About the continuity of the integral on the boundary of a ball

Question

I’m considering a $H^1$ function u on a open domain D. Is the integral:

$$ \int_{\partial B_r(x)} u \hspace{2pt}dH^{n-1}$$

continuous with respect to x?

I tried to prove that it’s differential by showing that the derivative can be written as the integration:

$$ \int_{\partial B_r(x)} Du \hspace{2pt}dH^{n-1}$$

But it’s just right a.e. From which we can deduce that it’s continuous a.e.. Is there any possibility to show that it’s continuous everywhere?

Answer 1

I would say so. Denote your integral by $b_u(x)=\int_{|x-y|=r}u(y)dH^{n-1}$. Approximate $u$ in $H^1$ with test functions $u_j$. The property is certainly true for $u_j$ thus it is enough to prove that $b_{u_j}\to b_u$ uniformly. You can estimate $|b_{u_j}(x)-b_u(x)|$ with the $L^2$ norm of the trace of $u-u_j$ at $\partial B_r(x)$, which is controlled by the $H^1$ norm of $u_j-u$ (for fixed $r$)