Littlewood Paley Theory and Square Functions — Examples

The pieces on the low-frequency ball. Since $\mathrm{supp}\widehat{f}\subset \{\mathrm{\mid }\xi \mathrm{\mid }\le 1\}$ and $\psi =1$ there, ${\phi }_0\widehat{f}=\psi \widehat{f}=\widehat{f}$. For $j\ge 1$, $\mathrm{supp}\widehat{f}$ is contained in the vanishing region $\{\mathrm{\mid }\xi \mathrm{\mid }\le 2^{j-1}\}$ of ${\phi }_j$ (because $2^{j-1}\ge 1$), so ${\phi }_j\widehat{f}=0$.

The square function. By step 1.1 and [L2], ${\mathrm{\Delta }}_{0}f={\mathcal{F}}^{-1}(\widehat{f})=f$ and ${\mathrm{\Delta }}_{j}f={\mathcal{F}}^{-1}(0)=0$ for every $j\ge 1$; hence only the $j=0$ term of the square function survives, $Sf=\mathrm{\mid }{\mathrm{\Delta }}_{0}f\mathrm{\mid }=\mathrm{\mid }f\mathrm{\mid }$, and therefore $\mathrm{\parallel }Sf{\mathrm{\parallel }}_p=\mathrm{\parallel }\mathrm{\mid }f\mathrm{\mid }{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$ for every $1\le p<\mathrm{\infty }$. For a nonzero function in this class, $c_p\mathrm{\parallel }f{\mathrm{\parallel }}_p\le \mathrm{\parallel }Sf{\mathrm{\parallel }}_p\le C_p\mathrm{\parallel }f{\mathrm{\parallel }}_p$ therefore forces $c_p\le 1\le C_p$.

Disjointness of the active levels. Suppose first that $i\ge 1$ and ${\mathrm{\Delta }}_i{f}_1\mathrm{\ne }0$. Since $\widehat{{\mathrm{\Delta }}_i{f}_1}={\phi }_i\widehat{f_1}$ by [L1] and the Fourier transform is injective, there is $\xi$ with ${\phi }_i(\xi )\mathrm{\ne }0$ and $2^{j-1}\le \mathrm{\mid }\xi \mathrm{\mid }\le 2^{j+1}$. By [L1], ${\phi }_i(\xi )\mathrm{\ne }0$ forces $2^{i-1}<\mathrm{\mid }\xi \mathrm{\mid }<2^{i+1}$, so $2^{j-1}<2^{i+1}$ and $2^{i-1}<2^{j+1}$, which imply $j-1\le i\le j+1$. If instead $i=0$ and ${\mathrm{\Delta }}_0{f}_1\mathrm{\ne }0$, then some $\xi$ in the packet support also lies in $\mathrm{supp}{\phi }_0\subset \{\mathrm{\mid }\xi \mathrm{\mid }<2\}$; since $2^{j-1}\le \mathrm{\mid }\xi \mathrm{\mid }<2$, this forces $j\le 1$, and therefore $0\in \{j-1,j,j+1\}$. Thus every active $i$ for $f_1$ belongs to $\{j-1,j,j+1\}\cap \{0,1,2,\dots \}$. The same argument shows every active $i$ for $f_2$ belongs to $\{k-1,k,k+1\}$; because $k\ge j+3$, the two nonnegative index sets are disjoint. Hence for every $i$ at most one of ${\mathrm{\Delta }}_i{f}_1$, ${\mathrm{\Delta }}_i{f}_2$ is nonzero.

Pointwise Euclidean-square identity. For every $x$ and every $i$, step 1.1 gives $\mathrm{\mid }{\mathrm{\Delta }}_{i}f(x){\mathrm{\mid }}^2=\mathrm{\mid }{\mathrm{\Delta }}_i{f}_1(x)+{\mathrm{\Delta }}_i{f}_2(x){\mathrm{\mid }}^2=\mathrm{\mid }{\mathrm{\Delta }}_i{f}_1(x){\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\Delta }}_i{f}_2(x){\mathrm{\mid }}^2$ (the cross term vanishes because one of the two numbers is zero), hence $Sf(x{)}^2={\sum }_i\mathrm{\mid }{\mathrm{\Delta }}_{i}f(x){\mathrm{\mid }}^2=S{f}_1(x{)}^2+S{f}_2(x{)}^2$; the rearrangement is legitimate because each packet has at most three active levels by step 1.1, so only finitely many indices contribute.

Norms and orthogonality of the packets. Because $f_1,f_2\in \mathcal{S}$, both square functions lie in $L^2$ by [L2], and integrating the identity of step 2.1 gives $\mathrm{\parallel }Sf{\mathrm{\parallel }}_2^2=\mathrm{\parallel }S{f}_1{\mathrm{\parallel }}_2^2+\mathrm{\parallel }S{f}_2{\mathrm{\parallel }}_2^2$; the $L^2$ almost orthogonality [L2] identifies each side with the sum of the squared dyadic-piece norms. Finally, $\widehat{f_1}\widehat{f_2}=0$ pointwise because the two Fourier supports are disjoint, so Plancherel gives $⟨f_1,f_2⟩=\int \widehat{f_1}\overline{\widehat{f_2}}=0$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{f}_2{\mathrm{\parallel }}_2^2$.

The inverse transform of the sharp cutoff. Since $\chi \in L^1\cap L^2$, [F1] gives ${\mathcal{F}}^{-1}\chi =R{\mathcal{F}}_2\chi$ as an $L^2$ class, and ${\mathcal{F}}_2\chi =\widehat{\chi }$ almost everywhere, so $K_0(x)=\widehat{\chi }(-x)={\int }_1^2{e}^{2\pi ix\xi }d\xi$ for almost every $x$, the last function being continuous and hence the correct representative. For $x\mathrm{\ne }0$, [F3] gives ${\int }_1^2{e}^{2\pi ix\xi }d\xi =(e^{4\pi ix}-e^{2\pi ix})\mathrm{/}(2\pi ix)=e^{3\pi ix}(e^{\pi ix}-e^{-\pi ix})\mathrm{/}(2\pi ix)=e^{3\pi ix}\sin (\pi x)\mathrm{/}(\pi x)$, using $e^{\pi ix}-e^{-\pi ix}=2i\sin (\pi x)$ from Euler's formula, and $K_0(0)={\int }_1^{2}d\xi =1$; consequently $\mathrm{\mid }{K}_0(x)\mathrm{\mid }=\mathrm{\mid }\sin (\pi x)\mathrm{\mid }\mathrm{/}(\pi \mathrm{\mid }x\mathrm{\mid })$ for $x\mathrm{\ne }0$.

The kernel is not in $L^1$. For $m\ge 1$ let $I_m:=[m+\frac{1}{4},m+\frac{3}{4}]$. On $I_m$ one has $2\pi x\in [2\pi m+\frac{\pi }{2},2\pi m+\frac{3\pi }{2}]$, so $\cos (2\pi x)\le 0$; since $\mathrm{\mid }{K}_0(x){\mathrm{\mid }}^2=\mathrm{\mid }{e}^{4\pi ix}-e^{2\pi ix}{\mathrm{\mid }}^2\mathrm{/}(4{\pi }^2{x}^2)=(2-2\cos (2\pi x))\mathrm{/}(4{\pi }^2{x}^2)\ge 1\mathrm{/}(2{\pi }^2{x}^2)$ on $I_m$, and $\mathrm{\mid }x\mathrm{\mid }\le m+\frac{3}{4}$, there holds ${\int }_{I_m}\mathrm{\mid }{K}_0\mathrm{\mid }\ge \frac{1}{\pi \sqrt{2}}\cdot \frac{1}{m+3\mathrm{/}4}\cdot \frac{1}{2}\ge \frac{c}{m+1}$ with $c:=1\mathrm{/}(4\pi \sqrt{2})$. The intervals $I_m$ are pairwise disjoint. For every $N\ge 1$ set $a_m:=1\mathrm{/}(\pi \sqrt{2}(m+3\mathrm{/}4))$ and $s_N:={\sum }_{m=1}^N{a}_m{\mathbf{1}}_{I_m}$. The pointwise bound just proved gives $s_N\le \mathrm{\mid }{K}_0\mathrm{\mid }{\mathbf{1}}_{[1,\mathrm{\infty })}$, and [F4] yields ${\int }_1^{\mathrm{\infty }}\mathrm{\mid }{K}_0\mathrm{\mid }\ge \int s_N={\sum }_{m=1}^N{a}_m\mathrm{/}2\ge c{\sum }_{m=1}^{N}1\mathrm{/}(m+1)$. These finite lower bounds are unbounded by the harmonic-series argument in [F4], so ${\int }_1^{\mathrm{\infty }}\mathrm{\mid }{K}_0\mathrm{\mid }=+\mathrm{\infty }$ and hence $\mathrm{\parallel }{K}_0{\mathrm{\parallel }}_{L^1}=+\mathrm{\infty }$.

No uniform bound and the failure of the sharp replacement. By [F2] and step 1.1, $\widehat{{\chi }_j}(x)=2^j\widehat{\chi }(2^{j}x)$ and hence $K_j(x)=\widehat{{\chi }_j}(-x)=2^j\widehat{\chi }(-2^{j}x)=2^j{K}_0(2^{j}x)$ for every $j\in \mathbb{Z}$; the nonnegative change of variables in [F2], with $b=2^j>0$, gives $\mathrm{\parallel }{K}_j{\mathrm{\parallel }}_{L^1}=\mathrm{\parallel }{K}_0{\mathrm{\parallel }}_{L^1}=+\mathrm{\infty }$ for every $j$, so in particular ${\sup }_j\mathrm{\parallel }{\chi }_j^{\vee }{\mathrm{\parallel }}_{L^1}=\mathrm{\infty }$. This contradicts the uniform bound $\mathrm{\parallel }{K}_j{\mathrm{\parallel }}_1\le C$ of the smooth partition [F5]: the sharp-cutoff family cannot replace the smooth annular cutoffs in the convolution estimates, and smoothness of the partition is used essentially, not cosmetically.

The values of the pieces on $A_j$. Let $\xi \in A_j$. If $j=0$ then $\mathrm{\mid }\xi \mathrm{\mid }\le 1$ and ${\phi }_0(\xi )=\psi (\xi )=1$; if $j\ge 1$ then $\mathrm{\mid }\xi \mathrm{\mid }\le 2^j$ gives $\psi (2^{-j}\xi )=1$, while $\mathrm{\mid }\xi \mathrm{\mid }>2^{j-1}(1+\epsilon )$ gives $\mathrm{\mid }{2}^{-(j-1)}\xi \mathrm{\mid }>1+\epsilon$, hence $\psi (2^{-(j-1)}\xi )=0$ and ${\phi }_j(\xi )=1$. For $i\mathrm{\ne }j$: if $1\le i\le j-1$ then the smaller argument has $\mathrm{\mid }{2}^{-i}\xi \mathrm{\mid }\ge \mathrm{\mid }{2}^{-(j-1)}\xi \mathrm{\mid }>1+\epsilon$, and so does the larger argument, so both $\psi$ values vanish and ${\phi }_i(\xi )=0$; for $i=0<j$, $\mathrm{\mid }\xi \mathrm{\mid }>1+\epsilon$ gives ${\phi }_0(\xi )=\psi (\xi )=0$; if $i\ge j+1$ then the larger argument has $\mathrm{\mid }{2}^{-(i-1)}\xi \mathrm{\mid }\le 2^{-j}\mathrm{\mid }\xi \mathrm{\mid }\le 1$, so both $\psi$ values are $1$ and again ${\phi }_i(\xi )=0$.

The pieces of $f$. By step 1.1, $\widehat{{\mathrm{\Delta }}_{j}f}={\phi }_j\widehat{f}=\widehat{f}$ and $\widehat{{\mathrm{\Delta }}_{i}f}={\phi }_i\widehat{f}=0$ for $i\mathrm{\ne }j$; since the Fourier transform is injective on tempered distributions, ${\mathrm{\Delta }}_{j}f=f$ and ${\mathrm{\Delta }}_{i}f=0$ for $i\mathrm{\ne }j$.

The Sobolev comparison. Applying the characterisation [L2] to the regular distribution $u_f$ of $f$, whose ${\mathrm{\Delta }}_i$-images are $u_{{\mathrm{\Delta }}_{i}f}$ by [L2], and using step 2.1, gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^s}^2\asymp {\sum }_{i\ge 0}{2}^{2is}\mathrm{\parallel }{\mathrm{\Delta }}_{i}f{\mathrm{\parallel }}_2^2=2^{2js}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$; moreover on $A_j$ the Japanese bracket satisfies $⟨\xi ⟩\asymp 2^j$ for $j\ge 1$ (as $2^{j-1}(1+\epsilon )<\mathrm{\mid }\xi \mathrm{\mid }\le 2^j$) and $⟨\xi ⟩\asymp 1=2^0$ for $j=0$, so the weight implicit in the comparison is exactly the dyadic weight $2^{js}$. Taking square roots gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^s}{\asymp }_s{2}^{js}\mathrm{\parallel }f{\mathrm{\parallel }}_2$ with constants depending only on $n,s$ and the fixed partition (through $\epsilon$).

Conclusion. Steps 1.1 to 3.1 verify the asserted values of the pieces, the identities ${\mathrm{\Delta }}_{j}f=f$, ${\mathrm{\Delta }}_{i}f=0$ ($i\mathrm{\ne }j$), and the two-sided Sobolev comparison, including the low-frequency case $j=0$ where the weight $2^0=1$ is the correct one.