Free Gaussian dispersive kernel

[Damped kernel] Fix $\epsilon >0$, put $z_{\epsilon }=\epsilon +it$, and consider the multiplier $g_{\epsilon }(\xi ):=e^{-4{\pi }^2{z}_{\epsilon }\mathrm{\mid }\xi {\mathrm{\mid }}^2}$, i.e. the operator $T_{\epsilon }:={\mathcal{F}}_2^{\ast }{M}_{g_{\epsilon }}{\mathcal{F}}_2$. By [F3] with $a=4{\pi }^2{z}_{\epsilon }$, applied fiberwise in the three coordinates through the product structure and [F2], its integral kernel is $(4\pi z_{\epsilon }{)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4z_{\epsilon })}$, where the power is the principal branch of [F2]; hence for Schwartz $\psi$ one has $(T_{\epsilon }\psi )(x)=(4\pi z_{\epsilon }{)}^{-3\mathrm{/}2}\int e^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4z_{\epsilon })}\psi (y)dy$.

[Removing the damping] Let $\psi$ be Schwartz and $x\in {\mathbb{R}}^3$. As $\epsilon \downarrow 0$ one has $z_{\epsilon }\to it$, $(4\pi z_{\epsilon }{)}^{-3\mathrm{/}2}\to (4\pi it{)}^{-3\mathrm{/}2}$ by continuity of the principal branch on the closed right half-plane away from $0$, and $e^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4z_{\epsilon })}\to e^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4it)}$ pointwise in $y$; the uniform domination $\mathrm{\mid }(4\pi z_{\epsilon }{)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(4z_{\epsilon })}\mathrm{\mid }\le (4\pi \mathrm{\mid }{z}_{\epsilon }\mathrm{\mid }{)}^{-3\mathrm{/}2}\le (4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}$ holds for all $y$ and $0<\epsilon <1$ because $\mathrm{\mid }{z}_{\epsilon }\mathrm{\mid }\ge \mathrm{\mid }t\mathrm{\mid }$ and the exponential factor has modulus at most $1$. Since $\psi \in L^1$, dominated convergence [F4] gives $(T_{\epsilon }\psi )(x)\to (K_t\psi )(x)$ for every $x$, and $\mathrm{\mid }(T_{\epsilon }\psi )(x)\mathrm{\mid }\le (4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$ for all $x,\epsilon$.

[Identification with free evolution] For Schwartz $\psi$, Plancherel gives $\mathrm{\parallel }{T}_{\epsilon }\psi {\mathrm{\parallel }}_2\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_2$, because $\mathrm{\mid }{g}_{\epsilon }\mathrm{\mid }\le 1$; the damped operator is a contraction, not an isometry. Moreover $g_{\epsilon }{\mathcal{F}}_2\psi \to e^{-itq}{\mathcal{F}}_2\psi$ in $L^2$ by dominated convergence with squared majorant $4\mathrm{\mid }{\mathcal{F}}_2\psi {\mathrm{\mid }}^2$. Hence $T_{\epsilon }\psi \to e^{-it{H}_0}\psi$ in $L^2$. For integers $n\ge 1$, define recursively $m_0=1$ and let $m_n$ be the least integer $m>m_{n-1}$ with $m\ge n$ and $\mathrm{\parallel }{T}_{1\mathrm{/}m}\psi -e^{-it{H}_0}\psi {\mathrm{\parallel }}_2^2<2^{-n}$. The $L^2$ convergence just proved makes the eligible integer set nonempty. Thus ${\epsilon }_n:=1\mathrm{/}{m}_n\downarrow 0$ and the squared errors are summable; this prescribed integer selection needs no choice principle. Integrating the sum of these squared errors makes that sum finite almost everywhere, so the selected sequence converges almost everywhere to $e^{-it{H}_0}\psi$. Its pointwise limit from step 2.1 is $K_t\psi$ everywhere; therefore $K_t\psi =e^{-it{H}_0}\psi$ almost everywhere and $\mathrm{\parallel }{K}_t\psi {\mathrm{\parallel }}_2=\mathrm{\parallel }\psi {\mathrm{\parallel }}_2$. The scalar unit-modulus free multiplier is unitary and preserves the dense Schwartz class, so this kernel operator on that class has the unique unitary extension $e^{-it{H}_0}$.

[The full intersection class and domain preservation] For $\psi \in L^1\cap L^2$, choose Schwartz ${\psi }_n\to \psi$ jointly in $L^1$ and $L^2$: first truncate the input in support and value, then convolve each bounded compactly supported truncation with a smooth compact approximate identity and choose a small enough radius for both norm errors. Translation continuity in each $L^p$, $p=1,2$, follows first for finite box-step functions by the volume of symmetric differences and then by their elementary density; thus the convolution errors tend to zero in both norms. Each approximant is compactly supported smooth and hence Schwartz; at most countable choices are used. The uniform kernel bound gives $\mathrm{\parallel }{K}_t{\psi }_n-K_t\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\le (4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}\mathrm{\parallel }{\psi }_n-\psi {\mathrm{\parallel }}_1\to 0$, while the unitary extension of step 3.1 converges in $L^2$ to $e^{-it{H}_0}\psi$. The squared-error subsequence argument of step 3.1 identifies these two limits almost everywhere, proving (a) on the full $L^1\cap L^2$ class. The absolutely convergent integral is everywhere defined and directly bounded by $(4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$, proving (b). Finally multiplication by $e^{-itq}$ preserves exactly the domain $q{\mathcal{F}}_2\psi \in L^2$ and its weighted norm, so $H_0{e}^{-it{H}_0}\psi =e^{-it{H}_0}{H}_0\psi$ and $\mathrm{\parallel }{H}_0{e}^{-it{H}_0}\psi {\mathrm{\parallel }}_2=\mathrm{\parallel }{H}_0\psi {\mathrm{\parallel }}_2$. This proves (c) and all assertions.