As part of a proof in complex analysis, I have encountered a claim that is presented as evident but that I don't understand and don't know how to prove. We have a function $h(z)$, and have just defined $L(z)=\log h(z)$, for reasons unclear as it doesn't seem like it is used later on in the proof. Right after this definition, comes this statement:

Since $h(z)$ has no zeroes, $L$ is an entire function, and it holds that $|Re\ h(z)|\leq\log M(r,h)$.

In this discussion, we consider:


I wonder whether there is a typo and the author actually meant $|Re\ L(z)|\leq M(r,h)$. My attempt at trying to prove the claim is the following:

$$M(r,h)=\underset{|z|=r}{max}|h(z)|=\underset{|z|=r}{max}\sqrt{[Re\ h(z)]^2+[Im\ h(z)]^2}\geq\underset{|z|=r}{max}|Re\ h(z)|\geq |Re\ h(z)|,$$

$$\forall z: |z|\leq r$$

So, I have proven that $|Re\ h(z)|\leq M(r,h)$. But how to get the desired result from here? We have that $\log M(r,h)\leq M(r,h)$, not the other way around.

Any help or insights into this mystery would be greatly appreciated!


We are trying to prove that if a meromorphic function $f(z)$ satisfies the following relation:

$$\liminf_{r\to\infty}\dfrac{T(r,f)}{\log r}<\infty$$

then $f(z)$ is a rational function.

The outline of the proof is as follows. According to the first fundamental theorem of Nevanlinna, $f$ can only have a finite number of zeros ($a_1,...,a_M$) and poles ($b_1,...,b_N$). Then, $R(z)$ is a rational function with the same zeros and poles as $f(z)$ and h(z) is a meromorphic function without zeros that are not poles, that is, an analytic function without zeros, which will be rational if and only if $f(z)$ is rational:


Using the Jensen-Poisson formula and several other results, we can prove that:

$$\liminf_{r\to\infty}\dfrac{\log M(r,h)}{\log r}=\liminf_{r\to\infty}\dfrac{T(r,h)}{\log r}<\infty$$

On the other hand, the function $L(z)=\log h(z)$ is analytical and uniform in the complex plane, and therefore is an entire function, since $h$ has no zeros. Moreover,

$$|Re\ h(z)|\leq\log M(r,h)$$

which means we can conclude

$$\liminf_{r\to\infty}\dfrac{Re\ h}{\log r}<\infty$$

and so, according to the generalized Liouville's Theorem, $h(z)$ is a constant functions, thus proving that $f(z)$ is rational.

  • $\begingroup$ $|Re\ h(z)|\leq\log M(r,h)$ is surely wrong. Can you show more context how this inequality is used? That might help to recognize what the author actually meant. $\endgroup$
    – Martin R
    Commented Feb 2 at 12:17
  • 1
    $\begingroup$ @MartinR I added a sketch of the proof where the author introduces his strange claim. I agree that surely there must be a typo, but I can't fully understand which one or how he concludes the proof. I tried to summarize the proof as much as possible, skipping details that didn't seem necesarry to elucidate what is going on... $\endgroup$ Commented Feb 2 at 13:51
  • $\begingroup$ What exactly is the “generalized Liouville's Theorem” that you are referring to? $\endgroup$
    – Martin R
    Commented Feb 2 at 14:10
  • $\begingroup$ Apparently the last statement should be $\liminf_{r\to\infty}\dfrac{\max_{|z|=r}Re\ L(z)}{\log r}<\infty$. $\endgroup$
    – Martin R
    Commented Feb 2 at 14:13
  • $\begingroup$ @MartinR The generalized Liouville's theorem states, according to these notes, that if an entire function $g(z)$ satisfies $\liminf_{r\to\infty} \frac{Re\ g(z)}{\log r}<\infty$, then $g$ is constant. $\endgroup$ Commented Feb 2 at 16:20

1 Answer 1


$|\operatorname{Re} h(z)|\leq\log M(r,h)$ is definitely wrong, as the example $h(z) = e^z$ shows: For $z = r > 0$ is $$ |\operatorname{Re} h(z)| = e^r > r = \log M(r,h) \, . $$

The proof should go as follows: $h$ is an entire function without zeros, so $h(z) = e^{L(z)}$ with an entire function $L$. Then $$ |h(z)| = e^{\operatorname{Re} L(z)} \\ \implies \log |h(z)| = \operatorname{Re} L(z) \\ \implies \log M(r, h) = \max_{|z|=r} \operatorname{Re} L(z) \, . $$ Then $$ \liminf_{r \to \infty} \frac{\max_{|z|=r} \operatorname{Re} L(z)}{\log r} =\liminf_{r \to \infty} \frac{\log M(r,h)}{\log r} < \infty \, . $$ It follows that $L$ is constant, say $L(z) = C \in \Bbb C$. Then $h(z) = e^C$ is also constant, and $f(z) = e^C \cdot R(z)$ is a rational function.


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