# Poisson modification of subharmonic function

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 $U\in H^2(\Omega)?$.

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My guess is no in general: Consider the one-dimensional case with $\Omega = \mathbb{R}$ (you can pick a bounded interval if you want) and $u(x)=x^2$, take $B(0,1)$ then $U(x)=x^2$ if $|x|\geq1$ and $U(x)=1$ if $|x|<1$. Then $U\notin H^2 (\Omega)$ (or any interval containing $[-1,1]$). – Jose27 Aug 13 '12 at 4:35
@Jose27 You are right, and the same example works in higher dimensions. Please post as an answer. To med: $H^1$ is preserved, but you probably already knew that. – user31373 Aug 13 '12 at 5:07
The question was (unnecessarily) cross-posted on MO: mathoverflow.net/questions/104593/… – user31373 Aug 13 '12 at 21:46

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.

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There's a little shorter argument based on the ACL property of Sobolev functions. Draw a square such that an arc of the unit circle separates its vertical sides. Observe that $\partial_1 U$ is discontinuous on every horizontal line segment within the square (according to your formula for $\partial_i U$). Hence, $\partial_1 U\notin H^1$ and $U\notin H^2$. – user31373 Aug 20 '12 at 2:05
@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. – Jose27 Aug 20 '12 at 3:25