if $x_n \rightharpoonup x$ in $X$, then $Tx_n \rightharpoonup Tx$ in $Y$ , for $T \in B(X, Y )$

Question

Let $X, Y$ be Banach spaces and $T \in B(X, Y )$ be linear operator. Show that if $x_n \rightharpoonup x$ in $X$, then $Tx_n \rightharpoonup Tx$ in $Y$ .

My attempt

since $T \in B(X, Y )$, $\|T\| < \infty$ , given that $x_n \rightharpoonup x$, choose $N$ such that $\|\langle x_n,z\rangle-\langle x,z\rangle\| \le \frac{\epsilon}{\|T\|} , \ \forall n>N , \forall z\in X. \ \ \ $ Then;

\begin{align} \|\langle Tx_n,z\rangle-\langle Tx,z\rangle\| & = \|\langle T(x_n-x),z\rangle\|\\ & \le \|T\|\|\langle (x_n-x),z\rangle\| \\ & = \|T\|\|\langle x_n,z\rangle-\langle x,z\rangle\| \\ & \le \|T\|\frac{\epsilon}{\|T\|} = \epsilon , \ \ \forall z \in X \end{align}

Answer 1

There are several problems with what you wrote:

• You seem to be treating $X$ as a Hilbert space, and assuming that $Y=X$. In a general Banach space there is no inner product nor Riesz Representation Theorem.

• You cannot choose $N$ uniformly on $z$, as you wrote.

• You write norms when applied to numbers, which is not necessarily wrong (the absolute value in $\mathbb C$ is actually a norm) but it's weird.

• Your first estimate makes no sense, even in a Hilbert space.

How to do the exercise:

In a Banach space, that $x_n\to x$ weakly means that (the numbers) $f(x_n)\to f(x)$ for all $f\in X^*$. The exercise is then trivial, since given $T\in B(X,Y)$ and $g\in Y^*$, then $g\circ T\in X^*$. Then $g(Tx_n)\to g(Tx_n)$ by hypothesis.