# How to prove that $|Re\ h(z)|\leq\log M(r,h)$

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:

$$M(r,h)=\underset{|z|=r}{max}|h(z)|$$

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!

Context:

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:

$$h(z)=\dfrac{f(z)}{R(z)}=f(z)\dfrac{\prod_{j=1}^N(z-b_j)}{\prod_{i=1}^M(z-a_i)}$$

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.

• $|Re\ h(z)|\leq\log M(r,h)$ is surely wrong. Can you show more context how this inequality is used? That might help to recognize what the author actually meant. Commented Feb 2 at 12:17
• @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... Commented Feb 2 at 13:51
• What exactly is the “generalized Liouville's Theorem” that you are referring to? Commented Feb 2 at 14:10
• Apparently the last statement should be $\liminf_{r\to\infty}\dfrac{\max_{|z|=r}Re\ L(z)}{\log r}<\infty$. Commented Feb 2 at 14:13
• @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. Commented Feb 2 at 16:20

## 1 Answer

$$|\operatorname{Re} h(z)|\leq\log M(r,h)$$ 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.