0
$\begingroup$

Let $u$ be a subharmonic function on $\mathbb{C}$. It is known that the convolution with standard mollifier gives a sequence $u_{\epsilon}$of subharmonic functions with the property: $u_{\epsilon}\in C^{\infty}$ and $u_{\epsilon}\searrow u$ (as $\epsilon\rightarrow0^{+}$) pointwise. However, I want one more extra property: $\Delta u_{\epsilon} \geq \Delta u$, $\forall \epsilon$ (or at least a subsequence). Here "$\geq$" means $\Delta u_{\epsilon}$ has more mass than $\Delta u$ on any compact set.

I think this is too much. So the better question may be that under what condition on $u$ that we can find a sequence $u_{\epsilon}$ (still understood as the convolution with standard mollifier) with this property.

I could find one condition. Let's assume $u$ is smooth enough and $\Delta u$ is also subharmonic. Then it follows easily that $\Delta u_{\epsilon}=(\Delta u)_{\epsilon}\geq \Delta u$. But this condition is so heavy.

I wonder whether we can do it under the condition that $\Delta u$ is a doubling measure.

$\endgroup$

1 Answer 1

1
$\begingroup$

This is not possible, even for the simplest function $u(z)=\log|z|$. The Riesz mass has an atom at $0$, so the condition $\Delta v\geq\Delta u$ implies that $\Delta v$ has an atom at $0$, but then $v$ cannot be smooth.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .