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Suppose $D$ is a bounded domain of $\mathbb{R}^m$ for $m>1$ and $\{u_n\}_{n\geq1}$ is a sequence of subharmonic functions on $D$. Assume $u_n\to u_0$ pointwise on $D$ and $u_0$ is subharmonic on $D$. Let $\mu_n$ be the Riesz measure associted to each $u_n$ for $n\geq0$. Suppose also that for a compact set $K\subset D$ we have $$\mu_n(K)=0$$ for all $n>0$. It is well-known that the sequence of measures $\{\mu_n\}$ has a subsequence that is vaguely convergent, and so $$\int_Kf(x)d\mu_{n_k}(x)\to \int_Kf(x)d\nu(x),$$ as $k\to\infty$, for all continuous functions $f$ and for some measure $\nu$.

My question is: can we conclude that the restrictions of $\nu$ and $\mu_0$? to $K$ coincide? In particular, do we have also $\mu_0(K)=0$?

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  • $\begingroup$ I don't understand. $\mu(K) = 0$ for each compact $K \subset D$ implies $\mu \equiv 0$ by inner regularity. What am I missing? $\endgroup$ Oct 3 '20 at 10:02
  • $\begingroup$ I think you can use Lebesgue Dominated convergence theorem $\endgroup$ Oct 3 '20 at 10:04
  • $\begingroup$ Plus, you should replace "compact $K$" by "open and bounded $K$", I believe. $\endgroup$ Oct 3 '20 at 11:20
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    $\begingroup$ Vague convergence would not imply that $\nu(K)=0$. $\endgroup$ Oct 3 '20 at 11:50
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The answer is no. Example (I use complex notation in dimension 2) $$u_n(z)=\max\{\log|z|-1/n,0\}\to\max\{\log|z|,0\}$$ $K=\{ z:|z|\leq 1\}$ is compact. $\mu_n(K)=0$ but he limit measure is supported on $K$. In fact, in this example, $\mu_n$ is the uniform measure on the circle $|z|=e^{1/n}>1$ while the limit is the uniform measure on the unit circle.

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  • $\begingroup$ Sorry, but I am confused! I have two problems. 1) How did you get the measures of $K$? 2) According to my understanding, the Riesz measure $\mu_n$ of $u_n$ is defined as $$\int_D \Psi d\mu_n =a\int_Du_n\Delta \Psi d\lambda$$ for all test function $\Psi$ in $D$ ($C^\infty $-smooth compactly supported in $D$). Now if you take for $D$ the ball of, say, radius 2 centered at the origin, then $u_n$'s are bounded and by dominated convergence theorem we get, as $n\to\infty$, $$\int_Du_0 \Psi d\lambda=\int_D \Psi d\mu_0=\int_D \Psi d\nu.$$ Isn't this a contradiction with what you are saying? $\endgroup$
    – M. Rahmat
    Oct 3 '20 at 21:00

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