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2 votes

Does the rate of decay of an entire function dictate the global growth rate?

The answer to your question is no. More generally, for functions of exponential type $b$ (which means $|f(z)|\leq Me^{b|z|}$), define the indicator: $$h(\theta)=\limsup_{r\to\infty}\frac{\log|f(re^{i\ …
Alexandre Eremenko's user avatar
3 votes
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Largest asymptotic growth for $2f(x)-f(2x)$

Let us discretise the problem by setting $a_n=2^{-n}f(2^n)$, $b_n=2^{-n-1}\Delta_f(2^n)$. Then your relation becomes, $$b_n=a_n-a_{n+1}.$$ since $a_n,b_n$ are non-negative, we conclude that $$\sum_{n= …
Alexandre Eremenko's user avatar