$W_0^{1,p} \cap W^{1,q} = W_0^{1,q}$?

Question

Let $\Omega \subset \mathbb R^n$ be an open set. We denote by $W^{1,p}(\Omega)$ the usual Sobolev spaces and $W_0^{1,p}(\Omega)$ is the closure of $C_c^\infty(\Omega)$ in $W^{1,p}(\Omega)$.

Let $1 \le p < q \le \infty$. Do we have $W_0^{1,p}(\Omega) \cap W^{1,q}(\Omega) = W_0^{1,q}(\Omega)$?

Of course, "$\supset$" is clear, but "$\subset$" is hard. I do not see that this inclusion can be tackled by applying the definitions directly.

If the boundary of $\Omega$ possesses some regularity (e.g. Lipschitz boundary), then we can define a trace operator (which is independent of the regularity exponent) and the conclusion follows from $$W_0^{1,r}(\Omega) = \{ u \in W^{1,r}(\Omega) \;\mid\; u = 0 \text{ on }\partial\Omega\}.$$

Is there some easier argument? In particular, I would like to see what happens in case that $\partial\Omega$ has low regularity (or "no regularity" at all).

Answer 1

The answer is negative for general open sets; to see this let $\Omega = B_1 \setminus \{0\} \subset \Bbb R^n$ be the punctured ball in $\Bbb R^n,$ and take $p < n < q.$ Fix any $\varphi \in C^{\infty}_c(B_1)$ such that $\varphi(0)=1.$

By Sobolev embedding convergence in $W^{1,q}$ implies uniform convergence, so $\varphi \in W^{1,q}(\Omega) \setminus W^{1,q}_0(\Omega).$ On the other hand I claim $\varphi \in W^{1,p}_0(\Omega)$ by considering $$ v_r(x) = 1 - \frac{\int_{|x|}^1 t^{(1-n)/(p-1)} \,\mathrm{d} t}{\int_r^1 t^{(1-n)/(p-1)} \,\mathrm{d} t} $$ for $0 < r < |x| < 1,$ setting $v_r(x) = 0$ for $|x| \leq 1.$ Then $v_r \in C^{1}_c(\Omega)$ and one can compute that $$ \int_{B_1} |\nabla v_r(x)|^p \,\mathrm{d} x = \omega_{n-1} \left(\frac{p-1}{n-p}\right)^{1-p} |1 - r^{(p-n)/(p-1)}|^{1-p} \to 0$$ as $r \to 0.$ Hence $v_r \to 1_{\Omega}$ in $W^{1,p}(\Omega),$ and by mollifying $v_r \varphi$ we see that $\varphi \in W^{1,p}_0(\Omega)$ as required.

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The above counterexample is furnished using the theory of Sobolev capacity, and I've adapted the calculations from 2.11 of the following reference.

Heinonen, Juha; Kilpeläinen, Tero; Martio, Olli, Nonlinear potential theory of degenerate elliptic equations, Oxford Mathematical Monographs. Oxford: Clarendon Press. v, 363 p. (1993). ZBL0780.31001.

More generally from chapters 2 and 4 of the text we have the following results, which provide some insight about when we may or may not expect results of this type to hold true.

Theorem (2.43): If $\Omega \subset \Bbb R^n$ is open and $E \subset \Omega$ is relatively closed, then we have $$ W_0^{1,p}(\Omega) = W^{1,p}_0(\Omega \setminus E) $$ if and only if $E$ has $p$-capacity zero.

Theorem (4.5): If $\Omega \subset \Bbb R^n$ is open, then $\varphi \in W^{1,p}(\Omega)$ lies in $W^{1,p}_0(\Omega)$ if and only if there is a $p$-quasicontinuous function $\tilde \varphi$ on $\Bbb R^n$ agreeing with $\varphi$ almost everywhere in $\Omega,$ such that $\tilde\varphi = 0$ $p$-quasieverywhere in $\Bbb R^n \setminus \Omega.$

Using Theorem 2.43 we can generate more counterexamples as follows: suppose $E \Subset \Omega$ has $p$-capacity zero but non-zero $q$-capacity, and choose $\varphi \in C_c^{\infty}(\Omega)$ such that $\varphi \equiv 1$ on $E.$ Then we see that $\varphi \not\in W^{1,q}_0(\Omega \setminus E),$ but $\varphi \in W^{1,p}_0(\Omega \setminus E) = W^{1,p}_0(\Omega).$ The above counterexample does this by noting the point $\{0\}$ has $p$-capacity zero if and only if $p < n,$ reproducing the necessary calculations.