I'm having trouble completing a homework question which will produce an alternative proof for Liouville's Theorem. The question reads
Let $f$ be an entire function. Evaluate, for $|a|<R,|b|<R,\int_{|z|=R}\frac{f(z)dz}{(z-a)(z-b)}$. When $f$ is bounded, let $R\to \infty$ and deduce another proof for Liouville's Theorem.
Liouville's Theorem: If a function f is entire and bounded in the complex plane, then $f(z)$ is constant throughout the plane.
So I begin by evaluating the integral. Using the Cauchy-Residue theorem (and skipping some methodic computation) I have that $\int_{|z|=R}\frac{f(z)dz}{(z-a)(z-b)}=2\pi i[\frac{f(a)-f(b)}{a-b}].$Now, it is given that $f$ is entire (analytic everywhere) and bounded. Thus all I must show is that it is constant. Since $f$ is bounded there exists some non-negative number $M$ such that $|f(z)|<M$.
I proceed by using the M-L Lemma in attempts to try and generate some ideas.
$|\frac{f(z)dz}{(z-a)(z-b)}|\leq\frac{M}{(R-a)(R-b)}\times 2\pi R$, by the conditions given in the question and the use of the reverse triangle inequality.
Hence, $|\int_{|z|=R}\frac{f(z)dz}{(z-a)(z-b)}|\leq\frac{M}{(R-a)(R-b)}\times 2\pi R$. Now as $R\to\infty$, the RHS of the inequality clearly approaches 0. Thus $\int_{|z|=R}\frac{f(z)dz}{(z-a)(z-b)}=0?$ I think I saw this in one of my text books. Does this lead on to the conclusion that $f(z)$ is constant? But then I realise I have not made use of the equation I established above.
Thank you to all in advanced for you help.
Best Answer
Fix $a$. Take $b$ arbitrarily and you showed that $$ \big| f(a) - f(b) \big| \leq \text{constant}\cdot \frac{M}{R} $$ taking $R \to \infty$ you get that $f(a) = f(b)$. Therefore for any $b \in \mathbb{C}$ you have $f(b) = f(a)$, implying that $f$ is constant.