I asked this question on MSE here some time ago, but I couldn't get an answer. There was a suggestion in the comments for a counterexample using a fat Cantor set, but I couldn't show a contradiction with the statement in my question.

Let $\Omega\subset \mathbb R^d$ ($d=2,3$) is a bounded Lipschitz domain.

Question: Is it true that for each function $g(x)\in L^2(\Omega)$ one can find a sequence $\{g_n\}_1^\infty$ of $H_0^1(\Omega)$ functions such that $g_n(x)\to g(x),\,a.e$ in $\Omega$ and $|g_n(x)|\leq |g(x)|+\epsilon,\,a.e,\,\forall n\ge 1$ for some $\epsilon>0$ ?

In the case when $g\in L^\infty(\Omega)$ and $m:=\|g\|_{L^\infty(\Omega)}$, then also $g\in L^p(\Omega),\,1\leq p<\infty$ so there is $g_n\in C_0^\infty(\Omega)$ s.t $g_n\to g$ in $L^p(\Omega)$ and from it we can extract a subsequence $g_{n_k}(x)\to g(x),\,a.e$. Finally, we construct the smooth function $\varphi:\mathbb R\to\mathbb R$ s.t $\varphi(t)=t,\,|t|\leq m+1$, $\varphi(t)=m+2,\,t>m+2$, $\varphi(t)=-m-2,\,t<-m-2$ and take the functions $\varphi\circ g_{n_k}$. These functions satisfy $\varphi\circ g_{n_k}(x)\to g(x),\,a.e$ and $|\varphi\circ g_{n_k}(x)|\leq m+2$.

Remark: It is easy to show (see for example Th. 4.9 in Brezis' book Functional Analysis) that for $g_n\to g$ in $L^p(\Omega)$ there exists a subsequence $g_{n_k}$ s.t $g_{n_k}(x)\to g(x),\,a.e$ and $|g_{n_k}(x)|\leq h(x)$ for some $h\in L^p(\Omega)$. As $C_0^\infty(\Omega)\subset H_0^1(\Omega)$ is dense in $L^2(\Omega)$ we can find $g_n\to g$ in $L^2(\Omega)$ and apply Th. 4.9, but then we get only $|g_{n_k}(x)|\leq h(x)$

  • $\begingroup$ You can try the 1D case to be convinced that the result cannot be true in general. $\endgroup$ – O.G. Apr 14 '16 at 14:53
  • $\begingroup$ @O.G. Yes, this was also the suggestion in the comments in MSE. But as I said, I cannot show a contradiction even in 1D. $\endgroup$ – Svetoslav Apr 14 '16 at 15:07
  • $\begingroup$ In 1D any $H^1$ function has a continuous representant. Consider $E_n=]1/n,1/n+1/n^4[$ and define $g$ as $n$ on $E_n$, and 0 elsewhere. I think a problem occurs in $n$ for $n$ sufficiently large ($n>2\varepsilon$ or $\varepsilon$) for $g_p$ ($p$ is the index of the sequence $g_p$ belonging to $H^1$ and converging to $g$). $\endgroup$ – O.G. Apr 14 '16 at 15:45
  • $\begingroup$ @O.G. Do you mean $g(x)=n$ on $E_n$ ? $\endgroup$ – Svetoslav Apr 14 '16 at 19:28
  • $\begingroup$ Yes and study near $1/n$ (not $n$ there is a typo). $\endgroup$ – O.G. Apr 14 '16 at 20:12

I just stumbled on the following fact in Conway's complex analysis book, vol. 2, Lemma 19.11.6: If $f\in H_0^1$, then, after modification on a null set, $f(x,y)$ will be (in fact: absolutely) continuous as a function of $x$ for fixed $y$. (Since it's a complex analysis book, it's done for $d=2$ there, but the proof works in any dimension.)

This means that the type of counterexample that was already suggested works. We can essentially work in one dimension: we take $\Omega\subseteq\mathbb R^d$ as the unit cube and $g(x,y)=h(x)$ with $h=0$ on a (dense open) set $A$ with $|A\cap I|>0$ for every open interval $I$ and $h=n$ on a set $C_n$ with $|C_n|=2^{-n}$ (use Cantor sets).

Then, by the continuity property reviewed above, a $g_n\in H_0^1$ with $|g_n|\le g+M$ a.e. will satisfy $|g_n|\le M$ a.e.

  • $\begingroup$ This is in contrast with the fact that a function in $H^1$ may be unbounded, right? $\endgroup$ – Piero D'Ancona Apr 16 '16 at 6:20
  • $\begingroup$ Sorry, I do not understand how a section of an unbounded function might be AC. E.g. $\log|x|$ in $R^3$ $\endgroup$ – Piero D'Ancona Apr 16 '16 at 7:29
  • $\begingroup$ @PieroD'Ancona: You did notice that I may have to modify on a null set? Try it perhaps for a standard example such as $f(x,y)=\log|\log |(x,y)||$: I'll modify by setting $f(x,0)=0$, and now every section is $AC$ (and bounded, though the bound is of course not uniform in $y$). $\endgroup$ – Christian Remling Apr 16 '16 at 7:32
  • $\begingroup$ @PieroD'Ancona: The idea of the proof is pretty simple really: $\partial_x f$ will be in $L^2(I)\subseteq L^1(I)$ for a.e. $y$, so $\widetilde{f}=\int_a^x \partial_t f(t,y)\, dt$ is as desired. We need to show that $\widetilde{f}=f$ a.e., which looks quite plausible right away. $\endgroup$ – Christian Remling Apr 16 '16 at 7:40
  • $\begingroup$ Ok you can make this work in one direction but not in both directions with the same modification $\endgroup$ – Piero D'Ancona Apr 16 '16 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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