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Let $(f_n)_n$ be a sequence. We say that it is a uniformly absolutely continuous sequence if given $\varepsilon>0$ there exists $\delta>0$ such that $$\left|\int_{A} f_n\, \mathrm{d}\mu\right|<\varepsilon$$ for all $n\in\mathbb{N}$ if $\mu(A)<\delta$.

Keeping in mind this definition, consider $\Omega\subset\mathbb{R}^N$ open and bounded subset. Let $g:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $\displaystyle\lim_{|t|\to +\infty} g(t) =0$ and let $(u_n)_n\subset W_0^{1, p}(\Omega)$ be a bounded sequence in $W_0^{1, p}(\Omega)$, $+\infty>p>1$.

Consider the sequence $(g(u_n) e^{|u_n|})_n$. How to show that it is a uniformly absolutely continuous sequence?

It seems very difficult to me using the definition above. Could anyone please help?

Thank you in advance!

${\bf Edit. \; MY \, ATTEMPT:}$ Since $\displaystyle\lim_{|t|\to +\infty} g(t) =0$, thus $$\int_{\Omega} |g(u_n)| e^{|u_n|} dx\leq \varepsilon e^{|u_n|} + c $$ with $c=c(\varepsilon)$. Although, I don’t know how to proceed by using the definition. Any hint?

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    $\begingroup$ When you say that $(u_n)_n$ are bounded, do you mean bounded in $||\cdot||_\infty$ norm? or in the sobolev norms? $\endgroup$ – Andrew McMillan Apr 3 at 0:42
  • $\begingroup$ Bounded in the Sobolev norm. $\endgroup$ – C. Bishop Apr 3 at 5:54
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Assumption: I interpret the statement "$(u_n)_n\subset W_0^{1,p}(\Omega)$ is a bounded sequence in $W_0^{1,p}$, $p\in (1,+\infty)$" to mean that $\sup_n||u_n||_{W_0^{1,p}}\leq M\,\,\forall p\in (1,\infty)$.

As a consequence of Sobolev's Embedding theorem, this assumption gives that $\sup_n||u_n||_{L^\infty}\leq M_2$ for some $M_2\in \mathbb{R}$.

Since $g(\cdot)$ is a continuous function that vanishes at infinity, $g\in C_0(\Omega)\subset C_b(\Omega)$, the space of continuous bounded functions equipped with the $||\cdot||_\infty$ norm. Therefore, $||g||_\infty:=\sup_{x\in \Omega}|g(x)|\leq M_3$ for some $M_3\in \mathbb{R}$.

We then have:

\begin{equation} \begin{split} \Big{|}\int_A g(u_n)e^{|u_n|}d\mu\Big{|}&\leq \int_A|g(u_n)|e^{|u_n|}d\mu\\ &\leq M_3e^{M_2}\int_Ad\mu\\ &=M_3e^{M_2}\mu(A) \end{split} \end{equation} Therefore, choose $\mu(A)$ appropriately so that the last term is $O(\epsilon)$.

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  • $\begingroup$ $W^{1,p}(\Omega)$ is not necessarily contained in $L^{2}(\Omega)$ if $p$ is small. Similarly, there's no hope to get the boundedness of $\int_{\Omega} \text{exp}(2 |u_{n}|) \, dx$. $\endgroup$ – Peter Morfe Apr 4 at 19:48
  • $\begingroup$ OP says $p>1$. In particular for $W^{1,2}(\Omega)$. Also, I'm not bounding $\int_\Omega \exp(2|u_n|)$, but instead $\int_A\exp(2|u_n|)$, where A is compact. $\endgroup$ – Andrew McMillan Apr 4 at 22:00
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    $\begingroup$ @C.Bishop The heart of the ideas are the so called "density arguments", which are ways to approximate bad functions with nice ones. There are many such results such as in Folland's Real Analysis book. Also, $C_b(\Omega)$ stands for the space of continuous bounded functions. $\endgroup$ – Andrew McMillan Apr 5 at 12:57
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    $\begingroup$ @AndrewMcMillan, if that's your assumption, then why not use the Sobolev embedding theorem to get $\sup_{n} \|u_{n}\|_{L^{\infty}} \leq C$ and write a much simpler proof? You might want to make it clear what your assumptions are in the answer. $\endgroup$ – Peter Morfe Apr 5 at 14:38
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    $\begingroup$ @C.Bishop If you look to the bottom of the link I sent, you get a bound for the $L^\infty$ norm via Sobolev Embedding. Also, for your particular problem, we are indeed working with Lebesgue measure, but the definition would still make sense for any abstract measure on a well-defined measure space. $\endgroup$ – Andrew McMillan Apr 8 at 13:41

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