Privalov operators

Privalov parameters

Operators that allow one to express the condition of harmonicity of a function without using partial derivatives (cf. Harmonic function). Let $u$ be a locally integrable function in a bounded domain $D$ of a Euclidean space ${\mathbf{R}}^n$, $n\ge 2$; let $\omega (h)$ denote the volume of the ball $B(x;h)$ of radius $h$ with centre $x\in D$, lying in $D$; and let

$${\mathrm{\Delta }}_{h}u(x)=\frac{1}{\omega (h)}\underset{B(x;h)}{\int }u(y)dy-u(x).$$

The upper and lower Privalov operators $\bar{\mathrm{\Delta }}{}^{\ast }u(x)$ and ${\underset{―}{\mathrm{\Delta }}}^{\ast }u(x)$ are defined, respectively, by the formulas

$$\bar{\mathrm{\Delta }}{}^{\ast }u(x)=\overline{lim}{}_{h\to 0}2(n+\frac{2)}{h^2}{\mathrm{\Delta }}_{h}u(x),$$

$${\underset{―}{\mathrm{\Delta }}}^{\ast }u(x)=\underset{\overline{h\to 0}}{lim}2(n+\frac{2)}{h^2}{\mathrm{\Delta }}_{h}u(x).$$

If the upper and lower Privalov operators coincide, then the Privalov operator ${\mathrm{\Delta }}^{\ast }u(x)$ is defined by

$${\mathrm{\Delta }}^{\ast }u(x)=\bar{\mathrm{\Delta }}{}^{\ast }u(x)={\underset{―}{\mathrm{\Delta }}}^{\ast }u(x)=\underset{h\to 0}{lim}2(n+\frac{2)}{h^2}{\mathrm{\Delta }}_{h}u(x).$$

If the function $u$ has continuous partial derivatives up to and including the second order at $x\in D$, then the Privalov operator ${\mathrm{\Delta }}^{\ast }u(x)$ exists at $x$ and is equal to the value of the Laplace operator: ${\mathrm{\Delta }}^{\ast }u(x)=\mathrm{\Delta }u(x)$. Privalov's theorem says: If a function $u$, continuous in a domain $D$, satisfies everywhere in $D$ the conditions

$${\underset{―}{\mathrm{\Delta }}}^{\ast }u(x)\le 0\le \bar{\mathrm{\Delta }}{}^{\ast }u(x),$$

then $u$ is harmonic in $D$. This implies that a function $u$, continuous in $D$, is harmonic if and only if at every point $x\in D$ one has ${\mathrm{\Delta }}_{h}u(x)=0$, from some sufficiently small $h$ onwards, or, in other words, if and only if

$$u(x)=\frac{1}{\omega (h)}\underset{B(x;h)}{\int }u(y)dy.$$

The average value over the volume of a sphere can be replaced by that over the surface area.

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More generally, if $u>-\mathrm{\infty }$ is lower semi-continuous, then $u$ is hyperharmonic if and only if ${\underset{―}{\mathrm{\Delta }}}^{\ast }u\le 0$ on $\{u<\mathrm{\infty }\}$( the theorem of Blaschke–Privalov).

Similar results hold if the average value over the surface area is used for the operators and $2(n+2)$ is replaced by $2n$.