I have been studying Reproducing Kernel Hilbert Spaces (RKHS). The definition I am using is as follows: An RKHS is a Hilbert space $\mathcal{H}$ of real-valued functions on a set $X$ such that for all $x \in X$, the evaluation functional $E_x : \mathcal{H} \to \mathbb{R}$ defined by $E_x(f) = f(x)$ is continuous. (Recall that continuity in this context means $\sup_{\|f\|\leq 1}|f(x)| < \infty$ for all $x$.)

As one of the first steps in my study I want to see some positive and negative examples of these spaces. So I want to find a non-trivial (meaning infinite dimensional) space of functions which is a Hilbert space but not an RKHS, meaning the evaluation functionals are not continuous.

The problem is that in all the references I find, the only example they give is the space of functions $L_{2}$. However, this example is trivial since $L_2$ fails for the simple reason that it is not a space of functions, rather it contains equivalence classes of functions.

Can someone please help me find an infinite dimensional Hilbert space of functions that is not an RKHS?

  • 1
    $\begingroup$ Why not just consider $\mathcal{L}_2$, i.e. $L_2$ before taking equivalence classes? That is a hilbert space of functions. $\endgroup$
    – user27182
    Commented Apr 25, 2021 at 9:04
  • $\begingroup$ @user27182 See both answers below. The problem is that $\mathcal{L}_2$ is not a space of functions; rather, elements of that space are equivalence classes of functions. $\endgroup$
    – ttb
    Commented May 14, 2021 at 15:17
  • 2
    $\begingroup$ $\mathcal{L}_2$ is a space of functions (it's the space of all square integrable functions, which you turn it into $L_2$ by considering equivalence classes), the problem with my earlier comment is that it is not an inner product space (so not a Hilbert space). $\endgroup$
    – user27182
    Commented May 14, 2021 at 16:14

1 Answer 1


There are no explicit examples of this kind. Indeed, if one could show you such a space $\mathcal H$, then point evaluation $f\mapsto f(x_0)$ would give an explicit example of a discontinuous linear functional on a Hilbert space. And there can be no explicit examples of this kind, because it's consistent with ZF that they do not exist. Only with the Axiom of Choice can one give a (non-constructive) proof of their existence. This is discussed in many places, such as

So, any time someone gives you a concrete Hilbert space of functions, you can be sure that it's either RKHS, or is not really a space of functions.

  • $\begingroup$ That's very interesting. So if I understand correctly, there can not be "constructive" example of a Hilbert space of functions and point $x_0$, such that the evaluation functional at that point is discontinuous? $\endgroup$
    – ttb
    Commented Mar 15, 2018 at 14:28

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