I'm learning Complex Analysis and need to verify my work to this problem since my textbook does not provide any solution:

If $f$ is entire and $\lim_\limits{z\to\infty} \dfrac{f(z)}{z} = 0$ show that $f$ is constant.

My work and thoughts:

From the $\varepsilon$ — $\delta$ definition of the limit we have that $$\forall{\varepsilon} > 0, \exists{n_0} \in \mathbb{N} : \forall{\left|z\right|} \geq n: \left| \frac{f(z)}{z} \right| < \varepsilon \iff \frac{\left| f(z) \right|}{\left| z \right|} < \varepsilon\iff \left| f(z) \right| < \varepsilon \left| z \right|.$$

Now let $C_R = \{z \in \mathbb{C} : \left| z \right| = R \}$.

For every $\left| z \right| < R$, by Cauchy's integral formula for derivatives we have that

$$ \left| f'(z) \right| = \frac{1}{2 \pi } \left| \int_{|\zeta|=R} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta \right|= \frac{1}{2 \pi } \left| \int_{0}^{2\pi} \frac{f(\zeta)}{(\zeta - z)^2} \, \zeta'(t) dt \right| \le$$

$$\le \frac{1}{2\pi} \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} 2\pi R = \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} R.$$

Thus, letting $R \rightarrow \infty$ yields the desired result, that is $$\left| f'(z) \right| \leq \lim_{R \to \infty} \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} R = 0 \implies f(z) = c \;\; \text{with} \; c \in \mathbb{C}.$$

Is my work correct? Are there parts of the proof that need improvements? I'm also looking for other (possibly quicker) solutions using the "big guns" theorems. The only one I'm familiar with is Picard's little theorem but it doesn't apply here.

  • $\begingroup$ Yet another method might be to consider the function $g(z) := f(\frac{1}{z})$ and prove that g is entire (only have to check that g has a removable singularity at 0). Now go back to the Taylor expansion of f at $\infty$ and conclude $\endgroup$ Apr 17, 2016 at 10:33

3 Answers 3


There is a flaw in the proof. Note that for the given $\epsilon>0$, $|f(z)|<\epsilon |z|$ for $|z|>n_0$.

Therefore, on $|\zeta|=R$, $|f(\zeta)|<\epsilon |\zeta|=\epsilon R$. Then, we can write for $R>z$

$$\begin{align} |f'(z)|&=\left|\frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta\right|\\\\ &\le \frac{1}{2\pi}\frac{\epsilon R}{(R-|z|)^2}\,(2\pi R)\\\\ &=\epsilon \frac{R^2}{(R-|z|)^2} \end{align}$$

As $R\to \infty$, we find that for any $\epsilon>0$, there exists a number $n_0$, such that whenever $|z|>n_0$, $|f'(z)|<\epsilon$. We can conclude from this only that

$$\lim_{z\to \infty}f'(z)=0$$

Another approach is to write $f(z)$ in terms of its Taylor series. Then, we see that

$$\begin{align} \lim_{z\to \infty}\frac{f(z)}{z}&=\lim_{z\to \infty}\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^{n-1}\\\\ &=\lim_{z\to \infty}\left(\frac{f(0)}{z}+f'(0)+\frac12 f''(0)z+\cdots \right)\\\\ &=0 \end{align}$$

only if all terms in the series are zero, except possibly $f(0)$. Therefore, $f(z)$ must be a constant.

  • $\begingroup$ I was looking back to your answer to prepare myself for an exam and I think there's a typo in the series expansion of $f(z)/z$. For $n=2$ the term in the series is $\frac12 f''(0)z^1$ (the power of $z$ should be $1$). $\endgroup$
    – glpsx
    Jul 2, 2016 at 19:52
  • $\begingroup$ Von, yes you're correct. I've edited. $\endgroup$
    – Mark Viola
    Jul 2, 2016 at 20:24
  • $\begingroup$ @MarkViola Sorry for a question about your answer from a year ago. Regarding the second approach, can you please explain how should one justify it? For example, if we write the taylor expansion of $e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^3}{3}+....$ then the limit as $x$ goes to infinity is zero, but non of the coeffiecients is zero... Thanks in advance for any comment! $\endgroup$
    – User3231
    May 4, 2018 at 17:06
  • $\begingroup$ @kiko One cannot say that $\lim_{z\to \infty}e^{-z}=0$ for $z\in \mathbb{C}$. $\endgroup$
    – Mark Viola
    May 4, 2018 at 17:27
  • 1
    $\begingroup$ @Kiko That suffices. $\endgroup$
    – Mark Viola
    May 4, 2018 at 17:37

A simpler solution with the use of maximum modulus theorem.

Define $g : \mathbb{C} \rightarrow \mathbb{C}$, $g(z)=\begin{cases} \frac{f(z)-f(0)}{z-0}, & \text{if }z \neq 0 \\ f'(0), & \text{if }z=0 \end{cases}$

So $g$ is also an entire function by addition and quotient of entire functions. $\lim_{z\to\infty}|g(z)|=\lim_{z\to\infty} |\frac{f(z)}{z}|=0$. So $\exists L$ such as $\forall z > L, |g(z)|< \epsilon$. In particular it is true on the circle of center $0$ and of radius $L$.

With the maximum modulus theorem we have that the $max$ of $g$ on the closed discus of center $0$ and of radius $L$ is equal to the $max$ of $g$ on the circle of center $0$ and of radius $L$. So $\forall z \in \mathbb{C}, |g(z)|< \epsilon$. So $\forall z \in \mathbb{C}, g(z)=0$. So $\forall z \in \mathbb{C},f(z)=f(0)$. So $f$ is constant.

  • $\begingroup$ Jennifer, I was proceeding under the assumption that the OP had not yet learned either the maximum modulus principle or Liouville's Theorem. And the series approach that I presented seems to be quite straightforward. ;-)) -Mark $\endgroup$
    – Mark Viola
    Apr 16, 2016 at 20:35
  • 1
    $\begingroup$ Oh my remark of simpler was not about your solution, but about OP's approach, actually I did not see your answer while I was writing mine. But indeed your approach is concise and more straightforward. :) $\endgroup$
    – Bérénice
    Apr 16, 2016 at 20:37
  • $\begingroup$ Jennifer, no worry at all please. You've posted a solid answer! Well done. -Mark $\endgroup$
    – Mark Viola
    Apr 16, 2016 at 20:38

As @Dr.Mv said, we have that $\lim_{z\rightarrow\infty}f'(z)=0$, which means that $f'$ is bounded. By Liouville's theorem $f'$ is constant and, since $\lim_{z\rightarrow\infty}f'(z)=0$, $f'=0$. Thus , $f$ is constant.

  • 1
    $\begingroup$ Chris, this is fine. I was proceeding under the assumption that the OP had not yet learned either the maximum modulus principle or Liouville's Theorem. -Mark $\endgroup$
    – Mark Viola
    Apr 16, 2016 at 20:33
  • $\begingroup$ @MarkViola How does one justify that $\lim_{z \to \infty} f'(z)=0$ imply that $f'$ is bounded? $\endgroup$
    – user330477
    Sep 16, 2020 at 7:40
  • $\begingroup$ @user330477 Can an entire function, $g(z)$ be unbounded on any bounded domain? Now if the entire function $g(z)$ also has the property that $\lim_{z\to\infty}g(z)=0$, then can $g(z)$ be unbounded? $\endgroup$
    – Mark Viola
    Sep 20, 2020 at 16:15

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