Trace defined in terms of integral averages

Question

It is known that if $u \in W^{1,1}(U)$ where $U\subset \mathbb{R}^n$ is bounded and $\partial U$ Lipschitz then $\mathcal{H}^{n-1}$ a.e we have $$\lim_{r\to 0} \frac{1}{|B(x,r)\cap U|}\int_{B(x,r)\cap U} \left|f(y) - Tf(x)\right|dy = 0 $$ where $f\mapsto Tf$ is the classical trace operator. Now I'm wondering if the following result is true : Take $f\in W^{1,p}(U)$, is it true that $$\lim_{r\to 0} \frac{1}{|B(x,r)\cap U|}\int_{B(x,r)\cap U} \left|f(y) - Tf(x)\right|^p dy = 0 $$ $\mathcal{H}^{n-1}$ a.e.

Answer 1

Yes, and you actually can say a bit more. Every bounded Lipschitz domain is a Sobolev extension domain (classical theorem of Calderón and Stein, referenced here). Thus we can assume that $u\in W^{1,p}(\mathbb R^n)$.

If $p>n$, then $u$ is Hölder continuous and the matter is trivial. So let's assume $1<p\le n$.

I consulted Fine Regularity of Solutions of Elliptic Partial Differential Equations by Malý and Ziemer (Google Books).

Theorem 2.20 (page 74) asserts that every $ W^{1,p}$ function has a $p$-quasicontinuous representative. There is no need to go into the definition of $p$-quasicontinuous here. It suffices to say that this is a representative of $u$, i.e., is equal to original function almost everywhere. Let's use this representative from now on.

Theorem 2.55 on page 90 is a strong Lebesgue-point result:

Suppose $u \in W_{\rm loc}^{1,p}(\mathbb R^n)$, $1<p\le n$, and $q$ is a finite exponent with $1\le q\le np/(n-p)$. Then any $p$-quasicontinuous representative of $u$ satisfies $$\lim_{t\to 0} \frac{1}{r^n}\int_{B(x,r)} |u(y)-u(x)|^q\,dy = 0 \tag{1}$$ at $p$-quasi every point of $\mathbb R^n$.

Here, "$p$-quasi every" means outside of set of $p$-capacity zero. By Theorem 2.53 (page 87), such a set has $s$-dimensional Hausdorff measure zero for every $s>n-p$. (In particular, for $s=n-1$.)

So, $\mathcal H^{n-1}$-a.e. on the boundary we have (1), and we also have $$\lim_{r\to 0} \frac{1}{|B(x,r)\cap U|}\int_{B(x,r)\cap U} \left|u(y) - Tu(x)\right|\,dy = 0$$ Hence $Tu(x)=u(x)$, $\mathcal H^{n-1}$-a.e. Conclusion: $$\lim_{r\to 0} \frac{1}{|B(x,r)\cap U|}\int_{B(x,r)\cap U} \left|u(y) - Tu(x)\right|^q dy = 0$$ for every finite $q\le np/(n-p)$, and for $\mathcal H^{n-1}$-a.e. $x\in \partial U$.

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Another book with similar results is Weakly Differentiable Functions by Ziemer: see Theorem 3.3.3 on page 120. However, the theorem is stated for $1<p<n$ only. Its conclusion is $$\lim_{t\to 0} \frac{1}{r^n}\int_{B(x,r)} |u(y)-u(x)|^p\,dy = 0$$ (outside of a set with zero capacity, as above).