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?

  • $\begingroup$ 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. $\endgroup$ Jun 30, 2022 at 16:56
  • $\begingroup$ 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 $\endgroup$
    – Rob Tan
    Jul 1, 2022 at 8:34
  • $\begingroup$ 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? $\endgroup$
    – Rob Tan
    Jul 1, 2022 at 8:37
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    $\begingroup$ 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. $\endgroup$
    – daw
    Jul 1, 2022 at 20:57
  • $\begingroup$ @daw Exactly I was just thinking to that! $\endgroup$
    – Rob Tan
    Jul 2, 2022 at 9:47


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