Schwartz functions are dense in the Fourier Bessel-potential norm

Put $g_R=g{\mathbf{1}}_{B(0,R)}$. By [F3], $\mathrm{\parallel }g-g_R{\mathrm{\parallel }}_{L_s^2}^2={\int }_{\mathrm{\mid }\xi \mathrm{\mid }>R}(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^s\mathrm{\mid }g(\xi ){\mathrm{\mid }}^{2}d\xi ⟶0(R\to \mathrm{\infty })\mathrm{.}$ Choose $R$ so this norm is less than a prescribed $\epsilon \mathrm{/}2$.

Let $f$ be any Schwartz function and define $\widehat{f}(\xi )=\int f(x)e^{-2\pi ix\cdot \xi }dx$. Polynomial multiples of derivatives of $f$ have arbitrarily rapid decay by [F5], so they and their derivatives are integrable, including on each coordinate line with the other coordinates fixed. Differentiation in $\xi$ is justified by dominated convergence [F6] using these integrable moments. Apply [F7] on each coordinate line to such a function and its Fourier exponential: the derivative products are integrable and the boundary product tends to zero. Absolute Fubini [F6] then combines the coordinate identities and shows that every ${\xi }^{\gamma }{D}_{\xi }^{\alpha }\widehat{f}(\xi )$ is a finite linear combination of Fourier transforms of derivatives of polynomial multiples of $f$. Each such function is in $L^1$ by [F5], so its Fourier transform is bounded; hence $\widehat{f}$ is Schwartz and in $L^1$. In particular $\int (1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^s\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi <\mathrm{\infty }$, by polar integration [F6].

On the bounded ball $O=B(0,R+1)$, the weight $(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^s$ is bounded above by a finite constant $C_R$. The restriction of $g_R$ to $O$ lies in $L^2(O)$, so [F2] gives $\varphi \in C_c^{\mathrm{\infty }}(O)$ with $\mathrm{\parallel }{g}_R-\varphi {\mathrm{\parallel }}_{L^2(O)}<\epsilon \mathrm{/}(2\sqrt{C_R})$. Extending $\varphi$ by zero is still smooth and compactly supported on ${\mathbb{R}}^d$ because its support is compactly contained in $O$, and $\mathrm{\parallel }{g}_R-\varphi {\mathrm{\parallel }}_{L_s^2}^2\le C_R\mathrm{\parallel }{g}_R-\varphi {\mathrm{\parallel }}_{L^2(O)}^2<{\epsilon }^2\mathrm{/}4.$

For $ϵ>0$, absolute Fubini [F6] and the Gaussian transform [F6] give $\int \widehat{f}(\xi )e^{2\pi ix\cdot \xi }{e}^{-\pi ϵ\mathrm{\mid }\xi {\mathrm{\mid }}^2}d\xi =\int f(y){ϵ}^{-d\mathrm{/}2}{e}^{-\pi \mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}ϵ}dy=(f\ast G_{ϵ})(x),$ where $G_{ϵ}(z)={ϵ}^{-d\mathrm{/}2}{e}^{-\pi \mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}ϵ}$. Since $\widehat{f}\in L^1$, dominated convergence sends the left side to $\int \widehat{f}(\xi )e^{2\pi ix\cdot \xi }d\xi$. The right side tends to $f(x)$: given $\epsilon >0$, continuity at $x$ gives $\delta >0$ such that $\mathrm{\mid }f(x-z)-f(x)\mathrm{\mid }<\epsilon$ for $\mathrm{\mid }z\mathrm{\mid }<\delta$. Since $G_{ϵ}$ has total mass one, the contribution from this ball is at most $\epsilon$; boundedness of $f$ and the Gaussian tail estimate from scaling make the contribution on $\mathrm{\mid }z\mathrm{\mid }\ge \delta$ tend to zero. Thus Fourier inversion holds and $f=U_{\widehat{f}}\in H^s$.

For any multi-indices $\alpha ,\beta$, differentiation under the integral is justified by the compactly supported integrable moments. For each coordinate, apply [F7] with the smooth compactly supported factor $(2\pi i\xi {)}^{\alpha }\varphi (\xi )$ (or a derivative of it) and the inverse Fourier exponential; both derivative products are integrable and the boundary product is zero outside a bounded interval. Repetition and absolute Fubini [F6] give $x^{\beta }{D}_x^{\alpha }{U}_{\varphi }(x)=(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }}(2\pi i{)}^{-\mathrm{\mid }\beta \mathrm{\mid }}\int {\mathrm{\partial }}_{\xi }^{\beta }((2\pi i\xi {)}^{\alpha }\varphi (\xi ))e^{2\pi ix\cdot \xi }d\xi \mathrm{.}$ The right side is uniformly bounded by the $L^1$ norm of a smooth compactly supported function. Hence every Schwartz seminorm of $U_{\varphi }$ is finite, so $U_{\varphi }$ is Schwartz.

Steps 1.1 and 2.1 show that $C_c^{\mathrm{\infty }}$ frequency functions are dense in the weighted Fourier norm. Step 3.1 shows their inverse transforms are Schwartz. Steps 1.2 and 2.2 show every Schwartz function belongs to $H^s$, and [F4] identifies the normed Fourier images injectively with the functions in $H^s$. Thus Schwartz functions are dense in $H^s({\mathbb{R}}^d)$.