is summable. Thus, 9x X such that: X X x (q (k) q (k+) ) y k q () lim lim q (k) + (x q () ) k! lim k! q(k) X y k Y (Y is a closed subspace) x q () k! q (k) Fially, sice f[q ]g is Cauchy ad has a coverget subsequece, the the etire sequece is coverget. Hece, the quotiet space is complete. y k

4 C5.4 Prove that every ite-dimesioal subspace of a ormed liear space is closed. (Hit: use the fact that all orms are equivalet o ite-dimesioal spaces to show that every ite-dimesioal subspace is complete). Solutio 6 Let X be a liear space, F X ite-dimesioal subspace. The, F spa(fe ; ::; e g) for some e i F. That is, 8x F 9!x R so that: x X x i e i i Now, we ca de e a orm o F that makes it isometrically isomorphic to R. For example, we ca de e kxk F k(x ; ::; x )k ;R ; ad the isomorphism (x) (x ; ::; x ). However, R is complete, ad completeess is a topological property (preserved uder isomorphism). Hece, (F; kk F ) is a complete space. Fially, sice all orms are equivalet o ite-dimesioal spaces, it follows that (F; kk X ) is a complete liear space. Hece, it must be a closed subspace of X (sice this implies every coverget sequece o F coverges to a poit i F ). C5.5 Show that the orms of examples (a); (c); (d) ad (e) are ot strictly subadditive. Solutio 7 For each space, we produce a example that shows these spaces are ot strictly subadditive: (a) l (C): v (; ; ; :::) ad v (; ; ; ; :::), the kv + v k + kv k + kv k. (c) L (S): Assumig S has at least two distict elemets, let s 6 s S. We de e f (s) fsg(s); ad f fs;s g(s) ad agai: kf + f k + kf k + kf k. (d) BC(Q; C): Let f ; ad f ay ocostat fuctio i BC. The, kf + f k + kf k kf k + kf k (e) C (Q; C): If Q is compact, the previous example also works. Otherwise, give K K K compact sets, if Q is locally compact ad Hausdor, by Tietze s Extesio Theorem we ca de e a cut-o fuctios f such that f o K ad p(f ) K ; ad f s.t. f o K ad p(f ) K, the: kf + f k + kf k + kf k (f + f i K ). C5.6 Show that the orms i examples (b) ad (f) are ot strictly subadditive for p : Solutio 8 Agai, we produce a example for each case: (b) l (C): Now, v (; ; ; :::) ad v (; ; ; ; :::), the kv + v k + kv k + kv k. (f) C (D): Here, let B B r (x ) ad B B r (x ) two disjoit ope balls i D, i the correspodig molli ers, that is: i (x) kr i exp ri ri kx x i k! Bi (x) k chose so that k i k. The, sice these two fuctios have disjoit ports, it follows that kv + v k + kv k + kv k. (I fact, this also proves this statemet for C k (D) 8k [; ]) Let R be a ope subset, for a fuctio u :! R, let kuk x ju(x)j : Deote: E fu :! R : kuk < ad u is uiformly cotiuousg Show that (E; kk ) is a Baach space. Solutio 9 Clearly, E is a subspace of the liear space of all cotiuous fuctios (C(); kk ); which we kow is a Baach space. Thus, all we eed to prove is that E is closed (that is, that the uiform limit of a sequece of uiformly cotiuous fuctios is uiformly cotiuous): Let fu g E so that u! u uiformly. This meas 8" > 9N(") > s.t. N ) ku uk < " 4

5 Now, sice u N is uiformly cotiuous, for this " there exists a (") > such that x; y ; jx yj < ) ju N (y) u N (x)j < ". But the, this also implies: ju(y) u(x)j ju(y) u N (y)j + ju N (y) u N (x)j + ju(x) u N (x)j < " Ad so, u is uiformly cotiuous (with (") N(") ): Hece, E is a closed subspace of a Baach space, ad thus, it is also a Baach space. 4 Let R be a bouded ope subset, (; ]. For ay fuctio u :! R; let: [u] x 6x ju(x ) u(x )j jx x j ad C () fu :! R : [u] < g space of Hölder cotiuous fuctios. (i) Show that kuk C () kuk + [u] is a orm o C (); uder which it becomes a Baach space. (ii) Is C () separable? Why? Solutio First of all, we otice that [u] < ) u is uiformly cotiuous (Hölder cotiuous), sice 8x ; x, ju(x ) u(x )j < jx x j (ad so, we ca always use (") " ). Also, sice u is cotiuous o a bouded, closed set (), we kow kuk <. Now, we rst eed to show kk C () is a orm: (i) Positive: We kow kuk ad [u] ; sice they are both de ed as remums of oegative quatities. Now, kuk C () ) ju(x ) u(x )j x 6x jx x ) ju(x j ) u(x )j 8x 6 x ) u is costat. However, the orm beig also implies kuk ; so this costat must be. (ii) Homogeeous: this follows immediately from homogeeity of kk ad jj: kuk + x6x ju(x jj kuk + jj ) u(x )j x6x jx x jj kuk j C (). (iii) Subadditive: This agai follows from the properties of kk ad jj: ku + vk + x6x ju(x kuk + kvk + ) u(x )j jv(x) v(x)j x6x jx x j + jx x kuk j C () + kvk C () ju(x ) u(x )j jx x j j(u+v)(x ) (u+v)(x )j jx x j Now, give a Cauchy sequece fu g C (); it is clear that it is also a Cauchy sequece i (C(); kk ) (sice kuk C () kuk ). This last space is complete, so u! u i (C(); kk ). So, all that is left to prove is that u C (), ad that u! u relative to the orm we have just de ed: (a) [u] < : For ", let N() s.t. ; m N ) ku u m k C () <. The: [u ] [u N ] + [u N u ] < [u N ] + [u] lim! [u ] < [u N ] + < (b) [u u]! : lim j(u u )(x ) (u u )(x )j! x 6x jx x j lim m;! x 6x That is: 8" > ; by de itio of remum, 9x () 6 x () such that: j(u u )(x ) (u u )(x )j x 6x jx x j < (u The, sice u m! u uiformly, 9P (") such that p P ) (u j(u m u )(x ) (u m u )(x )j jx x j u )(x () ) (u u )(x () ) x () + " x () u )(x () ) (u u )(x () ) (u+p < u )(x () ) (u +p u )(x () ) " + x () x () 5

6 so: j(u u )(x ) (u u )(x )j (u +p x 6x jx x j < u )(x () ) (u +p u )(x () ) x () + " x () Fially, sice our sequece is Cauchy, 9N(") such that N ) [u +p j(u u )(x ) (u u )(x )j x 6x jx x j < " u ] < ". But this implies: Hece, [u u]! ; ad so the space of Hölder cotiuous fuctios is complete. Fially, we d that this space is ot separable. To see this, let h t (x ; x d ) (x d t) [;) (x d ). We ca see that h t C () 8t, ad although kh t h s k! (s!t) ; the distace i the [] semiorm is uiformly bouded below. Here is a plot of h t h for decreasig values of t (t ; :5; :5; :) i R: y So, we have: j(x t) (y t) j [h t ] x6y jx yj (sice jx yj jx y j for (; ); )ad it is achieved whe y t) Also, for ay s < t ; [h t h s ] (sice h t h s h s for x [s; t]) Hece, we have a ucoutable family of fuctios i C () such that kh t h s k C () 8s 6 t ) C () is ot separable. x 5 Let C be a ope subset, for p [; ], let: (i) Show that (A p (); kk p ) is a Baach space. (ii) Is A p () separable? Why? A p () ff :! C : f is holomorphic ad f L p ()g Solutio First of all, we idetify A p () as a subspace of L p (). Therefore, all that remais is to show that it is closed to prove it is a Baach space. Let ff g A p ; kf fk p! : For p, it is immediate that f A, sice (by Weierstrass Theorem) a uiform limit of aalytic fuctios is aalytic. For p < ; let z ; B r (z ). By the Cauchy Itegral formula: f (z ) Z f (z + re i )d 6

7 However, we wat to have a result that liks f (z ) with a itegral over a disc. Let R > such that the Cauchy Itegral formula applies for r R. The, we ca itegrate ad average over the whole disc: f (z ) R R R Z f (z + re i )ddr Z R Z R Z Z B R (z ) f (z + re i )(rdrd) f (z)da Hece, jf (z ) f(z )j R Z B R (z ) jf (z) R jf (z) f(z )j dz p f(z )j p dza (R ) q! as! Ad so f A p (f is aalytic, sice f coverges to f uiformly o compact sets i ). Fially, A p () is separable for p < ; sice it is a subspace of L p (); which we kow is separable. However, A () is ot separable. To see this, we take D the uit disc, ad we start by geeratig harmoic fuctios u t that solve the Dirichlet problem: ut x D u t (; ) [;t) () Now, for each t; u t has a cojugate harmoic fuctio v t ; ad f u t + iv t is a aalytic fuctio. However, we ca t cotrol or boud v t, so we take h t exp(u t + iv t ) 8t [; ). We see that: (i) kh t k e: jh t j exp(u t ) [; e]; ad the maximum is attaied i the boudary ( [; t)) (by maximum priciple for holomorphic fuctios). (ii) kh t h s k e : We look i the boudary, o the iterval [s; t). There kh t h s k jexp(u t ) exp(u s )j e >. So, we have foud a ucoutable family of fuctios i A (D) s.t. kh t h s k is uiformly bouded below, ad therefore, A () is ot separable. To traslate this to the geeral case, we could perhaps compose this result with a coformal map from the uit disc to ; or equivaletly, solve a similar Dirichlet problem directly i. 7