# Holomorphic function $f: \Bbb C \rightarrow \Bbb C$ such that $|f(z)|\leq C|\negthinspace\cos(z)|$ for all $z \in \Bbb C$ [duplicate]

The Question:

Let $f:\Bbb C \rightarrow \Bbb C$ be a holomorphic function (i.e. an entire function) such that $|f(z)|≤C|\negthinspace\cos(z)|$ for all $z \in \Bbb C$, where $C \in \Bbb R$ is a constant.

What can you say about $f$?

My Thoughts:

I know that if $|f(z)|≤C|z^n|$ then $f$ must be a polynomial of degree $≤n$ (as I have proven in a previous part of the question), but I don't see how it generalizes to $|\cos(z)|$.

Also, this seems somehow related to Liouville's Theorem.

Certainly, functions of the form $f(z)=C' \cos (z)$ would work, provided $|C'|≤C$, and I can't seem to come up with any other function.

Any hints?

• Show that $f/\cos$ extends to an entire function which is bounded.
– WimC
Commented May 11, 2018 at 2:38

The condition on $f$ immediately implies that when $cos(z)$ vanishes, so does $f$. Let $z_0$ be a zero of order $k$ for $cos(z)$ and therefore also a zero of $f$, for some other order $l$. Now, in an open disc $D$ around $z$, $cos(z)= (z-z_0)^k\phi(z)$ and $f(z) = (z-z_0)^l\psi(z)$, with $\phi,\psi \in \mathcal{H}(D)$ never zero in the disc. Since $|f(z)| \leq C|cos(z)|$,

$$|z-z_0|^{l-k} \leq C\frac{|\phi(z)|}{|\psi(z)|}$$

Taking limits, we have that

$$\lim_{z\to z_0}|z-z_0|^{l-k} \leq C\frac{|\phi(z_0)|}{|\psi(z_0)|}$$

and in particular, for this function to be bounded near $z_0$ we shall have that $k\leq l$. Therefore,

$$\frac{f(z)}{cos(z)} = (z-z_0)^{l-k}\frac{\psi(z)}{\phi(z)}$$

which is holomorphic in $D$. Since this can be done for each zero of $cos(z)$, the discontinuities of $g(z) = \frac{f(z)}{cos(z)}$ are all avoidable and therefore the function is entire. By the inequality of the problem we also know that $g$ is bounded, so by Liouville's theorem, it has to be constant. Finally, this shows that in effect, the family

$$\mathcal{F} = \{\lambda \cos(z) : |\lambda| \leq C , \ \lambda \in \mathbb{C}\}$$

consists of all the functions that satisfy this property.

• Aren't all zeros of $\cos z$ of order 1? Commented May 11, 2018 at 5:19
• @KaviRamaMurthy In hindsight, this should simplify the proof (aesthetically, at least). The idea is the same though, right? Commented May 11, 2018 at 16:39
• Yes, that it was I wanted to say. In your proof $k=1$ necessarily. Commented May 12, 2018 at 11:43
• It does not really matter what the zeros are: they are isolated, and since $|f(z)|/|\cos z|$ is bounded, they are removable singularities.
– user357151
Commented May 12, 2018 at 19:39