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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?

el_tenedor
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  • See http://math.stackexchange.com/q/301745/49437 – Martin Mar 21 '13 at 17:57
  • @Martin Thank you for your link. But it isn't sparking any ideas! –  Mar 21 '13 at 18:05
  • 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). – Martin Mar 21 '13 at 18:34
  • 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? –  Mar 21 '13 at 18:38
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    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. – Martin Mar 21 '13 at 18:58
  • Sorry, I made a typo in my question. I am wanting to consider a linear function/map between two arbitrary Banach spaces. (I.e the map is not a functional.) Then it does not directly follow from the definition does it? –  Mar 21 '13 at 19:13
  • @Martin Write it as an answer? – Julien Mar 21 '13 at 23:05

2 Answers2

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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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$T:(X,\lVert \cdot \rVert)\rightarrow (Y,\lVert \cdot \rVert)$ is continuous if and only if $ T: (X,\sigma(X,X^*))\rightarrow (Y,\sigma(Y,Y^*)) $ is continuous.

$(\Rightarrow) $ $ T: (X,\sigma(X,X^*))\rightarrow (Y,\sigma(Y,Y^*)) $ will be continuous if and only if $y^*\circ T:(X,\sigma(X,X^*))\rightarrow \mathbb{R} $ is continuous for every $y^*\in Y^*$. However, because $T$ is continuous, $y^*\circ T\in X^*$ and this direction follows.

$(\Leftarrow)$ This is considerably harder. Let us prove some lemmas first.

Lemma 1: There are no extra continuous linear functionals in the weak topology. Precisely: $$(X,\sigma (X,X^*))^*=X^*$$

Proof: $X^*\subseteq (X,\sigma (X,X^*))^*$ by definition of the weak topology. Now take $g\in (X,\sigma (X,X^*))^*$. If $g\equiv 0$, we are done. Suppose there is $x_o\in (\ker g)^C$. Because $\ker g$ is closed in the weak topology we have:

$$x_o\in \cap_{i=1}^n \{x| \:|f_i(x)-f_i(x_o)|<\varepsilon\}\subseteq (\ker g)^C$$

If it doesn't hold that $\cap_{i=1}^n \ker f_i \subseteq \ker g$ there is a certain $x$ satisfying $f_i(x)=0$ for every $i$ and $g(x) \not=0$. But this means that $x_o+\lambda x\in (\ker g)^C$ for every $\lambda$, which is absurd, say $\lambda=-g(x_o)/g(x)$. Hence $\cap_{i=1}^n \ker f_i \subseteq \ker g$ and by this result, we have, $g=\sum_i \lambda_i f_i\in X^*$ and we are done.

Lemma 2: weak convergence implies norm boundedness.

Proof: This has been done in other MSE posts. Brezis also does this as a consequence of the uniform boundedness principle.

Finally let us prove our result. Because $y^*\circ T$ is continuos with respect to $\sigma(X,X^*)$, by Lemma 1, we have $y^*\circ T \in X^*$. Suppose by way of contradiction that there are $x_n$ in the circle with $\lVert T(x_n)\rVert \geq n$. Clearly:

$$\frac{x_n}{\sqrt{\lVert T(x_n) \rVert}}\rightarrow 0$$

But because $y^*\circ T\in X^* $, this implies:

$$y^*\left(\frac{T(x_n)}{\sqrt{\lVert T(x_n) \rVert}}\right)\rightarrow 0\quad \text{but $y\in Y^*$ is arbitrary, so:}\quad \frac{T(x_n)}{\sqrt{\lVert T(x_n)\rVert}}\rightharpoonup 0$$

But by Lemma 2 this means:

$$\sqrt{n}\leq \left\lVert \frac{T(x_n)}{\sqrt{\lVert T(x_n)\rVert}}\right \rVert \leq C$$

This contradiction implies $T$ must be bounded.

Kadmos
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