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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.

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2 Answers 2

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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.

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  • $\begingroup$ I'm not quite on track. So I may understand that $|f(z)-f(b)|\leq 0$ and $R \to \infty$, but the whole idea behind fixing $a$, $b$ and the equality established is a little confusing. Can you please elaborate? Thanks. @Tom $\endgroup$ Commented Oct 31, 2013 at 13:09
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    $\begingroup$ Sure! Let $a \in \mathbb{C}$ and set $P = f(a)$. Now take any arbitrary $z \in \mathbb{C}$ and as you have written you have shown that for $R$ large enough $|f(z) - f(a)|\leq \text{constant}\cdot R^{-1}$. But since this is true for any $R$ large enough, then you can take it as large as you want, which forces the inequality $|f(z) - f(a)| \leq 0$ to be true for any $z$. Since $0 \leq |f(z)-f(a)| \leq 0$ then $f(z)=f(a)$ for every $z \in \mathbb{C}$. That is $f(z) = P$ for every $z \in \mathbb{C}$, meaning that $f$ is constant. $\endgroup$
    – Tom
    Commented Oct 31, 2013 at 13:17
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    $\begingroup$ Brilliant, @Tom. Thanks a lot mate! That made perfect sense. $\endgroup$ Commented Oct 31, 2013 at 13:19
  • $\begingroup$ @eXtremiity Glad to help! $\endgroup$
    – Tom
    Commented Oct 31, 2013 at 13:27
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You just computed the result in two ways so you have that for all a,b complex numbers $$f(a)=f(b)$$ This happens only when $f$ is constant

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