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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 $|\{|Tf|>\lambda \}|\le ((4\sqrt{d}{)}^d+2^{d+2}{A}^2+4C)\Vert f{\Vert }_1/\lambda$.

Let $c$ be the center and $\ell$ the side length of $Q$. The zero integral of $f$ lets us subtract a constant kernel value:Choose $L=4\sqrt{d}$. For $x$ outside the concentric cube $\hat{Q}$ of side length $L\ell$, and for $y\in Q$,Thus the domain of integration in $x$ is contained in $\{|x-c|>2|y-c|\}$. After the translations $X=x-c$ and $Y=y-c$, the Hörmander integral kernel condition givesIf $Y=0$ the difference is zero and the same inequality holds. Apply the triangle inequality and Tonelli theorem to obtain the cancellation estimate outside a dilated cube:In particular the constant on the right is the same $C$ as in the hypothesis; only the dilation $L$ depends on the dimension.