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Let $X,Y$ be normed spaces. A bounded operator $T: X \to Y$ is arbitrary. $B(X,Y)$ denotes the set of bounded operators. How to prove that if ${x_n}$ converges weakly to $x_0$, then for $T \in B(X,Y)$, $T(x_n)$ converge weakly to $T(x_0)$?

I know it has to be true in the case that weak convergence implies strong convergence. Also, when $T$ is not only bounded but also linear, Riesz representation theorem comes in and the proof becomes trivial. However, if $T$ is only bounded, I have no clue.

This is an exercise from the book. I seem to find counter a example of this though. Can anyone at least show me an example of weak convergence where the limit is not 0?

Johnny Chen
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Assume $x_n \xrightarrow{w} x_0$. For any $f \in Y^*$ we have that $f \circ T \in X^*$ so

$$f(Tx_n) = (f \circ T)(x_n) \xrightarrow{n\to\infty} (f \circ T)(x_0) = f(Tx_0)$$

Therefore $Tx_n \xrightarrow{w} Tx_0$.

mechanodroid
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  • Thank you first. May I ask why f($Tx_n$) $\in$ $X^*$? If T is linear, then I understand why. However, if T is non-linear, and only bounded, it can not be in the dual of X right? I mean, what will happen if T maps the whole space X into a ball of radius R. – Johnny Chen May 25 '18 at 21:41
  • @JohnnyChen The set $B(X,Y)$ is usually defined as all bounded linear maps $X \to Y$, and the term bounded operator usually refers to bounded linear maps. Boundedness for general functions is defined differently than for linear maps. – mechanodroid May 25 '18 at 21:45
  • Thank you. If that is the case, I have no more confusion. – Johnny Chen May 25 '18 at 22:48