I am reading a lecture on RKHS of kernels and here is an exercise I have not been able to solve
Given a translation invariant kernel $$ k(x,y) = \varphi (x-y) $$
The theorem states that once $\varphi$ is symmetric and positive definite, and that $\varphi$ and $\hat\varphi$ are integrable, the RKHS would be $$ \mathcal H = \left\{ f \in \mathcal C(\mathbb R) \cap L^2 (\mathbb R) \middle\vert \int \frac{\vert\hat f(\omega) \vert^2}{\hat \varphi(\omega)} \mathrm d \omega < \infty\right\} $$ Where $\hat \ast$ denotes the Fourier transform of $\ast$.
Can anyone please show me why this is a Hilbert space, typically by showing that it is closed in $L^2(\mathbb R)$? Thank you
Since $K(x,y) = \varphi (x-y)$, Bochner theorem tell us that there exists a nondecreasing measure $d\sigma(t)=\rho(t)\,dt$, with $\rho(t)>0$ and continuos, such that $$\varphi(x-y) = \frac{1}{2\pi}\int_{\mathbb{R}} e^{it(x-y)}d\sigma(t)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{it(x-y)}\rho(t)\,dt,$$ and we denote $\rho=\hat{\phi}$.
Let $F(x)$ being such that $g(x)=F(x)\sqrt{\rho(x)}$ is in $L^2(\mathbb{R})$, that is $$\int_{\mathbb{R}}|F(x)|^2\rho(x)\,dx<\infty.$$ Then the function $$f(x)=\frac{1}{2\pi}\int_{\mathbb{R}} F(t)\rho(t)e^{-ixt}dt$$ is in the reproducing kernel Hilbert space $\mathcal{H}_K$, and you can see that $$|f(x)|=\frac{1}{2\pi}\left|\int_{\mathbb{R}}\left(F(t)\sqrt{\rho(t)}\right)\left(e^{-ixt}\sqrt{\rho(t)}\right)dt\right|\leq \frac{1}{2\pi}\|g\|_{L^2}\sqrt{K(x,x)} .$$ The inversion formula to Fourier transform gives us that $$F(t)\rho(t)=\int_{\mathbb{R}} f(x)e^{ixt}dx,$$ in $L^2(\mathbb{R})$. It follows that $$\|f\|_{\mathcal{H}_K}^2=\frac{1}{2\pi}\int_{\mathbb{R}}|F(t)|^2\rho(t)\,dt=\frac{1}{2\pi}\int_{\mathbb{R}}\left|F(t)\rho(t)\right|^2\frac{1}{\rho(t)}dt=\frac{1}{2\pi}\int_{\mathbb{R}}\left|\int_{\mathbb{R}} f(x)e^{ixt}dx\right|^2\frac{1}{\rho(t)}dt.$$
Please see Theorem 2.1.2 in the book Integral Transforms, Reproducing Kernels and Their Applications, by S. Saitoh.
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