Calderón–Zygmund operators need not map L∞ to L∞

The function $q$ is unbounded above and below on every punctured neighbourhood of the endpoints: for $0<x<1$, $\frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}=\frac{x}{1-x}$, which tends to $0$ as $x\to 0^{+}$ and to $+\mathrm{\infty }$ as $x\to 1^{-}$; since $\log$ is continuous, strictly increasing, with ${\lim }_{u\downarrow 0}\log u=-\mathrm{\infty }$ and ${\lim }_{u\to \mathrm{\infty }}\log u=+\mathrm{\infty }$, one has $q(x)\to -\mathrm{\infty }$ as $x\to 0^{+}$ and $q(x)\to +\mathrm{\infty }$ as $x\to 1^{-}$. Hence for every $M>0$ the sets $\{q>M\}$ and $\{q<-M\}$ contain nondegenerate intervals, so both have positive Lebesgue measure and $q$ is not essentially bounded.

Suppose $T:L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})\to L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$ were bounded and agreed with the $L^2$ Hilbert transform on $L^{\mathrm{\infty }}\cap L^2$. Since $f\in L^{\mathrm{\infty }}\cap L^2$ by [F1], the class $Tf$ would equal the class $Hf=q$; but $q$ is not essentially bounded by step 1.1, whereas every class in $L^{\mathrm{\infty }}$ is essentially bounded. This contradiction shows that no bounded $L^{\mathrm{\infty }}\to L^{\mathrm{\infty }}$ extension compatible on $L^{\mathrm{\infty }}\cap L^2$ exists, which is the asserted failure; nothing here concerns a BMO-valued estimate.