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In the proof of the statement that weak convergence is equivalent to strong convergence in finite dimensional normed vector space, we need to use the fact that in finite dimensional space convergence in coordinates is equivalent to convergence in norm.

Since in infinite dimensional normed vector space weak convergence is generally different from strong convergence, I guess in infinite dimensional case convergence in coordinates is NOT equivalent to convergence in norm. But I have no example. Could someone give me an example of “convergence in coordinates is not equivalent to convergence in norm”?

Thanks so much.

Update: I think the infinite dimensional space must be separable, for otherwise it does not make sense to talk about “convergence in coordinates”

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  • $\begingroup$ Related $\endgroup$ – Giuseppe Negro Jun 24 '18 at 9:00
  • $\begingroup$ @GiuseppeNegro Thank you Negro. $\endgroup$ – Sam Wong Jun 24 '18 at 9:11
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Consider $\ell^2:=\{(x_1,x_2,\dots):\sum_{n\geq 0}x_n^2<\infty\}$ with norm $\Vert(x_1,x_2,\dots)\Vert=\sqrt{\sum_{n\geq 0}x_n^2}$. Now consider two sequences $(x_n)$ and $(y_n^{(k)})$ where $x_n=0$ for all $n$ and $y_n^{(k)}=\mathbb{1}[n=k]$. Notice that we have $(x_n),(y_n^{(k)})\in\ell^2$ for all $k$. Furthermore, $y_n^{(k)}\to x_n$ as $k\to\infty$ since $y_n^{(k)}=0$ whenever $k>n$. However, for every $k$, $\Vert (x_n)-(y_n^{(k)})\Vert=1$, so $(y_n^{(k)})$ does not converge to $(x_n)$ in the norm.

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  • $\begingroup$ Good example. Thank you Munchhausen. $\endgroup$ – Sam Wong Jun 24 '18 at 9:10
  • $\begingroup$ What is $y_n^{(k)}$ for $n\neq k$? $\endgroup$ – Sigur Aug 13 '19 at 17:53
  • $\begingroup$ The notation $\mathbf{1}[n=k]$ is the indicator of the event "$n=k$". So $y_n^{(k)}$ is $1$ when $n=k$ and $0$ otherwise $\endgroup$ – munchhausen Aug 14 '19 at 18:12

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