Cotlar's inequality for maximal truncations

Fix the auxiliary bump once as a function of dimension only: let $a(t)=e^{-1\mathrm{/}t}$ for $t>0$ and $a(t)=0$ otherwise, put $\rho (y)=a(1-16\mathrm{\mid }y{\mathrm{\mid }}^2)$ and $\phi (y)=c_n\rho (y)$ with $c_n=(\int \rho {)}^{-1}$. The derivatives of $a$ on $t>0$ have the form $P_k(1\mathrm{/}t)e^{-1\mathrm{/}t}$, with $P_{k+1}(s)=s^2(P_k(s)-P_k^{\mathrm{\prime }}(s))$; [F6] makes each derivative and its difference quotient tend to zero at $t=0$, proving smoothness across that point. Thus $\rho$ is smooth, supported in the closed radius-$1\mathrm{/}4$ ball, nonnegative and radially nonincreasing since $a^{\mathrm{\prime }}(t)\ge 0$. Its mass is finite by boundedness and compact support, and positive since it is bounded below by $a(3\mathrm{/}4)>0$ on the radius-$1\mathrm{/}8$ ball, which has positive measure by [F6]. This gives the required unit-mass bump with support inside $B(0,1\mathrm{/}2)$. All its derivative bounds and $c_n$ depend only on $n$. For fixed $\epsilon >0$ put $k^{(\epsilon )}=k{\mathbf{1}}_{\{\mathrm{\mid }\cdot \mathrm{\mid }\ge \epsilon \}}$ and $R_{\epsilon }=k^{(\epsilon )}-W\ast {\phi }_{\epsilon }$. By [F2,F3], $W\ast {\phi }_{\epsilon }$ is smooth and equals the principal-value limit ${\lim }_{j\to \mathrm{\infty }}{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge {\delta }_j}k(z){\phi }_{\epsilon }(x-z)dz$. This need not be an absolutely convergent integral near $z=0$. If $\mathrm{\mid }x\mathrm{\mid }\ge 2\epsilon$, the support condition $\mathrm{\mid }x-z\mathrm{\mid }\le \epsilon \mathrm{/}2$ implies $\mathrm{\mid }z\mathrm{\mid }\ge 3\epsilon \mathrm{/}2$, so in that region the same formula is an ordinary absolutely convergent integral. Near the origin retain the principal-value limit and use the cancellation estimate in the next step.

Case $\mathrm{\mid }x\mathrm{\mid }<2\epsilon$. Write $(W\ast {\phi }_{\epsilon })(x)={\lim }_{j\to \mathrm{\infty }}(I_1^{(\delta )}+I_2^{(\delta )}+I_3^{(\delta )})$ along $\delta ={\delta }_j$ for all sufficiently large $j$ such that $0<{\delta }_j<\epsilon \mathrm{/}4$, with the three pieces obtained by inserting ${\phi }_{\epsilon }(y)={\phi }_{\epsilon }(x)+({\phi }_{\epsilon }(y)-{\phi }_{\epsilon }(x))$ and splitting at $\mathrm{\mid }x-y\mathrm{\mid }=\epsilon \mathrm{/}4$: $I_1^{(\delta )}={\int }_{\mathrm{\mid }x-y\mathrm{\mid }>\epsilon \mathrm{/}4}k(x-y){\phi }_{\epsilon }(y)dy$, $I_2^{(\delta )}={\int }_{\delta \le \mathrm{\mid }x-y\mathrm{\mid }\le \epsilon \mathrm{/}4}k(x-y)({\phi }_{\epsilon }(y)-{\phi }_{\epsilon }(x))dy$, and $I_3^{(\delta )}={\phi }_{\epsilon }(x){\int }_{\delta \le \mathrm{\mid }x-y\mathrm{\mid }\le \epsilon \mathrm{/}4}k(x-y)dy$; the three pieces are absolutely convergent and their sum is the truncation ${\int }_{\mathrm{\mid }x-y\mathrm{\mid }\ge \delta }k(x-y){\phi }_{\epsilon }(y)dy$. Here $\mathrm{\mid }{I}_1^{(\delta )}\mathrm{\mid }\le A_1(4\mathrm{/}\epsilon {)}^n\int {\phi }_{\epsilon }=A_1{4}^n{\epsilon }^{-n}$ because $\mathrm{\mid }k(x-y)\mathrm{\mid }\le A_1\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}$ on the domain; $\mathrm{\mid }{I}_2^{(\delta )}\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }{\phi }_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}{A}_1{\int }_{\mathrm{\mid }x-y\mathrm{\mid }\le \epsilon \mathrm{/}4}\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}\mathrm{\mid }x-y\mathrm{\mid }dy\le C_{\phi }{A}_1{\epsilon }^{-n}$ by the mean value theorem, the bound $\mathrm{\mid }k\mathrm{\mid }\le A_1\mathrm{\mid }\cdot {\mathrm{\mid }}^{-n}$ and polar coordinates of the punctured ball; and $\mathrm{\mid }{I}_3^{(\delta )}\mathrm{\mid }\le \mathrm{\parallel }{\phi }_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}{A}_3\le C_{\phi }{A}_3{\epsilon }^{-n}$ by the cancellation bound after the substitution $z=x-y$. Finally $\mathrm{\mid }{k}^{(\epsilon )}(x)\mathrm{\mid }\le A_1\mathrm{\mid }x{\mathrm{\mid }}^{-n}{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }\ge \epsilon \}}\le A_1{\epsilon }^{-n}$. Hence $\mathrm{\mid }{R}_{\epsilon }(x)\mathrm{\mid }\le C_{n,\phi }(A_1+A_3){\epsilon }^{-n}\le C_{n,\delta ,\phi }(A_1+A_2^{\mathrm{\prime }}+A_3){\epsilon }^{\delta }(\epsilon +\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }$, the last inequality because $\mathrm{\mid }x\mathrm{\mid }<2\epsilon$ gives $(\epsilon +\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }\ge (3\epsilon {)}^{-n-\delta }$.

Case $\mathrm{\mid }x\mathrm{\mid }\ge 2\epsilon$ of the error bound. Since ${\phi }_{\epsilon }$ is supported in $\mathrm{\mid }y\mathrm{\mid }\le \epsilon \mathrm{/}2$, for $\mathrm{\mid }x\mathrm{\mid }\ge 2\epsilon$ one has $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }$ on the support, so $k^{(\epsilon )}(x)=k(x)$ and, substituting $z=x-y$ in the formula of step 1.1, $R_{\epsilon }(x)=k(x)-\int k(x-y){\phi }_{\epsilon }(y)dy=\int [k(x)-k(x-y)]{\phi }_{\epsilon }(y)dy,$ whence the Hölder bound gives $\mathrm{\mid }{R}_{\epsilon }(x)\mathrm{\mid }\le A_2^{\mathrm{\prime }}\int \mathrm{\mid }y{\mathrm{\mid }}^{\delta }{\phi }_{\epsilon }(y)dy\mathrm{\mid }x{\mathrm{\mid }}^{-n-\delta }=A_2^{\mathrm{\prime }}{C}_{\phi }{\epsilon }^{\delta }\mathrm{\mid }x{\mathrm{\mid }}^{-n-\delta }\le A_2^{\mathrm{\prime }}{C}_{\phi }{2}^{n+\delta }{\epsilon }^{\delta }(\epsilon +\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }$.

The convolution identity is $T_{\epsilon }f=((Tf)\ast {\phi }_{\epsilon })+f\ast R_{\epsilon }$. Indeed, by the bounds in steps 1.2 and 2.1, $R_{\epsilon }$ is integrable and its convolution with $f$ converges absolutely; $k^{(\epsilon )}\ast f=T_{\epsilon }f$ also converges absolutely by the size bound and Schwartz decay. For the remaining term use the distribution pairing rather than interchange nonabsolute kernel integrals. The compactly supported integral $\int {\phi }_{\epsilon }(y)f(x-y-\cdot )dy$ converges in every Schwartz seminorm: all derivatives of $f$ decay rapidly, uniformly over $y$ in the fixed compact support. Continuity of the tempered distribution $W$ therefore permits its pairing to pass through that integral. This gives $(W\ast f)\ast {\phi }_{\epsilon }=W\ast (f\ast {\phi }_{\epsilon })$. Applying the same argument to $\int f(x-z){\phi }_{\epsilon }(z-\cdot )dz$ gives $f\ast (W\ast {\phi }_{\epsilon })=W\ast (f\ast {\phi }_{\epsilon })$: its Schwartz seminorms are bounded by integrals of $\mathrm{\mid }f(x-z)\mathrm{\mid }(1+\mathrm{\mid }z\mathrm{\mid }{)}^N$, finite for every $N$. Hence $f\ast (k^{(\epsilon )}-R_{\epsilon })=(W\ast f)\ast {\phi }_{\epsilon }$, proving the identity.

Steps 2.1 and 1.2 together show that for every $\epsilon >0$ and every $x\in {\mathbb{R}}^n$, $\mathrm{\mid }{R}_{\epsilon }(x)\mathrm{\mid }\le C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3){\epsilon }^{\delta }(\epsilon +\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }=C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3){\omega }_{\epsilon }(x),$ where $\omega (x):=(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }$, ${\omega }_{\epsilon }(y)={\epsilon }^{-n}\omega (y\mathrm{/}\epsilon )$, and $\omega$ is nonnegative, radially nonincreasing and integrable; note ${\epsilon }^{\delta }(\epsilon +\mathrm{\mid }x\mathrm{\mid }{)}^{-n-\delta }={\epsilon }^{-n}(1+\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\epsilon {)}^{-n-\delta }$.

First term bound: $\mathrm{\mid }((Tf)\ast {\phi }_{\epsilon })(x)\mathrm{\mid }\le \int \mathrm{\mid }Tf(x-y)\mathrm{\mid }{\phi }_{\epsilon }(y)dy\le \mathrm{\parallel }{\phi }_{\epsilon }{\mathrm{\parallel }}_{1}M(Tf)(x)=M(Tf)(x)$, using that ${\phi }_{\epsilon }$ is radially nonincreasing with $\int {\phi }_{\epsilon }=1$ and applying the domination lemma [F4] with $g=Tf$ (a smooth function of polynomial growth, hence locally integrable).

Second term bound: $\mathrm{\mid }(f\ast R_{\epsilon })(x)\mathrm{\mid }\le \int \mathrm{\mid }f(x-y)\mathrm{\mid }\mathrm{\mid }{R}_{\epsilon }(y)\mathrm{\mid }dy\le C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3)\int \mathrm{\mid }f(x-y)\mathrm{\mid }{\omega }_{\epsilon }(y)dy\le C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3)\mathrm{\parallel }\omega {\mathrm{\parallel }}_{1}Mf(x)$ by step 3.2 and the domination lemma [F4] applied to ${\omega }_{\epsilon }$, which is radially nonincreasing, integrable with $\mathrm{\parallel }{\omega }_{\epsilon }{\mathrm{\parallel }}_1=\mathrm{\parallel }\omega {\mathrm{\parallel }}_1$.

For every $\epsilon >0$ and every $x$, steps 3.1, 4.1 and 4.2 give $\mathrm{\mid }{T}_{\epsilon }f(x)\mathrm{\mid }\le M(Tf)(x)+C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3)\mathrm{\parallel }\omega {\mathrm{\parallel }}_{1}Mf(x)$; taking the supremum over $\epsilon >0$ and using [F1] yields the asserted inequality with $C_{n,\delta }$ redefined to absorb $\mathrm{\parallel }\omega {\mathrm{\parallel }}_1=\int (1+\mathrm{\mid }y\mathrm{\mid }{)}^{-n-\delta }dy<\mathrm{\infty }$, in particular for almost every $x$.