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Jensen's formula bounds the number of zeros in a smaller disc

Statement

Let $f$ be holomorphic on a neighbourhood of the closed disc ${∣ z ∣ \leq R}$ , assume $f (0) \neq 0$ , and let $n (r)$ denote the number of zeros of $f$ in $∣ z ∣ \leq r$ , counted with multiplicity, for $0 < r < R$ . Then

$n (r) \log \frac{R}{r} \leq \frac{1}{2 π} \int_{0}^{2 π} \log ∣ f (R e^{i t}) ∣ d t - \log ∣ f (0) ∣ .$

Facts & Assumptions

Given: A holomorphic function $f$ on a neighbourhood of the closed disc ${∣ z ∣ \leq R}$ with $f (0) \neq 0$ , and a radius $0 < r < R$ .

[F1]

Jensen's formula gives $\log ∣ f (0) ∣ = \frac{1}{2 π} \int_{0}^{2 π} \log ∣ f (R e^{i t}) ∣ d t - \sum_{k = 1}^{N} \log \frac{R}{∣ a_{k} ∣}$ for the zeros $a_{k}$ of $f$ in $∣ z ∣ < R$ (Jensen's formula on a disc).

Proof

technique · direct

1.1givenalgebra

If $a_{k}$ is a zero with $∣ a_{k} ∣ \leq r$ , then $\log (R / ∣ a_{k} ∣) \geq \log (R / r)$ . There are exactly $n (r)$ such zeros, counted with multiplicity.

2.1F1step 1.1algebra∎

Therefore the Jensen sum in [F1] satisfies $\sum_{k = 1}^{N} \log (R / ∣ a_{k} ∣) \geq n (r) \log (R / r)$ . Substituting this lower bound into [F1] and rearranging gives the stated inequality.

Depends on

Jensen's formula on a disc

Used by

The exponent of convergence of the zeros of an entire function does not exceed its order Theorem

Dependency tree · two levels

5 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 §1 (standard reference, not scraped)