Weak-type estimate

Remarks

Equivalently, $\norm{Tf}_{q, \infty} \le C \norm{f}_p$, where $\norm{\cdot}_{q,\infty}$ is the quasinorm of the weak Lp space $L^{q,\infty}(\nu)$. The operator is of strong type $(p, q)$ if $\norm{Tf}_q \le C \norm{f}_p$. By Chebyshev's inequality $\nu(\set{\abs{g} > \lambda}) \le \lambda^{-q} \norm{g}_q^q$, strong type $(p, q)$ implies weak type $(p, q)$ with the same constant. By convention, weak type $(p, \infty)$ means strong type $(p, \infty)$.

Example: let $X = (0, 1)$ and $Y = (0, \infty)$ with Lebesgue measure, and let $Tf(y) = c_f / y$ with $c_f = \int_0^1 f$. Then $\set{y : \abs{Tf(y)} > \lambda} = (0, \abs{c_f}/\lambda)$, of measure $\abs{c_f}/\lambda \le \norm{f}_1/\lambda$. So $T$ is of weak type $(1, 1)$ with $C = 1$, but not of strong type $(1, 1)$, because $\int_0^\infty dy/y = \infty$ when $c_f \neq 0$.

Non-example: on the same spaces, with $Y = (0, 1)$, let $Tf(y) = c_f / y^2$. For $f = 1$ one has $c_f = 1$ and $\set{Tf > \lambda} = (0, \lambda^{-1/2})$ for $\lambda > 1$. The ratio $\lambda^{-1/2} / (C/\lambda) = \lambda^{1/2}/C$ is unbounded as $\lambda \to \infty$, so $T$ is not of weak type $(1, 1)$.