Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $g(u)u = 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.