Japanese-bracket and Laplacian Bessel-potential operators

Assume Countable Choice and let $n\ge 1$. Throughout, $⟨\xi ⟩=(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ is the Japanese bracket and $\mathcal{F}$ is the negative-sign $2\pi$-normalized Fourier transform, an automorphism of ${\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ with inverse ${\mathcal{F}}^{-1}$ (Fourier transform is a topological automorphism of tempered distributions).

Japanese-bracket operator. For real $t$ and $u\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ define $⟨D{⟩}^{t}u:={\mathcal{F}}^{-1}(⟨\xi {⟩}^t\mathcal{F}u)\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n),$ where $⟨\xi {⟩}^t\mathcal{F}u$ is multiplication of the tempered distribution $\mathcal{F}u$ by the smooth symbol $⟨\xi {⟩}^t$.

Laplacian Bessel-potential operator. Independently, define $(I-\mathrm{\Delta }{)}^{t\mathrm{/}2}u:={\mathcal{F}}^{-1}((1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{t\mathrm{/}2}\mathcal{F}u)\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)\mathrm{.}$

Well-definedness and invertibility. The symbols $w_t(\xi )=⟨\xi {⟩}^t$, $w_{-t}(\xi )=⟨\xi {⟩}^{-t}$, $a_t(\xi )=(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{t\mathrm{/}2}$ and $a_{-t}(\xi )=(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-t\mathrm{/}2}$ are smooth, and every derivative has polynomial growth. For $w_{\pm t}$ this is Real powers of the Japanese bracket act on Schwartz space, which also states that these two multipliers act continuously and inversely on $\mathcal{S}$ and, by transposition, on ${\mathcal{S}}^{\mathrm{\prime }}$. For $a_{\pm t}$ the chain rule and induction on $\mathrm{\mid }\alpha \mathrm{\mid }$ write ${\mathrm{\partial }}^{\alpha }{a}_t(\xi )={\sum }_j{P}_{\alpha ,j}(\xi )(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{t\mathrm{/}2-j}$ with polynomials $P_{\alpha ,j}$ of degree at most $\mathrm{\mid }\alpha \mathrm{\mid }$; since $1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$ is bounded above and below by positive constant multiples of $⟨\xi {⟩}^2$, each term is $O(⟨\xi {⟩}^{t+\mathrm{\mid }\alpha \mathrm{\mid }})$ on ${\mathbb{R}}^n$, so every derivative of $a_t$ has polynomial growth, and the same holds for $a_{-t}$. Hence Smooth polynomially bounded multipliers on schwartz space makes multiplication by either symbol a continuous endomorphism of $\mathcal{S}({\mathbb{R}}^n)$ whose transpose is a continuous endomorphism of ${\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ for both dual topologies. Since $a_t{a}_{-t}=1$ pointwise, the two transposed maps are inverse: evaluated on a Schwartz test $\phi$ one has $⟨a_t(a_{-t}u),\phi ⟩=⟨u,a_{-t}(a_t\phi )⟩=⟨u,\phi ⟩$, and likewise with the factors exchanged. Thus both $⟨D{⟩}^t$ and $(I-\mathrm{\Delta }{)}^{t\mathrm{/}2}$ are continuous bijections of ${\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ with inverses $⟨D{⟩}^{-t}$ and $(I-\mathrm{\Delta }{)}^{-t\mathrm{/}2}$. No self-adjointness, spectral-theorem or positivity assertion is made here; the operators are defined by their Fourier symbols on tempered distributions.

Consistency at integer order. For every nonnegative integer $m$, $\mathcal{F}((I-\mathrm{\Delta }{)}^{m}u)=(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^m\mathcal{F}u,$ because the published Fourier differentiation identity gives $\mathcal{F}({\mathrm{\partial }}_j^{2}u)=(2\pi i{\xi }_j{)}^2\mathcal{F}u=-4{\pi }^2{\xi }_j^2\mathcal{F}u$, summation over $j$ gives $\mathcal{F}(\mathrm{\Delta }u)=-4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathcal{F}u$, and iterating $m$ times (Fourier differentiation and multiplication identities on tempered distributions). So the symbol $(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^m$ really is the integer power of $I-\mathrm{\Delta }$ under this normalization.

The two symbols differ. For $t\mathrm{\ne }0$ the symbols $⟨\xi {⟩}^t$ and $(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{t\mathrm{/}2}$ differ at every $\xi \mathrm{\ne }0$: equality would give $1+\mathrm{\mid }\xi {\mathrm{\mid }}^2=1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$, hence $\mathrm{\mid }\xi \mathrm{\mid }=0$, since $4{\pi }^2\mathrm{\ne }1$ and $x\mapsto x^{t\mathrm{/}2}$ is injective on $(0,\mathrm{\infty })$ for $t\mathrm{\ne }0$. They agree at $\xi =0$, a Lebesgue-null set of frequencies. Consequently $⟨D{⟩}^t$ and $(I-\mathrm{\Delta }{)}^{t\mathrm{/}2}$ are different operators for $t\mathrm{\ne }0$, and in particular $⟨D{⟩}^t$ is not the Bessel potential $(I-\mathrm{\Delta }{)}^{t\mathrm{/}2}$; the bracket symbol uses the $2\pi$-independent weight, while the Laplacian symbol carries the factor $4{\pi }^2$ from $\mathcal{F}({\mathrm{\partial }}_j)=2\pi i{\xi }_j\mathcal{F}$.