# Superposition operator in Sobolev spaces

While working on an elliptic problem in $\mathbb{R}^N$, I met an issue that I cannot work out clearly. Assume that we have a continuous function $g \colon \mathbb{R} \to \mathbb{R}$ such that

1. $\lim_{s \to 0}\frac{g(s)}{s}=0$;
2. $\lim_{s \to +\infty} \frac{g(s)}{s^{2^*-1}}=0$, where $2^*=\frac{2N}{N-2}$ is the critical Sobolev exponent in dimension $N\geq 3$.

Now assume I have a sequence $\{u_n\}_n$ of functions that converges strongly to some $u$ in $L^q(\mathbb{R}^N)$, where $q<2^*$. My question is: is it true that $\{g(u_n)u_n\}_n$ converges in $L^1$ to $g(u)u$?

I am in trouble because there is no precise growth condition on $g$, just some asymptotic behavior. It would be standard if $|g(s)| \leq |s|^q$ or something alike.

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For $q < 2^*$ (as written), the answer is no.
Let $q' = \frac12(q + 2^*) > q$. Let $$u_n = n^{1/q'}\chi_{[n,n+1/n]}$$ then $$\|u_n\|_q = n^{(q - q')/(qq')} \searrow 0$$ and we see that $u_n$ converges strongly to 0 in $L^q$.
Let $g(s)$ be roughly $s^{q'-1}\chi_{[1,\infty)}$, modified to be continuous. This satisfies both conditions 1 and 2, as $q' - 1 < 2^* - 1$. We have that $$\| g(u_n) u_n \|_1 = 1 \qquad \| g(u_n) u_n - g(u_m) u_m \|_1 = 2(1- \delta_{mn})$$ so does not converge strongly to $g(u)u = 0$.
I agree, actually there is no information about the behavior of $g$ far from zero and infinity. In most cases, a pointwise decay of $u_n$ at infinity is added, but I was trying to avoid it. – Siminore Sep 10 '12 at 14:09
BTW, if $q \geq 2^*$, then note that for any continuous function $F(s)$ on $[0,\infty)$ with limit $$-\infty < \liminf_{s\to\infty} F(s) \leq \limsup_{s\to\infty} F(s) < \infty$$ is necessarily bounded, so you can reduce to the case $g(s) \leq C s^q$. – Willie Wong Sep 10 '12 at 14:10
Well, as you can imagine, the convergence of $u_n$ follows from a compact embedding. It is almost impossible to choose $q \geq 2^*$... – Siminore Sep 10 '12 at 14:11
Umm... I see, the problem is that even with $u_n \to u$ for all $q < 2^*$ you may not have enough: I can imagine $g(s) \approx s^{2^* - 1 - \frac{1}{\log\log s}}$ or something similar... – Willie Wong Sep 10 '12 at 14:20
Exactly. I either need a uniform decay at infinity or a precise bound on the growth of $g$. – Siminore Sep 10 '12 at 14:23