Introduction to Functional Analysis

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1 MIT OpeCourseWare Itroductio to Fuctioal Aalysis Sprig 009 For iformatio about citig these materials or our Terms of Use, visit:

2 LECTURE OTES FOR 18.10, SPRIG Lecture 1. Tuesay, Mar 17: Compactess ad weak covergece A subset i a geeral metric space is oe with the property that ay sequece i it has a coverget subsequece, with its limit i the set. You will recall with pleasure o doubt the equivalece of this coditio to the (more geeral sice it makes good sese i a arbitrary topological space) equivalece of this with the coverig coditio, that ay ope cover of the set has a fiite subcover. So, i a separable Hilbert space the otio of a compact set is already fixed. We wat to characterize it i the problems this week you will be asked to prove several characterizatios. A geeral result i a metric space is that ay compact set is both closed ad bouded, so this must be true i a Hilbert space. The Heie-Borel theorem gives a coverse to this, R or C (ad hece i ay fiite dimesioal ormed space) ay closed ad bouded set is compact. Also recall that the covergece of a sequece i C is equivalet to the covergece of the sequeces give by its compoets ad this is what is used to pass first from R to C ad the to C. All of this fails i ifiite dimesios ad we eed some coditio i additio to beig bouded ad closed for a set to be compact. To see where this might come from, observe that a set, S, cosistig of the poits of a coverget sequece, s : M, together with its limit, s, i ay metric space is always compact. The set here is the image of the sequece, thought of as a map from the itegers ito the metric space, together with the limit (which might of course already be i the image). Certaily this set is bouded, sice the distace from the itial poit is certaily bouded. Moreover it is closed, although you might eed to thik about this for a miute. A sequece i the set which is the image of aother sequece cosists of elemets of the origial sequece i ay order ad maybe repeated at will. Sice the origial sequece may itself have reapeated poits, the labellig of poits is by o meas uique. However S is closed sice M \ S is ope a poit i p M \ S is at a fiite o-zero distace, d(p, s) from the limit so B(p, d(p, s)/) ca cotai oly fiitely may elemets of S hece a smaller ope ball does ot meet it. Lemma 6. The image of a coverget sequece i a Hilbert space is a set with equi-small tails with respect to ay orthoormal sequece, i.e. if e k is a othoormal sequece ad u u is a coverget sequece the give ɛ > 0 there exists such that (1.1) (u, e k ) < ɛ. k> Proof. Bessel s iequality shows that for ay u H, (1.) (u, e k ) u. k The covergece of this series meas that (1.1) ca be arraged for ay sigle elemet u or the limit u by choosig large eough, thus give ɛ > 0 we ca choose so that (1.3) (u, e k ) < ɛ /. k>

3 7 LECTURE OTES FOR 18.10, SPRIG 009 I fact, for ay orthoormal sequece such as e k whether complete or ot, (1.4) P : H u P u = (u, e k )e k H is cotiuous ad i fact has orm at most oe. Ideed from Bessel s iequality, P u u. ow, applyig this to (1.5) P u = (u, e k )e k k> the covergece u u implies the covergece i orm P u P u ad so (1.6) (u, e k ) < ɛ. k> So, we have arraged (1.1) for > with =. Of course, this estimate remais valid if is icreased, ad we may arrage it for by chossig large eough. Thus ideed (1.1) holds for all if is chose large eough. This suggest oe useful characterizatio of compact sets i a separable Hilbert space. Propositio 19. A set K H i a separabel Hilbert space is compact if ad oly if it is bouded, closed ad has equi-small tails with respect to ay oe orthoormal basis. Proof. We already kow that a compact set is closed ad bouded. Suppose the equi-smalless of tails coditio fails with respect to some orthoormal basis e k. This meas that for some ɛ > 0 ad all there is a elemet u K such that (1.7) (u, e k ) ɛ. k> The the sequece {u } ca have o coverget subsequece, sice this would cotradict the Lemma we have just proved, hece K is ot compact i this case. Thus we have proved the equi-smalless of tails coditio to be ecessary for the compactess of a closed, bouded set. So, it remais to show that it is sufficiet. So, suppose K is closed, bouded ad satisfies the equi-small tails coditio with respect to a orthoormal basis e k ad {u } is a sequece i K. We oly eed show that {u } has a Cauchy subsequece, sice this will coverge (H beig complete) ad the limit will be i K (sice it is closed). ow, cosider each of the sequeces of coefficiets (u, e k ) i C. Here k is fixed. This sequece is bouded: (1.8) (u, e k ) u C by the boudedess of K. So, by the Heie-Borel theorem, there is a subsequece of u l such that (u l, e k ) coverges as l. We ca apply this argumet for each k = 1,,.... First extractig a subsequece of {u } so that the sequece (u, e 1 ) coverges alog this subsequece. The extract a subsequece of this subsequece so that (u, e ) also coverges alog this sparser subsequece, ad cotiue iductively. The pass to the diagoal subsequece of {u } which has kth etry the kth term i the kth subsequece. It is evetually a subsequece of each of the subsequeces previously costructed meaig it coicides with a subsequece for some poit oward (amely the k

4 LECTURE OTES FOR 18.10, SPRIG kth term oward for the kth subsquece). Thus, for this subsequece each of the (u l, e k ) coverges. ow, let s relabel this subsequece v for simplicity of otatio ad cosider Bessel s idetity (the orthoormal set e k is complete by assumptio) for the differece v v +l H = (v v +l, e k ) + (v v +l, e k ) k (1.9) (v v +l, e k ) + (v, e k ) + (v +l, e k ) Proof. Let e k be a orthoormal basis. The if u is weakly coverget it follows immediately that (u, e k ) (u, e k ) coverges for each k. Coversely, suppose this is true for a bouded sequece, just that (u, e k ) c k i C for each k. The orm boudedess ad Bessel s iequality show that (1.1) c k = lim (u, e k ) C sup u k> k k> k> where the parallelogram law o C has bee used. To make this sum less tha ɛ we may choose so large that the last two terms are less tha ɛ / ad this may be doe for all ad l by the equi-smalless of the tails. ow, choose so large that each of the terms i the first sum is less tha ɛ /, for all l > 0 usig the Cauchy coditio o each of fiite umber of sequece (v, e k ). Thus, {v } is a Cauchy subsequece of {u } ad hece as already oted coverget i K. Thus K is ideed compact. It is coveiet to formalize the idea that each of the (u, e k ), the sequece of coefficiets of the Fourier-Bessel series, should coverge. Defiitio 6. A sequece, {u }, i a Hilbert space, H, is said to coverge weakly to a elemet u H if it is bouded i orm ad (u j, v) (u, v) coverges i C for each v H. This relatioship is writte (1.10) u u. I fact as we shall see ext time, the assumptio that u is bouded ad that u exists are both uecessary. That is, a sequece coverges weakly if ad oly if (u, v) coverges i C for each v H. Coversely, there is o harm i assumig it is bouded ad that the weak limit u H exists. ote that the weak limit is uique sice if u ad u both have this property the (u u, v) = lim (u, v) lim (u, v) = 0 for all v H ad settig v = u u it follows that u = u. Lemma 7. A (strogly) coverget sequece is weakly coverget with the same limit. Proof. This is the cotiuity of the ier product. If u u the (1.11) (u, v) (u, v) u u v 0 for each v H shows weak covergece. ow, there is a couple of thigs I will prove here ad leave some more to you for the homework. Lemma 8. For a bouded sequece i a separable Hilbert space, weak covergece is equivalet to compoet covergece with respect to a orthoormal basis. k p k p

5 74 LECTURE OTES FOR 18.10, SPRIG 009 for all p. Thus i fact {c k } l ad hece (1.13) u = w k e k H Proof. Choose a orthoormal basis e k ad observe that (1.16) u, e k = lim u, e k. ow the sequece o the right is bouded by u idepedetly of p so (1.17) u, e k lim if u k p k by the completeess of H. Clearly (u, e k ) (u, e k ) for each k. It remais to show thta (u, v) (u, v) for all v H. This is certaily true for ay fiite liear combiatio of the e k ad for a geeral v we ca write (1.14) (u, v) (u, v) = (u, v p ) (u, v p ) + (u, v v p ) (u, v v p ) = where v p = (u, v) (u, v) = (u, v p ) (u, v p ) + C v v p (v, e k )e k is a fiite part of the Fourier-Bessel series for v ad C is a k p boud for u. ow the covergece v p v implies that the last term i (1.14) ca be made small by choosig p large, idepedet of. The the secod last term ca be made small by choosig large sice v p is a fiite liear combiatio of the e k. Thus ideed, (u, v) (u, v) for all v H a it follows that u coverges weakly to u. Propositio 0. Ay bouded sequece {u } i a separable Hilbert space has a weakly coverget subsequece. This ca be thought of as a aalogue i ifiite dimesios of the Heie-Borel theorem if you say a bouded closed subset of a separable Hilbert space is weakly compact. Proof. Choose a orthoormal basis e k ad apply the procedure i the proof of Propositio 19 to extract a subsequece of the give bouded sequece such that (u p, e k ) coverges for each k. ow apply the preceedig Lemma to coclude that this subsequece coverges weakly. Lemma 9. For a weakly coverget sequece u u (1.15) u lim if u. k p by the defiitio of lim if. ow, take p to coclude that (1.18) u lim if u from which (1.15) follows.

6 LECTURE OTES FOR 18.10, SPRIG Problems 6: Due 11AM Tuesday, 31 Mar Hit: Do t pay too much attetio to my hits, sometimes they are a little offthe-cuff ad may ot be very helpfult. A example beig the old hit for Problem 6.! Problem 6.1 Let H be a separable Hilbert space. Show that K H is compact if ad oly if it is closed, bouded ad has the property that ay sequece i K which is weakly coverget sequece i H is (strogly) coverget. Hit:- I oe directio use the result from class that ay bouded sequece has a weakly coverget subsequece. Problem 6. Show that, i a separable Hilbert space, a weakly coverget sequece {v }, is (strogly) coverget if ad oly if the weak limit, v satisfies (1.19) v H = lim v H. Hit:- To show that this coditio is sufficiet, expad (1.0) (v v, v v) = v Re(v, v) + v. Problem 6.3 Show that a subset of a separable Hilbert space is compact if ad oly if it is closed ad bouded ad has the property of fiite dimesioal approximatio meaig that for ay ɛ > 0 there exists a liear subspace D H of fiite dimesio such that (1.1) d(k, D ) = sup if {d(u, v)} ɛ. u K v D Hit:- To prove ecessity of this coditio use the equi-small tails property of compact sets with respect to a orthoormal basis. To use the fiite dimesioal approximatio coditio to show that ay weakly coverget sequece i K is strogly coverget, use the covexity result from class to defie the sequece {v } i D where v is the closest poit i D to v. Show that v is weakly, hece strogly, coverget ad hece deduce that {v } is Cauchy. Problem 6.4 Suppose that A : H H is a bouded liear operator with the property that A(H) H is fiite dimesioal. Show that if v is weakly coverget i H the Av is strogly coverget i H. Problem 6.5 Suppose that H 1 ad H are two differet Hilbert spaces ad A : H 1 H is a bouded liear operator. Show that there is a uique bouded liear operator (the adjoit) A : H H 1 with the property (1.) Au 1, u H = u 1, A u H1 u 1 H 1, u H.

7 76 LECTURE OTES FOR 18.10, SPRIG 009 Solutios to Problem set 5 You should be thikig about usig Lebesgue s domiated covergece at several poits below. Problem 5.1 Let f : R C be a elemet of L 1 (R). Defie { (1.3) f L (x) = f(x) x [ L, L] 0 otherwise. 1 Show that fl L ( R) ad that fl f 0 as L. Solutio. If χl is the characteristic fuctio of [, ] the fl = fχl. If to f the f χ L is absolutely summable, sice f χ L f ad coverges a.e. to f L, so f 1 L L (R). Certaily f L (x) f(x) 0 for each x as L ad f L (x) f(x) f l (x) + f(x) f(x) so by Lebesgue s domiated covergece, f f L 0. f is a absolutely summable series of step fuctios covergig a.e. Problem 5. Cosider a real-valued fuctio f : R R which is locally itegrable i the sese that { (1.4) g L (x) = f(x) x [ L, L] 0 x R \ [ L, L] is Lebesgue itegrable of each L. (1) Show that for each fixed L the fuctio ( ) (1.5) g L (x) = g L (x) if g L (x) [, ] if g L (x) > if g L (x) < is Lebesgue itegrable. () Show that gl g L 0 as. (3) Show that there is a sequece, h, of step fuctios such that (1.6) h (x) f(x) a.e. i R. (4) Defiig 0 x [ L, L] h (x) if h (x) [, ], x [ L, L] (1.7) h,l =. if h (x) >, x [ L, L] if h (x) <, x [ L, L] Show that h,l g L 0 as. Solutio: (1) By defiitio g L = max( χ L, mi(χ L, g L )) where χ L is the characteristic fucito of [L, L], thus it is i L 1 (R). () Clearly g L (x) g L (x) for every x ad g L (x) g L (x) so by Dom iated Covergece, gl g i L 1 L, i.e. gl g L 0 as sice the sequece coverges to 0 poitwise ad is bouded by g(x). (3) Let S L, be a sequece of step fuctios covergig a.e. to g L for example the sequece of partial sums of a absolutely summable series of step fuctios covergig to g L which exists by the assumed itegrability.

8 LECTURE OTES FOR 18.10, SPRIG The replacig S L, by S L, χ L we ca assume that the elemets all vaish outside [, ] but still have covergece a.e. to g L. ow take the sequece { S k, k o [k, k] \ [(k 1), (k 1)], 1 k, (1.8) h (x) = 0 o R \ [, ]. This is certaily a sequece of step fuctios sice it is a fiite sum of step fuctios for each ad o [ L, L] \ [ (L 1), (L 1)] for large itegral L is just S L, L g L. Thus h (x) f(x) outside a coutable uio of sets of measure zero, so also almost everywhere. (4) This is repetitio of the first problem, h,l (x) g L almost everywhere ad h χ L so g 1 (R) ad h,l L L,L g L 0 as. Problem 5.3 Show that L (R) is a Hilbert space sice it is rather cetral to the course I wated you to go through the details carefully! First workig with real fuctios, defie L (R) as the set of fuctios f : R R which are locally itegrable ad such that f is itegrable. (1) For such f choose h ad defie g L, g L ad h by (1.4), (1.5) ad (1.7). () Show usig the sequece h for fixed ad L that g ad (g,l L L ) are i L 1 (R) ad that (h,l ) (g L ) 0 as. (3) Show that ( g ) L ) L 1 (R) ad that (g L (g L ) 0 as. (4) Show that (g L ) f 0 as L. (5) Show that f, g L (R) the fg L 1 (R) ad that (1.9) fg fg f L g L, f L = f. (6) Use these costructios to show that L (R) is a liear space. (7) Coclude that the quotiet space L (R) = L (R)/, where is the space of ull fuctios, is a real Hilbert space. (8) Exted the argumets to the case of complex-valued fuctios. Solutio: (1) Doe. I thik it should have bee h,l. () We already checked that g L L 1 (R) ad the same argumet applies to ), ) (g L amely (h,l g L almost everywhere ad both are bouded by χ L so by domiated covergece (h ) g ) χ L a.e. = g ) 1,L L L L (R) ad (1.30) h ) ),L g L 0 a.e., h,l ) g L ) χ L = h,l ) g L ) 0. (3) ow, as, (g ) (g L ) a.e. ad (g L ) (g L ) so by L f domiated covergece, (g L ) L 1 ad (g L ) (g L ) 0 as. (4) The same argumet of domiated covergece shows ow that g f 1 L ad g L f 0 usig the boud by f L (R).

9 78 LECTURE OTES FOR 18.10, SPRIG 009 (5) What this is all for is to show that fg L 1 (R) if f, F = g L (R) (for easier otatio). Approximate each of them by sequeces of step fuctios as above, h for f ad H,L,L for g. The the product sequece is i L 1 beig a sequece of step fuctios ad (1.31) h ( ) (x)h (x) g (x)g (x),l,l L L almost everywhere ad with absolute value bouded by χ L. Thus by domiated covergece g G L L L 1 (R). ow, let ; this sequece coverges almost everywhere to g L (x)g L (x) ad we have the boud ( ) 1 (f + F ) (1.3) g L (x)g L (x) f(x)f (x) so as always by domiated covergece, the limit g L G L L 1. Fially, lettig L the same argumet shows that ff L 1 (R). Moreover, ff L 1 (R) ad (1.33) ff ff f L F L where the last iequality follows from Cauchy s iequality if you wish, first for the approximatig sequeces ad the takig limits. (6) So if f, g L (R) are real-value, f + g is certaily locally itegrable ad (1.34) (f + g) = f + fg + g L 1 (R) by the discussio above. For costats f L (R) implies cf L (R) is directly true. (7) The argumet is the same as for L 1 versus L 1. amely f = 0 implies that f = 0 almost everywhere which is equivalet to f = 0 a@ė. The the orm is the same for all f + h where h is a ull fuctio sice fh ad h are ull so (f + h) = f + fh + h. The same is true for the ier product so it follows that the quotiet by ull fuctios (1.35) L (R) = L (R)/ is a prehilbert space. However, it remais to show completeess. Suppose {[f ]} is a absolutely summable series i L (R) which meas that f L <. It follows that the cut-off series f χ L is absolutely summable i the L 1 sese sice 1 (1.36) f χ L L ( f 1 ) by Cauchy s iequality. Thus if we set F = f k the F (x)χ L coverges k 1 almost everywhere for each L so i fact (1.37) F (x) f(x) coverges almost everywhere. We wat to show that f L (R) where it follows already that f is locally itegrable by the completeess of L 1. ow cosider the series (1.38) g 1 = F 1, g = F F 1.

10 LECTURE OTES FOR 18.10, SPRIG The elemets are i L 1 (R) ad by Cauchy s iequality for > 1, (1.39) g = F F 1 F F 1 L F + F 1 L f L f k L where the triagle iequality has bee used. Thus i fact the series g is absolutely summable i L 1 (1.40) g ( f L ). So ideed the sequece of partial sums, the F coverge to f L 1 (R). Thus f L (R) ad moroever (1.41) (F f) = F + f F f 0 as. Ideed the first term coverges to f ad, by Cauchys iequality, the series of products f f is absulutely summable i L 1 with limit f so the third term coverges to f. Thus i fact [F ] [f] i L (R) ad we have proved completeess. (8) For the complex case we eed to check liearity, assumig f is locally itegrable ad f L 1 (R). The real part of f is locally itegrable ad the approximatio F L discussed above is square itegrable with (F L ) f so by domiated covergece, lettig first ad the L the real part is i L (R). ow liearity ad completeess follow from the real case. Problem 5.4 Cosider the sequece space (1.4) h,1 = c : j c j C; (1 + j ) c j <. (1) Show that (1.43) h,1 h,1 (c, d) c, d = (1 + j )c j d j is a Hermitia ier form which turs h,1 ito a Hilbert space. () Deotig the orm o this space by,1 ad the orm o l by, show that (1.44) h,1 l, c c,1 c h,1. Solutio: (1) The ier product is well defied sice the series defiig it coverges absolutely by Cauchy s iequality: c, d = (1 + j 1 ) c j (1 + j 1 ) d j, j (1.45) (1 + j ) c j (1 + j 1 ) d j ( (1 + j ) c j ) ( (1 + j ) d j ). j j j j j k

11 80 LECTURE OTES FOR 18.10, SPRIG 009 It is sesquiliear ad positive defiite sice (1.46) c,1 = ( (1 + j ) c j ) (1.51) w i C. () Coclude that {w i } l ad that (1.5) w = w i e i H. j oly vaishes if all c j vaish. Completeess follows as for l if c () is a Cauchy sequece the each compoet c () j coverges, sice (1 + j) 1 c () j is Cauchy. The limits c j defie a elemet of h,1 sice the sequece is bouded ad 1 (1.47) (1 + j ) cj = lim (1 + j ) c () j A j=1 j=1 where A is a boud o the orms. The from the Cauchy coditio c () c i h,1 by passig to the limit as m i c () c (m),1 ɛ. () Clearly h, l sice for ay fiite (1.48) c j (1 + j) c j c j=1 j=1 ad we may pass to the limit as to see that (1.49) c l c, 1. Problem 5.5 I the separable case, prove Riesz Represetatio Theorem directly. Choose a orthoormal basis {e i } of the separable Hilbert space H. Suppose T : H C is a bouded liear fuctioal. Defie a sequece (1.50) w i = T (e i ), i. (1.53) (1) ow, recall that T u C u H for some costat C. Show that for every fiite, (3) Show that j=1 i Solutio: (1) The fiite sum w = w i e i is a elemet of the Hilbert space with orm i=1 w w i = by Bessel s idetity. Expadig out 1,1 T (u) = u, w H u H ad T = w H. i=1 (1.54) T (w ) = T ( w i e i ) = w i T (e i ) = w i i=1 i=1 i=1 ad from the cotiuity of T, (1.55) T (w ) C w = = C H w H C w H w

12 LECTURE OTES FOR 18.10, SPRIG which is the desired iequality. () Lettig it follows that the ifiite sum coverges ad (1.56) w i C = w = w i e i H i sice w w w i teds to zero with. j> (3) For ay u H u = u, e i e i by the completess of the {e i } so from the i=1 cotiuity of T (1.57) T (u) = lim T (u ) = lim u, e i T (e i ) i=1 = lim u, w i e i = lim u, w = u, w i=1 where the cotiuity of the ier product has bee used. From this ad Cauchy s iequality it follows that T = sup u H =1 T (u) w. The coverse follows from the fact that T (w) = w H. i