Universal approximation of Gaussians Can gaussian kernels reproduce non continuous L2 integrable functions? ( Do non continuous L2 integrable functions lie in the RKHS constructed by a Gaussian Kernel?)
Edit:
I think my question is being misconceived. I do not intend to say that the RKHS is constructed using this argument. My question is that due to the reproducing property, every function that belongs to a Hilbert space must satisfy the reproducing property give the kernel is a reproducing kernel. Would a constant function over a compact support satisfy this for gaussian kernels?

More generally, a Hilbert space $\mathcal F$ of functions from $\mathcal X$ to $\mathbb R$ is a Reproducing Kernel Hilbert Space if and only if for all $x\in\mathcal X$, the map :
$$L_x : f\in\mathcal F\mapsto f(x)$$
is bounded. That is, if and only if there exists for all $x\in\mathcal X$ a finite $M_x$ such that
$$|f(x)|\le M_x \|f\|_{\mathcal F}\quad \forall f\in\mathcal F \tag1$$
So even if you forget for a minute that for $\mathcal F=\mathcal L^2(\mathcal X)$, $f(x)$ is not well-defined (as rightfully mentioned in @frank's answer), you can easily see that the uniform bound $(1)$ doesn't hold. For instance, let $x_0\in\mathcal X$ be arbitrary and consider the family of functions
$$f_n(x) =\begin{cases} n \text{ if } x=x_0,\\
 0 \text{ otherwise}\end{cases}$$
It is clear that all the $f_n$'s are in $\mathcal F =\mathcal L^2$ but the evaluation functional $L_{x_0}$ is unbounded by construction. Hence we can conclude that $\mathcal L^2$ is not a RKHS for any kernel, and in particular for the Gaussian kernel.
By a similar argument you can show that even a "well-behaved" space such as the set of all continuous functions on an interval $[a,b]$ is not an RKHS either.
