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Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this imply that my operator T must be bounded?

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    No problem. In the meantime I noticed that the Wikipedia article on weak topology mentions your original notation, but it was something I wasn't used to. http://en.wikipedia.org/wiki/Weak_topology#Weak_convergence2010-10-01

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In my original answer I only mentioned that it works for $Y$ complete, but as Nate pointed out in a comment, I never actually used completeness of $Y$.

The answer is yes. Weakly convergent sequences in a normed space are bounded, as a consequence of the uniform boundedness principle applied to the dual space (which is a Banach space) and the fact that a convergent sequence of real (or complex) numbers is bounded. If $T$ is unbounded, then there is a sequence $x_1,x_2,\ldots$ in $X$ converging in norm (and hence weakly) to 0 such that $\|T(x_n)\|\to\infty$, so by the previous sentence this implies that $T(x_1),T(x_2),\ldots$ does not converge weakly.

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    @Svetoslav: Unboundedness of$T$implies that for each M>0 there exists x with norm 1 such that \|Tx\|>M. Thus for each positive integer n take $y_n$ with norm 1 such that \|Ty_n\|>n^2, and take $x_n=\frac1n y_n$.2016-02-15