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In Erwin Kreyszig's "Introduction to Functional Analysis" there is the following statement:enter image description here

It seems to me that if every entry in $x_n=(\xi_j^{(n)})$ converges to $\xi_j$ then $x_n\longrightarrow x$, which is the definition of strong convergence. If this is the case, the statement reduces to

Weak convergent if and only if strong convergent.

Edit: Anna has pointed out that my mistake lies in the statement: "if every entry in $x_n=(\xi_j^{(n)})$ converges to $\xi_j$ then $x_n\longrightarrow x$". Can someone help explain why this is untrue?

Edit2: Finally, the intuition hits. Thank you Anna.

David Raveh
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    Does this answer your question? Prove that weak convergence does not necessarily imply strong convergence without counterexample. The proof by counterexample given there for your "thought that in $l^2$ weak convergence $\nRightarrow$ strong convergence" should help you to realize that your "if every entry in $x_n=(\xi_j^{(n)})$ converges to $\xi_j$ then $x_n\longrightarrow x$" was wrong. – Anne Bauval Jan 01 '23 at 08:40
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    @AnneBauval I have seen the proof in the link before; I am trying to understand why what I said is untrue. I will edit my question. – David Raveh Jan 01 '23 at 08:43
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    What don't you understand in that counterexample? (a) the fact that every entry of $(e_n)$ converges to 0 or (b) the fact that $(e_n)$ does not (strongly) converge to 0? – Anne Bauval Jan 01 '23 at 08:53
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    @AnneBauval I understand that the sequence $(e_n)$ converges weakly to 0 because of Riesz representation and Bessel's inquality. Clearly it doesn't converge strongly. I believe the theorem. I guess I am just unused to thinking about sequences like this. Thank you for your help. – David Raveh Jan 01 '23 at 09:12

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