1
$\begingroup$

Let $u\in C^2(\Omega)$ be such that $\Delta u \ge 0$ on $\Omega\supset \overline{B(a,r)}$. We consider the Poisson modification $U$ of $u$ for the ball $B(a,r),$ that is $U$ equals $u$ on $\Omega-B(a,r)$ and that on $B=B(a,r)$ equals the solution to Direchlet problem with boundary data $u|_{\partial B}$, which is given by the Poisson kernel classically denoted by $P(x,y)$. It is known that $U$ is subharmonic in the sense that it verifies an inequality mean property. My question : Do we have $U\in H^2(\Omega)?$.

  • 0
    The question was (unnecessarily) cross-posted on MO: http://mathoverflow.net/questions/104593/poisson-modification-of-subharmonic-function2012-08-13

1 Answers 1

1

In fact a slight variant from the one dimensional case works in general: Consider $\Omega =B(0,2)=:B_2$ and $u(x)=|x|^2$ and take $B_1:=B(0,1)\subset \Omega$ then

$ U(x)=\left\{ \begin{array}{lr} |x|^2 & \text{if }|x|\geq 1\\ 1 & \text{if } |x|<1 \end{array} \right. $

We want to prove that $U\notin H^2(\Omega)$, for this we note that

$ \partial_iU(x)=\left\{ \begin{array}{lr} 2x_i & \text{if }|x|\geq 1\\ 0 & \text{if } |x|<1 \end{array} \right. $

and so, since this is almost smooth, a possible second derivative has only one candidate

$ \partial_{ij}U(x)=\left\{ \begin{array}{lr} 0 & \text{if }|x|\geq 1\\ 0 & \text{if } |x|<1 \end{array} \right. $ and so, for the contradiction just note that integration by parts gives $ 0=\int_{B_2\setminus B_1} \partial_{i} U\partial_j \phi =2\int_{B_2\setminus B_1} x_i\partial_j\phi=2 \int_{\partial(B_2\setminus B_1)} x_i\phi\nu^j = 2\int_{\partial B_1} x_i\phi\nu^j $ for all $\phi\in C_c^\infty (B_2)$, which is clearly not true.

  • 0
    @LVK: Thanks, I like that argument, it makes it easier to see (and prove!) that Sobolev functions can't have jumps over $n-1$-submanifolds.2012-08-20