3

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?

Andrew McMillan
  • 1,005
C. Bishop
  • 3,418

1 Answers1

1

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 $O(\epsilon)$.

Andrew McMillan
  • 1,005
  • $W^{1,p}(\Omega)$ is not necessarily contained in $L^{2}(\Omega)$ if $p$ is small. Similarly, there's no hope to get the boundedness of $\int_{\Omega} \text{exp}(2 |u_{n}|) , dx$. –  Apr 04 '21 at 19:48
  • 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. – Andrew McMillan Apr 04 '21 at 22:00
  • Perhaps I should have said "$W^{1,p}(\Omega)$ is not necessarily contained in $L^{2}(\Omega)$ if $p$ is very close to $1$." –  Apr 04 '21 at 23:29
  • @AndrewMcMillan Thank you for your answer. Could you please tell me if it is a "standard" way to proceed? If yes, could you please give me a reference? Thank you in advance! – C. Bishop Apr 05 '21 at 09:09
  • @AndrewMcMillan sorry, but what do you mean by $C_b(\Omega)$? – C. Bishop Apr 05 '21 at 11:52
  • @PeterMorfe Yes, I agree. However, I interpreted the claim "$(u_n)n$ is uniformly bounded for $p\in (1,\infty)$ to mean that $\sup_n||u_n||{W^{1,p}}\leq M,,\forall p$. – Andrew McMillan Apr 05 '21 at 12:55
  • 1
    @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. – Andrew McMillan Apr 05 '21 at 12:57
  • @AndrewMcMillan Thank you! Just a last thing: where I could find the relation $C_0(\Omega)\subset C_b(\Omega)$ and $C_b(\Omega)$ is dense in Lipschitz function? – C. Bishop Apr 05 '21 at 13:31
  • Here's a discussion about $C_0(\Omega)\subset C_b(\Omega)$: https://math.stackexchange.com/questions/2059085/functions-which-are-zero-at-infinity-are-closed-subset-of-bounded-functions-eve ,but you should try to prove it yourself first! – Andrew McMillan Apr 05 '21 at 13:36
  • Here is the density of Lipschitz functions https://math.stackexchange.com/questions/665587/how-to-show-that-the-set-of-all-lipschitz-functions-on-a-compact-set-x-is-dense – Andrew McMillan Apr 05 '21 at 13:37
  • 1
    @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. –  Apr 05 '21 at 14:38
  • @PeterMorfe Ah yes! You are right. I will correct the answer. – Andrew McMillan Apr 05 '21 at 14:56
  • Thank you @AndrewMcMillan for your answer and thank you Peter Morfe, too. – C. Bishop Apr 05 '21 at 17:13
  • @AndrewMcMillan sorry, I have just another question. How to obtain $\sup_n | u_n|_{\infty}\leq M_2$? Sobolev embedding theorem states that $W_0^{1,p}(\Omega)\hookrightarrow L^q(\Omega)$ for $q\in [1, p^*]$. (Sorry if my question is trivial). – C. Bishop Apr 05 '21 at 18:54
  • @AndrewMcMillan, sorry again. If you take $(\mathbb{R}^N, \mu)$ where $\mu$ denotes the Lebesgue measure, does the property holds for any Lebesgue integrable $f$, isn'it? It should be the so called absolute continuity of Lebesgue integral. – C. Bishop Apr 06 '21 at 13:20
  • 1
    @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. – Andrew McMillan Apr 08 '21 at 13:41