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Plane subharmonic functions are locally integrable

Statement

Every subharmonic function on a complex domain belongs to $L_{l o c}^{1}$ .

Facts & Assumptions

Given: A subharmonic function $u$ on a complex domain $Ω$ .

[L1]

On every closed disc inside the domain, a harmonic function that dominates $u$ on the boundary dominates $u$ throughout the disc (Subharmonicity is equivalent to harmonic comparison on compactly contained discs).

[L2]

On a compact circle, the boundary values of an upper semicontinuous function are Borel measurable and bounded above, so the circle averages in the submean inequality are finite or $- \infty$ (Upper semicontinuous functions are Borel and their circle averages are defined).

[L3]

Every open connected subset of $R^{2}$ is polygonally connected (For an open subset of $R^{n}$ , connectedness, path-connectedness and polygonal connectedness are equivalent).

[L4]

For a nonnegative Borel function $F$ on a disc, the polar-coordinate identity $\int_{D (c, s)} F d A = \int_{0}^{s} \int_{0}^{2 π} F (c + t e^{i θ}) t d θ d t$ follows first for indicators of annular sectors and nonnegative simple functions, and then for general $F$ by monotone convergence.

Proof

technique · direct

1.1

Fix a compact disc $\overline{D (a, R)} \subseteq Ω$ and a smaller concentric disc $\overline{D (a, ρ)}$ with $0 < ρ < R$ . The function $u$ is upper semicontinuous on the compact circle $\partial D (a, R)$ , so [L2] gives a finite upper bound there. By the harmonic-comparison theorem [L1], $u$ is therefore bounded above on $\overline{D (a, ρ)}$ by the harmonic majorant obtained from any continuous boundary majorant on $\partial D (a, R)$ .

L1L2given

1.2

The set $A = {z \in Ω : u (z) = - \infty}$ has empty interior. Suppose instead that $D (c_{0}, R) \subseteq A$ for some $R > 0$ , and fix any point $x$ in the connected component of $Ω$ containing $c_{0}$ . By [L3], choose a polygonal path $γ$ in that component from $c_{0}$ to $x$ . Because $γ ([0, 1])$ is compact and lies in the open set $Ω$ , choose $r > 0$ with $3 r < R$ and $\overline{D (y, 4 r)} \subseteq Ω$ for every $y \in γ ([0, 1])$ . Subdivide the path by points $c_{0}, c_{1}, \dots, c_{N} = x$ on $γ$ with $∣ c_{j + 1} - c_{j} ∣ < r / 2$ for every $j$ . We claim inductively that $D (c_{j}, 3 r) \subseteq A$ for all $j$ . The case $j = 0$ holds because $3 r < R$ . If $D (c_{j}, 3 r) \subseteq A$ and $b \in D (c_{j}, 7 r / 2)$ , choose $ρ$ with $\max (0, ∣ b - c_{j} ∣ - 3 r) < ρ < r / 2$ . Then the circle $∣ z - b ∣ = ρ$ lies in $D (c_{j}, 4 r) \subseteq Ω$ and meets $D (c_{j}, 3 r)$ in an open arc, so $u = - \infty$ on a set of positive arc-length measure on that circle. By [L2], the circle values are Borel measurable and bounded above, hence the circle average is $- \infty$ . The submean inequality therefore gives $u (b) = - \infty$ , proving $D (c_{j}, 7 r / 2) \subseteq A$ . Since $∣ c_{j + 1} - c_{j} ∣ < r / 2$ , one has $D (c_{j + 1}, 3 r) \subseteq D (c_{j}, 7 r / 2) \subseteq A$ , completing the induction. In particular $x \in A$ . Because $x$ was arbitrary in the component, this contradicts the subharmonic convention that $u$ is not identically $- \infty$ there. Thus $A$ has empty interior, so every open subdisc contains a point where $u$ is finite.

L2L3givenalgebra

2.1

Fix $a \in Ω$ and choose $r > 0$ with $\overline{D (a, 3 r)} \subseteq Ω$ . Step 1.1, applied with outer radius $3 r$ and inner radius $2 r$ , gives a finite upper bound $M$ for $u$ on $D (a, 2 r)$ . By step 1.2 choose $c \in D (a, r / 2)$ with $u (c) > - \infty$ , and then choose $r + ∣ c - a ∣ < s < 2 r - ∣ c - a ∣ .$ These inequalities give $D (a, r) \subseteq D (c, s) \subseteq D (a, 2 r)$ .

step 1.1step 1.2choosealgebra

3.1

For every $0 < t < s$ , the submean inequality at $c$ gives $\int_{0}^{2 π} (M - u (c + t e^{i θ})) d θ \leq 2 π (M - u (c)) .$ The integrand is nonnegative and Borel by [L2]. Multiply by $t$ , integrate from $0$ to $s$ , and apply [L4] to obtain $\int_{D (c, s)} (M - u) d A \leq π s^{2} (M - u (c)) < \infty .$ Thus the negative part of $u$ is integrable on $D (c, s)$ , while its positive part is bounded there by $M$ . Hence $u \in L^{1} (D (c, s))$ , and therefore $u \in L^{1} (D (a, r))$ .

step 2.1L2L4algebra

4.1

Every point $a \in Ω$ admits such a disc $D (a, r)$ , so $u \in L_{l o c}^{1} (Ω)$ .

step 3.1∎

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Boris Khoruzhenko, Potential Theory lecture notes (standard reference, not scraped)