# $\forall g\in L^2(\Omega)$ exists $g_n\in H_0^1(\Omega)$ and $\epsilon>0$ s.t $g_n(x)\to g(x),\,a.e$ and $|g_n(x)|\leq |g(x)|+\epsilon$

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 $|g_{n_k}(x)|\leq h(x)$

• You can try the 1D case to be convinced that the result cannot be true in general. – O.G. Apr 14 '16 at 14:53
• @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. – Svetoslav Apr 14 '16 at 15:07
• 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 $g$). – O.G. Apr 14 '16 at 15:45
• @O.G. Do you mean $g(x)=n$ on $E_n$ ? – Svetoslav Apr 14 '16 at 19:28
• Yes and study near $1/n$ (not $n$ there is a typo). – 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.
• This is in contrast with the fact that a function in $H^1$ may be unbounded, right? – Piero D'Ancona Apr 16 '16 at 6:20
• Sorry, I do not understand how a section of an unbounded function might be AC. E.g. $\log|x|$ in $R^3$ – Piero D'Ancona Apr 16 '16 at 7:29
• @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 $y$). – Christian Remling Apr 16 '16 at 7:32
• @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. – Christian Remling Apr 16 '16 at 7:40