Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel

Normalisation and scaling of the kernel. The construction below is uniform in the scale: for a fixed parameter $s_0>0$ the annuli $\{\mathrm{\mid }\xi \mathrm{\mid }\asymp s_0^{-1}{2}^j\}$ play the role of the annuli $\{\mathrm{\mid }\xi \mathrm{\mid }\asymp 2^j\}$ at $s_0=1$, and every estimate keeps the same form with constants depending on $s_0$; in particular the same argument run at a smaller parameter gives the statement for every smaller scale, the constants changing by a fixed factor. Since $\widehat{\phi }(0)=\int \phi \mathrm{\ne }0$ and $\widehat{\phi }$ is continuous, after multiplying $\phi$ by the nonzero complex multiple $(1\mathrm{/}\int \phi )$ and then replacing it by a suitable positive dilation ${\delta }^n\phi (\delta \cdot )={\phi }_{1\mathrm{/}\delta }$, we may assume ${\int }_{{\mathbb{R}}^n}\phi =1,\mathrm{\mid }\widehat{\phi }(\xi )\mathrm{\mid }\ge \frac{1}{2}\text{for }\mathrm{\mid }\xi \mathrm{\mid }\le 2.$ We prove the lemma in this normalisation with $s_0=1$; undoing the dilation replaces the scale $1$ by the fixed positive number $1\mathrm{/}\delta$ and does not change the form of the estimates.

A smooth dyadic partition of unity. With the smooth step $\sigma$ of The standard smooth step function, take $\zeta (\xi )=\sigma ((9\mathrm{/}4-\mathrm{\mid }\xi {\mathrm{\mid }}^2)\mathrm{/}(5\mathrm{/}4))$; it equals one on $B(0,1)$ and its support is contained in the closed ball of radius $3\mathrm{/}2$, hence in $B(0,2)$, and put ${\zeta }_0=\zeta$ and ${\zeta }_j(\xi )=\zeta (2^{-j}\xi )-\zeta (2^{-j+1}\xi )$ for $j\ge 1$. Then for every $J\ge 0$, ${\sum }_{j=0}^J{\zeta }_j(\xi )=\zeta (2^{-J}\xi ),$ so ${\sum }_{j\ge 0}{\zeta }_j(\xi )=1$ for every $\xi$: at $\xi =0$ both sides equal one because $\zeta (0)=1$, and for $\xi \mathrm{\ne }0$ the limit $\zeta (2^{-J}\xi )\to \zeta (0)=1$ as $J\to \mathrm{\infty }$ gives the identity. If ${\zeta }_j(\xi )\mathrm{\ne }0$, then $2^{-j}\xi \in \mathrm{supp}\zeta$ or $2^{-j+1}\xi \in \mathrm{supp}\zeta$, so $\mathrm{\mid }{2}^{-j}\xi \mathrm{\mid }\le 2$ in either case; by step 1.1, $\mathrm{\mid }\widehat{\phi }(2^{-j}\xi )\mathrm{\mid }\ge \frac{1}{2}$ on $\mathrm{supp}{\zeta }_j$.

The deconvolution coefficients and convergence. For $\psi \in \mathcal{S}$ and $j\ge 0$ define the compactly supported smooth function $\widehat{{\eta }^j}(\xi )=\frac{{\zeta }_j(\xi )}{\widehat{\phi }(2^{-j}\xi )}\widehat{\psi }(\xi )\mathrm{.}$ The quotient is well defined and smooth on a neighbourhood of $\mathrm{supp}{\zeta }_j$ by step 2.1, and $\widehat{{\eta }^j}\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)\subseteq \mathcal{S}$; hence ${\eta }^j:={\mathcal{F}}^{-1}\widehat{{\eta }^j}\in \mathcal{S}$ by [F1]. On the Fourier side, for every $J\ge 0$, $\mathcal{F}({\sum }_{j=0}^J{\eta }^j\ast {\phi }_{2^{-j}})(\xi )=\widehat{\psi }(\xi ){\sum }_{j=0}^J{\zeta }_j(\xi )=\widehat{\psi }(\xi )\zeta (2^{-J}\xi ),$ where we used $\widehat{{\phi }_{2^{-j}}}(\xi )=\widehat{\phi }(2^{-j}\xi )$ from [F1]. To prove convergence in $\mathcal{S}$, put ${\chi }_J(\xi )=\zeta (2^{-J}\xi )-1$. The multiplier ${\chi }_J$ itself is not Schwartz and does not converge to zero in $\mathcal{S}$; instead we prove $\widehat{\psi }{\chi }_J\to 0$ in every Schwartz seminorm. Fix multi-indices $\alpha ,\beta$. Leibniz's rule writes ${\mathrm{\partial }}^{\beta }(\widehat{\psi }{\chi }_J)$ as a finite sum of terms $C_{\gamma ,\beta }({\mathrm{\partial }}^{\gamma }\widehat{\psi })({\mathrm{\partial }}^{\beta -\gamma }{\chi }_J)$, $\gamma \le \beta$. For the term with $\gamma =\beta$, the cutoff is undifferentiated: ${\chi }_J=0$ on $\mathrm{\mid }\xi \mathrm{\mid }\le 2^J$ because $\zeta =1$ on $B(0,1)$, and $\mathrm{\mid }{\chi }_J\mathrm{\mid }\le 1+\mathrm{\parallel }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}$ everywhere. Thus its $p_{\alpha \beta }$ contribution is bounded by $(1+\mathrm{\parallel }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}){\sup }_{\mathrm{\mid }\xi \mathrm{\mid }\ge 2^J}\mathrm{\mid }{\xi }^{\alpha }{\mathrm{\partial }}^{\beta }\widehat{\psi }(\xi )\mathrm{\mid }$, which tends to zero by Schwartz decay. For every term with $\gamma <\beta$, let $\delta =\beta -\gamma \mathrm{\ne }0$. The chain rule gives ${\mathrm{\partial }}^{\delta }{\chi }_J(\xi )=2^{-J\mathrm{\mid }\delta \mathrm{\mid }}({\mathrm{\partial }}^{\delta }\zeta )(2^{-J}\xi )$, supported in the annulus $2^J\le \mathrm{\mid }\xi \mathrm{\mid }\le 2^{J+1}$ since $\zeta$ is constant on $B(0,1)$ and vanishes outside $B(0,2)$. Its contribution is therefore bounded by $C_{\delta }{2}^{-J\mathrm{\mid }\delta \mathrm{\mid }}{\sup }_{\mathrm{\mid }\xi \mathrm{\mid }\ge 2^J}\mathrm{\mid }{\xi }^{\alpha }{\mathrm{\partial }}^{\gamma }\widehat{\psi }(\xi )\mathrm{\mid }$, which also tends to zero. There are only finitely many terms for each $\alpha ,\beta$, so $p_{\alpha \beta }(\widehat{\psi }{\chi }_J)\to 0$. This proves $\widehat{\psi }\zeta (2^{-J}\cdot )\to \widehat{\psi }$ in $\mathcal{S}$. Since Fourier transformation is a homeomorphism of $\mathcal{S}$ [F1], the partial sums converge to $\psi$ in $\mathcal{S}$, and (with ${\sum }_{j\ge 0}$ denoting that limit) $\psi ={\sum }_{j\ge 0}{\eta }^j\ast {\phi }_{2^{-j}}$.

The rapid norm decay. Fix $L,N>0$; all constants below depend on $\phi ,n,L,N$ only. For $j\ge 1$, on $\mathrm{supp}{\zeta }_j$ one has $(1+\mathrm{\mid }\xi \mathrm{\mid })\asymp 2^j$: if ${\zeta }_j(\xi )\mathrm{\ne }0$ then $2^{-j}\mathrm{\mid }\xi \mathrm{\mid }\le 2$, while $2^{-j+1}\mathrm{\mid }\xi \mathrm{\mid }\ge 1$ because otherwise both $\zeta (2^{-j}\xi )$ and $\zeta (2^{-j+1}\xi )$ would equal one, so $\frac{1}{2}\cdot 2^j\le 1+\mathrm{\mid }\xi \mathrm{\mid }\le 3\cdot 2^j$. For $j=0$, the support lies in a fixed ball and all the following estimates hold by enlarging the constant, since the target factor is $2^{-0nL}=1$. The quotient $\xi \mapsto {\zeta }_j(\xi )\mathrm{/}\widehat{\phi }(2^{-j}\xi )$ is smooth on the neighbourhood $\{\mathrm{\mid }\xi \mathrm{\mid }<2^{j+1}\}$ of $\mathrm{supp}{\zeta }_j$, and its derivatives of order at most $N+n+1$ are bounded by a constant $C_0=C_0(n,N,\phi )$ independent of $j$: the chain rule contributes the factors $2^{-j\mathrm{\mid }\beta \mathrm{\mid }}\le 1$ to the derivatives both of ${\zeta }_j$ and of the composition of $1\mathrm{/}\widehat{\phi }$ with $\xi \mapsto 2^{-j}\xi$, and $1\mathrm{/}\widehat{\phi }$ is smooth with bounded derivatives on the fixed ball $\mathrm{\mid }z\mathrm{\mid }\le 2$, where $\mathrm{\mid }\widehat{\phi }\mathrm{\mid }\ge \frac{1}{2}$ by step 1.1. Multiplying by $\widehat{\psi }$ with the higher product rule, $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }\widehat{{\eta }^j}(\xi )\mathrm{\mid }\le C_1{\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le N+n+1}\mathrm{\mid }{\mathrm{\partial }}^{\beta }\widehat{\psi }(\xi )\mathrm{\mid },\mathrm{\mid }\alpha \mathrm{\mid }\le N+n+1,\xi \in \mathrm{supp}{\zeta }_j\mathrm{.}$ Multiplying by $(1+\mathrm{\mid }\xi \mathrm{\mid }{)}^{N+n+1}$ and using the definition of the $S_M$ norm with $M\ge N+n+1$ together with the lower bound just proved, $\mathrm{\parallel }\widehat{{\eta }^j}{\mathrm{\parallel }}_{S_{N+n+1}}\le C_1{\sup }_{\xi \in \mathrm{supp}{\zeta }_j}(1+\mathrm{\mid }\xi \mathrm{\mid }{)}^{N+n+1-M}\mathrm{\parallel }\widehat{\psi }{\mathrm{\parallel }}_{S_M}\le C_2{2}^{-j(M-N-n-1)}\mathrm{\parallel }\widehat{\psi }{\mathrm{\parallel }}_{S_M},$ so with $M\ge N+n+1+nL$ the right-hand side is at most $C_2{2}^{-jnL}\mathrm{\parallel }\widehat{\psi }{\mathrm{\parallel }}_{S_M}$. Finally [F2] applied with the roles of a function and its transform interchanged gives $\mathrm{\parallel }{\eta }^j{\mathrm{\parallel }}_{S_N}\le C_3\mathrm{\parallel }\widehat{{\eta }^j}{\mathrm{\parallel }}_{S_{N+n+1}}$ (the Fourier transform of $\widehat{{\eta }^j}$ is ${\eta }^j(-\cdot )$, which has the same seminorms), and [F2] gives $\mathrm{\parallel }\widehat{\psi }{\mathrm{\parallel }}_{S_M}\le C_M\mathrm{\parallel }\psi {\mathrm{\parallel }}_{S_{M+n+1}}$. Hence $\mathrm{\parallel }{\eta }^j{\mathrm{\parallel }}_{S_N}\le C{2}^{-jnL}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{S_{M+n+1}}$ with $C=C_2{C}_3{C}_M$ independent of $\psi$ and $j$, so the statement holds with $M$ replaced by $M+n+1$.

Conclusion. Steps 2.1 and 3.1 construct ${\eta }^j\in \mathcal{S}$ with $\psi ={\sum }_{j\ge 0}{\eta }^j\ast {\phi }_{2^{-j}}$ in $\mathcal{S}$, and step 4.1 gives the estimate $\mathrm{\parallel }{\eta }^j{\mathrm{\parallel }}_{S_N}\le C{2}^{-jnL}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{S_M}$ with constants independent of $\psi$. Undoing the normalisation of step 1.1 replaces the scale family $2^{-j}$ by $s_0{2}^{-j}$ with the fixed $s_0=1\mathrm{/}\delta$ and does not affect the convergence or the estimates, and the same construction run at a smaller parameter gives the statement there. This proves the lemma.