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.

  • 1
    $\begingroup$ 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 $n\ne m$. $\endgroup$ – DisintegratingByParts Mar 25 '14 at 17:43
  • $\begingroup$ I don't know if maybe he is asking why 'convergent implies weakly convergent'...? $\endgroup$ – Frank Mar 25 '14 at 17:56
  • $\begingroup$ I can't tell precisely what is meant by the question, especially "every sequence converges in a given space." $\endgroup$ – DisintegratingByParts Mar 25 '14 at 18:28

Strong convergence implies weak convergence.

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 $x_n = 1_{[n,n+1]}\in L^2(\Bbb R)$.


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