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

Question

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 $$\left| f'(z) \right| \leq \lim_{R \to \infty} \frac{\varepsilon \left| z \right|}{(R - \left| z \right|)^2} R = 0 \implies f(z) = c \;\; \text{with} \; c \in \mathbb{C}.$$

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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.

Answer 1

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.

Answer 2

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.

Answer 3

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.