Calderón–Zygmund theorem

Remarks

The proof has three steps. The three conditions together give a uniform bound on the Fourier transforms of the truncated kernels, hence a uniform $L^2$ bound by the Plancherel theorem. The Calderón–Zygmund lemma splits $f$ into a bounded good part and a bad part with mean zero on disjoint cubes. Condition 2 controls the operator on the bad part away from the cubes, and this gives the weak type $(1, 1)$ bound. The Marcinkiewicz interpolation theorem then gives $1 < p \le 2$, and duality gives $2 \le p < \infty$.

Example: on $\R$, $K(x) = \frac{1}{\pi x}$ satisfies condition 1 with $A = 1/\pi$, and condition 3 because $K$ is odd. For condition 2, $\abs{K(x - y) - K(x)} = \frac{\abs{y}}{\pi \abs{x}\abs{x - y}} \le \frac{2\abs{y}}{\pi x^2}$ when $\abs{x} \ge 2\abs{y}$, whose integral is $2/\pi$. So the truncated Hilbert transforms are bounded on $L^p(\R)$ uniformly in $\varepsilon$, which recovers the M. Riesz theorem.

The cancellation condition 3 is needed. For $K(x) = \abs{x}^{-n}$, conditions 1 and 2 hold, but $T_\varepsilon f(x)$ grows like $\log(1/\varepsilon)$ for $f$ equal to $1$ near $x$. The conclusion fails for $p = 1$ and $p = \infty$, as already for the Hilbert transform.

Going deeper

For a student

This is the central theorem of the theory of singular integral operators. Its method, an $L^2$ bound, the Calderón–Zygmund decomposition and interpolation, also proves the Hörmander–Mikhlin multiplier theorem and the bounds for maximal functions. It extends to non-convolution operators, the Calderón–Zygmund operators, once an $L^2$ bound is known; the T(1) theorem gives a test for that $L^2$ bound.