Evaluate $f^{prime prime}(z)$ using Cauchy's inequality.

Question

Suppose an entire function $f$ satisfies $|f(z)| leq pi|z|$ for all $z in mathbb{C}$.
(a) Evaluate $f^{prime prime}(z)$ for each $z in mathbb{C}$ using Cauchy's inequality.
By applying the Cauchy's estimate
($|f^n(a)|leq frac{Mn!}{R^n}$), we have
$|f''(0)|leq frac{2pi R}{R^2}$ and by analyticity we can let $Rrightarrow infty$ to get $f''(0)=0$, but what about other points? How to get the bound $M$ of $|f(z)|$.

Martin R · Answer

You can apply Cauchy's estimate for $f''$ to a disk of radius $R$ with arbitrary center $a$, that gives
$$
 |f''(a)| le frac{2 pi(|a| + R)}{R^2}
$$
and with $R to infty$ you can conclude that $f''(a) = 0$ for all $a in Bbb C$.

Kavi Rama Murthy · Answer

$|f(z)| leq pi R$ when $|z|=R$. Hence $|f^{(n)}(0)| leq frac {pi Rn!} {R^{n}}$ and letting $R to infty$ gives $f^{(n)}(0)=0$ for all $n >1$. The power series expansion now shows that $f(z)=f(0)+f'(0)z$ for all $z$. Hence $f''(z)=0$ for all $z$.

Answered by Kavi Rama Murthy on November 19, 2021

Evaluate $f^{prime prime}(z)$ using Cauchy's inequality.

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