Finding an entire function satisfying the given condition

Question

Find all entire functions that satisfy $Re(f(z)) \leq 2/{|z|}$ for $|z| \geq 1.$ Justify your answer.

My approach: I tried to go along the lines of Cauchy Integral Formula by taking disk $D(z_0,r)$ in $|z| \geq 1$ and writing $$f(\zeta)=\frac{1}{2 \pi i} \int_{C(z_0, r)} \frac{f(z)}{(z-\zeta)} d\zeta.$$ I tried the absolute values of both sides but since $Re(z) \leq |z|$ for any $z,$ I cannot use the given upper bound for $Re(f(z)).$ How should I go by in solving problems like this? If the problem says the function is a 'constant' I would have tried to show $f'(z) \equiv 0.$ Any help in solving this is much appreciated. Thank you.

Answer 1

Integrating over the circle $C(z_0,r)$ of radius r centered at $z_0$ we obtain, by the Cauchy integral formula, $$ f(z_0)=\frac1{2\pi i}\int_{C(z_0,r)}\frac{f(z)\,dz}{z-z_0}= \frac1{2\pi}\int_0^{2\pi} f(z_0+re^{i\phi})d\phi $$ and hence $$ \operatorname{Re}f(z_0)=\frac1{2\pi}\int_0^{2\pi}\operatorname{Re} f(z_0+re^{i\phi})d\phi\le \frac2{r-|z_0|}. $$ By letting $r\to\infty$, we obtain $\operatorname{Re}f(z_0)\le 0$. A semibounded harmonic function is constant by Liouville's theorem. Thus, $\operatorname{Re}f(z)$ is a nonpositive constant, and $\operatorname{Im}f(z_0)$ is a constant (being a harmonic function conjugate to a constant harmonic function). Thus, the answer is, $f(z)$ is a constant function with nonpositive real part.

Answer 2

$\text{Re}(f(z)) \le 2$ for $|z| \ge 1$, and since $f$ is continuous it is bounded on $|z| \le 1$. Thus there is some $M$ such that $\text{Re}(f(z)) \le M$ for all $z \in \mathbb C$. If $N > M$, $1/(N-f(z))$ is a bounded entire function, and therefore constant by Liouville's theorem.