Hilbert transform does not map L-infinity to L-infinity

The indicator $f$ is measurable with $0\le f\le 1$, so $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, and ${\int }_{\mathbb{R}}\mathrm{\mid }f{\mathrm{\mid }}^2={\lambda }_1((0,1))=1$ by [F4]; hence $f\in L^{\mathrm{\infty }}(\mathbb{R})\cap L^2(\mathbb{R})$.

$q$ is not essentially bounded. Indeed, fix $M>0$; by [F1] and [F3], for $x\in (0,1)$ one has $q(x)>M$ exactly when $\log \frac{x}{1-x}>\pi M$, i.e. $\frac{x}{1-x}>e^{\pi M}$, i.e. $x>\frac{1}{1+e^{-\pi M}}$. Hence the set $E_M:=\{x\in (0,1):q(x)>M\}$ is the interval $(\frac{1}{1+e^{-\pi M}},1)$, which by [F4] has measure $\frac{e^{-\pi M}}{1+e^{-\pi M}}>0$. Since $M$ was arbitrary, no real number bounds $q$ from above almost everywhere, so $q\mathrm{\notin }{L}^{\mathrm{\infty }}(\mathbb{R})$.

Let $\phi$ be the unit-mass bump of [F5] and for $j\in \mathbb{N}$ put $f_j:=f\ast {\phi }_{1\mathrm{/}(j+1)}$. Then $f_j$ is smooth with support in $[-2\mathrm{/}(j+1),1+2\mathrm{/}(j+1)]$, hence $f_j\in C_c^{\mathrm{\infty }}(\mathbb{R})\subseteq \mathcal{S}(\mathbb{R})$; and $0\le f_j\le 1$ because $0\le f\le 1$ and ${\phi }_{1\mathrm{/}(j+1)}\ge 0$ has integral one. By [F5], $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_2\to 0$.

$\mathrm{\parallel }H{f}_j-q{\mathrm{\parallel }}_2\to 0$: by [F1] $q=Hf$ and by [F2] $H$ is a linear isometry, so $\mathrm{\parallel }H{f}_j-q{\mathrm{\parallel }}_2=\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_2\to 0$ by step 2.1.

Suppose, for contradiction, that $T:L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})\to L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$ is bounded and linear with $Tg=Hg$ almost everywhere for every $g\in L^{\mathrm{\infty }}(\mathbb{R})\cap L^2(\mathbb{R})$. Each $f_j$ of step 2.1 lies in this intersection, so $T{f}_j=H{f}_j$ almost everywhere; by [F7] and $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, $\mathrm{\parallel }H{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }T{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }T\mathrm{\parallel }$.

By step 3.1 and [F6] there is a subsequence $(H{f}_{j_k}{)}_k$ converging almost everywhere to $q$. The sets where $T{f}_{j_k}\mathrm{\ne }H{f}_{j_k}$ are null, the sets where $\mathrm{\mid }H{f}_{j_k}\mathrm{\mid }>\mathrm{\parallel }T\mathrm{\parallel }$ are null by step 3.2, and the set where the subsequence fails to converge to $q$ is null; their countable union is null by [F6]. Off that union one has $\mathrm{\mid }H{f}_{j_k}\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }$ for every $k$ by step 3.2 and $H{f}_{j_k}\to q$, so $\mathrm{\mid }q\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }$ almost everywhere. Hence $q\in L^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }T\mathrm{\parallel }$.

Step 4.1 contradicts step 1.2, so no such bounded linear operator $T$ exists. The compatibility required of $T$ was only on $L^{\mathrm{\infty }}\cap L^2$, hence also holds for every Schwartz function; therefore no bounded $L^{\mathrm{\infty }}$ action agreeing with the $L^2$ Hilbert transform on the intersection exists. A $BMO$-valued endpoint is a different assertion and is not addressed.