FUNCTIONAL ANALYSIS LECTURE NOTES: WEAK AND WEAK* …

1 FUNCTIONAL ANALYSIS LECTURE NOTES: WEAK AND WEAK* CONVERGENCECHRISTOPHER and Weak* convergence of VectorsDefinition a normed linear space, and letxn,x We say thatxnconverges,converges strongly, orconverges in normtox, and writexn x, iflimn kx xnk= We say thatxnconverges weaklytox, and writexnw x, if X ,limn hxn, i=hx, Show that strong convergence implies weak Show that weak convergence does not imply strong convergence in general (look for aHilbert space counterexample).If our space is itself the dual space of another space, then there is an additional mode ofconvergence that we can consider, as a normed linear space, and suppose that n, X.

2 Then wesay that nconverges WEAK* to , and write nw* , if x X,limn hx, ni=hx, that weak* convergence is just pointwise convergence of the operators n!Remark * convergence only makes sense for a sequence that lies in a dualspaceX . However, if we do have a sequence{ n}n NinX , then we can consider threetypes of convergence of nto : strong, weak, and weak*. By definition, these are: n limn k nk= 0, nw T X ,limn h n,Ti=h ,Ti, nw* x X,limn hx, ni=hx, 2007 by Christopher AND WEAK* CONVERGENCEE xercise n, X , show that n = nw = nw*.

3 ( )IfXis reflexive, show that nw nw* .In general, however, the implications in ( ) do not hold inthe reverse Weak* limits are Weak limits are thatXis a normed linear space, and that we had both nw* and nw* inX . Then, by definition, x X,hx, i= limn hx, ni=hx, i,so = .b. Suppose that we have bothxnw xandxnw yinX. Then, by definition, X ,hx, i= limn hxn, i=hy, , by Hahn Banach,kx yk= supk k=1|hx y, i|= 0,sox=y. It is trivial to show that strongly convergent sequences arebounded. However, we needsome fairly sophisticated machinery (the Uniform Boundedness Principle) to show thatweakly convergent and weak* convergent sequences are likewise Show that weak* convergent sequences in the dual of a Banach space an example of an unbounded but weak* convergence sequence in the dual of anincomplete normed.

4 The dual space ofc00under the norm is (c00) = Show that weakly convergent sequences in a normed space are , we will show that strong convergence is equivalent to weak convergence in finite-dimensional a finite-dimensional vector space, then strong convergence is equivalentto weak AND WEAK* first the case thatX=Fdunder the Euclidean normk k2. Suppose thatxnw xinFd. Then for each standard basis vectorek, we havexn ek x ek, k= 1,.., is, weak convergence implies componentwise convergence . But since there are onlyfinitely many components, this implies norm convergence , sincekx xnk22=d k=1|x ek xn ek|2 0 asn.

5 For the general case, choose any basisB={e1,..,ed}forX, and use the fact that allnorms onXare equivalent to define an isomorphism betweenXandFd. Often, there exists a connection between componentwise or pointwise convergence andweak convergence . This is related to the question of whether point evaluation are contin-uous linear functionals on a given 1 p , and consider the space p. As usual, givenx pandy p , writehx,yi= k=1xk standard basis vectorsekbelong to every p, and henceek p ( p) (with equalityifp < ).

6 Therefore, if we havexn= (xn(k))k Nandy= (y(k))k Nand we know thatxnw y, then we have for eachk Nthatxn(k) =hxn,eki hy,eki=y(k).Thus,weak convergence in p= componentwise convergence in converse is not true in general, but the following resultgives necessary and sufficientconditions, at least for 1< p < , and letxn,y pbe given. Prove that the following twostatements are (k) y(k) for eachk(componentwise convergence ) and supkxnkp< .What happens ifp= 1 orp= ?Exercise thatxn,y 1are given. Since 1 =c 0, we can consider WEAK* convergence ofxntoy.

7 Prove that the following two statements are * AND WEAK* (k) y(k) for eachk(componentwise convergence ) and supkxnk1< .Exercise ,f C0(R) be given. SinceC0(R) =Mb(R), we can consider weakconvergence offntof. Prove that the following two statements are (x) f(x) pointwise for eachx, and supkfnk < .Exercise n, Mb(R) be given. Show that nw* does not implyk nk k 1< p < , and letfn Lp(R) be given. Prove that the following twostatements are Efn 0 for everyE Rwith|E|< , and supkfnkp< . a. Suppose that statement b holds, and letR= supkfnkp.

8 Choose anyg Lp (R). Since the step functions are dense inLp (R), we can find a function of the norm =M k=1ck Fk,with eachFka measurable subset ofR, such thatkg kp < Lp (R), there exists a compact setKsuch that if we set = K, then we havek kp < , note that is a step function, since =M k=1ck Fk K=M k=1ck Ek,whereEk=Fk k=1|ck|,and assume for now thatC >0. Since eachEkhas finite measure, by hypothesis we canfind an integerNsuch thatn > N= Ekfn < AND WEAK* CONVERGENCE5 Hence forn > Nwe have|hg,fni| |hg ,fni|+|h ,fni|+|h ,fni|<kg kkfnk|+k kkfnk+ M k=1ck Ekfn < 4RR+ 4RR+M k=1|ck| Ekfn < 4+ 4+M k=1|ck| 2C=.

9 IfC= 0 then we still obtain|hg,fni|< . This shows that0 lim supn |hg,fni| .Since this is true for every , we conclude thatlimn |hg,fni|= since this is true for everyg Lp (R), we havefnw 0. Exercise a Hilbert space. Show thatfn f fnw fandkfnk Hilbert spaces, and letT B(H,K) be a compact thatfnw f= Tfn , a compact operator maps weakly convergent sequences to strongly convergent AND WEAK* of OperatorsWe can apply similar notions to convergence of ,Ybe normed linear spaces, and letAn,A B(X,Y) be We say thatAnconverges in operator norm toA, or thatAnis uniformly operatorconvergent toA, and writeAn A, iflimn kA Ank= the definition of operator norm, this is equivalent tolimn (supkxk=1kAx Anxk)

10 = We say thatAnconverges in the strong operator topology (SOT) toA, or thatAnisstrongly operator convergent toA, if x X, Anx Ax(strong convergence inY).Equivalently, this holds if x X,limn kAx Anxk= We say thatAnis weakly operator convergent toA, if x X, Anxw Ax(weak convergence inY).Equivalently, this holds if x X, Y ,limn hAnx, i=hAnx, particular, consider the caseY=F, , the operatorsAnare boundedlinear functionals onX. SinceY=Y , strong and weak convergence inYare equiva-lent. Hence for this case, strong operator convergence and weak operator convergence areequivalent, and in fact, they are simply weak* convergence of the operatorsAninX.