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If $X$ and $Y$ are banach spaces and T map from $X$ to $Y$. If every sequence $x_n$ in weak topology in $X$ converges to $0$ , and $T(x_n)$ converges to $0$ in weak topology , does that imply that $T$ is bounded ?

From this question i would like to understand how convergence in weak topology differs from other topology. What would be the case if the topology defined was not weak ?

Thank you for your kind help . Keenly Looking forward to get some good ideas about weak topologies .

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    Do you mean to require that for *every* sequence $x_n$ which converges weakly to 0, that $T(x_n)$ converges weakly to 0?2012-11-20
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    @NateEldredge : Yes Sir .2012-11-20
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    The fact that $T$ is bounded follows from closed graph theorem. Let $\{x_n\}$ a sequence which converges strongly to $0$, and $\lVert Tx_n-y\rVert\to 0$. Let $f\in Y^*$. Then we have that $\{Tx_n\}$ converges weakly to $0$. As $\{Tx_n\}$ converges strongly hence weakly to $y$, $y=0$. Now let $x_n\to x$, $Tx_n\to y$. Then $x_n-x\to 0$ and $T(x_n-x)\to y-Tx$ so $y=Tx$.2012-11-20
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    Now I guess the question is, for example when $X=Y$, what are the topologies $\cal T$ on $X$ such that if $T\colon (X,\cal T)\to (X,\cal T)$ is continuous, then $T$ is strongly continuous (and conversely)?2012-11-20
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    @DavideGiraudo : Sir, thats exactly what i want to know. And also how to connect the continuity and other properties of norm with the weak topologies . I am kind of mixing up everything and not able to see clearly how to think connect norm and weak topology . I hope i am asking a sensible question :P2012-11-21
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    @DavideGiraudo : Can you tell me why u said $f\in Y^*$2012-11-22
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    I don't know myself, maybe a copy-paste error.2012-11-22

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