(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.

  • $\begingroup$ 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

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 ....

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  • $\begingroup$ 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

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

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