(I am relatively new to the field of Reproducing Kernel Hilbert Spaces (RKHS) and functional analysis, and I have come across a conceptual discrepancy during my exploration of these subjects.)
Established Results in Textbooks
Let $\mathcal{X}$ be a compact set inside $\mathbb{R}^n$ and $\kappa: \mathcal{X} \times \mathcal{X} \to \mathbb{R}$ a continuous Mercer kernel. We can construct a reproducing kernel Hilbert space $\mathcal{H}$ via the kernel $\kappa$. Based on Reference[1, Corollary 4.31] and [2, Theorem 1. (c)(d)]:
The inclusion embedding $\imath: \mathcal{H} \hookrightarrow C(\mathcal{X})$ is compact and more strongly, for any closed and bounded subset $B \subseteq \mathcal{H}$, $\imath (B)$ is a compact subset of $C(\mathcal{X})$. [Claim $(\star)$]
The norm for $C(\mathcal{X})$ in Reference [2] is the infinity norm $\|f\|_{C(\mathcal{X})} = \|f\|_\infty$.
Construction
Now, let's consider $\mathcal{X} = [0, \pi] \subseteq \mathbb{R}$ and the RKHS $\mathcal{H}$ constructed using the absolute exponential kernel, i.e., \begin{align} \kappa(x, x') = \frac{1}{2}\exp\big( -|x - x'| \big). \end{align} This kernel is not differentiable but continuous in $\mathcal{X} \times \mathcal{X}$, and we can leverage the claim above to this case. It has been established in Reference [3, Theorem 1.3] that such an RKHS $\mathcal{H}$ shares the same vector space with the Sobolev space $W^{1,2}[0, \pi]$. Additionally, the RKHS norm induced by $\kappa$ can be explicitly expressed as follows: for any $f \in \mathcal{H}$ \begin{align} \|f\|_{\mathcal{H}} = \sqrt{\langle f, f\rangle_{\mathcal{H}}} = \sqrt{\Big(\int_{0}^{\pi} |f(x)|^2 + |f'(x)|^2 dx\Big) + |f(0)|^2 + |f(\pi)|^2}. \end{align}
Let consider the following sequence of functions $(f_n)_{n \in \mathbb{N}_+}, \mathbb{N}_+ = \{1, 2, 3, \ldots\}$: \begin{align} f_n(x) = \alpha_n \cdot \frac{1}{n}\sin(nx), x\in [0, \pi],\text{ where } \alpha_n = \frac{\varepsilon}{\sqrt{\frac{\pi}{2} + \frac{\pi}{2n^2}}}. \end{align} The parameter $\varepsilon > 0$ above can be any arbitrary positive constant. By some straightforward computation in undergraduate calculus, we have \begin{align} & f_n(0) = \alpha_n \cdot\frac{1}{n} \sin(n \cdot 0) = 0, f_n(\pi) = \alpha_n \cdot\frac{1}{n} \sin(n \cdot \pi) = 0, \\ & \int_0^\pi \frac{1}{n^2} \sin^2(nx) dx = \frac{1}{n^2}\int_0^\pi \frac{1}{2}\big(1 - \cos(2nx)\big)dx = \frac{1}{4n^3} \int_{0}^{2n\pi}(1 - \cos u)du = \frac{\pi}{2n^2} \\ & \int_0^\pi \cos^2(nx) dx = \int_0^\pi \frac{1}{2}\big(1 + \cos(2nx)\big) dx = \frac{1}{4n} \int_{0}^{2n\pi}\big(1 + \cos u\big) du = \frac{\pi}{2} \\ & \|f_n\|_{\mathcal{H}} = \alpha_n \cdot \sqrt{\frac{\pi}{2n^2} + \frac{\pi}{2}} = \varepsilon. \end{align} Hence, the sequence $(f_n)_{n \in \mathbb{N}_+}$ sits on the sphere $\mathbb{S}_\varepsilon(0) = \{f \in \mathcal{H}: \|f\|_{\mathcal{H}} = \varepsilon\}$. Additionally, considering the infinity norm defined as $\|f\|_\infty = \sup_{x \in \mathcal{X}}|f(x)| = \max_{x \in [0, \pi]}|f(x)|$, we also have $\|f_n\|_\infty \leq \frac{\varepsilon}{\sqrt{\pi/2}} \cdot \frac{1}{n} \searrow 0$ as $n \to \infty$. In other words, $(f_n)_{n \in \mathbb{N}_+}$ converge to the zero function $f_0 \equiv 0$ in the $\|\cdot\|_\infty$-norm.
Conflicts
For one thing, for the zero element $f_0 \equiv 0$, we have $\|f_0\|_{\mathcal{H}} = 0$ and hence $f_0 \notin \mathbb{S}_\varepsilon(0)$.
For another thing, I think $\mathbb{S}_\varepsilon(0)$ is closed and bounded in the $\|\cdot\|_{\mathcal{H}}$-norm. By [Claim $(\star)$], after the inclusion embedding, $\imath(\mathbb{S}_\varepsilon(0))$ should be a compact subset in $\big(C(\mathcal{X}), \|\cdot\|_\infty\big)$. My understanding is that the inclusion $\imath$ keeps every element as it is (without adding any additional elements) but changes the norm used from $\|\cdot\|_{\mathcal{H}}$ to $\|\cdot\|_\infty$. We now have the following facts:
$\imath(\mathbb{S}_\varepsilon(0))$ is closed in $\big(C(\mathcal{X}), \|\cdot\|_\infty\big)$;
the constructed sequence $(f_n)_{n \in \mathbb{N}_+}$ converges strongly to the zero element $f_0 \equiv 0$ in the $\|\cdot\|_\infty$-norm;
every convergent sequence in a closed set $F$ has its limit sit inside $F$.
Altogether, the limit $f_0 \equiv 0$ of $(f_n)_{n \in \mathbb{N}_+}$ in the $\|\cdot\|_\infty$-norm satisfies $f_0 \in \imath(\mathbb{S}_\varepsilon(0))$.
The aforementioned statements appear to result in a contradiction. I acknowledge that there may be an error/errors in the provided reasoning, yet I find myself unable to pinpoint it/them. I would sincerely welcome any insights or guidance from the community to help clarify this matter.
Reference
[1] Christmann, A., & Steinwart, I. (2008). Support vector machines.
[2] Zhou, D. X. (2008). Derivative reproducing properties for kernel methods in learning theory. Journal of computational and Applied Mathematics, 220(1-2), 456-463.
[3] Saitoh, S., & Sawano, Y. (2016). Theory of reproducing kernels and applications. Singapore: Springer Singapore.