# Is every weak sequence in a separable Hilbert space, a strong sequence?

I'm concluding something that is not convincing me.

Imagine a sequence of elements $$x_k$$ in an Hilbert space $$\mathcal{H}$$, then if it is weakly convergent to $$x$$ that means $$\lim_{k\to +\infty} \langle x_k - x, y\rangle = 0 \,\forall y\in\mathcal{H}$$ In particular this holds for all the elements $$y=e_j\in\mathcal{H}$$ where $$\{e_j\}$$ constitutes a complete orthonormal sequence in a separable Hilbert space, so we have $$\lim_{k\to +\infty} \langle x_k - x, e_j\rangle = \left\langle \lim_{k\to +\infty} x_k - x, e_j\right\rangle = 0 \,\forall j$$ but Hilbert space is indeed separable so we have $$\lim_{k\to +\infty} x_k - x = \sum\limits_j \left\langle \lim_{k\to +\infty} x_k - x, e_j\right\rangle e_j = 0$$ (see for example "Hilbert spaces with applications - Debnath, Mikusinski" third edition pag. 115 theorem 3.4.14) and this means $$\left\lVert \lim_{k\to +\infty} x_k - x\right\rVert = \lim_{k\to +\infty} \lVert x_k - x\rVert = 0$$ so the sequence $$\{x_k\}$$ is strongly convergent to its weak limit. What am I doing wrong?

• You're using the expression $\lim_{k\to\infty}x_k-x$ as though it had a meaning, but your assumptions don't guarantee that. Jun 30, 2022 at 16:56
• Ok, so in other words I can define by force a quantity like $\lim_{k\to +\infty}(x_k-x)$ on itself, only if it is strongly convergent? Norm $\lVert\cdot\rVert$, but I can call it $A$, is a continuos application and that means $\lim_{k\to +\infty} A(X_k)=A(\lim_{k\to +\infty} X_k)$ where $\lim_{k\to +\infty}X_k$ is determined, by definition of continuity, by strong convergence. So in this case $X_k= x_k - x$ and defining $\lim_{k\to +\infty} x_k - x\doteq 0$ we get by continuity of the norm $\lim_{k\to +\infty}\lVert x_k-x\Vert=\lVert\lim_{k\to +\infty}x_k-x\rVert=0$, so the definition works Jul 1, 2022 at 8:34
• But scalar product is a continuos application with the induced norm, so if I fix $y$ and I call $\langle x, y\rangle\doteq A x$ I can say $A$ is a continuos application and I can "enter the limit" defined above, only if the sequence converges strongly first. Is my interpretation right? Am I brute-forcing too much, or becoming tautological? Jul 1, 2022 at 8:37
• You cannot pull the limit in the inner product. Because the limit of $x_k$ might not exists. Test your argument for the standard non-strongly but weakly converging sequence: $(e_k)$ an orthonormal sequence.
– daw
Jul 1, 2022 at 20:57
• @daw Exactly I was just thinking to that! Jul 2, 2022 at 9:47