TITLE: $\int_\Omega F(u_n)\to0$ implies $\int_\Omega F(au_n)\to 0$ for $a\in [0,\infty)$?
QUESTION [5 upvotes]: Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $F:[0,\infty)\to [0,\infty)$ be a convex, strictly increasing and continuous function satisfying $F(0)=0$. Suppose that $u_n\in L^\infty(\Omega)$ (Lebesgue measure) is a sequence of non-negative functions satisfying $$\int_\Omega F(u_n)\to0$$
Can I conclude that for every $a\in [0,\infty)$ the following convergence is true: $$\int_\Omega F(au_n)\to 0$$
Remark 1: Convexity and $F(0)=0$ implies that $F(ax)\leq aF(x)$ for $a\leq 1$, hence the only case that needs to be treat is the case $a>1$.
Update 1: In the circumstances described abode, $u_n\to 0$ in measure. Maybe this can help...
Thank you

REPLY [2 votes]: You cannot conclude that without the extra assumption that $F(a x) / F(x)$ is bounded as $x \to \infty$.
Suppose there is a sequence $\{x_n\}$ such that
$$
\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{F(a x_n)}{F(x_n)} = \infty.
$$
Define $u_n(t) = x_n\chi(t)$ where $\chi$ is the characteristic function of a set of measure $1/F(a x_n)$. Then 
$$
\int_\Omega F(u_n) = \frac{F(x_n)}{F(a x_n)} \to 0
$$
but
$$
\int_\Omega F(a u_n) = \frac{F(a x_n)}{F(a x_n)} = 1.
$$
So, as long as $F$ is bounded by a polynomial it will work, but for e.g. $F(x) = e^x - 1$ it fails.