Is $H_0^1(\Omega)\subset C(\Omega)?$

Solution 1:

As mentioned in the comments, if your integrability exponent is high enough then you have Hölder continuity (in one dimension the threshold is $p=1$ so everything's fine there). However, if your exponent is low, then you can't expect something like that.

In what follows we'll assume $n>2$ for simplicity; in the case $n=2$ use a variant of $\ln\ln(|x|)$ instead of $\ln(|x|)$.

Consider $f(x)=-\eta(x)\ln(|x|)$, where $0\leq \eta\in C_c^\infty(B(0,1/2))$ with $\eta \equiv 1$ in $B(0,1/4)$. Let $(p_k)$ be an enumeration of the rationals in $\Omega$ and define $$ u(x)=\sum_{k=1}^\infty 2^{-k}f(x-p_k). $$ Notice that $u$ is well defined, as an element of $H^1$, since the series is absolutely convergent in $H^1$ by translation invariance and the triangle inequality. You can also check that the series converges absolutely a.e. but clearly such a function can't be continuous a.e. or a.e. equal to a continuous function (it's unbounded on every open set of $\Omega$!).

As a fun sanity check: Given $\varepsilon>0$ what is the set $E$ given by Lusin's theorem? i.e. such that $|\Omega\setminus E|<\varepsilon$ and $u|_E$ is continuous.

If you're into this stuff, you can improve Lusin's theorem for Sobolev functions by replacing the Lebesgue measure in the statement above by a quantity that we call, in this case, the $2$-capacity of $\Omega\setminus E$. In this case we say that $u$ has a quasi-continuous representative. But again, even this improved Lusin's theorem is not enough to give the a.e. continuity.