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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:
$$h(\theta)=\limsup_{r\to\infty}\frac{\log|f(re^{i\ …
3
votes
Accepted
Largest asymptotic growth for $2f(x)-f(2x)$
Let us discretise the problem by setting $a_n=2^{-n}f(2^n)$, $b_n=2^{-n-1}\Delta_f(2^n)$. Then your relation becomes,
$$b_n=a_n-a_{n+1}.$$
since $a_n,b_n$ are non-negative, we conclude that
$$\sum_{n= …