If $T: X \to Y$ is norm-norm continuous then it is weak-weak continuous

Question

Let $X,Y$ be normed linear spaces (or Banach spaces if necessary) and let $T: X \to Y$ be linear. We call $T$ norm-norm continuous if $X,Y$ are endowed with the norm topology and similarly, weak-weak continuous if $X,Y$ are endowed with the weak topology.

I am trying to show that if $T$ is norm norm continuous then it is weak-weak continuous. My idea was to use the sequential definition of continuity and to show that if $x_n \to x$ weakly then $Tx_n \to Tx$ weakly. That was easy enough but to complete my proof I would now have to show that this implies that $T$ is continuous and I can't seem to prove it. It would be easy if the topologies were the norm topologies but with both spaces carrying the weak topology I don't see how to proceed.

My question is: Is it true that if $T$ is linear and $x_n \to x$ weakly implies $Tx_n \to Tx$ weakly then $T$ is continuous? If yes, could someone please show me a proof, I can't seem to work it out.

Answer 1

It is of course true that norm-continuous linear maps are weakly continuous. This follows from the fact that the weak topology is the initial topology w.r.t. to all continuous linear functionals, i.e. $\sigma(Y,Y^*)$ is the coarsest topology on $Y$ such that all $f\in Y^*$ are continuous. Then, by abstract nonsense, a map $T:E \to (Y,\sigma(Y,Y^*))$ (where $E$ is an arbitrary topological space) is continuous if (and only if) all compositions $f\circ T$ are continuous. For $E=(X,\sigma(X,X^*))$ you have that continuity because $f\circ T$ is norm-continuous and hence $\sigma(X,X^*)$-continuous.

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You should be very careful with your sequential proof (the weak topologies are not metrizable if the spaces are infinite dimensional). There are Banach spaces where the weakly convergent sequences are always norm-convergent, the space $\ell^1$ of absolutely summable sequences is the most prominent example of such Schur-spaces. This means that the identity $(\ell^1,\sigma(\ell^1,\ell^\infty)) \to (\ell^1,$norm-topology$)$ is sequentially continuous but NOT continuous.

Answer 2

Here is a very simple and elementary answer:

Ps: The completeness of the spaces are not needed at all here. The first implication is obvious.

Assume $T$ is weakly continuous. If $T$ is not norm continuous, then in particular $T$ is not continuous at the $x=0$. Thence there is $r>0$ such that for all $n\geq 1$ there is $x_n$ such that $\|y_n\|\leq \frac{1}{n^2}$ and $\|Ty_n\|\geq r$

Letting $x_n= ny_n$ we find that $\|x_n\|\leq\frac1n\to 0$ and $\|Tx_n\|= n\|Ty_n\|\geq rn\to\infty.$

In particular $x_n\rightharpoonup 0$ weakly and hence the weak continuity of $T$ implies $Tx_n\rightharpoonup 0$ weakly, which in turn implies that $(Tx_n)_n$ is norm bounded. This is clearly a contradiction since $\|Tx_n\|\to\infty.$ Necessarily $T$ is norm bounded.