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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!

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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?

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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.

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  • $\begingroup$ I do not think your $A$ is weakly sequentially continuous. Modify the example above. Let $a(t)=\max(0,t)$ and $u_n(x)=\sin(2n\pi x)$. Then $u_n\rightharpoonup 0$ weakly in $L^2(0,1)$, but $A(u_n)$ can not converge weakly to $0=A(u)$ in $L^2(0,1)$. $\endgroup$ – teh Sep 23 '13 at 16:29
  • $\begingroup$ @teh What I say is if you want $A$ to be weakly sequentially continuous then $a$ has to be an affine function. That's why your example does not work. $\endgroup$ – user37238 Sep 23 '13 at 17:30
  • $\begingroup$ So, you meant for a sublinear function $a$, in order $(Au)(x)=a(u(x))$ to be weakly sequentially continuous, we must have $a$ is affine? If this is the case, the question above may not be true. Then how to show $a$ is an affine function? Could you show this? $\endgroup$ – teh Sep 23 '13 at 19:49
  • $\begingroup$ Yes it is what I meant. $\endgroup$ – user37238 Sep 23 '13 at 20:46
  • $\begingroup$ @teh Just added a sketch of the proof. $\endgroup$ – user37238 Sep 23 '13 at 21:00

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