The Matérn native space has its weighted Fourier norm and strict Gram positivity

Since $w=A(a^2+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-s}$, the two positive functions $(a^2+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^s$ and $(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^s$ have ratio bounded above and below by positive constants: compare $a^2+4{\pi }^2{r}^2$ with $1+r^2$ separately using the minimum and maximum of $a^2$ and $4{\pi }^2$, and use monotonicity of the positive power [F7]. Multiplying by $A^{-1}$ and taking square roots gives the displayed norm equivalence. Thus the weighted $L^2$ norms defining ${\mathcal{H}}_k$ and $H^s$ are equivalent and have the same representatives by injectivity [F1,F2]; completeness of $H^s$ implies completeness of ${\mathcal{H}}_k$.

For a complex coefficient vector $c=(c_j)$, the Hermitian quadratic form of the Gram matrix is $c^{\ast }{K}_{X}c={\int }_{{\mathbb{R}}^d}w(\xi ){\mid {\sum }_{j=1}^N{c}_j{e}^{-2\pi i{x}_j\cdot \xi }\mid }^{2}d\xi \mathrm{.}$ If it is zero, [F6] says $w$ times the squared modulus of the continuous exponential sum vanishes almost everywhere. Since $w>0$, the sum is zero almost everywhere. A continuous function zero almost everywhere is zero everywhere, because any point of nonzero value would have a neighborhood on which its modulus is bounded below, and such a neighborhood contains a positive-measure box [F5].

The spectral representative of $k(\cdot -y)$ is $w(\xi )e^{-2\pi iy\cdot \xi }$. Its ${\mathcal{H}}_k$ norm squared is $\int w=1$, and for $U_g\in {\mathcal{H}}_k$, $⟨U_g,k(\cdot -y){⟩}_k=\int g(\xi )e^{2\pi iy\cdot \xi }d\xi =U_g(y)\mathrm{.}$ The integral exists because Cauchy--Schwarz bounds $\int \mathrm{\mid }g\mathrm{\mid }$ by $\mathrm{\parallel }g{\mathrm{\parallel }}_k(\int w{)}^{1\mathrm{/}2}$. Hence point evaluation is continuous and the stated kernel sections reproduce it.

If $U$ is orthogonal to every section centered at $y\in {\mathbb{R}}^d$, step 2.1 gives $U(y)=0$ for every $y$. Fourier injectivity [F1] then makes $U=0$ in ${\mathcal{H}}_k$. By [F3] the span of all sections is dense.

Choose $v=(1,t,\dots ,t^{d-1})$ so that the numbers $v\cdot x_j$ are pairwise distinct. For each pair $i\mathrm{\ne }j$, the polynomial $(x_i-x_j)\cdot (1,t,\dots ,t^{d-1})$ is nonzero because $x_i-x_j\mathrm{\ne }0$; [F4] says its zero set is finite. The ordered field $\mathbb{R}$ has $0<1$, so the natural multiples $0,1,1+1,\dots$ are strictly increasing and $\mathbb{R}$ is infinite. Thus a real $t$ exists outside the finite union of root sets. Restrict the vanishing exponential sum to $\xi =uv$ and differentiate at $u=0$ for orders $r=0,\dots ,N-1$. This gives ${\sum }_j{c}_j(v\cdot x_j{)}^r=0$. For each $j$, the Lagrange polynomial $L_j(t)={\prod }_{i\mathrm{\ne }j}\frac{t-v\cdot x_i}{v\cdot x_j-v\cdot x_i}$ has degree at most $N-1$ and satisfies $L_j(v\cdot x_i)=1$ if $i=j$ and $0$ otherwise. Taking the same linear combination of the moment equations as the coefficients of $L_j$ yields $c_j=0$. Thus all $c_j=0$, and the real Gram matrix is strictly positive definite.