2016
DOI: 10.1134/s0001434616030317
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The logarithm of the modulus of a holomorphic function as a minorant for a subharmonic function

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Cited by 7 publications
(3 citation statements)
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“…then, for each function (16) with (19) and with a number P ∈ R + if D := C and for any number a > 0, there exists a function f ∈ Hol(D) such that f = 0, f (Z) = 0 and…”
Section: Main Results On Sufficient Conditions For Zero Subsets In Do...mentioning
confidence: 99%
“…then, for each function (16) with (19) and with a number P ∈ R + if D := C and for any number a > 0, there exists a function f ∈ Hol(D) such that f = 0, f (Z) = 0 and…”
Section: Main Results On Sufficient Conditions For Zero Subsets In Do...mentioning
confidence: 99%
“…The first one is to restrict to inequalities for points z lying outside some (small, if possible) exceptional set E ⊂ C. The second approach is to obtain lower bounds for integral averages of the form (2.10) or (2.7) over discs or circles of variable small radius r : C → R + \ 0. The latter option is often preferable because, first, it is additive, unlike the most commonly used lower bounds for the supremum of the function u on discs D z r(z) , and second, this approach is also capable of delivering lower bounds outside the exceptional sets E, as reflected in Theorem 2 of [50]. In this regard, we have the following result.…”
Section: Shifts and Balayage Of Amentioning
confidence: 92%
“….3)If, for this subharmonic function u ̸ ≡ −∞, its Riesz mass distribution ∆ u is of finite order ord[∆ u ] < +∞, then, for each d ∈ (0, 2], there exists an entire function f ̸ ≡ 0 satisfying (8.2),(8.3) and such that Proof. Using[51], § 1.3, Corollary 2, with a comment after this corollary, which is based on[51], § 2.4, the proof of Corollary 2, see also assertion 4 of Lemma 5.1 in[52] and Theorem 1 in[50], we find that, for each subharmonic function u ̸ ≡ −∞ on C and any number P ∈ R + with particular function P ̸ ≡ 0 such that ln |f P (z)| ⩽ u •p (z) for all z ∈ C. Hence, for the function r, with lower bounds (2.24) from Remark 2, we have r(z) ⩾ p(z) and ln |f P (z)| ⩽ u •r (z) for all z ∈ C \ RD with sufficiently large R ∈ R + . At the same time, by inequality (2.24) from Remark 2 we have r(z) ⩾ c for all z ∈ RD for some number c ∈ R + \ 0.…”
mentioning
confidence: 99%