I would like to know is the following result true ?

"Each Hilbert space of holomorphic functions on $\mathbb C$ is a reproducing kernel Hilbert space".

Where a Hilbert space of holomorphic functions on $\mathbb C$ is a Hilbert space $H \subseteq O(\mathbb C)$ such that the inclusion mapping $H \hookrightarrow O(\mathbb C)$ is continuous.

i.e., If $H \subseteq O(\mathbb C)$ is a Hilbert space of holomorphic functions, then the point evaluation map, $f \mapsto f(z)$, is continuous for all $z\in \mathbb C$ .

If so, where can I find its proof?

Thank you in advance

  • $\begingroup$ @Yaddle: The boundedness of the functions is not assumed. E.g. you could take the space of polynomials of degree at most $n$. $\endgroup$
    – gerw
    Dec 5, 2018 at 10:48
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    $\begingroup$ Is the topology on $O(\mathbb{C})$ the one of uniform convergence on compacta? If it is, your claim follows immediately from the continuity of the point evaluation maps on $O(\mathbb{C})$. $\endgroup$
    – MaoWao
    Dec 5, 2018 at 13:43
  • $\begingroup$ I don't know anything about "reproducing kernels", but judging from the name, a reproducing kernel should be a representation of the identity operator as an integral. For holomorphic functions, this is Cauchy's formula; $$f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)\, d\zeta}{\zeta - z}.$$ $\endgroup$ Dec 5, 2018 at 13:52
  • $\begingroup$ I think the non-trivial idea is that $f_n$ analytic on $U$ converges to an analytic function on $U$ iff for every compact $K\subset U$ it converges uniformly on $K$ iff (Cauchy integral formula) for some sequence of curves (enclosing every compacts) its converges uniformly on those curves. So that your normed space of analytic functions is closed means the norm must be stronger than the previous ones. $\endgroup$
    – reuns
    Dec 5, 2018 at 14:55
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    $\begingroup$ @GiuseppeNegro Cauchy's formula does not show that any such $H$ is a reproducing-kernel Hilbert space, because the kernel is not an element of $H$. $\endgroup$ Dec 5, 2018 at 15:51

1 Answer 1


If the inclusion $i: H \to O(\mathbb{C})$ is continuous, then of course the evaluations are continuous, since they are continuous on $O(\mathbb{C})$ so that $f(z) = (if)(z)$ is a composition of two continuous maps. I'm assuming here that $O(\mathbb{C})$ is supplied with the usual topology of uniform convergence on compacta.

A perhaps more interesting case happens if the continuity of this inclusion is dropped. Without this assumption, the continuity of evaluations certainly should not be true, although an explicit counterexample might be difficult to construct. If you allow the use of the axiom of choice, then the following construction I think will work.

Let $E$ be the vector space of all entire functions. It should not be difficult to show by example that a linearly independent sequence of functions $f_n \in E$ exists such that $|f_n(0)| \to \infty$ as $n \to \infty$. By the axiom of choice, we can complete this sequence to a vector space basis $\{f_\alpha\}_{\alpha \in A}$ for $E$.

There certainly exists a Hilbert space $H$ with a a vector space basis of the same cardinality as $A$, say this basis is $\{h_\alpha\}_{\alpha \in A}$. By scaling, we can assume that $\|h_\alpha\|_{H} = 1$. Now, define the norm on $E$ in the following way: $$\|\sum_{i=1}^n c_if_{\alpha_i}\|_{E} = \|\sum_{i=1}^n c_ih_{\alpha_i}\|_{H}.$$ This makes $E$ into a Hilbert space of holomorphic functions on $\mathbb{C}$. Certainly $\|f_n\|_E = 1$, yet $|f_n(0)| \to \infty$ by construction, so the evaluation at $z = 0$ is discontinuous.


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