Let $u$ be a real-valued harmonic function on $\mathbb{D}$, which extends continuously to the boundary. I wonder how to prove the inequality $$\int_\mathbb{D} e^{2u} dxdy\leq\dfrac{1}{4\pi}\left(\int_{\partial\mathbb{D}}e^uds\right)^2.$$

There are a lot of properties with harmonic function $u$ (like maximum principle and mean value equality), but I did not know how to handle with this $e^u$.


1 Answer 1


This is the isoperimetric inequality in disguise. Define the Riemannian metric with length element $e^{u(z)}|dz|$. The curvature of this metric is $$-e^{-2u}\Delta u=0,$$ so the metric is flat. Then your inequality says that the area of the disk is at most $1/4\pi$ times the square of the length of the boundary, the usual isoperimetric inequality for the flat metric.

Remark. It is sufficient to assume that curvature is non-positive, that is $\Delta u\geq 0$.

Ref. MR0936419 Burago, Yu. D.; Zalgaller, V. A. Geometric inequalities. Translated from the Russian by A. B. Sosinskiĭ. Springer-Verlag, Berlin, 1988.

Isoperimetric inequality for arbitrary surfaces is usually called Fiala-Aleksandrov inequality. The original publications are:

MR0006422 Fiala, F. Le problème des isopérimètres sur les surfaces ouvertes à courbure positive, Comment. Math. Helv. 13 (1941), 293–346, and

MR0052136 Aleksandrov, A. D. An isoperimetric problem, Doklady Akad. Nauk SSSR (N.S.) 50, (1945). 31–34.


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