3
$\begingroup$

Hörmander in Analysis of linear partial differential equations II proves Proposition 16.1.2 on page 304 to the effect that a sequence of subharmonic functions converging to another subharmonic function in the distributional sense necessarily converges in the $L^p_{loc}$ sense. Note that the proposition assumes the existence of a limiting subharmonic function. Is there a version of the Arzela-Ascoli theorem in this context that would guarantee the existence of a limit for a suitable subsequence, and under what hypotheses? I am mainly interested in the real 2-dimensional case.

As a follow-up, can one get control over the set of points where the limiting function is $-\infty$?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ There is such a thing, at least for plurisubharmonic functions. More precisely, if $(u_n)$ is a sequence of psh functions such that the $L^1$ norm of $(u_n)$ is bounded on each compact set, then there exists a convergent subsequence – in $L^p_{\rm loc}$ norm for any $p \ge 1$. Moreover if $(u_n)$ is only assumed to be locally uniformly bounded above, then either $(u_n)\to -\infty$ locally uniformly or it has the compactness property above. $\endgroup$
    – Henri
    Commented Mar 13, 2017 at 17:21
  • 2
    $\begingroup$ For a reference, you can look at Hörmander's book Notions of Convexity, Thm 3.2.12 $\endgroup$
    – Henri
    Commented Mar 13, 2017 at 17:22
  • $\begingroup$ @Henri, thanks! I forgot to mention that I am mainly interested in the real 2-dimensional case, so perhaps this is easier than it appears. $\endgroup$
    – Mikhail Katz
    Commented Mar 13, 2017 at 17:26
  • $\begingroup$ @Henri, Hormander's theorem 3.2.12 seems to be the right result. Notice that it is in general dimension and talks about subharmonic functions. I wonder if in the 2-dimensional case one can get control over the set of points where the limiting function is $-\infty$. Does that ring a bell? $\endgroup$
    – Mikhail Katz
    Commented Mar 13, 2017 at 17:48
  • $\begingroup$ Polar sets of subharmonic functions can be pretty nasty (e.g. they can be dense). In general they are always $G_\delta$'s. $\endgroup$
    – YangMills
    Commented Mar 13, 2017 at 20:14

1 Answer 1

Reset to default
5
$\begingroup$

Yes, there is such an analog. Let $U$ be a family of subharmonic functions in a region $D$ subject to the following conditions:

a) For every compact $K\subset D$ there is a constant $C(K)$ such that $u(z)\leq C(K)$ for $z\in K$ and $u\in U$. And

b) For every $u\in U$ we have $u(z_0)>-C$ for some $z_0\in D$ and some real $C$.

Then there is a subsequence which converges to a subharmonic function. The convergence is in $L^1_{\mathrm{loc}}$ and in $D'$ and in many other senses. In particular, Laplacians converge in the weak topology of measures.

If you remove the condition b), one has to add the possibility that the subsequence converges to $-\infty$ (uniformly on compact subsets). Condition b) can be modified in various ways. For example, you can make $z_0$ depending on $u$ but restricted to some fixed compact $K$. All this is true in any dimension.

All this is contained in Theorem 4.1.9 (Hormander, vol. I), or with more details about modes of convergence, in his book Notions of convexity, Birkhauser, Boston, 1994.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, Alexandre. Do conditions a) and b) come equipped with a universal quantifier on $u\in U$? $\endgroup$
    – Mikhail Katz
    Commented Mar 14, 2017 at 7:52
  • $\begingroup$ @Mikhail Katz: Yes, I edited. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 14, 2017 at 13:01
  • $\begingroup$ The theorem you mention seems to be the same as the one mentioned in the comments by @Henri, from Notions of convexity. $\endgroup$
    – Mikhail Katz
    Commented Mar 19, 2017 at 14:54
  • $\begingroup$ Yes, it is the same. The formulation is Notions of convexity is somewhat more complete than in Linear differential operators. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 19, 2017 at 15:23

You must log in to answer this question.

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