complex inequality (solution verification)

Question

Let $f$ be entire, $|f(z)|\leq 3|z|^4+1$ prove $f$ is a polynomial with deggre $\leq 4$

Solution:

$f^{(4)}(z_0)=\frac{4!}{2\pi i} \int_c\frac{f(z)}{(z-z_0)^5}dz$, $\Rightarrow$ $|f^{(4)}(z_0)|\leq \frac{4!}{2\pi i} \int_c |\frac{f(z)}{(z-z_0)^5}|dz \leq \frac{4!}{2\pi i} \int_c \frac{|f(z)|}{|(z-z_0)^5|}dz \leq$ $$\leq \frac{4!}{2\pi i} \int_{0}^{2\pi}\frac{|f(z_0+re^{it})ire^{it}|}{|(re^{it})^5|}dt \leq \frac{4!}{2\pi i} \int_{0}^{2\pi}\frac{(3(|z_0|+r)^4+1)r}{r^5}dt \leq \frac{4!}{2\pi i} 2\pi\frac{(3(|z_0|+r)^4+1)}{r^4}$$

that means $f^{(4)}(z)$ is bounded so from Liouville's theorem its constant and therefore $f$ is a polynomial with deggre $\leq 4$

Is my solution correct ?

Answer 1

Your solution is almost correct. There are a few minor errata. Note that

$$f^{(4)}(z)=\frac{4!}{2\pi i}\oint_C \frac{f(w)}{(w-z)^5}\,dw$$

where $C$ is a rectifiable closed contour that encloses the point $z$.

We may choose $C$ to be the circular contour $|w-z|=r$. Then, we have

$$\begin{align} |f^{(4)}(z)|&=\left|\frac{4!}{2\pi i}\oint_C \frac{f(w)}{(w-z)^5}\,dw\right|\\\\ &\le \frac{4!}{2\pi \color{red}{\underbrace{|i|}_{=1}}}\left|\int_0^{2\pi } \frac{|f(z+re^{i\phi})|}{|r|^5}|\color{red}{\underbrace{ire^{i\phi}}_{=r}}|\,d\phi\right|\\\\ &\le\frac{4!}{r^4}\max_{\phi}|f(z+re^{i\phi})|\\\\ &\le \frac{4!(3\max_{\phi}(|z+re^{i\phi}|^4)+1)}{r^4}\\\\ &=\frac{4!(3(r+|z|)^4+1)}{r^4} \end{align}$$

Letting $r\to \infty$ shows that

$$|f^{(4)}(z)|\le 72$$

whereby appealing to Liousville's Theorem implies that $f^{(4)}(z)$ is constant from which we conclude that $f(z)$ is a polynomial of order $4$.