Dyadic Mihlin pieces sum to an off-support kernel representation

Transposition of the inverse Schwartz transform is the inverse distribution transform: composing either way tests against $\mathcal{F}{\mathcal{F}}^{-1}\phi =\phi$. Thus the inverse in [F1] uses ${\mathcal{F}}^{-1}$ on tests. The partial sums converge in ${\mathcal{S}}^{\mathrm{\prime }}$: for $\phi \in \mathcal{S}$, using [F1] and [F2], $⟨{\sum }_{\mathrm{\mid }j\mathrm{\mid }\le N}{K}_j,\phi ⟩={\int }_{{\mathbb{R}}^n}({\sum }_{\mathrm{\mid }j\mathrm{\mid }\le N}{m}_j(\xi ))({\mathcal{F}}^{-1}\phi )(\xi )d\xi ⟶{\int }_{{\mathbb{R}}^n}m(\xi )({\mathcal{F}}^{-1}\phi )(\xi )d\xi =⟨W,\phi ⟩,$ by dominated convergence with majorant $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mathcal{F}}^{-1}\phi \mathrm{\mid }\in L^1$ and pointwise convergence ${\sum }_{\mathrm{\mid }j\mathrm{\mid }\le N}{m}_j(\xi )=m(\xi ){\sum }_{\mathrm{\mid }j\mathrm{\mid }\le N}\zeta (2^{-j}\xi )\to m(\xi )$ for $\xi \mathrm{\ne }0$; the limit pairing is $⟨m^{\vee },\phi ⟩=⟨W,\phi ⟩$.

Two elementary estimates. First, from $K_j(x)=\int m_j(\xi )e^{2\pi ix\cdot \xi }d\xi$, which is absolutely convergent because $m_j$ is supported in the annulus $2^{j-1}\le \mathrm{\mid }\xi \mathrm{\mid }\le 2^{j+1}$ and bounded by $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$, one has the pointwise bound $\mathrm{\mid }{K}_j(x)\mathrm{\mid }\le 2^{jn}{C}_{\zeta }A$ for every $j$ and every $x$. Second, for every $\delta >0$ and $j>0$, [F3] gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_j(x)\mathrm{\mid }dx\le C_{n}A(1+2^j\delta {)}^{-1\mathrm{/}4}$, so ${\sum }_{j>0}{\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_j\mathrm{\mid }$ is finite for every fixed $\delta >0$, and for $j\le 0$ the pointwise bound gives ${\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }{K}_j\mathrm{\mid }\le 2^{jn}{C}_{\zeta }A\mathrm{\mid }B(0,2\delta )\mathrm{\mid }\le C_{n}A{\delta }^n{2}^{jn}$, whose sum over $j\le 0$ is at most $C_{n}A{\delta }^n$, finite for every fixed $\delta$.

Almost everywhere convergence. By step 1.2, ${\sum }_{j\le 0}\mathrm{\mid }{K}_j(x)\mathrm{\mid }<\mathrm{\infty }$ for every $x$; for each fixed $\delta >0$, ${\sum }_{j>0}{\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }\mathrm{\mid }{K}_j\mathrm{\mid }<\mathrm{\infty }$ implies ${\sum }_{j>0}\mathrm{\mid }{K}_j(x)\mathrm{\mid }<\mathrm{\infty }$ for almost every $x$ with $\mathrm{\mid }x\mathrm{\mid }\ge \delta$. Taking the union of the exceptional sets over $\delta =1\mathrm{/}m$, $m\ge 1$, the series ${\sum }_{j\in \mathbb{Z}}{K}_j(x)$ converges absolutely for almost every $x\in {\mathbb{R}}^n\setminus \{0\}$; denote its sum by $k(x)$, a measurable function on ${\mathbb{R}}^n\setminus \{0\}$.

On every compact $K\subset {\mathbb{R}}^n\setminus \{0\}$ the dominating function ${\sum }_j\mathrm{\mid }{K}_j\mathrm{\mid }$ is integrable: $K$ lies in some annulus $\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2^m\delta$, and the two estimates of step 1.2 (the second applied after covering the outer annulus by finitely many dyadic annuli of the same type) give ${\int }_K{\sum }_j\mathrm{\mid }{K}_j\mathrm{\mid }<\mathrm{\infty }$. Hence for $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n\setminus \{0\})$, dominated convergence with the partial sums bounded by ${\sum }_j\mathrm{\mid }{K}_j\mathrm{\mid }$ gives $⟨W,\phi ⟩={\lim }_N⟨{\sum }_{\mathrm{\mid }j\mathrm{\mid }\le N}{K}_j,\phi ⟩=\int k\phi$ by step 1.1, so $k$ coincides with $W$ on ${\mathbb{R}}^n\setminus \{0\}$.

Annular size bound. Fix $\delta >0$. Split ${\sum }_j{\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }{K}_j\mathrm{\mid }$ according to whether $2^j\delta >1$. For $2^j\delta \le 1$ the pointwise bound of step 1.2 gives ${\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }{K}_j\mathrm{\mid }\le 2^{jn}{C}_{\zeta }A\mathrm{\mid }B(0,2\delta )\mathrm{\mid }$, so the sum over these $j$ is at most $C_{n}A{\delta }^n{\sum }_{2^j\le 1\mathrm{/}\delta }{2}^{jn}\le C_{n}A$. For $2^j\delta >1$, ${\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }{K}_j\mathrm{\mid }\le (2^j\delta {)}^{-1\mathrm{/}4}\int \mathrm{\mid }{K}_j\mathrm{\mid }(1+2^j\mathrm{\mid }x\mathrm{\mid }{)}^{1\mathrm{/}4}\le C_{n}A(2^j\delta {)}^{-1\mathrm{/}4}$ by [F3], Put $j_0=\min \{j\in \mathbb{Z}:2^j\delta >1\}$, so $1<2^{j_0}\delta \le 2$ by minimality. The high-frequency sum is therefore bounded by $C_{n}A{\sum }_{j\ge j_0}(2^j\delta {)}^{-1\mathrm{/}4}=C_{n}A(2^{j_0}\delta {)}^{-1\mathrm{/}4}\mathrm{/}(1-2^{-1\mathrm{/}4})\le C_{n}A\mathrm{/}(1-2^{-1\mathrm{/}4})$, independent of $\delta$. Hence ${\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }k\mathrm{\mid }\le {\sum }_j{\int }_{\delta \le \mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }{K}_j\mathrm{\mid }\le C_{n}A$, uniformly in $\delta$.

Hörmander's condition. Fix $y\mathrm{\ne }0$ and choose $k\in \mathbb{Z}$ with $2^{-k}\le \mathrm{\mid }y\mathrm{\mid }\le 2^{1-k}$. For $j>k$ the triangle inequality and [F3] give ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{K}_j(x-y)-K_j(x)\mathrm{\mid }dx\le 2{\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{K}_j(x)\mathrm{\mid }dx\le 2C_{n}A(1+2^j\mathrm{\mid }y\mathrm{\mid }{)}^{-1\mathrm{/}4}$, and summing over $j>k$ yields at most $C_{n}A$, since $2^j\mathrm{\mid }y\mathrm{\mid }\ge 2^{j-k}$. For $j\le k$, the mean value theorem and translation of the integral give ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{K}_j(x-y)-K_j(x)\mathrm{\mid }dx\le \mathrm{\mid }y\mathrm{\mid }{\int }_{{\mathbb{R}}^n}\mathrm{\mid }\mathrm{\nabla }{K}_j(u)\mathrm{\mid }du\le C_{n}A\mathrm{\mid }y\mathrm{\mid }{2}^j$ by [F3]. Hence the sum over $j\le k$ is bounded by $C_{n}A\mathrm{\mid }y\mathrm{\mid }{\sum }_{j\le k}{2}^j=2C_{n}A\mathrm{\mid }y\mathrm{\mid }{2}^k\le 4C_{n}A$, using the upper dyadic inequality $\mathrm{\mid }y\mathrm{\mid }\le 2^{1-k}$. Summing in $j$ yields the asserted Hörmander bound, uniformly in $y\mathrm{\ne }0$.

Steps 1.1, 3.1, 3.2 and 3.3 are the four assertions: ${\mathcal{S}}^{\mathrm{\prime }}$ convergence, the a.e. convergent series defining $k$, its coincidence with $W$ off the origin, and the two kernel bounds with constant $C_{n}A$.