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Let $l^{2}=\left\{x=(x^{(1)},x^{(2)},...):\sum_{i=1}^{\infty }\left\vert x ^{(i)}\right\vert ^{2}<\infty \right\} $. Would you help me to prove that $({\vert|x_n |\vert})$ is bounded sequence and $(x_n)$ converge componentwise ($x_n^{(i)}\rightarrow x^{(i)}$ for all $i$) if and only if $\langle x_n,y\rangle\rightarrow \langle x,y\rangle$ for all $y\in l^2$

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For one direction, put $y = e_k$, $e_k$ being the standard basis vector with 1 in the $k$th position and zero everywhere else. – Jas Ter Oct 15 '12 at 14:05
up vote 1 down vote accepted

Assume that $\{\lVert x_n\rVert\}$ is bounded (say by $M$) and we have componentwise convergence. Fix $y\in \ell^2$. We have $$|\langle x_n-x,y\rangle|\leq \sum_{j=1}^N|x_n^{(i)}-x^{(i)}||y^{(i)}|+(M+\lVert x\rVert)\sqrt{\sum_{j\geq N+1}|y^{(j)}|^2}.$$ We get $$\limsup_{n\to+\infty}|\langle x_n-x,y\rangle|\leq (M+\lVert x\rVert)\sqrt{\sum_{j\geq N+1}|y^{(j)}|^2}$$ for each $N$, which gives what we want as $y\in\ell^2$.

Conversely, to see componentwise convergence take $y=e^{(i)}$ where $e^{(i)}_j=1$ if $i=j$ and $0$ otherwise. To see boundedness, we apply Baire categories theorem to $F_k:=\bigcap_{n\geq 1}\{y\in\ell^2,|\langle x_n-x,y\rangle|\leq k\}$, which are closed set in a complete metric space.

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Elementary proof without Baire bla2 ? – beginner Oct 15 '12 at 14:39

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