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.

  • $\begingroup$ Yes. See this. $\endgroup$ Commented Aug 17, 2014 at 10:59
  • $\begingroup$ @DavidMitra This is very helpful to me, thank you. $\endgroup$
    – user167889
    Commented Aug 17, 2014 at 11:09
  • 1
    $\begingroup$ My previous comment addressed the question in the post body. I assumed (probably incorrectly) that you were referring to norm-continuity. For the question in your title, you could show that if a net $(x_\alpha)$ converges weakly to $x$, then $(Tx_\alpha)$ converges weakly to $Tx$. Towards this end, note $y^*\circ T\in X^*$ if $y^*\in Y^*$ (I presume you already did this for sequences; the proof for nets is the same). $\endgroup$ Commented Aug 17, 2014 at 11:14

2 Answers 2


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.

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.

  • $\begingroup$ You used a characterisation of initial topologies for the proof of the implication " norm-norm $\implies$ weak-weak ". Is this because for weak topologies on Banach spaces there is no closed-graph theorem? I asked a similar question (still no answer), trying to prove facts of this type for a more general setting (locally convex spaces): math.stackexchange.com/questions/2131594 $\endgroup$
    – el_tenedor
    Commented Feb 9, 2017 at 16:25
  • $\begingroup$ This post answers the same question, using the closed graph theorem (for Banach spaces) only for the counterpart of the implication that you showed: math.stackexchange.com/questions/337103 $\endgroup$
    – el_tenedor
    Commented Feb 9, 2017 at 16:27

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.


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