# Banach-Alaoglu Theorem to Prove Linear Operator is Bounded

I'm attempting to answer the following:

Let X be a reflexive Banach space, and Y be a normed space. Suppose T : X → Y is a linear operator with the property that if a sequence $x_n$ converges weakly to 0 in X, then $T(x_n)$ converges weakly to 0 in Y . Prove that T is bounded.

I feel like I can use the Banach-Alaoglu Theorem to get there, namely that it guarantees that every bounded sequence in X has a weakly convergent subsequence, but I'm not sure how to proceed.

Any help is much appreciated!

• I think X must be a reflexive space – Guy Fsone Dec 11 '17 at 19:33

By Contradiction: If T is not bounded then for all n there eixst $x_n\in X$ with $\|x_n\|=1$ such that $$\|T(x_n)\|\ge n$$
the sequence $x_n$ is bounded then by Banach-Alaoglu theorem, we can extract a sub-sequence $x_{n_j}$ which converges weakly in X. And by assumption $(Tx_{n_j})$ converges weakly in $Y$ and Hence $(T(x_{n_j}))_j$ is a bounded sequence in $Y$. This goes in contradiction with the fact that $$\|T(x_{n_j})\|\ge n_j$$
• it is necesary if you take $x_n-x\to 0$ instead of $x_n\to x$ – Guy Fsone Dec 11 '17 at 21:57
• OK, so then by Banach-Alaoglu $x_{n_k} \rightarrow x$ weakly, so taking $y_n = x_{n_k} - x$, we have that $y_n \rightarrow 0$ weakly? – OGBerglemir Dec 11 '17 at 22:03
Assume $T$ is not bounded, then there is a sequence $(x_n)$ such that $\|x_n\|=1$ and $Tx_n\to\infty$. Then we can extract a subsequence (denoted the same) such that $x_n \rightharpoonup x$, and by the properties of $T$, $Tx_n\rightharpoonup Tx$. Weakly converging sequences are bounded, contradiction.