Is a Reproducing Kernel Hilbert Space just a Hilbert space equipped with an "indexed basis"?

Question

I haven't studied any functional analysis yet. My linear algebra is pretty good, I think.

Consider the tuple $(H, I, \phi)$ where $H$ is a Hilbert space, $I$ is an abstract set, and $\phi:I \to H$ is a injective mapping whose image is an unconditional Schauder basis of $H$. We can make any element $v$ of $H$ into a function by defining $v(i\in I)$ to mean $\langle v, \phi(i)\rangle$. Now observe that $|v(i)| =|\langle v, \phi(i)\rangle| \leq \|v\| \|\phi(i)\|$ which means that the function space I've described is an RKHS.

Going in the other direction, I believe every RKHS space is of the form above. Is this true?

The nice thing about thinking of RKHSes in this way, is that it means that an RKHS is the same thing as a Hilbert Space $H$ with a chosen "indexed basis". By indexed basis, I'm talking about the $\phi$ above. This shows that every Hilbert space can be made into an RKHS by "blessing" one of its bases. It also suggests that operations between vector spaces like direct sum ($\oplus$) and tensor product ($\otimes$) and $L(H,K)$ (forming the space of linear maps) all make sense over RKHSes; it's just a matter of keeping track of what happens to their bases.

Answer 1

No, not every RKHS arises this way.

Consider "the" basic example, the Bergmann space $H=A_2(\Bbb D)$, consisting of all functions $f$ holomorphic in the unit disk $\Bbb D$ such that $\int_{\Bbb D}|f|^2<\infty$. If $z\in\Bbb D$ and $K_z$ is the Bergmann kernel then $f(z)=\int_{\Bbb D}f\overline K_z$, but the $K_z$ certainly do not form a basis; for example $H$ is separable...

Answer 2

Let $H$ be a Hilbert space of functions from $[0,1]$ to $\mathbb R$ which are constant everywhere.

$H$ satisfies Wikipedia's definition of an RKHS because its evaluation functional is bounded. By the Riesz representation theorem, the representive of $\operatorname{ev}_x$ is simply the function $t \mapsto 1$. We will denote this function $1$. $1$ forms a basis of $H$. The construction in the question would make all the elements of $H$ a function over a single element, forgetting the fact that the elements of $H$ are functions over $[0,1]$.

Notice that the resulting kernel $k(x,y)=\langle\phi(x), \phi(y) \rangle = \langle 1, 1\rangle = 1$ is not positive definite. This shows that an RKHS doesn't need to have a positive definite kernel.