1
$\begingroup$

Consider a compact subset $E\subset\mathbb{C}$, holomorphic functions $f_j:V\to \mathbb{C}$, $1\leq j\leq k$, defined in a neighbourhood $V$ of $E$, and set $u:V\to\mathbb{R}\cup\{-\infty\}$, $u(z)=\max_{1\leq j\leq k} \log|f_j(z)|$.

Is there an explicit formula for the potential part $$ \int \log |z-w|\,d\mu(w) $$ of the Riesz decomposition of the subharmonic function $u(z)$?

If not, is at least the support of $\mu$ contained in the set $\Gamma=\Gamma_0\cup \cup_{i<j} \Gamma_{i,j}$, where $\Gamma_0=\{z\in E: f_j(z)=0,\,\forall j\}$ and $\Gamma_{i,j}=\{z\in E: |f_i(z)|=|f_j(z)|=e^{u(z)}\}$?

And can we relate the density of $\mu$ with respect to Lebesgue on $\Gamma_{i,j}$ with $|f_i'(z)-f_j'(z)|$?

$\endgroup$
2
$\begingroup$

The "formula" is this. Let $j(z)$ be that $j$ for which $|f_j(z)|$ is maximal among all $|f_k(z)|$. (There can be several $j$ with this property, choose one, or just remove from your set several functions, to make this $j(z)$ unique). Then $j(z)$ is locally constant on some open set, the boundary of this open set consists of analytic curves which can meet at isolated points. If $m$ and $k$ are two values of $j(z)$ on the right and the left of such a curve, which means that $|f_n(z)|=|f_m(z)|$ on the curve, while $|f_m(z)|$ is the largest one on the left, and $|f_n|$ on the right, then the density of the Riesz measure on this curve (with respect to the arclength) is equal to $$\frac{1}{2\pi}|(d/dn)\log|f_n(z)|-(d/dn)\log|f_m(z)||$$ where $d/dn$ is the derivative along the normal to the curve. The direction of the normal in this formula can be chosen arbitrarily, but the same in both summands. In other words, this is the jump of the directional derivative of $\max_k\log|f_k|$, in the direction normal to the curve on which this $\max$ is not smooth.

All this is easy to prove, and people frequently use this, see, for example, Proc. Amer. Math. Soc., 140, (2012) 1397-1402, formula (7), but there is no convenient reference where this fact is stated. People either prove this, in the situation where they need it, or say that this is evident, or well-known. I've seen this formula as an exercise in some Russian book, but I do not remember which book. (There was no proof anyway).

Of course, the gradient of $\log|f|$ is $|f'/f|$ but here we deal not with the gradient but with the derivative in the normal direction of the curve. But of course you can produce from the above formula a formula which is more convenient for your purpose.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.