Suppose $f$ is entire and $|f(z)| \leq A + B |z|^{3/2}$. Show that $f$ is a linear polynomial.

Question

Suppose $f$ is entire and $|f(z)| \leq A + B |z|^{3/2}$. Show that $f$ is a linear polynomial.

My attempt: Since $f$ is entire, we know that it has a Taylor series expansion on the circle $|z| = R$, that is $f(z) = \sum_{n = 0}^\infty a_n z^n$. By using Cauchy's estimate, we can write

$|f^n(0)| \leq \frac{n! M}{R^n}$. Since this holds for any $R>0$, it follows that for $n\geq 2$, $|f^n(0)| \leq 0$. This implies that $f^n(0) = 0$ for $n\geq 2$. Hence $f$ is a polynomial of degree at most $1$.

Edit 1 : On the circle $|z| = R$, $|f(z)|\leq A + B R^{3/2}$.

Thank you very much.

Kindly suggest me if I am wrong? Is there any other apporoach to solve such kind of problems?

Answer 1

A comment :

Note that if $f(z) = \sum_{n= 0}^\infty a_n z^n$ has a radius of convergence $> R+1$, then $$\int_{|z| = R} \frac{f(z)}{z^{k+1}}dz = \int_0^{2\pi} f(R e^{it}) e^{-i k t}R^{-k}dt = \int_0^{2\pi}\sum_{n=0}^\infty a_n e^{i (n-k) t}R^{n-k}dt $$ $$ = \left(\sum_{n=0}^N\underbrace{\int_0^{2\pi} a_n e^{i (n-k) t}R^{n-k}dt}_{= 0 \text{ if } k\ne n}\right)+\int_0^{2\pi}\underbrace{\sum_{n=N+1}^\infty a_n e^{i (n-k) t}R^{n-k}}_{o((R+1)^{-N})}dt= 2i \pi a_k$$ is obvious.

That is to say, the Cauchy integral formula is non-trivial only for $f(z)$ holomorphic (before we know that holomorphic $\implies $ analytic) and/or for a non-circular contour $\gamma$ instead of $|z| = R$.

Answer 2

Here's another proof. Write $f(z)=\sum_{n=0}^{\infty}a_nz^n.$ Using the orthogonality of the exponentials, we get

$$2\pi (A^2 + 2ABr^{3/2} +A^2r^3)\ge \int_0^{2\pi} |f(re^{it})|^2 \, dt = 2\pi \sum_{n=0}^{\infty}|a_n|^2r^{2n}.$$

That can't be true unless $a_n = 0, n = 2,3,\dots$ (think of $r\to \infty$). Thus $f(z) = a_0 + a_1z$ as desired.