Do the holomorphic or meromorphic functions on a domain $D \subseteq \mathbb{C}$ form a Hilbert space $\mathcal{H}$? In a physics paper I am reading that the meromorphic functions on $\mathbb{C}$ with $f(x) = f(\overline{x})$ form a Hilbert space.
$$ \mathcal{H} = \{ f(x) : f(x) = f(\overline{x})  \}$$
Even let $f(x)$ just be holomorphic. This looked plausible for a time but I am wondering if that is still correct.


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*if they were using the Schwartz reflection principle the relation should be $f(\overline{x}) = \overline{f(x)}$ (Ex)


Even without the reflections, do holomorphic or meromorphic functions form a Hilbert space?
$$ \mathcal{H} = \{ f(z): \text{ holomorphic} \} $$
The paper does not specify a norm - we might be tempted to guess the paper doesn't say.  In any case, they claim this Hilbert space is spanned by the poles.  For any holo/mero-morphic $f(z) \in \mathcal{H}$ 
$$ f(z) = \sum \frac{1}{z - A_i} $$
for some numbers $A_i \in \mathbb{C}$.  After reading this enough times I do not even remember anymore:


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*Do holomorphic or meromorphic functions form a Hilbert space?

*Is that Hilbert space spanned by functions of the type $\frac{1}{z-A}$ with $ A \in \mathbb{C}$.


I have to understand better how this example can be made into a Hilbert space or why it simply just can't.
 A: Do the entire functions $f$ such that $f(x)=f(\overline x)$ form a Hilbert space? Amusingly, the strictly speaking mathematically correct answer is "yes". But in the real world of fallible human beings saying yes would be very misleading; the right answer in practical terms is "What? Of course not; if you're quoting the paper correctly the author is very confused!" Note the "if"...
First the  yes: In fact if $f$ is an entire function and $f(x)=f(\overline x)$ then $f$ is constant. And so the space of such functions is a Hilbert space; a one-dimensional Hilbert space. 
Correct but awesomely silly; that cannot be what the author meant.
The space of entire functions is definitely not a Hilbert space. Now, there are plenty of Hilbert spaces of entire functions out there; such a space does not contain all the entire functions, just those satisfying some extra growth condition.
(There are nuances in the English language: If someone said something about "a Hilbert space of entire functions" that would be no problem. If someone said something about "the Hilbert space of entire functions" that would indicate he was confused.)
Just for the sake of inserting some mathematical content, I'm going to give a proof that the entire functions do not form a Hilbert space; that is, a proof that it's impossible to define a suitable norm.
Some background: A (complex) topological vector space is a complex vector space with a topology so that addition and scalar multiplication are jointly continuous. Say $V$ is a TVS and $B\subset V$. We say that $V$ is bounded if for every neighborhood $O$ of $0$ there exists $t>0$ such that $B\subset tO$. Note that if the topology on $V$ is given by a metric then this notion of boundedness is definitely not the same as saying that the metric is bounded on $B$. We may refer to the notion we just defined as TVS-boundedness to disambiguate things. Note as well that if the topology on $V$ is given by a norm then saying that $B$ is TVS-bounded is the same as saying that the norm is bounded on $B$.
Now say $D$ is an open subset of the plane and let $H(D)$ denote the holomorphic functions in $D$. Say $(K_n)$ is an exhaustion of $D$ by compact sets ($K_n$ is compact, $D=\bigcup K_n$, and $K_n$ is contained in the interior of $K_{n+1}$.) In general say $$||f||_K=\sup_{z\in K}|f(z)|.$$Then $$d(f,g)=\sum_{n=1}^\infty2^{-n}\frac{||f-g||_{K_n}}{1+||f-g||_{K_n}}$$defines a metric on $H(D)$ making $H(D)$ into a (complete, locally convex) TVS. This gives the standard topology on $H(D)$; it's easy to see that $d(f_n,f)\to0$ if and only if $f_n\to  f$ uniformly on compact sets.
And now the proof that $H(D)$ is not a Hilbert space: Montel's theorem on normal families shows that a subset of $H(D)$ is compact if and only if it is closed and TVS-bounded (hint: show that here TVS boundedness is equivalent to uniform boundeness on compact sets). A Hilbert space certainly does not have this property.
