Hörmander integral kernel condition (${\sup }_{y\ne 0}{\int }_{|x|>2|y|}|K(x-y)-K(x)|dx<\infty$)

This integral smoothness condition on a convolution kernel controls how translations change the kernel away from the translation scale. It need not imply differentiability. Coupled with an $L^2$ operator bound, it yields a weak type (1,1) bound through the Calderón–Zygmund decomposition. It is distinct from the bracket-generating condition for differential operators also named after Hörmander.

A measurable convolution kernel satisfying the displayed size and cancellation properties together with the Hörmander integral kernel condition defines uniformly bounded Fourier multipliers after deletion of a ball at the origin. The deleted-ball kernels belong to $L^2$, although they need not belong to $L^1$ at infinity. Cancellation controls low-frequency integration near the deleted ball; translation by half an oscillation controls the far tail through the oscillatory tail estimate for a singular kernel.

For $\xi \ne 0$ set $y=\xi /(2|\xi {|}^2)$, and take $R>r\ge 4|y|$. Translation by $y$ reverses the oscillatory exponential's sign. Comparing the original annulus with its translate gives twice the integral as an integrable kernel difference plus boundary-shell errors. The shell errors are bounded by $C_{d}A|y|(r^{-1}+R^{-1})$. The displayed estimate follows by enlarging the constant. Its right-hand side tends to zero as $r\to \infty$, proving convergence of the Fourier integral's improper far tail without absolute integrability of $K$ itself.

Suppose a convolution operator has $L^2$ operator norm at most $A$, and its kernel satisfies the Hörmander integral kernel condition with constant $C$. Under hypotheses making its integrals well-defined, the Calderón–Zygmund decomposition gives weak type (1,1). Indeed, the bounded part has squared $L^2$ norm at most $2^d\lambda \Vert f{\Vert }_1$. The bad cubes occupy at most $\Vert f{\Vert }_1/\lambda$, while outside their $4\sqrt{d}$ dilations the cancellation estimate outside a dilated cube bounds the bad image in $L^1$. Applying distributional estimates to these three regions gives $|\{|Tf|>\lambda \}|\le ((4\sqrt{d}{)}^d+2^{d+2}{A}^2+4C)\Vert f{\Vert }_1/\lambda$.

If $K$ obeys the Hörmander integral kernel condition with constant $C$, and an integrable $b$ is supported in a cube $Q$ with $\int b=0$, then ${\int }_{(LQ{)}^c}|K\ast b|\le C\Vert b{\Vert }_1$ for $L=4\sqrt{d}$. Here $LQ$ is the concentric cube with multiplied side length. Subtract the kernel value at the cube center inside the convolution; outside $LQ$, every source displacement is smaller than half the observation displacement. Tonelli theorem and the integral kernel condition give the bound.