By Vladimir S. Azarin

In this e-book an account of the expansion concept of subharmonic features is given, that's directed in the direction of its purposes to complete capabilities of 1 and a number of other complicated variables.

The presentation goals at changing the noble artwork of creating a whole functionality with prescribed asymptotic behaviour to a handicraft. For this one should still in basic terms build the restrict set that describes the asymptotic behaviour of the full functionality.

All priceless fabric is constructed in the ebook, accordingly will probably be most precious as a reference booklet for the development of whole functions.

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Let N > 0. Set GN (x, y) := max(G(x, y), −N ), a truncation of the function G(x, y). The functions GN are continuous in Rm × Rm and GN (x, y) ↓ G(x, y) for every (x, y) when N → ∞. Set ΠN (x, μ, D) := − GN (x, y, D)μ(dy). The functions ΠN are continuous and ΠN (x, •) ↑ Π(x, •) by the B. 2). 9). Let us prove GPo5). 4 (D and C ∗ convergences) lim ΠN (x, μn , D) = ΠN (x, μ, D). n→∞ Further Π(x, μn , D) ≥ ΠN (x, μn , D), hence lim inf Π(x, μn , D) ≥ ΠN (x, μ, D). n→∞ Passing to the limit while N → ∞, we obtain GPo5).

Auxiliary Information. 3) holds in D (Ω). Let f (x) be a continuous function on ∂Ω. , [Vl, Ch. 4) ∂Ω is the only harmonic function that coincides with f on ∂Ω. The unique solution of the Poisson equation Δu = p, u|∂Ω = f for a continuous function p is given by the formula u(x, f, p) := f (y) ∂ −1 G(x, y, Ω)dsy + θm ∂ny G(x, y, Ω)p(y)dy. We can define a G(x, y, D) in the following way. Consider a sequence Ωn of a Lipschitz domain such that Ωn ↑ D. The sequence of the corresponding Green functions G(x, y, Ωn ) monotonically decreases.

O. ρ1 (r) such that (k) rk log rρ1 (r) → 0, k = 1, 2, . . when r → ∞. 8. Scale of growth. Growth characteristics of subharmonic functions Proof. 3). 5)dt n for x ∈ [n, n + 1). The function po1 (x) is continuous and infinitely differentiable due to properties of α and po1 (n) = po(n) for n = 1, 2, . . o. we have (n + 1)|po(n + 1) − po(n)| ≤ n+1 max |y · po (y)| → 0 n y∈[n,n+1] as n → ∞. Thus max y∈[n,n+1] (k) |y · po1 (y)| ≤ const ·(n + 1)|po(n + 1) − po(n)| → 0 as n → ∞. o. 8). Let us show that it is equivalent to ρ(r).

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