Assume that \(f\) is an entire function. Let \(M(r) = \sup_{|z|=r} |f(z)|\). Assume that the discrete sequence \(\{a_n\}\) are zeros of \(f\), indexed so that \(|a_1|\le |a_2|\le\cdots\), and \(f(0) = 1\). Show that \[k\log 2 \leq \log(M(2|a_k|)).\]
The Fall 2006 source does not state the ordering of \(\{a_n\}\). The displayed inequality requires the standard convention that the zeros are indexed by nondecreasing modulus; without it, an arbitrary re-enumeration makes the claim false.
Solution
Jensen’s formula for \(f(0)=1\) gives, for every \(R>0\) avoiding zeros on the circle, \[ \sum_{|a_j|<R}\log\frac{R}{|a_j|} =\frac1{2\pi}\int_0^{2\pi}\log|f(Re^{it})|\,dt \le \log M(R). \] The same inequality follows for arbitrary \(R\) by a limiting argument.
Take \(R=2|a_k|\). By the ordering, \[ |a_j|\le |a_k|\qquad(1\le j\le k), \] so for each of the first \(k\) zeros, \[ \log\frac{2|a_k|}{|a_j|}\ge\log2. \] Therefore \[ k\log2 \le \sum_{j=1}^k\log\frac{2|a_k|}{|a_j|} \le \log M(2|a_k|), \] which is the required estimate.