Reproducing kernel Hilbert space of set functions

Question

Let $\Omega$ be a finite set. Can we construct a reproducing kernel Hilbert space (RKHS) of real-valued functions $2^\Omega \to \mathbb{R}$? If so, how can we construct one and how is the kernel defined?

Thank you!

Answer 1

If you let $X = 2^\Omega$, then the space of real valued functions on $X$ is basically $\mathbb{R}^n$ for $n = |2^\Omega|$. Indeed, you equip it with the standard dot product to get a Hilbert space. From there, you quickly see that the evaluation functionals $T_x(f) := f(x)$ are all continuous (linear maps from finite dimensional vector spaces are always continuous). Hence $\mathbb{R}^n$ is a RKHS. Note also that $T_x(f)$ is basically a projection. So, for example, if $\Omega = \{1\}$ and we allow ourselves to forgo function notation to instead work directly with $\mathbb{R}^2$, we get $T_\emptyset((x,y)) = x$ and $T_{\{1\}}((x,y)) = y$. It's probably time to note the notational convenience of using natural numbers as the base set $X$. Indeed, $T_1$ looks better than $T_\emptyset$.

So, how does the kernel of standard Euclidean $n$-space look like?

Simple, $K(x,y) = \langle K_x, K_y \rangle$, where $K_x$ is the vector/function in our space that represents $T_x$ in the sense that $T_x(f) = \langle f, K_x \rangle$. In our case, $K$ is a function from $\{1,2\}^2$ to $\mathbb{R}$ so that we must define $K(1,1), K(1,2), K(2,1),$ and $K(2,2)$. For that, note $\langle (1,0), (x,y) \rangle = x$ and $\langle (0,1), (x,y) \rangle = y$ so that the canonical vectors $e_1$ and $e_2$ are, respectively, $K_1$ and $K_2$. From here, you only need to compute four (three if you are smart) inner products, and you're done.