Calderon reproducing pair and the telescoping identity in ${\mathcal{S}}^{\mathrm{\prime }}$

Support and moments. Since $\mathrm{supp}\phi \subseteq B(0,1)$, both $\mathrm{supp}(2^n\phi (2\cdot ))$ and $\mathrm{supp}\phi$ lie in $B(0,1)$; hence $\mathrm{supp}\psi ,\mathrm{supp}\tilde{\psi }\subseteq B(0,1)$ and, after dilation, $\mathrm{supp}{\psi }_k,\mathrm{supp}{\tilde{\psi }}_k\subseteq B(0,2^{-k})$ for every $k$. For $0<\mathrm{\mid }\alpha \mathrm{\mid }\le K$ one has $\int x^{\alpha }\psi (x)dx=2^{-\mathrm{\mid }\alpha \mathrm{\mid }}\int x^{\alpha }\phi (x)dx-\int x^{\alpha }\phi (x)dx=(2^{-\mathrm{\mid }\alpha \mathrm{\mid }}-1)\int x^{\alpha }\phi =0$ by [F1] and the flatness of $\widehat{\phi }$; the case $\alpha =0$ gives $\int \psi =0$ as well.

Telescoping. For every $k\in \mathbb{Z}$ the definitions give ${\psi }_k={\phi }_{k+1}-{\phi }_k$ and ${\tilde{\psi }}_k={\phi }_{k+1}+{\phi }_k$: indeed $h=2^n\phi (2\cdot )$ has $h_k={\phi }_{k+1}$ and ${\phi }_k={\phi }_k$. Therefore ${\psi }_k\ast {\tilde{\psi }}_k\ast f=({\phi }_{k+1}\ast {\phi }_{k+1}-{\phi }_k\ast {\phi }_k)\ast f$, and the finite sums telescope: ${\phi }_j\ast {\phi }_j\ast f+{\sum }_{k=j}^N{\psi }_k\ast {\tilde{\psi }}_k\ast f={\phi }_{N+1}\ast {\phi }_{N+1}\ast f=(\phi \ast \phi {)}_{N+1}\ast f$ for every $N\ge j$. By [F2], $(\phi \ast \phi {)}_{N+1}\ast f\to f$ in ${\mathcal{S}}^{\mathrm{\prime }}$ as $N\to \mathrm{\infty }$, so the partial sums converge to $f-{\phi }_j\ast {\phi }_j\ast f$ and the displayed one-sided identity holds for every $f\in {\mathcal{S}}^{\mathrm{\prime }}$ and $j\in \mathbb{Z}$.

The two-sided identity for $H^p$ elements. Let $f\in H^p$ and set $g=M_{\phi }^{0}f\in L^p$, so $\mathrm{\mid }{\phi }_j\ast f\mathrm{\mid }\le g$ pointwise for every $j$. If $0<p<1$, choose an admissible integer order $N$ with $G=M_{N}f\in L^p$ by [F5], and let $C_N=3^N{P}_N(\phi )$. For $r\in \mathbb{Z}$ put ${\mathrm{\Omega }}_r=\{x:G(x)>2^r\}$; by [F7] it is measurable, and ${\mathrm{\Omega }}_r\subseteq \{G^p\ge 2^{rp}\}$, so [F8] applied to $G^p$ at threshold $2^{rp}$ gives $\lambda ({\mathrm{\Omega }}_r)\le 2^{-rp}\mathrm{\parallel }G{\mathrm{\parallel }}_p^p<\mathrm{\infty }$. Fix $r$. By [F6], $\lambda (B(y,2^{-j+1}))=c_n{2}^{(-j+1)n}\to \mathrm{\infty }$ as $j\to -\mathrm{\infty }$, uniformly in $y$. Thus for all sufficiently negative $j$ and every $y\in {\mathbb{R}}^n$ there is $x\in B(y,2^{-j+1})\setminus {\mathrm{\Omega }}_r$; otherwise this ball would be contained in ${\mathrm{\Omega }}_r$ and have measure at most $\lambda ({\mathrm{\Omega }}_r)$. Then $\mathrm{\mid }y-x\mathrm{\mid }<2\cdot 2^{-j}$, so [F5] gives $\mathrm{\mid }({\phi }_j\ast f)(y)\mathrm{\mid }\le C_{N}G(x)\le C_N{2}^r$. Hence $\mathrm{\parallel }{\phi }_j\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_N{2}^r$, and [F6] gives $\mathrm{\parallel }{\phi }_j\ast {\phi }_j\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{\phi }_j{\mathrm{\parallel }}_1\mathrm{\parallel }{\phi }_j\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_1{C}_N{2}^r\mathrm{.}$ Given $\epsilon >0$, choose $r$ so negative that the right-hand side is less than $\epsilon$, and then choose $j$ sufficiently negative. Thus ${\phi }_j\ast {\phi }_j\ast f\to 0$ uniformly, hence in ${\mathcal{S}}^{\mathrm{\prime }}$ because every Schwartz test function is integrable by [F4]. If $p\ge 1$, Hölder instead gives $\mathrm{\mid }{\phi }_j\ast {\phi }_j\ast f(x)\mathrm{\mid }\le \mathrm{\parallel }{\phi }_j{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\mathrm{\parallel }{\phi }_j\ast f{\mathrm{\parallel }}_p\le 2^{jn\mathrm{/}p}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\mathrm{\parallel }g{\mathrm{\parallel }}_p⟶0$ as $j\to -\mathrm{\infty }$. For $p>1$, [F6] gives this norm scaling since $\int \mathrm{\mid }{\phi }_j{\mathrm{\mid }}^{p^{\mathrm{\prime }}}=2^{jn{p}^{\mathrm{\prime }}}{2}^{-jn}\int \mathrm{\mid }\phi (u){\mathrm{\mid }}^{p^{\mathrm{\prime }}}du$, so $n(1-1\mathrm{/}{p}^{\mathrm{\prime }})=n\mathrm{/}p$; for $p=1$ it is the supremum scaling $\mathrm{\parallel }{\phi }_j{\mathrm{\parallel }}_{\mathrm{\infty }}=2^{jn}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$. The other bound uses $\mathrm{\mid }{\phi }_j\ast f\mathrm{\mid }\le g$ and [F3]. Hence in every case ${\phi }_j\ast {\phi }_j\ast f\to 0$ in ${\mathcal{S}}^{\mathrm{\prime }}$. Passing to the limit $j\to -\mathrm{\infty }$ in the one-sided identity of step 2.1 (with the series understood as ${\lim }_{j\to -\mathrm{\infty }}{\sum }_{k\ge j}$, whose partial sums are ${\phi }_{N+1}\ast {\phi }_{N+1}\ast f-{\phi }_j\ast {\phi }_j\ast f$) gives $f={\sum }_{k\in \mathbb{Z}}{\psi }_k\ast {\tilde{\psi }}_k\ast f$ in ${\mathcal{S}}^{\mathrm{\prime }}$.

Conclusion. Step 1.1 gives the support and moment properties of $\psi$; step 2.1 gives the one-sided telescoping identity for every tempered distribution; step 3.1 gives the two-sided identity for $H^p$ elements. This proves the lemma.