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I have the forward direction: $(\Longrightarrow)$ Let $T\in\mathcal{L}(X,X)$ and let $f\in X^*$. Since both $T$ and $f$ are bounded, then both $T$ and $f$ are continuous in the norm topology. Then $f\circ T$ is continuous with respect to the norm topology, and therefore bounded. Thus $f\circ T\in X^*$. Since $x_n\rightharpoonup x$, then $(f\circ T)(x_n)\to (f\circ T)(x)\Longleftrightarrow f(T(x_n))\to f(T(x))$. Since our choice of $f$ was arbitrary, we conclude that $T(x_n)\rightharpoonup T(x)$.

The reverse direction is going to require the use of reflexivity, but I have no idea where to begin with this. Any help would be appreciated.

Laars Helenius
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Suppose $T$ were an unbounded operator; then there is a norm-bounded sequence $\{x_n\}$ in $X$ such that $\{T(x_n)\}$ is not norm-bounded; we can actually choose the sequence in such a manner that $\|T(x_n)\|$ is strictly increasing.

Now in a reflexive Banach space, every norm-bounded sequence has a weakly-convergent subsequence, so we can extract a subsequence $\{x_{n_k}\}$ with weak limit $x$, so that $T(x_{n_k}) \rightharpoonup T(x)$. But every weakly-convergent sequence has to be norm-bounded, which is a contradiction.

  • Couldn't we just cut to the chase and say since $T(x_n)\rightharpoonup T(x)$ in a reflexive space, then we have to have $T$ bounded? – Laars Helenius Mar 31 '14 at 03:30