Mihlin endpoints: weak (1,1), but no general strong endpoint bounds

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). The displayed estimate in The Mihlin–Hörmander Fourier multiplier theorem gives strong $L^p$ bounds for $1<p<\mathrm{\infty }$. Its proof also identifies $T_m$ as a Calderón–Zygmund operator with Hörmander constant at most $C_{n}A$ and $L^2$ norm $\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$. Consequently Calderón–Zygmund operators are of weak type (1,1) gives the weak endpoint $\mathrm{\mid }\{\mathrm{\mid }{T}_{m}f\mathrm{\mid }>\lambda \}\mathrm{\mid }\le C_n(A+\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1(f\in L^1,\lambda >0),$ where $A$ is the derivative-bound constant in the Mihlin theorem. This is a bound for the multiplier operator; it does not require the stronger principal-value and pointwise kernel hypotheses used for maximal truncations.

No general strong $L^1$ or $L^{\mathrm{\infty }}$ bound follows. In dimension one, $m(\xi )=-i\mathrm{sgn}\xi$ satisfies the Mihlin hypotheses despite its jump at the origin, and its Hilbert transform has the weak endpoint above while the interval-indicator counterexamples refute compatible strong $L^1$ and $L^{\mathrm{\infty }}$ bounds (Strong type (1,1) fails for the Hilbert transform ↗, Calderón–Zygmund operators need not map L∞ to L∞ ↗). A jump at a nonzero frequency violates the Mihlin smoothness hypothesis; boundedness of a symbol alone does not imply a strong $L^1$ bound.