# Weakly continuous function implies some sort of triviality

So, I asked a similar question at math stackexchange and was directed here. Please let me know if this question is better suited elsewhere.

Let $$U$$ and $$V$$ be (infinite-dimensional) Banach spaces. Assume we have a sequence $$(u_n)$$ in $$U$$ converging weakly to $$u_0$$ and a nonlinear function $$F:U \to V$$, where we can assume for instance that $$\|F(u)\|_V \leq C \|u\|_U$$, it is not really important for the question.

Moving on to the question, in $$L^p(0,1)$$, weak continuity of a function $$\psi: \mathbb{R} \to \mathbb{R}$$ implies that that $$\psi$$ is affine (see 2.10 in these lecture notes )

I am wondering if there are similar results for other Banach (or Hilbert spaces), that weak continuity implies some sort of triviality? I suspect there is no general theorem, but I am particularly concerned with spaces such as $$L^2([a,b],V)$$ where $$V$$ is a reflexive, separable Banach space (perhaps even finite-dimensional), or $$C^1([a,b],V)$$, $$W^{k,p}([a,b],V)$$. In the first case my guess is that it does carry over and weakly continuous functions are affine, in the other cases, I am not so sure.

• It seems that you're misinterpreting the reference you quote: the result there is not for general mappings on $L^p$, but only for mappings that are given by composition with a function $\mathbb{R} \to \mathbb{R}$. Commented Jun 27, 2021 at 21:04
• (It is very easy to give examples of non-affine but weakly continuous mappings on $L^p$.) Commented Jun 27, 2021 at 21:06
• Ohhh, so in general, saying something like "let F be nonlinear weakly continuous map from U to V" will be completly fine and not equivalent to "let F be trivial"?
– ejk
Commented Jun 27, 2021 at 22:27
• Yes, indeed. (But of course, when saying something like that, one should typically have some actual examples of such a map in mind.) Commented Jun 28, 2021 at 5:20
• Thank you for your answer. Is it possible to write down an example from for instance L^2((a,b),U) to L^2((a,b),V)? I struggle a bit to find good examples... I suspect that in general, even if I have a weakly continuous mapping from U to V, it won't help me?
– ejk
Commented Jun 28, 2021 at 14:02

From the question and from the OP's related question on Mathematics StackExchange, I infer that the OP is in general interested in the weak continuity of nonlinear mappings. So here are two general facts the seem to be relevant:

Let $$X$$, $$Y$$ be Banach spaces (over the same field). Assume that $$X$$ is infinite-dimensional and that $$Y$$ is non-zero. Then:

(a) There exists a norm-continuous nonlinear mapping $$F: X \to Y$$ that is not weakly continuous.

(b) There exists a non-linear mapping $$G: X \to Y$$ that is weakly continuous.

Proof. (a) Choose a non-zero vector $$y \in Y$$ and set $$F(x) = \|x\| \cdot y$$ for each $$x \in X$$. Obviously, $$F$$ is norm-continuous. But if $$F$$ were weakly continuous, then the preimage of the set $$\{y\}$$ under $$F$$ - which is the unit sphere in $$X$$ - would be weakly closed. However, the unit sphere an an infinite-dimensional Banach space is never weakly closed.

(b) Let $$x' \in X'$$ be a non-zero bounded linear functional on $$X$$ and fix a non-zero vector $$y \in Y$$. We set $$G(x) = \langle x', x \rangle^2 \cdot y$$ for each $$x \in X$$. Then $$G$$ is not linear, but it is weakly continuous since it is the composition of the weakly continuous mappings $$X \overset{x'}{\longrightarrow} \mathbb{F} \; \overset{s \mapsto s^2}{\longrightarrow} \; \mathbb{F} \; \overset{t \mapsto t \cdot y}{\longrightarrow} \; Y$$ (where $$\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$$ denotes the underlying scalar field of $$X$$ and $$Y$$). $$\square$$

Remarks.

(1) The mapping $$F$$ from (a) is not even sequentially weakly continuous. This follows from the fact that, in an infinite-dimensional Banach space $$X$$, there always exists a sequence $$(x_n)$$ in the unit sphere that converges weakly to $$0$$.

(2) It is maybe worthwhile to recall that, for a linear mapping $$T$$ between two Banach spaces $$X$$ and $$Y$$, the following are equivalent:

(i) $$T$$ is norm continuous (i.e., bounded).

(ii) $$T$$ is weakly continuous.

(iii) $$T$$ is sequentially weakly continuous.

(Assertion (i) implies (ii) due to the existence of the dual operator $$T'$$, (ii) obviously implies (iii), and (iii) implies (i) due to the closed graph theorem.)

Disclaimer. Actually, I think that this would be a better fit for Mathematics StackExchange, but since the OP was directed to MathOverflow from there, it thought would be a bit unfair to send them back without an answer.