# Dimension of a space of holomorphic functions

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.

• Weird, I recall someone else asking the same question somewhat recently, and I commented on it, but I cannot find it again. Anyway, this could help you. Jun 4 at 23:48
– user1181358
Jun 5 at 0:45

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 $$\lfloor r \rfloor + 1$$.

• Thank you! What is the sg(a)∞ notation?
– user1181358
Jun 5 at 1:30
• It's "sign of $a$ infinity". Jun 5 at 1:34

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 $$\lfloor r \rfloor$$.

• The condition $\sup_{|z|\geq1}|f(z)/z^r|<\infty$ does not imply that $f(z)/z^r = const$, all polynomials of degree at most a fixed integer $n$ satisfy $|f(z)| \leq M |z|^{n+1}$ for some constant $M$ for example. Moreover, $z \mapsto z^r$ is holomorphic iff $r$ is an integer anyway. Jun 5 at 0:12
• @user5684138 I was assuming that $r$ was an integer. If $r$ is not an integer, the same condition holds if you replace $r$ by the ceiling of $r$, as Bruno B explains. The reason why if a holomorphic function bounded by a polynomial $C|z|^n$ has to be a polynomial of degree $\le n$ itself is a consequence of Liouville's theorem applied to the $n$-th derivative of $f$, see the link supplied by Bruno B. And yes, a basis of the space of functions satisfying the given property is $\{1,z,z^2,\dots z^n\}$ where $n=\lfloor r \rfloor$. Its dimension is $n+1$ as you have $n+1$ free coefficients. Jun 5 at 18:15