Given a sequence of real numbers $c_k\to-\infty$, is there always a $C^\infty$ subharmonic function $f$ on $\mathbb R^2$ and a sequence $z_k\to\infty$ with $|z_k|<k$ such that $$\displaystyle\limsup_{z\to\infty} \frac{f(z)}{\log |z|}<\infty\ \ \text{and}\ \ f(z_k)<c_k\ ?$$

I do not even know the answer for non-smooth $f$. In Hayman's "Subharmonic functions", volume 2, in example 7.26 on page 449 there is a weaker result that there is subharmonic $f$ of sub-logarithmic growth (in the sense above) and $z_k\to\infty$ such that $f(z_k)<-\log|z_k|$.

The question came up in research but it seems like an interesting question by itself.


1 Answer 1


Yes, of course. Take any convergent series with positive terms $a_k$. Consider the function $u(z)=\sum a_k\log|1-z/z_k|$. This is a subharmonic function, $u(z_k)=-\infty$, and satisfies $u(z)=O(\log|z|)$ because the Riesz mass is finite. It is continuous (in the extended sense, $-\infty$ is allowed).

To make it smooth, take a symmetric positive cap with bounded support, and take a convolution with this cap. When the support does not contain $z_k$ this convolution does not change the function (because it is harmonic there and has the average property). This allows to to take the cap support smaller and smaller with $k$.

More formally, let $\phi_k$ be a positive symmetric $C^\infty$ cap supported in the disk of radius less than $1/4$ of the distance from $z_k$ to the rest of $z_j$. In a neighborhood of $z_j$, replace $u$ by $u*\phi_k$. The result is the $C^\infty$ function you need.

  • $\begingroup$ Wonderful! I was thinking of how to do precisely this and was stuck. The two points I was missing: (1) the Riesz mass equals $\sum_k a_k$ by e.g. [Berenstein-Gay's book, Prop 4.4.22], and also equals the limsup from my question [Hayman's book, Theorem 6.32]. (2) $\sum_k a_k\log|1-z/z_k|$ is subharmonic as a locally uniform limit of its partial sums. To check that the limit exists fix a compact set and drop several terms in the sum to ensure that $|z/z_k|\ll 1$ so that each summand is negative. $\endgroup$ May 15, 2015 at 2:51
  • $\begingroup$ To bound the partial sums from below consider $\log |P_n^m|$ where $m$ is any integer $>\sup_k a_k$ and $P_n$ is the product of $(1-z/z_k)$'s, $k=1,\dots, n$. Convergence of $P_n$'s follows because $|z/z_k|$ is bounded away from $1$, see e.g. [Berenstein-Gay, Prop 4.6.12]. $\endgroup$ May 15, 2015 at 2:52

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