# Relation between RKHS and space of continuous functions

Consider a Mercer Kernel $K\colon \mathcal{X}\times \mathcal{X}\to \mathbb{R}$, $\mathcal{X}$ being a compact subset of $\mathbb{R}^m$, and its (unique) associated Reproducing Kernel HIlbert Space $H_K$. Then $H_k\subset \mathcal{C}^0(\mathcal{X})$, where $\mathcal{C}^0(\mathcal{X})$ is the space of continuous function on $\mathcal{X}$, i.e., there exist some functions of $\mathcal{C}^0(\mathcal{X})$ which are not in $H_K$.

Could you give me an explanation of the latter statement? And/or could you give me an example of functions in $\mathcal{C}^0(\mathcal{X})$ which do not belong to $H_K$?

Thank you.

• What is the definition of a Mercer kernel? Commented Mar 16, 2020 at 22:51

What I know is that for some popular kernels such as SE kernels or Matern kernels, $$H_k$$ is dense in $$C^{0}(\mathcal{X})$$ with respect to sup-norm if $$\mathcal{X}$$ is compact (ref: Miccheli et al., 2006). This result can also be expanded to $$L^p$$-denseness (ref: Sriperumbudur et al., 2011). Thus it might be helpful for you to read these references.

Also, in Kanagawa et al., 2018, there is an explicit representation of RKHS $$H_k$$ using Fourier transform (Theorem 2.4) for stationary kernels. According to this reference (Remark 2.12), roughly speaking, a function $$f$$ is in $$H_k$$ if and only if $$f \in C(\mathcal{X}) \cap L^2 (\mathcal{X})$$ and $$f$$ is smooth enough (i.e., $$\hat{f}(\omega)$$ is small enough for large $$\omega$$) compared to the kernel $$k$$.

However, neither do I haven't seen explicit example of function $$f$$ that is not in RKHS generated by popular kernels such as SE kernels, but continuous.