A positive semidefinite kernel has a canonical reproducing kernel Hilbert space under countable choice

Let $E_0$ be the finitely supported functions on $X$ and define $B(a,b)={\sum }_{x,z}a(x)b(z)k(x,z)$, where each sum is over the finite supports. Symmetry and bilinearity follow from symmetry of $k$ and finite-sum algebra. For any $a$, listing its finite support gives $B(a,a)\ge 0$ by [F1].

For $a,b\in E_0$ and $t\in \mathbb{R}$, $0\le B(a+tb,a+tb)=B(a,a)+2tB(a,b)+t^{2}B(b,b)$. If $B(b,b)>0$, nonnegativity of this quadratic for all $t$ gives $B(a,b{)}^2\le B(a,a)B(b,b)$; if $B(b,b)=0$, its linear coefficient must vanish, so $B(a,b)=0$.

It follows that every $a\in N:=\{a:B(a,a)=0\}$ satisfies $B(a,b)=0$ for every $b\in E_0$. Hence $N$ is a subspace and the form descends to a well-defined positive definite inner product on $Q:=E_0\mathrm{/}N$: changing a representative by an element of $N$ does not change any pairing, and a quotient class has zero squared norm exactly when it is the zero class.

The induced inner product makes $Q$ a real inner-product space. By [F2] and [A1], its norm metric has a metric completion with compatible Banach structure, hence a norm completion; the inner product extends uniquely and makes the completion $\widehat{Q}$ a real Hilbert space.

For each $x\in X$, the image $k_x$ of $[{\delta }_x]\in Q$ satisfies $\mathrm{\parallel }{k}_x{\mathrm{\parallel }}^2=B({\delta }_x,{\delta }_x)=k(x,x)\ge 0$. Thus the finite kernel sections are elements of $\widehat{Q}$, and their span is dense because the image of $Q$ is dense in its completion.

For $a\in E_0$, the quotient pairing gives $⟨[a],k_x⟩={\sum }_{z\in \mathrm{supp}(a)}a(z)k(z,x)$, independently of the representative by [step 3.1]. Given $h\in \widehat{Q}$, density in [step 4.1] provides an approximating sequence $[a_m]\to h$. By [F3], the real sequence $⟨[a_m],k_x⟩$ is Cauchy, and if $[b_m]\to h$ is another such sequence then $\mathrm{\mid }⟨[a_m-b_m],k_x⟩\mathrm{\mid }\le \mathrm{\parallel }[a_m-b_m]\mathrm{\parallel }\mathrm{\parallel }{k}_x\mathrm{\parallel }\to 0$, so its limit is independent of the approximants. The extended inner product also satisfies [F3], hence this limit equals $⟨h,k_x⟩$.

Define the associated function of $h\in \widehat{Q}$ by $h(x):=⟨h,k_x⟩$. Applying the extended pairing to $k_z$ gives $k_z(x)=⟨k_z,k_x⟩=B({\delta }_z,{\delta }_x)=k(z,x)=k(x,z)$, so $k_x$ is exactly the section $k(\cdot ,x)$ and $⟨h,k_x⟩=h(x)$ for every $h$.

If $h(x)=0$ for every $x\in X$, then $h$ is orthogonal to every section $k_x$ and therefore to their dense span from [step 5.1]. Continuity of the inner product gives $⟨h,h⟩=0$, so positive definiteness gives $h=0$. Thus the associated-function map is injective and identifies $\widehat{Q}$ with a Hilbert space of functions having reproducing kernel $k$.