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Let me first introduce the restricted setting in which this question has a nice answer. I came up with this when messing around with a homework problem in a PDE class a couple years back.

Let $\phi \in C^2[0,1]$. It induces a function $\phi_*: L^2[0,1] \to L^2[0,1]$, $f \mapsto \phi\circ f$. One can ask whether $\phi_*$ is continuous with respect to the weak topology on $L^2[0,1]$: Does $f_n \rightharpoondown f \Rightarrow \phi(f_n) \rightharpoondown \phi(f)$?.

By looking at the function $f(x)=r\cos(n\pi x)+x_0$, one can get a necessary condition on $\phi$, that $\phi''(x_0)=0$ for all $x_0$. I.e. $\phi(x)=ax+b$. This is obviously sufficient too. Here's how you get the condition:

Assuming the continuity of $\phi_*$ with respect to the weak topology: $$\int_{-r+x_0}^{r+x_0}\frac{\phi(z)}{\sqrt{r^2-(z-x_0)^2}}=\int_{[0,1]} \phi(x_0+r\cos(n\pi x))dx$$

$$=\langle \text{weak limit of } \phi(x_0+r\cos(n \pi x)),1 \rangle $$

$$=\langle \phi(\text{weak limit of } (r\cos(n \pi x)+x_0)),1\rangle=\langle \phi(x_0),1\rangle=\phi(x_0).$$ So if $\phi''(x_0)>0$ for some $x_0$, one can find a small ball $B_r(x_0)$ where $\phi''>0$ and the extreme LHS won't be the extreme RHS. Similarly $\phi''$ is not less than 0 anywhere.

My question for stackexchange is whether this is something that is well studied? It has to be. It is such a simply phrased question. If that is too vague, my question for you is what happens when one removes the differentiability conditions on $\phi$, and one looks at the induced function on $L^2[\text{non compact euclidean space}]$. This argument doesn't even work in higher dimensions so I am at a loss to see how to generalize what I did above.

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    $\begingroup$ Note that the weak topology of a infinite dimensional Hilbert space $H$ is not metrizable (it is only its restriction on the unit ball), and for a map $H\to H$ continuity wrto the weak topology is not a priori equivalent to the sequential continuity ($f_n \rightharpoondown f \Rightarrow \phi(f_n) \rightharpoondown \phi(f)$) . $\endgroup$ – Pietro Majer Oct 2 '16 at 9:07
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    $\begingroup$ Also, for the composition $f\to \phi \circ f$ to be defined on $L^2[0,1]$, you need $\phi$ to be defined on the whole $\mathbb{R}$ (not just $[0,1]$), together with some $|\phi(t)|\le a|t|+b$ for all $t$. $\endgroup$ – Pietro Majer Oct 2 '16 at 9:19
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Consider the sequence $f_k:[0,1]^n\to\mathbb{R}$ defined by $f_k(x_1,\dots,x_n):=(-1)^{\sum_{j=1}^n \lfloor kx_j \rfloor}$: it converges weakly to $0$. For any $\phi:\mathbb{R}\to\mathbb{R}$, and for any $a$ and $b$ in $\mathbb{R}$, there holds: $$\phi\circ (a+bf_k)=\frac{\phi(a+b)+\phi(a-b)}{2}+\frac{\phi(a+b)-\phi(a-b)}{2}\, f_k$$ so assuming $f\to \phi\circ f$ is sequentially continuous wrto the weak topology of $L^2[0,1]^n$ and taking weak limits we get $$\phi (a)=\frac{\phi(a+b)+\phi(a-b)}{2}.$$

Thus, testing the weak continuity along all sequences $a+bf_k$, we conclude that $\phi$ must be an affine function.

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Adding to the answer of Pietro:

Exercise 4.20 in Brezis' "Functional Analysis, Sobolev Spaces and Partial Differential Equations" also tells you that weakly continuous Nemytskii operators on $L^p(0,1)$ are already necessarily affine linear. In fact, the proof uses so-called Rademacher's functions, which are closely related to the sequence $f_k$ defined by Pietro in his answer (luckily, Brezis provides solutions for his exercises :-)).

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