14
$\begingroup$

(To clarify, I would just like a hint. Please do not give me the answer to this problem. ) The solution to the following problem has really evaded me here:

Problem: Assume that $f$ is entire and that $\lim_{z \to \infty} f(z)/z = 0.$ Prove that $f(z)$ is constant.

My Thoughts and Work So Far: We know that proving that $f'(z) = 0$ or that for some fixed $c \in \mathbb{C}$, $f(z) = c$ for all $z\in \mathbb{C}$. My first approach was to use Liouville's Theorem; If I could could show that $f$ is bounded then I am done. Since $\lim_{z \to \infty} f(z)/z = 0$, for all $\varepsilon > 0$ there exists a $N \in \mathbb{C}$ so large that if $z \geq N$ then $|f(z)/z| \leq \varepsilon$. Thus, if $C_R$ is the circle of radius $R$ centered at the point $z$, then, as long as z is large enough by Cauchy's Inequality
$$ |f'(z)| \leq \frac{1}{2\pi i} \oint_{C_R} \frac{f(\zeta)}{(\zeta - z)^2}\ d\zeta \leq \bigg | \frac{1}{2\pi i} \bigg | \oint_{C_R} \bigg | \frac{f(\zeta)}{(\zeta - z)^2} \bigg | d \zeta \leq \frac{1}{2\pi} \frac{|\zeta|\varepsilon 2\pi R}{R^2} = \frac{|\zeta|\varepsilon}{R}. $$

Now taking the limit as $R \to \infty$ (which to me says, "let our circle around our point z dilate to an infinite radius so that it covers all of $\mathbb{C}$) $$|f'(z)| \leq \lim_{R \to \infty} \frac{|\zeta|\varepsilon}{R} = 0.$$

Thus $f'(z) = 0$ and $z$ was arbitrary, so $f$ must be constant.

Why I Think Im Wrong: I say $z$ was arbitrary, but really it is "any $z \geq N$" which really isn't all that arbitrary.

This is where I am stuck. Am I right, wrong, close, or totally lost? Any hints would be great.

Edit: I am very sorry but I accidentally posted this before I was done typing the problem.

$\endgroup$
  • $\begingroup$ You take a circle $C_R$ centered at $0$. Then you use that to estimate $\lvert f'(z)\rvert$ for arbitrary $\lvert z\rvert \leqslant R/2$ (or $\lvert z\rvert \leqslant 1$, by the identity theorem, that's enough). You get an estimate $\lvert f'(z)\rvert \leqslant K\cdot \sup \left\{\left\lvert \frac{f(\zeta)}{\zeta}\right\rvert : \lvert \zeta\rvert \geqslant R\right\}$. $\endgroup$ – Daniel Fischer Oct 16 '13 at 15:03
  • $\begingroup$ What is $K$ in that last inequality? $\endgroup$ – Eric Oct 16 '13 at 15:05
  • $\begingroup$ Some constant. If you constrain $\lvert z\rvert \leqslant R/2$, then $K = 4$. More generally, for $\lvert z\rvert \leqslant R/c$, you have $K = \left(\frac{c}{c-1}\right)^2$ (or anything larger). It must be a bound for $\left(\frac{R}{R - \lvert z\rvert}\right)^2$, if you don't allow your $z$ to come too close to the circle, there is such a constant. $\endgroup$ – Daniel Fischer Oct 16 '13 at 15:10
  • $\begingroup$ Of course we could be lazy and say that $\dfrac{f(z)}{z}$ has a removable singularity in $\infty$, and the value is $0$, hence $f(z)$ is holomorphic in $\infty$, thus it is a holomorphic mapping $\widehat{\mathbb{C}} \to \mathbb{C}$, hence constant (since the image is compact, hence not open). $\endgroup$ – Daniel Fischer Oct 16 '13 at 15:14
  • $\begingroup$ Well we haven't covered what holomorphic is, so I don't think that I would be allowed to do that. $\endgroup$ – Eric Oct 16 '13 at 15:16
8
$\begingroup$

Hint: Use the Cauchy Integral Formula for the second derivative and show that $f'' \equiv 0$. So $f$ is a polynomial of degree at most __, and ....

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If $f$ was a polynomial function then of degree $\geq 1$ then wouldn't $z$ simply divide out an one woulde be left with a limit that is constant or tending to infinity? $\endgroup$ – Eric Oct 17 '13 at 6:50
  • $\begingroup$ @Eric $f$ is a polynomial of degree at most $1$. $\endgroup$ – user61527 Oct 17 '13 at 6:51
  • $\begingroup$ If thats true, then how could $f$ be constant? Wouldn't $f(z) = Az + B$ for some constants A,B? $\endgroup$ – Eric Oct 17 '13 at 6:55
  • $\begingroup$ @Eric If it's degree $1$, write $f(z) = az + b$. Then away from $0$, $f(z) / z = a + b/z$ tends to what as $z \to \infty$? $\endgroup$ – user61527 Oct 17 '13 at 6:56
  • $\begingroup$ then $f(z)/z$ would then tend to $a$. Which seems to be a contradiction given $\lim_{z \to \infty} f(z)/z =0$. $\endgroup$ – Eric Oct 17 '13 at 7:00
7
$\begingroup$

Hint: Consider the Taylor expansion of $f$ and the Cauchy integral formula. How can you combine the two to make the desired conclusion?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.