## sequence in Hilbert space,example

Hi. Im working on some exercises but I couldt find any clue for this one:

Find a bounded sequence (as like the norm) in l^2 Hilbert space,that weakly converges to 0 (as like the weak topology)
but doesn`t have any convergent subsequences (as in strong topology).
Could someone help me?

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 Recognitions: Science Advisor I think the unit cube in Hilbert space l^2:{ei:=δij}, i.e., ei has a 1 in the i-th position and is 0 otherwise, is an example: for one thing, it does not have a finite ε-net for ε <1/2, say, since the l2-distance is √2, so that it cannot have convergent subsequences -- the √2 distance is a barrier to being Cauchy.
 Recognitions: Science Advisor I forgot the second part, about the sequence converging weakly: first , in l2: We need to show: Limn→∞ =<0,an>=0 But notice that the product on the left equals the n-th term of a sequence an in l2. Since an is square-summable, it is Cauchy, so that its n-th term goes to zero as n→ ∞. But this sequence {en}=(δij)j=1,...,∞ converges weakly to zero in _any_ Hilbert space: Let h be an element in any Hilbert space H. Then, since {en} is a maximal orthonormal set, it is a Hamel basis for H . Then h is a finite linear combination of the basis elements in {en}. Let ek be the largest index in the linear combination. Then, when n>k , < en,h>=0. h=Ʃ

 Tags hilbert space, weak convergence