Continuity+differentiability imply weak sequential continuity? Let $E$ and $F$ be two Banach spaces. We know that, if $f:E\rightarrow F$ is a nonlinear continuous operator, then $f$ may fail to send weakly convergent sequences to weakly convergent sequences, i.e., $u_n \rightharpoonup u$ weakly in $E$  does not necessarily imply $f(u_n)\rightharpoonup f(u)$ weakly in $F$. Even through $f$ has very mild non-linearity,  this weak sequential continuity may fail. For example, let $u_n(x)=\sin(nx), x\in (0,2\pi)$. Then Riemann-Lebesgue lemma shows that $u_n\rightharpoonup u= 0$ in $L^p(0,2\pi)$ for any $p\geq 1$. 
Now, for $f(u)=\max(0,u)=u^+$, one has 
$$f(u_n)\rightharpoonup \frac{1}{\pi}\neq f(u)=0  \text{ in } L^1(0,2\pi). 
$$
Also,
$$
\int_0^{2\pi}f(u_n(x))dx=\int_0^{2\pi}\sin^+(nx)dx=\frac{1}{n}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}\sin (y)dy=2, 
$$
which implies $f(u_n)$ can not converge weakly to  $0=f(u)$ in $L^2(0,2\pi)$. 
If $g(u)=|u|$, then 
$$g(u_n)\rightharpoonup \frac{2}{\pi}\neq g(u)=0  \text{ in } L^1(0,2\pi). 
$$
From these two examples, we see there is no general hope that a continuous map  preserves weak sequential continuity. We are wondering what kind of conditions imposed on $f$ so that it preserves weak sequential continuity.  For the two examples above, we note that both $f$ and $g$ are sub-linear ($|f(u)|\leq c|u|$ for some  constant $c>0$),  continuous but not differentiable. Now, a natural question arises: whether a sub-linear ($|f(u)|\leq a|u|+b$ for some positive constant $a$ and non-negative constant $b$), continuous and differentiable map $f:E\rightarrow F$ preserves weak sequential continuity? Here, we emphasize the non-linearity of $f$, since, for linear map $f$, it is known from functional analysis this question holds. For this question, one can choose $E$ and $F$ to be Hilbert spaces, for convenience. As concrete examples, one may test: 1. $u_n\rightharpoonup u$ with $u_n\geq 0$ in $L^2(\Omega)$ and $f(u)=\frac{u}{1+u}$, do we have $f(u_n)\rightharpoonup f(u)$ in $L^2(\Omega)$? 2. $u_n\rightharpoonup u$ in $L^2(\Omega)$ and $g(u)=\frac{u}{1+u^2}$, do we have $g(u_n)\rightharpoonup g(u)$ in $L^2(\Omega)$? Thank you for your effort! 
 A: Thank user 37238 first. Actually, the example listed in the last sentence  has already illustrated that the question is not true. Here, I make it more clear. Consider the sequence $u_n(x)=\sin(nx)$. Then $u_n\rightharpoonup 0:=u$ in $L^2(0,2\pi)$. Now, by a known result about weak limits of rapidly oscillating periodic functions, we have 
$$
f(u_n(x)):=\frac{u_n(x)}{2+u_n(x)}=\frac{\sin(nx)}{2+\sin(nx)}\rightharpoonup \frac{1}{2\pi}\int_0^{2\pi}\frac{\sin(x)}{2+\sin(x)}dx\neq 0=f(u) \text{ in } L^2(0,2\pi). 
$$
Of course,, $f$ is sub-linear, continuous and differentiable. As answered by User37238, it seems that, if $u_n\rightharpoonup u$ implies $f(u_n)\rightharpoonup f(u)$, then we must necessarily have $f$ is affine? 
A: Here is an example (from Brézis' book Functional Analysis, Sobolev Spaces and Partial Differential Equations exercise 4.20) that shows that it is much more restrictive than you think :
Take $\Omega = (0,1)$ and $a : \mathbb{R}\to \mathbb{R}$ a continous function such that
$$ |a(t)|\le a|t|+b$$
and define $A : L^2(\Omega)\to L^2(\Omega)$ by $(Au)(x) = a(u(x))$ for almost every $x\in \Omega$. Then you can show that $A$ is strongly continuous. 
Suppose now that for every sequence $(u_n)$ such that $u_n \to u$ weakly in $L^2(\Omega)$, we have $Au_n \to Au$ weakly in $L^2(\Omega)$. 
Then you can show that $a$ is necessarily an affine function.
Sketch of the proof :
First define $f$ a function $1$ periodic such that 
$$f = \left\{\begin{array}{rcl}
\alpha &\text{on}& (0,1/2) \\
\beta &\text{on}& (1/,2) \\
\end{array} \right. $$
where $(\alpha,\beta)\in\mathbb{R}^2$. Then define $u_n : x\in (0,1) \mapsto f(nx)$ and show that $u_n$ converges weakly in $L^2$ towards $\frac{\alpha+\beta}{2}$.
Using this $u_n$, you can show that 
$$\forall(\alpha,\beta)\in \mathbb{R}^2,\ \frac{a(\alpha)+a(\beta)}{2} = a\left( \frac{\alpha+\beta}{2}\right) $$
Finally, you get $a$ affine from the previous equation.
