Dyadic Mihlin pieces: uniform L1 and first-difference bounds

The difference $\zeta =\chi (\xi )-\chi (2\xi )$ vanishes for $\mathrm{\mid }\xi \mathrm{\mid }\le 1\mathrm{/}2$, since there $\chi (\xi )=\chi (2\xi )=1$, and vanishes for $\mathrm{\mid }\xi \mathrm{\mid }\ge 2$, since there $\chi (\xi )=\chi (2\xi )=0$; hence $\zeta$ is supported in the annulus $1\mathrm{/}2\le \mathrm{\mid }\xi \mathrm{\mid }\le 2$, and $0\le \zeta \le 1$ for the fixed smooth-step cutoff: its construction is $\chi (\xi )=\sigma ((4-\mathrm{\mid }\xi {\mathrm{\mid }}^2)\mathrm{/}3)$, $\sigma (t)=a(t)\mathrm{/}(a(t)+a(1-t))$, with $a(t)=e^{-1\mathrm{/}t}$ for $t>0$ and $a(t)=0$ otherwise. On $0<t<1$ one has ${\sigma }^{\mathrm{\prime }}(t)=(a^{\mathrm{\prime }}(t)a(1-t)+a(t)a^{\mathrm{\prime }}(1-t))\mathrm{/}(a(t)+a(1-t){)}^2\ge 0$; on the constant regions its derivative is zero. Thus $\chi$ decreases with radius and $\chi (\xi )\ge \chi (2\xi )$, proving the asserted nonnegativity. For every $N\ge 1$ the sum telescopes: ${\sum }_{j=-N}^N\zeta (2^{-j}\xi )={\sum }_{j=-N}^N[\chi (2^{-j}\xi )-\chi (2^{1-j}\xi )]=\chi (2^{-N}\xi )-\chi (2^{N+1}\xi )$. For $\xi \mathrm{\ne }0$ one has $2^{-N}\mathrm{\mid }\xi \mathrm{\mid }\le 1$ and $2^{N+1}\mathrm{\mid }\xi \mathrm{\mid }\ge 2$ for all large $N$, so the last expression equals $1-0=1$ there; this gives the asserted partition of unity.

For every $j$ the function $m_j=m\zeta (2^{-j}\cdot )$ is supported in the annulus $2^{j-1}\le \mathrm{\mid }\xi \mathrm{\mid }\le 2^{j+1}$, where it agrees almost everywhere with $m_0(\xi )\zeta (2^{-j}\xi )$; we use this representative in the derivative estimates. Since $m_0$ is $C^{n_0}$ on that annulus and $\zeta$ is compactly supported and smooth, $m_j$ is represented by a compactly supported $C^{n_0}$ function, so $m_j\in L^1\cap L^2$ and $u_{m_j}$ is a well-defined regular tempered distribution. By [F3] the object $K_j={\mathcal{F}}^{-1}(u_{m_j})$ is the regular distribution of the $L^2$ function ${\mathcal{F}}_2^{-1}{m}_j$; we use $K_j$ to denote that $L^2$ class, so that $\mathcal{F}{K}_j=u_{m_j}$ and $\mathrm{\parallel }{K}_j{\mathrm{\parallel }}_2\le \mathrm{\parallel }{m}_j{\mathrm{\parallel }}_2$. It has the smooth integral representative $G_j(x)=\int m_j(\xi )e^{2\pi ix\cdot \xi }d\xi$: for every Schwartz test $\phi$, Fubini applies with absolute bound $\mathrm{\parallel }{m}_j{\mathrm{\parallel }}_1\mathrm{\parallel }\phi {\mathrm{\parallel }}_1$, giving $\int G_j\phi =\int m_j{\mathcal{F}}^{-1}\phi =⟨{\mathcal{F}}^{-1}{u}_{m_j},\phi ⟩$. Thus [F6] identifies $G_j$ with the $L^2$ class. Every ${\xi }^{\beta }{m}_j$ is integrable on the fixed compact frequency support. Put $G_{j,\beta }(x)=\int (2\pi i\xi {)}^{\beta }{m}_j(\xi )e^{2\pi ix\cdot \xi }d\xi$. The bounds $\mathrm{\mid }{e}^{it}-1\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$ and $\mathrm{\mid }{e}^{it}-1-it\mathrm{\mid }\le t^2\mathrm{/}2$ give continuity of $G_{j,\beta }$ and a coordinate difference-quotient remainder bounded uniformly in $x$ by $C_{j,\beta }\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }{m}_j{\mathrm{\parallel }}_1$ as $h\to 0$. Hence ${\mathrm{\partial }}_r{G}_{j,\beta }=G_{j,\beta +e_r}$, proving $G_j\in C^{\mathrm{\infty }}$. These derivatives are bounded; repeated integration by parts against rapidly decaying Schwartz tests therefore has no boundary term and identifies each classical derivative with its regular distributional derivative as defined in [F6]. We henceforth use this smooth representative for $K_j$ and its gradients.

Claim: for every multi-index $\gamma$ with $\mathrm{\mid }\gamma \mathrm{\mid }\le n_0$, the product $x^{\gamma }{K}_j$ is (the regular distribution of) an $L^2$ function and $\parallel x^{\gamma }{K}_j{\parallel }_2=(2\pi {)}^{-\mathrm{\mid }\gamma \mathrm{\mid }}\parallel {\mathrm{\partial }}^{\gamma }{m}_j{\parallel }_2\mathrm{.}$ Indeed, applying the first identity of [F4] to $u=K_j$ and using $\mathcal{F}{K}_j=u_{m_j}$ gives $\mathcal{F}(x^{\gamma }{K}_j)=(-1\mathrm{/}(2\pi i){)}^{\mathrm{\mid }\gamma \mathrm{\mid }}{\mathrm{\partial }}^{\gamma }{u}_{m_j}=c_{\gamma }{u}_{{\mathrm{\partial }}^{\gamma }{m}_j}$ for the scalar $c_{\gamma }=(-1\mathrm{/}(2\pi i){)}^{\mathrm{\mid }\gamma \mathrm{\mid }}$, the last equality because ${\mathrm{\partial }}^{\gamma }{m}_j$ is continuous and compactly supported, hence a regular distribution, and differentiation of a regular distribution of a $C^{\mathrm{\mid }\gamma \mathrm{\mid }}$ function is the regular distribution of its classical derivative. Since ${\mathrm{\partial }}^{\gamma }{m}_j\in C_c\subset L^2$, [F3] applied to $h=c_{\gamma }{\mathrm{\partial }}^{\gamma }{m}_j$ identifies $x^{\gamma }{K}_j$ with the regular distribution of ${\mathcal{F}}_2^{-1}(c_{\gamma }{\mathrm{\partial }}^{\gamma }{m}_j)$, and Plancherel gives $\mathrm{\parallel }{x}^{\gamma }{K}_j{\mathrm{\parallel }}_2=\mathrm{\parallel }{c}_{\gamma }{\mathrm{\partial }}^{\gamma }{m}_j{\mathrm{\parallel }}_2=(2\pi {)}^{-\mathrm{\mid }\gamma \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\gamma }{m}_j{\mathrm{\parallel }}_2$.

Claim: there is $C_{n,\chi }$ with $\mathrm{\parallel }{\mathrm{\partial }}^{\gamma }{m}_j{\mathrm{\parallel }}_2\le C_{n,\chi }A{2}^{j(n\mathrm{/}2-\mathrm{\mid }\gamma \mathrm{\mid })}$ for all $j\in \mathbb{Z}$ and all $\mathrm{\mid }\gamma \mathrm{\mid }\le n_0$. Leibniz's rule on $m_j=m_0\zeta (2^{-j}\cdot )$ gives ${\mathrm{\partial }}^{\gamma }{m}_j={\sum }_{\delta \le \gamma }{C}_{\delta ,\gamma }{\mathrm{\partial }}^{\gamma -\delta }(\zeta (2^{-j}\cdot )){\mathrm{\partial }}^{\delta }{m}_0$; the chain rule bounds the factor by $2^{-j\mathrm{\mid }\gamma -\delta \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\gamma -\delta }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}$, and on the support of $m_j$ one has $\mathrm{\mid }{\mathrm{\partial }}^{\delta }{m}_0(\xi )\mathrm{\mid }\le C_{\delta }\mathrm{\mid }\xi {\mathrm{\mid }}^{-\mathrm{\mid }\delta \mathrm{\mid }}\le C_{\delta }{2}^{-j\mathrm{\mid }\delta \mathrm{\mid }}\cdot 2^{\mathrm{\mid }\delta \mathrm{\mid }}$ since $\mathrm{\mid }\xi \mathrm{\mid }\ge 2^{j-1}$. Taking $L^2$ norms and bounding the support measure by $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }(2^n-2^{-n})2^{jn}$ yields $\mathrm{\parallel }{\mathrm{\partial }}^{\gamma }{m}_j{\mathrm{\parallel }}_2\le {\sum }_{\delta \le \gamma }{C}_{\delta ,\gamma }{2}^{-j\mathrm{\mid }\gamma -\delta \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\gamma -\delta }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}{C}_{\delta }{2}^{-j\mathrm{\mid }\delta \mathrm{\mid }}{2}^{jn\mathrm{/}2}\cdot c_n$, that is, $C_{n,\chi }({\max }_{\mathrm{\mid }\delta \mathrm{\mid }\le n_0}{C}_{\delta })2^{j(n\mathrm{/}2-\mathrm{\mid }\gamma \mathrm{\mid })}\le C_{n,\chi }A{2}^{j(n\mathrm{/}2-\mathrm{\mid }\gamma \mathrm{\mid })}$, because $\mathrm{\mid }\gamma -\delta \mathrm{\mid }+\mathrm{\mid }\delta \mathrm{\mid }=\mathrm{\mid }\gamma \mathrm{\mid }$ for $\delta \le \gamma$.

Proof of (1). Fix $j$ and write $w(x):=(1+2^j\mathrm{\mid }x\mathrm{\mid }{)}^{1\mathrm{/}4}$, $W(x):=(1+2^j\mathrm{\mid }x\mathrm{\mid }{)}^{n_0}$. Since $-2n_0+1\mathrm{/}2<-n$, the substitution $u=2^{j}x$ gives ${\int }_{{\mathbb{R}}^n}W(x{)}^{-2}w(x{)}^{2}dx={\int }_{{\mathbb{R}}^n}(1+2^j\mathrm{\mid }x\mathrm{\mid }{)}^{-2n_0+1\mathrm{/}2}dx=2^{-jn}{c}_n$ for a constant $c_n=\int (1+\mathrm{\mid }u\mathrm{\mid }{)}^{-2n_0+1\mathrm{/}2}du$. Cauchy–Schwarz and the elementary bound $W(x)\le C(n){\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le n_0}{2}^{j\mathrm{\mid }\gamma \mathrm{\mid }}\mathrm{\mid }{x}^{\gamma }\mathrm{\mid }$ give $\int \mathrm{\mid }{K}_j\mathrm{\mid }w\le (\int W^2\mathrm{\mid }{K}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}(\int W^{-2}{w}^2{)}^{1\mathrm{/}2}\le C(n){\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le n_0}{2}^{j\mathrm{\mid }\gamma \mathrm{\mid }}\mathrm{\parallel }{x}^{\gamma }{K}_j{\mathrm{\parallel }}_2\cdot 2^{-jn\mathrm{/}2}{c}_n^{1\mathrm{/}2}$. By steps 2.1 and 2.2 this is at most $C_{n,\chi }A{\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }\le n_0}{2}^{j\mathrm{\mid }\gamma \mathrm{\mid }}{2}^{j(n\mathrm{/}2-\mathrm{\mid }\gamma \mathrm{\mid })}{2}^{-jn\mathrm{/}2}=C_{n,\chi }^{\mathrm{\prime }}A$, uniformly in $j$.

Proof of (2), one coordinate at a time. Fix $r\le n$ and put ${\zeta }_r(\xi ):={\xi }_r\zeta (\xi )$ and ${\tilde{m}}_j(\xi ):=m(\xi ){\zeta }_r(2^{-j}\xi )=2^{-j}{\xi }_r{m}_j(\xi )$, so that ${\tilde{m}}_j$ is again compactly supported and $C^{n_0}$. The second identity of [F4] gives $\mathcal{F}({\mathrm{\partial }}_r{K}_j)=(2\pi i{\xi }_r)\mathcal{F}{K}_j=2\pi i{u}_{{\xi }_r{m}_j}$, hence $\mathcal{F}(2^{-j}{\mathrm{\partial }}_r{K}_j)=2\pi i{u}_{{\tilde{m}}_j}$. Identifying $2^{-j}{\mathrm{\partial }}_r{K}_j$ with the regular distribution of ${\mathcal{F}}_2^{-1}(2\pi i{\tilde{m}}_j)$ as in step 1.2 and repeating steps 2.1, 2.2 and 3.1 with the fixed cutoff ${\zeta }_r$ in place of $\zeta$ (whose support and derivatives are again bounded by constants $C_{n,\chi }$) yields ${\sup }_j\int \mathrm{\mid }{2}^{-j}{\mathrm{\partial }}_r{K}_j\mathrm{\mid }w\le C_{n,\chi }A$. Summing these $n$ estimates over $r\le n$ and using $\mathrm{\mid }\mathrm{\nabla }{K}_j\mathrm{\mid }\le {\sum }_r\mathrm{\mid }{\mathrm{\partial }}_r{K}_j\mathrm{\mid }$ gives (2).

Steps 3.1 and 4.1 are exactly the two asserted estimates, with constants depending only on $n$ and the fixed cutoff $\chi$; the auxiliary claim of step 1.2 supplies the $L^2$ reading of $K_j$ used throughout. This proves the lemma.