The Banach Gaussian covariance and RKHS construction is well-defined

For $\mathrm{\ell }\in B^{\ast }$, the map $\omega \mapsto \mathrm{\ell }(W(\omega ))W(\omega )$ is Borel as the composition of $W$ with the continuous map $x\mapsto \mathrm{\ell }(x)x$. It is strongly measurable by [F2], and $\mathrm{\parallel }\mathrm{\ell }(W)W\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{\parallel }W{\mathrm{\parallel }}^2$ is integrable by [F1]. Hence [F3] defines $C\mathrm{\ell }=E[\mathrm{\ell }(W)W]$ as a Bochner integral.

By [F1], $E\mathrm{\mid }\mathrm{\ell }(W){\mathrm{\mid }}^2\le \mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}^{2}E\mathrm{\parallel }W{\mathrm{\parallel }}^2<\mathrm{\infty }$ for every $\mathrm{\ell }\in B^{\ast }$; Cauchy–Schwarz [F6] therefore makes $E\mathrm{\mid }m(W)\mathrm{\ell }(W)\mathrm{\mid }$ finite. Applying $m\in B^{\ast }$ to the Bochner integral using [F4] gives $m(C\mathrm{\ell })=E[m(W)\mathrm{\ell }(W)]$. Thus $Q(\mathrm{\ell },m)=m(C\mathrm{\ell })$ is a finite symmetric bilinear form, nonnegative on the diagonal. Again by [F6], $\mathrm{\mid }Q(\mathrm{\ell },m)\mathrm{\mid }\le Q(\mathrm{\ell },\mathrm{\ell }{)}^{1\mathrm{/}2}Q(m,m{)}^{1\mathrm{/}2}$, so $Q(\mathrm{\ell },\mathrm{\ell })=0$ implies $Q(\mathrm{\ell },m)=0$ for every $m$; therefore $N$ is the radical of $Q$ and the quotient form is an inner product.

If $\mathrm{\ell }\in N$, step 2.1 gives $m(C\mathrm{\ell })=0$ for every $m$, so $C\mathrm{\ell }=0$ by [F5]. Conversely, if $C\mathrm{\ell }=0$, taking $m=\mathrm{\ell }$ in step 2.1 gives $Q(\mathrm{\ell },\mathrm{\ell })=0$, hence $\mathrm{\ell }\in N$. Therefore $[\mathrm{\ell }]\mapsto C\mathrm{\ell }$ is well-defined and injective on $B^{\ast }\mathrm{/}N$.

By [F3] and [F6], $\mathrm{\parallel }C\mathrm{\ell }\mathrm{\parallel }\le E(\mathrm{\mid }\mathrm{\ell }(W)\mathrm{\mid }\mathrm{\parallel }W\mathrm{\parallel })\le Q(\mathrm{\ell },\mathrm{\ell }{)}^{1\mathrm{/}2}(E\mathrm{\parallel }W{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\mathrm{.}$ So the map on the quotient is bounded and extends uniquely to its Hilbert completion because $B$ is complete. For every $m\in B^{\ast }$, the identity $m(C\mathrm{\ell })=⟨[\mathrm{\ell }],[m]{⟩}_H$ from step 2.1 extends by continuity to $m(iu)=⟨u,[m]{⟩}_H$. If $iu=0$, then $u$ is orthogonal to the dense subspace $B^{\ast }\mathrm{/}N$, so $u=0$. The extension is injective, proving all claims.