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.

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