I'm working on some practice exams and in one I am looking at the following question:
Let $f$ be a function holomorphic on $\mathbb{C}$. Suppose that there exist [real] constants $A$ and $B$ such that:
$$\lvert f(z) \rvert \leq A + B \lvert z \rvert ^{1/2} ~ ~ ~ (z \in > \mathbb{C})$$
Prove $f$ is constant on $\mathbb{C}$.
This is what I did, can someone check if it is correct and the steps follow properly? If not, what needs to be fixed? Thanks!
We have:
$$\lvert f(z) \rvert \leq A + B \lvert z \rvert ^{1/2} ~ ~ ~ \iff ~ ~ ~ \frac{\lvert f(z) \rvert - A}{\lvert z \rvert^{1/2}} \leq B$$
And by the triangle inequality and properties of the complex absolute value:
$$\left \lvert \frac{f(z) - A}{z^{1/2}} \right \lvert \leq \frac{\lvert f(z) \rvert - A}{\lvert z \rvert^{1/2}} \leq B$$
By Liouville's theorem, it follows the function:
$$\frac{f(z) - A}{z^{1/2}}$$
is constant. Therefore, for some constant $M \in \mathbb{C}$ we have:
$$\frac{f(z) - A}{z^{1/2}} = M ~ ~ ~ \implies ~ ~ ~ f(z) = A + M z^{1/2}$$
But $f$ is holomorphic on $\mathbb{C}$, whereas $z^{1/2}$ cannot be holomorphic on $\mathbb{C}$. Hence $M = 0$ and so:
$$f(z) = A$$
No, this is not correct, The inequality
$$|f(z) - A| \le |f(z)| - A$$
is quite false, and does not follow from the triangle inequality.
For a slightly different way to proceed, try using the Cauchy integral formula to $f'$, since $f$ is entire. For any $z$, we have
\begin{align*} |f'(z)| &= \left|\frac{1}{2\pi i} \int_{\gamma} \frac{f(s)}{(s - z)^2} ds\right| \end{align*}
where $\gamma = z + R e^{it}$, $0 \le t \le 2\pi$ is a large circle with center $z$. Now the numerator can be bounded away in terms of $R^{1/2}$, the path length is $R$, but the denominator contributes a factor of $R^2$. So
$$|f'(z)| \lesssim \frac{1}{R^{1/2}}$$
for all large $R$, so....