F. Riesz Theorem on harmonic and subharmonic functions

Question

In the book "Uniform Algebras and Jensen Measures" by T.W. Gamelin, p.39, says:

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By the F.Riesz Theorem, any subharmonic function $u$ in a neighborhood of a compact set $K$ in $C$ can be expressed in the form

$$u(z)= v(z) + \int log|z- \zeta|d\tau(\zeta),\quad z \in K$$

where $\tau$ is a positive measure supported on a compact neighborhood of $K$, and $v$ is harmonic in a neighborhood of $K$.

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Where can I find a reference for this result, especially for a proof? Can anyone provide a proof? Thank you.

score 1 · Answer 1 · edited Dec 10 '23 at 22:14

1

A proof for the fundamental theorem on subharmonic functions may be found in Masatsugu Tsuji's "On F. Riesz' fundamental theorem on subharmonic functions" (Tohoku Math. J. (2) 4(2): 131-140 (1952)).

edited Dec 10 '23 at 22:14

Alex M.

35,207

answered Nov 28 '16 at 12:32

Fred

77,394

Fantastic. Thank you. Meanwhile I should read the proof...do you think the analogous result holds in $R^d$, $d \ge 3$, where $log|z-\zeta|$ is replaced by the potential function on $R^d$ in the statement? – LDS Nov 28 '16 at 13:33

score 1 · Answer 2 · answered Jul 13 '19 at 19:22

This question was posed a long time ago so I will answer it from another point of view.

If you're willing to accept some things about distributions you can see this by looking at the potential as a convolution: if $\tau$ is seen as a distribution with compact support and then we have

\begin{equation} p_\tau(z) = \int \log |z - w|\ d\tau(w) = \log|z|*\tau \end{equation}

Now using the fact that $\Delta\log|z| = 2\pi\delta$, we have that $\Delta p_\tau = \Delta(\log|z|*\tau) = 2\pi(\delta*\tau) = 2\pi\tau$.

We need to accept some facts:

For a subharmonic function $f$ we have $\Delta f$ is a positive distribution.
Positive distributions come from positive Radon measures (also see this ).
$\Delta$ is a hypoelliptic operator meaning that if $\Delta h = 0$ for any distribution $h$, then $h$ is a harmonic (classical) function.

If we accept this, then the result follows since $\tau=\frac{1}{2\pi}\Delta f$ is a radon measure and $h = f - p_\tau$ is a harmonic function such that

\begin{equation} f = h + p_\tau. \end{equation}

PS: We also have to check that $\operatorname{supp} \tau$ is compact, so you actually need to change $f$ a little, but it is not too problematic.

PSS: I think you wanted a more classical answer but since 2 years had passed, i figured why not.

Also this result should work in $\mathbb{R}$ by convolutioning with $F$ such that $\Delta F = \delta$. All you would need to check is that $\Delta f$ is positive for any sub-harmonic function on dimension $n$. — Thiago, Jul 13 '19 at 19:24

F. Riesz Theorem on harmonic and subharmonic functions

2 Answers2