Claim : a linear function $T$ between Banach spaces is weakly continuous iff norm continuous?

Okay, So I think I have realised weakly continuous implies norm continuous. As weakly continuous implies weakly sequentially continuous. Now suppose that $T$ is unbounded. But we also know that 'weakly convergent implies weakly bounded', which implies norm bounded. But this would then imply that '$T(x_{n})$ converges weakly implies that $\|T(x_{n})\|$ is bounded'.

Hence $T(x_{n})$ is not weakly convergent, so $T$ cannot be weakly continuous. Contradiction! Hence T is bounded.

Any ideas on the converse? I.e How do I show norm continuous is weakly continuous?

  • $\begingroup$ See math.stackexchange.com/q/301745/49437 $\endgroup$ – Martin Mar 21 '13 at 17:57
  • $\begingroup$ @Martin Thank you for your link. But it isn't sparking any ideas! $\endgroup$ – user58514 Mar 21 '13 at 18:05
  • $\begingroup$ By definition norm-continuous functionals are weakly continuous. I show the converse: if a functional is weakly continuous then it is a norm-continuous linear functional (take $F = X^\ast$, $w_F$ is the weak topology and $s$ is the norm topology in the notation of that answer). $\endgroup$ – Martin Mar 21 '13 at 18:34
  • $\begingroup$ Why does it follow from the definition that norm continuous implies weakly continuous? Take any weakly open set, this is norm open, so the pre-image is norm open, but not necessarily weakly open? $\endgroup$ – user58514 Mar 21 '13 at 18:38
  • 1
    $\begingroup$ Yes, norm-open sets are not necessarily weakly open (e.g. the unit ball), but: By definition the weak topology is the weakest (coarsest) topology for which all norm-continuous functionals are continuous. $\endgroup$ – Martin Mar 21 '13 at 18:58

For definiteness consider the linear operator $T:X\to Y$. If $T$ is norm continuous, then for each $f \in Y^*$ we have $f \circ T \in X^*$. Thus $f\circ T$ is continuous w.r.t. the strong topology on $X$ and hence also w.r.t. the weak topology on $X$. But if $f \circ T$ is continuous from $X$ with the weak topology to $\mathbb{F}$ for every $f \in Y^*$ then $T$ is weak-weak continuous from $X$ to $Y$ since the weak topology on $Y$ is the induced topology of the continuous linear functionals on $Y$.

The other direction follows since if $T$ is weak-weak continuous then the graph of $T$ is a weakly closed convex set in $X \times X$ and hence a strongly closed set. The Closed Graph Theorem implies that $T$ is norm continuous.


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