Functional Analysis: Assignment Set # 10 Spring Professor: Fengbo Hang April 22, 2009

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1 Eduardo Coroa Fuctioal Aalysis: Assigmet Set # 0 Srig 2009 Professor: Fegbo Hag Aril 22, 2009 Theorem Let S be a comact Hausdor sace. Suose fgg is a collectio of comlex-valued fuctios o S satisfyig (i) The collectio fgg is equicotiuous (ii) fgg is uiformly bouded: jg(s)j M for all s 2 S ad every g i the collectio. The the collectio fgg is recomact i the maximum orm o S. 22. Show i Theorem that the coditios are ecessary: Solutio 2 So, we wat to show that fgg recomact i (C(S); kk ) imlies the coditios of Arzela- Ascoli. We kow that 8" > 0 9 fb " (g i )g N i ite esilo cover of fgg. Hece, (i) 8" > 0; let i (") the cotiuity delta for each g i ; ad mi in f i g: The, 8g 2 fgg; g 2 B " (g i ) for some i; ad: jg(s) g(t)j jg(s) g i (s)j + jg i (t) g i (s)j + jg(t) g i (t)j < 3" Hece, fgg is equicotiuous. (ii) Now, let M max in fkg i k g. The, 8g 2 fgg; g 2 B " (g i ) for some i; ad Hece, fgg is uiformly bouded. kgk kg g i k + kg i k < M + " 8" > Show that if Q is a oe bouded set i R such that ay two oits of Q ca be coected by a ath of legth l; the a family of fuctios fgg o Q is recomact i the maximum orm if the fuctios g ad their rst artial derivatives are uiformly bouded o Q : jg(s)j M j@ i g(s)j M 8i 8s 2 Q Solutio 3 As stated, this is robably false: we ca easily thik of examles where two oits could be arbitrarily close i sace, ad yet the shortest ath betwee them would ot shrik i size. However, we roose the followig alterative: let Q such that ay two oits ca be coected by a ath such that the ath legth l( x!y ) l jx yj (or eve l( x!y ) l jx yj ). The, for ay g 2 fgg; we have that: g(y) g(x) jg(y) g(x)j 0 0 d dt g( x!y(s))ds Jg( x!y (s)) 0 x!y(s) ds Cl( x!y ) Cl jx yj The fgg is uiformly Lischitz / Hölder cotiuous, ad as such, it is equicotiuous. Hece, by Arzela- Ascoli, fgg is recomact.

2 22.3 Formulate a versio of Theorem for fuctios whose values lie i a metric sace. Solutio 4 Let S be a comact Hausdor sace, Y a metric sace. Suose fgg C(S; Y ) satisfyig (i) The collectio fgg is equicotiuous (ii) fgg is oitwise recomact, that is: fg(s) : g 2 fggg is recomact i Y for all s 2 S. The the collectio fgg is recomact i the maximum orm o S Costruct a examle of a itegral oerator whose kerel satis es the Holmgre coditio (Theorem 3 i Chater 6) that is ot comact. That is, T K bouded from L 2 (X; )! L 2 (X; ) sucht that: kt K k su s 2 jk(s; t)j d(s) su t 2 jk(s; t)j d(t) Solutio 5 We cosider the itegral oerator T K : L 2 (R)! L 2 (R) with kerel: K(x; t) [0;] (x t) That is, we use a covolutio oerator with the idicator of the iterval [0; ]: Verifyig the Holmgre coditio is immediate, sice: [0;] (x t)dt [0;] (x t)dx However, this oerator is clearly ot comact, sice R is ot comact. This we kow sice su( [0;] g) [0; ] + su(g) Ad so, if we take g [2 ;2) ; su( [0;] g ) [2 ; 2 + ) 8. Sice the g all have the same L 2 orm, ad [0;] g have disjoit suorts, it follows that this oerator is ot comact Show that every Hakel oerator is comact. Solutio 6 I Lemma 8; Lax shows that for ay cotiuous s; C P + s s is a comact oerator. However, if we call P 0 the rojectio to the sace of costats, we have that: H s s P + s + P 0 s P 0 s C Hece, we have writte H s as the sum of a degeerate oerator (rak ) ad a comact oerator. It follows that H s is comact Show that the orm of a Hakel oerator H s satis es: kh s k if ks qk where q rages over all aalytic fuctios i the uit disc that are cotiuous o the uit circle S ad zero at z 0. Accordig to a theorem by Nehari, the sig of equality also holds. 2

3 Solutio 7 Let q 2 A fq: q is aalytic o the disc, q(0) 0g. The, it follows that q j S 2 H + ; ad has costat coe ciet equal to 0: This is also true for qu for all u, ad hece, H q P s 0. So, we have that H s H s q for all q 2 A. Fially, by the same reasoig as for T s ; we have that kh s k kh s q k ks qk 8q 2 A kh s k if q2a ks qk 28.7 Show that the ositive square root of A is uique, i.e, that there is o other ositive oerator whose square is A. Solutio 8 Sice A is ositive ad comact, it has a comlete set of orthogoal eigevectors, ad the corresodig eigevalues are all real ad oegative. As is oted i Corollary 6, A is the a ositive square root of A, ad ( A) (A). Now, suose S is a square root of A, 2 (S). The, for all v 2 N(S I); we have that Sv v ) S 2 v 2 v ) v 2 N(A 2 I). Hece, for some 2 (A), ad N(S I) N(A I) N( A I). Hece, S has the exact same sectrum as A; ad its eigesaces are cotaied i those of A. Moreover, the actio o each v 2 N(S I) by S ad A is the same, ad sice S has a comlete set of eigevectors, this imlies that S A. Alteratively, we could have also argued that, both beig ositive square roots of A; S ad A commute with A; ad hece commute with each other. Hece, they have the same ivariat subsaces (i articular, the eigesaces of S are ivariat uder the actio of A ad viceversa), ad sice they re both square roots of A it follows that they must have the same basis of eigevectors, with the same corresodig eigevalues Show without recourse to Theorem 3 (existece of a orthoormal base of eigevectors for comact symmetric oerators o Hilbert saces) that eigevectors of a symmetric oerator belogig to distict eigevalues are orthogoal. Solutio 9 Let A be a symmetric oerator, Av v ad Aw w. The hv; wi hav; wi hv; Awi hv; wi Hece, v ad w must be orthogoal. ( ) hv; wi Let U be a uitary oerator of form I + C; C comact. Show that U has a comlete set of orthoormal eigevectors, ad that all eigevectors have absolute value. Solutio 0 Sice U is uitary, we have that: (I + C)(I + C ) I (I + C )(I + C) CC C C Hece, C is ormal, ad by Corollary 2 we have that it has a comlete set of orthoormal eigevectors. Fially, each of these eigevectors is trivially a eigevector of U. Now, let v be a eigevector of u; the corresodig eigevalue. We the have that: huv; Uvi hv; Uvi jj 2 hv; vi Hece, all eigevalues have modulo. jj 2 3

4 . Let be a bouded oe set i R with smooth boudary, 0 < < <. Deote where C () fu 2 C() : [u] < g [u] su x6y2 ju(x) jx u(y)j yj ad the orm of the sace above is kuk C () kuk + [u].. Prove that the ijective ma C ()! C() is comact. 2. Prove that the ijective ma C ()! C () is comact. Solutio. Let B be the uit ball i C (). The, we have that u 2 C () are uiformly bouded by : kuk kuk C (), ad equicotiuous: ju(x) u(y)j jx yj ju(x) u(y)j jx yj Sice all u o the uit ball are uiformly Hölder cotiuous, it follows immediately that they are equicotiuous. Hece, by Arzela-Ascoli, the image of the uit ball uder this ma is recomact, ad hece the ma itself is comact. 2. We agai cosider the uit ball B ad its image uder the ijective ma. We already kow by. that B is recomact i C; ad so for every sequece u 2 B there exists a subsequece fu k g such that it coverges i the i ity orm to a fuctio u 2 C. Now, we have that: [u k u l ] su x6y2 jx yj su < su su jx yj + su + su yj< yj< j(u k u l )(x)j + j(u k u l )(y)j + [u k u l ] jx yj j(u k u l )(x) jx (u k yj u l )(y)j Now, sice > ; ad [u k u l ] 2; 8" > 0 we ca d such that the secod term i this boud is smaller tha "2. Also, sice ku k u l k! 0 as k; l! ; ; it follows that give this ; I ca d L such that k; l > L ) ku k u l k < " 4. Hece, these imly [u k u l ] < "; ad so this subsequece is coverget i C ; ad the ijectio ma is comact. 2 For a fuctio f de ed o R ad h 2 R ; we deote the traslatio of f by ( h f)(x) f(x h): Assume < ; show that a subset A L () is recomact if ad oly if (i) su f2a fkfk L (R ) g < (ii) For every " > 0; there exists R > 0 such that for every f 2 A; R jxj>r jf(x)j dx < " (iii) For every " > 0; there exists > 0 such that for ay f 2 A ad ay h 2 R with khk < ; k h f fk L (R ) < " Solutio 2 First, we cosider B R (0). By (i), the restrictio of A to B R is bouded, ad hece A is bouded i L (B R ). Hece, A e o ef j f 2 A the set of extesios by zero is bouded i L (R ) ad L (R ). Now, let f" m g m 4

5 R + such that " m # 0; ad m the corresodig sequece of molli ers. For each e f 2 e A; usig Hölder we have: f)(x) e f(x) e e f(x y) e f(x) (y)dy R 8 < : R f(x e y) f(x) e (y)dy 9 ; For " > 0; usig (iii) we choose m such that " m < ( " 4 ), Let H m e f)(x) e f(x) 0;;B R ) B (0) 8 < (y) : R f(x e y) f(x) e dx f)(x) e f(x) e 0;;BR < " 4 8f 2 A 9 ; dy < ( " 4 ) f e o j BR j f 2 A : We have that, 8f 2 F, by covolutio iequality 9C m > 0 such that: f e 0;;R 0;;R f e 0;;R C m Ad for all x; y 2 R ; f)(x) e f)(y) e 0;;R f e jx yj C0 m jx yj 0;;R Hece, for each xed m, H m is bouded ad equicotiuous i C(B R ); ad by Arzela-Ascoli, it is recomact. Hece, it is also recomact i L (B R ). o k Fially, for " > 0; we ick m such that " m < ("4). The, we d a ite "4-cover of H m : B L (B R ) "4 (f i ) o However, by the aroximatio roerties of these fuctios, this imlies that B L (B R ) "2 (f i ) k is a ite esilo cover for A j BR. Hece, A j BR is recomact. Sice this alies for all R > 0; by (ii) we ca choose such R > 0 such that theo -orm outside of B R is less o tha "2. The, sice A j BR is recomact, we ca d a cover B L (B R ) "2 (f i ) k ; ad hece B L (R ) " (f i ) k is a ite esilo cover for A. : 5