Hilbert transform of an interval indicator

Let $x\mathrm{\notin }\{0,1\}$ and $\epsilon >0$. Substituting $t=x-y$ in [F1] and using ${\mathbf{1}}_{(0,1)}(x-t)=1$ exactly for $t\in (x-1,x)$ gives $H_{\epsilon }f(x)=\frac{1}{\pi }{\int }_{(x-1,x)\cap \{\mathrm{\mid }t\mathrm{\mid }>\epsilon \}}\frac{dt}{t}$, the integrand being continuous on each piece because $t=0$ is either excluded by the truncation or avoided.

Construction of approximants. Put $\phi :=\chi \mathrm{/}\int \chi$ with $\chi$ as in [F3]. The bounds $0\le \chi \le 1$ and [F4] give $2={\lambda }_1([-1,1])\le \int \chi \le {\lambda }_1([-2,2])=4$, so $0<\int \chi <\mathrm{\infty }$ and $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ is nonnegative with $\int \phi =1$. Let $({\phi }_{\epsilon })$ be its mollifier family and put $f_j:=f\ast {\phi }_{1\mathrm{/}j}$ for $j\ge 1$. By [F7] each $f_j$ is smooth, and since $\mathrm{supp}(f)\subseteq [0,1]$ and $\mathrm{supp}({\phi }_{1\mathrm{/}j})\subseteq [-2\mathrm{/}j,2\mathrm{/}j]$, the support inclusion gives $\mathrm{supp}(f_j)\subseteq [-2\mathrm{/}j,1+2\mathrm{/}j]$. Also, if $\mathrm{dist}(y,\{0,1\})>2\mathrm{/}j$, the bump samples only where $f$ is constant, so $f_j(y)=f(y)$; hence $g_j=f_j-f$ is supported within distance $2\mathrm{/}j$ of the endpoints, and $f_j\in C_c^{\mathrm{\infty }}(\mathbb{R})$. Since $0\le \phi$ and $\int {\phi }_{1\mathrm{/}j}=1$, moreover $0\le f_j\le 1$ pointwise: $f_j(x)=\int f(x-y){\phi }_{1\mathrm{/}j}(y)dy\in [0,1]$.

Case $x>1$. For $0<\epsilon <x-1$ one has $(x-1,x)\subseteq (\epsilon ,\mathrm{\infty })$, so $H_{\epsilon }f(x)=\frac{1}{\pi }{\int }_{x-1}^x\frac{dt}{t}=\frac{1}{\pi }(\log x-\log (x-1))=\frac{1}{\pi }\log \frac{x}{x-1}$ by [F8] and [F9].

Case $x<0$. For $0<\epsilon <-x$ one has $(x-1,x)\subseteq (-\mathrm{\infty },-\epsilon )$, so $H_{\epsilon }f(x)=\frac{1}{\pi }{\int }_{x-1}^x\frac{dt}{t}=\frac{1}{\pi }(\log \mathrm{\mid }x\mathrm{\mid }-\log \mathrm{\mid }x-1\mathrm{\mid })=\frac{1}{\pi }\log \frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}$ by [F8], the antiderivative of $1\mathrm{/}t$ on the negative axis being $\log \mathrm{\mid }t\mathrm{\mid }$.

Case $0<x<1$. For $0<\epsilon <\min (x,1-x)$ the set $(x-1,x)\cap \{\mathrm{\mid }t\mathrm{\mid }>\epsilon \}$ is $(x-1,-\epsilon )\cup (\epsilon ,x)$, so by [F9] $H_{\epsilon }f(x)=\frac{1}{\pi }[\log \mathrm{\mid }-\epsilon \mathrm{\mid }-\log \mathrm{\mid }x-1\mathrm{\mid }+\log x-\log \epsilon ]=\frac{1}{\pi }[\log x-\log (1-x)]=\frac{1}{\pi }\log \frac{x}{1-x}$, the two $\log \epsilon$ terms cancelling exactly because $\log \mathrm{\mid }-\epsilon \mathrm{\mid }=\log \epsilon$; since $\mathrm{\mid }x-1\mathrm{\mid }=1-x>0$ this is $\frac{1}{\pi }\log \frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}$.

By [F6] applied with $p=1$ and $p=2$, the sequence of 1.2 satisfies $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_1\to 0$ and $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_2\to 0$; consequently $f_j\to f$ in $L^2$, and the $L^2$ boundedness of [F2] gives $\mathrm{\parallel }H{f}_j-Hf{\mathrm{\parallel }}_2\to 0$, where $H{f}_j$ is both the $L^2$ transform of $f_j$ and the pointwise principal value of [F2].

Fix $x\mathrm{\notin }\{0,1\}$ and put $\delta :=\frac{1}{2}\mathrm{dist}(x,\{0,1\})>0$; let $c\in \{0,1\}$ be the constant value of $f$ on $(x-\delta ,x+\delta )$ and set $r:=\delta \mathrm{/}4$. The mollifier is supported in $[-2\mathrm{/}j,2\mathrm{/}j]$, so for $j>4\mathrm{/}\delta$ its convolution samples only points of $(x-\delta ,x+\delta )$ when the argument lies in $(x-\delta \mathrm{/}2,x+\delta \mathrm{/}2)$; hence 1.2 gives $f_j=c$ there. Thus for $0<\eta <r$ the part of $H_{\eta }{f}_j(x)$ over $\eta <\mathrm{\mid }x-y\mathrm{\mid }<r$ is the integral of $c\mathrm{/}(\pi (x-y))$ over a symmetric annulus, hence is zero. The remaining integral is absolutely convergent because $f_j$ has compact support and $\mathrm{\mid }x-y\mathrm{\mid }\ge r$ there. Letting $\eta \downarrow 0$ in [F2] gives $H{f}_j(x)=\frac{1}{\pi }{\int }_{\mathrm{\mid }x-y\mathrm{\mid }>r}\frac{f_j(y)}{x-y}dy\mathrm{.}$

By 2.1, 2.2 and 2.3, for every $x\mathrm{\notin }\{0,1\}$ and every $0<\epsilon <r(x)$, where $r(x):=x-1$ for $x>1$, $r(x):=-x$ for $x<0$ and $r(x):=\min (x,1-x)$ for $0<x<1$, one has $H_{\epsilon }f(x)=q(x)=\frac{1}{\pi }\log \frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}$. Since $r(x)>0$, the symmetric principal value exists at every $x\mathrm{\notin }\{0,1\}$ and equals $q(x)$; this proves assertion 1.

The function $g_j=f_j-f$ is supported in $\{y:\mathrm{dist}(y,\{0,1\})\le 2\mathrm{/}j\}$ by 1.2, so for $j>4\mathrm{/}\delta$ and $y\in \mathrm{supp}(g_j)$ one has $\mathrm{\mid }x-y\mathrm{\mid }\ge 2\delta -\frac{2}{j}\ge \frac{\delta }{2}$; hence $\mid \frac{1}{\pi }{\int }_{\mathbb{R}}\frac{g_j(y)}{x-y}dy\mid \le \frac{\mathrm{\parallel }{g}_j{\mathrm{\parallel }}_1}{\pi }\cdot \frac{2}{\delta }\to 0$ as $j\to \mathrm{\infty }$ by 2.4.

Since $f=c$ on $\mathrm{\mid }x-y\mathrm{\mid }<r$, the same symmetric cancellation shows that for every $0<\eta <r$, $H_{\eta }f(x)=\frac{1}{\pi }{\int }_{\mathrm{\mid }x-y\mathrm{\mid }>r}\frac{f(y)}{x-y}dy$. This outer integral is absolutely convergent because $f$ has compact support and its denominator is bounded away from zero. By 3.1 its value is $q(x)$.

Combining 2.5, 3.2 and 4.1, $H{f}_j(x)\to q(x)$ for every fixed $x\mathrm{\notin }\{0,1\}$.

By 2.4, $H{f}_j\to Hf$ in $L^2$; by [F10] a subsequence converges almost everywhere to a representative of the class $Hf$, while 5.1 makes that same subsequence converge to $q$ at every point of the full-measure set $\mathbb{R}\setminus \{0,1\}$. Therefore $q=Hf$ almost everywhere: $q$ represents the $L^2$ multiplier extension of $f$, which is assertion 2.