Dimension of a space of holomorphic functions

Question

I would like to solve the following: for $r$ a positive real number, determine the dimension of the following vector space over $\mathbb{C}$ in terms of $r$:

$$[\text{holomorphic } f:\mathbb{C}\rightarrow\mathbb{C}\text{ where } f<\infty].$$

My idea is to show that all such holomorphic functions are really nice in some way, like polynomials. I've tried playing around with the power series expansion and showing this directly, but I'm not making much progress. Is this the right idea? Any help is appreciated.

Answer 1

Consider a function $f$ from the desired space, which we'll call $V_r$.
For $|z| \geq 1$, we have $|z|^r \leq |z|^{\lceil r \rceil}$ where $x \mapsto \lceil x \rceil$ is the ceiling function.
We thus get, for all $|z| \geq 1$, $|f(z)| \leq M |z|^{n}$ with $M := \sup_{|z|\geq1}\left|\frac{f(z)}{z^r}\right|$ and $n := \lceil r \rceil$.
Following this post: Entire function bounded by a polynomial is a polynomial, such $f$s are therefore polynomials of degree at most $n$.

However, they cannot be of degree exactly $n$ if $r$ is not an integer due to the fact that $z \mapsto \frac{f(z)}{z^r}$ is bounded (by $M$) since, if $f$ were of degree $n$, we would have for example $\frac{f(x)}{x^r} \underset{x \to +\infty}{\sim} \frac{az^n}{z^r} \xrightarrow[x \to +\infty]{} \operatorname{sg}(a)\infty$ along the real axis, where $a$ is the leading coefficient of $f$.
Hence, when $r$ is not an integer, $V_r$ is made up of polynomials of degree at most $\lceil r \rceil - 1 = \lfloor r \rfloor$ where $x \mapsto \lfloor x \rfloor$ is the floor function, and $V_r$ is therefore, in all cases ($r$ integer or not integer), comprised of polynomials of degree at most $\lfloor r \rfloor$.

Finally, it suffices to show that all polynomials of degree at most $\lfloor r \rfloor$ are in $V_r$, but that's fairly easy because, for $|z| \geq 1$: $$\left|\sum_{k = 0}^{\lfloor r \rfloor} a_k z^k\right| \leq \sum_{k = 0}^{\lfloor r \rfloor} |a_n||z^k| \leq \left(\sum_{k = 0}^{\lfloor r \rfloor} |a_n|\right) \cdot |z|^r$$

In conclusion, $V_r$ is exactly the space of all polynomials of degree at most $\lfloor r \rfloor$, thus of dimension $\lfloor r \rfloor + 1$.

Answer 2

Observe that, if $r=0$, your function has to be a constant by Liouville’s Theorem. The space of constants over $\mathbb{C}$ has dimension $=1$. If $r<0$ your condition is stronger. Apart from being a constant, it has to be zero, otherwise the limit at infinity would be infinity. In this case the dimension is obviously zero. If $r>0$, $f(z)/z^r= const$ at infinity implies $f(z)=P_r(z)$, where $P$ is a polynomial of degree $\le r$ and your space has dimension $=r+1$. I am assuming that $r$ is an integer, otherwise you have to replace by $\lfloor r \rfloor$.