Hilbert transform is not strong type (1,1)

The indicator $f$ is measurable with $0\le f\le 1$ and $\{f\mathrm{\ne }0\}=(0,1)$ of measure one, so $\mathrm{\mid }f{\mathrm{\mid }}^p\le 1$ on a set of measure one and vanishes elsewhere; hence $f\in L^p(\mathbb{R};\mathbb{C})$ for every $1\le p\le \mathrm{\infty }$, with $\mathrm{\parallel }f{\mathrm{\parallel }}_p\le 1$.

On $(1,\mathrm{\infty })$ one has $q(x)=\frac{1}{\pi }\log \frac{x}{x-1}=\frac{1}{\pi }\log (1+\frac{1}{x-1})>0$ because $\log$ is strictly increasing and $\log 1=0$ by [F3].

For $0\le u\le 1$ one has $\log (1+u)={\int }_1^{1+u}\frac{dt}{t}\ge {\int }_1^{1+u}\frac{dt}{1+u}=\frac{u}{1+u}\ge \frac{u}{2}$, using monotonicity of the integral and $1+u\le 2$.

$q\mathrm{\notin }{L}^1(\mathbb{R})$: for $x\ge 2$ put $u=\frac{1}{x-1}\in (0,1]$; steps 1.2 and 1.3 give $q(x)\ge \frac{1}{2\pi }\cdot \frac{1}{x-1}\ge \frac{1}{2\pi x}\ge 0$. Hence for every $R>2$, using additivity over the interval and [F8] with the antiderivative of [F3], ${\int }_{(2,R)}q\ge \frac{1}{2\pi }{\int }_{(2,R)}\frac{dx}{x}=\frac{1}{2\pi }(\log R-\log 2)$, which tends to $+\mathrm{\infty }$; monotone convergence [F4] gives ${\int }_{(2,\mathrm{\infty })}q=+\mathrm{\infty }$, and monotonicity in the domain gives ${\int }_{\mathbb{R}}\mathrm{\mid }q\mathrm{\mid }\ge {\int }_{(2,\mathrm{\infty })}q=+\mathrm{\infty }$. So $q$ is not an $L^1$ class.

Let $\phi$ be the unit-mass bump of [F5] and for $j\in \mathbb{N}$ put $f_j:=f\ast {\phi }_{1\mathrm{/}(j+1)}$. Each $f_j$ is smooth with $\mathrm{supp}{f}_j\subseteq [-2\mathrm{/}(j+1),1+2\mathrm{/}(j+1)]$, so $f_j\in C_c^{\mathrm{\infty }}(\mathbb{R})\subseteq \mathcal{S}(\mathbb{R})$; and $0\le f_j\le 1$ because $0\le f\le 1$ and ${\phi }_{1\mathrm{/}(j+1)}\ge 0$ has integral one. Since $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_p\to 0$ for $p=1$ and $p=2$, each $f_j$ lies in $L^1\cap L^2$ and the sequence converges to $f$ in both norms.

$\mathrm{\parallel }H{f}_j-q{\mathrm{\parallel }}_2\to 0$: by [F1] $q=Hf$, and by [F2] $H$ is a linear isometry, so $\mathrm{\parallel }H{f}_j-q{\mathrm{\parallel }}_2=\mathrm{\parallel }H(f_j-f){\mathrm{\parallel }}_2=\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_2\to 0$ by step 2.2.

Suppose, for contradiction, that $T:L^1(\mathbb{R};\mathbb{C})\to L^1(\mathbb{R};\mathbb{C})$ is bounded and linear with $Tg=Hg$ for every $g\in \mathcal{S}(\mathbb{R})$. Since $f_j\in \mathcal{S}$, one has $T{f}_j=H{f}_j$ as $L^1$ classes, and $\mathrm{\parallel }T{f}_j-Tf{\mathrm{\parallel }}_1\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_1\to 0$ by step 2.2. Fix $\psi \in C_c^{\mathrm{\infty }}(\mathbb{R})$. Then $\mathrm{\mid }\int (T{f}_j-Tf)\psi \mathrm{\mid }\le \mathrm{\parallel }T{f}_j-Tf{\mathrm{\parallel }}_1\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ and, by step 3.1, $\mathrm{\mid }\int (H{f}_j-q)\psi \mathrm{\mid }\le \mathrm{\parallel }H{f}_j-q{\mathrm{\parallel }}_2\mathrm{\parallel }\psi {\mathrm{\parallel }}_2\to 0$; since the $j$-th integrals of $T{f}_j$ and $H{f}_j$ agree, it follows that $\int (Tf)\psi =\int q\psi$, i.e. $\int (Tf-q)\psi =0$ for every $\psi \in C_c^{\mathrm{\infty }}(\mathbb{R})$.

For $R>8$ set ${\psi }_R:={\mathbf{1}}_{[4,R]}\ast {\phi }_1$ with ${\phi }_1$ as in [F5]. Then ${\psi }_R\in C_c^{\mathrm{\infty }}(\mathbb{R})$, its support is contained in $[2,R+2]$, where $q\ge 0$ by step 1.2, $0\le {\psi }_R\le 1$, and ${\psi }_R=1$ on $[6,R-2]$: indeed ${\psi }_R(x)={\int }_{x-R}^{x-4}{\phi }_1(y)dy$, and for $x\in [6,R-2]$ the interval $[x-R,x-4]$ contains the support of ${\phi }_1$, which is contained in $[-2,2]$. Step 4.1 gives $\int (Tf){\psi }_R=\int q{\psi }_R$ for every $R>8$. As $R\to \mathrm{\infty }$ the functions ${\psi }_R$ converge pointwise to the bounded function $x\mapsto {\int }_{-\mathrm{\infty }}^{x-4}{\phi }_1$, so dominated convergence [F7] with majorant $\mathrm{\mid }Tf\mathrm{\mid }\in L^1$ shows that the left-hand sides converge to a finite limit; but step 2.1 and ${\psi }_R=1$ on $[6,R-2]$ give $\int q{\psi }_R\ge {\int }_{[6,R-2]}q\ge \frac{1}{2\pi }(\log (R-2)-\log 6)\to \mathrm{\infty }$. A sequence cannot converge to a finite limit while equalling terms that tend to $+\mathrm{\infty }$, so no such $T$ exists.

The compatibility hypothesis in step 4.1 was imposed only on Schwartz functions, which lie in $L^1\cap L^2$; hence there is no bounded linear $L^1\to L^1$ operator agreeing with the $L^2$ Hilbert transform on $L^1\cap L^2$ either. The witness $f\in L^1$ with transform $q\mathrm{\notin }{L}^1$ therefore refutes strong type $(1,1)$. Nothing here addresses weak type $(1,1)$, which is a different assertion.