# Why for the sequence in a Hilbert space to be convergent we need the given sequence to be weakly convergent?

Why do we need a weak convergence for the sequence in a Hilbert space to be convergent? A Hilbert space is a complete pre-Hilbert space, so, every sequence converges in a given space.

Where is my confusion? Many thanks.

• Weak convergence and convergence are not the same for a Hilbert space. For example, if $\{ e_{n}\}_{n=1}^{\infty}$ is an orthonormal set of vectors, then this sequence converges weakly to 0, but it doesn't converge in norm because $\|e_{n}-e_{m}\|=\sqrt{2}$ for $n\ne m$. – DisintegratingByParts Mar 25 '14 at 17:43
• I don't know if maybe he is asking why 'convergent implies weakly convergent'...? – Frank Mar 25 '14 at 17:56
• I can't tell precisely what is meant by the question, especially "every sequence converges in a given space." – DisintegratingByParts Mar 25 '14 at 18:28

Proof: if $x_n\to x$ then
$$\langle x_n, y \rangle - \langle x, y \rangle =\langle x_n-x, y \rangle\\ |\langle x_n-x, y \rangle|\le |x_n-x||y|\to 0$$
Anyway, the other implication if false: consider $x_n = 1_{[n,n+1]}\in L^2(\Bbb R)$.