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<h1>Approximation of the Hilbert transform <br />in the Lebesgue spaces</h1>
<p class="authors">
<span class="author">Rashid A. Aliev\(^\ast \) Lale Sh. Alizade\(^\ddag \)</span>
</p>
<p class="date">March 23, 2023; accepted: October 8, 2023; published online: December 22, 2023.</p>
<p class="thanks">\(^\ast \)Baku State University, Baku, Azerbaijan; Institute of Mathematics and Mechanics, Baku, Azerbaijan, e-mail: <span class="tt">aliyevrashid@mail.ru</span>. <br />\(^\ddag \)Baku State University, Baku, Azerbaijan, e-mail: <span class="tt">lale-alizade-98@mail.ru</span>. </p>
</div>



<div class="abstract"> The Hilbert transform plays an important role in the theory and practice of signal processing operations in continuous system theory because of its relevance to such problems as envelope detection and demodulation, as well as its use in relating the real and imaginary components, and the magnitude and phase components of spectra. The Hilbert transform is a multiplier operator and is widely used in the theory of Fourier transforms. The Hilbert transform is the main part of the singular integral equations on the real line. Therefore, approximations of the Hilbert transform are of great interest. Many papers have dealt with the numerical approximation of the singular integrals in the case of bounded intervals. On the other hand, the literature concerning the numerical integration on unbounded intervals is by far poorer than the one on bounded intervals. The case of the Hilbert Transform has been considered very little. This article is devoted to the approximation of the Hilbert transform in Lebesgue spaces by operators which introduced by V.R. Kress and E. Mortensen to approximate the Hilbert transform of analytic functions in a strip. In this paper, we prove that the approximating operators are bounded maps in Lebesgue spaces and strongly converges to the Hilbert transform in these spaces. </div>
<p><b class="bfseries">MSC. </b>44A15, 65R10, 65D30. </p>
<p><b class="bfseries">Keywords. </b>Hilbert transform, singular integral, approximation, Lebesgue space. </p>
<h1 id="a0000000002">1 Introduction</h1>
<p> The Hilbert transform of a function \(u \in L_{p}({\mathbb R})\), \(1\le p{\lt}\infty \) is defined as the Cauchy principle value integral <span class="cite">
	[
	<a href="index.html#18" >18</a>
	]
</span> </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  (Hu) (t)=\tfrac {1}{\pi }\int _{R}\tfrac {u(\tau )}{t-\tau }d\tau , \  \   t \in {\mathbb R}, \end{equation*}
</div>
<p> where the integral is understood in the Cauchy principal value sense. It is well known (see <span class="cite">
	[
	<a href="index.html#14" >14</a>
	, 
	<a href="index.html#18" >18</a>
	, 
	<a href="index.html#32" >32</a>
	]
</span>) that the Hilbert transform of the function \(u \in L_{p}({\mathbb R})\), \(1\le p{\lt}\infty \), exists for almost all values of \(t \in {\mathbb R}\) . In case \(1{\lt}p{\lt}\infty \), the Hilbert transform is a bounded map in the space \(L_{p}({\mathbb R})\) and satisfies the equation: </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  H^{2}=-I. \end{equation*}
</div>
<p>The Hilbert transform plays an important role in the theory and practice of signal processing operations in continuous system theory because of its relevance to such problems as envelope detection and demodulation, as well as its use in relating the real and imaginary components, and the magnitude and phase components of spectra. The Hilbert transform is a multiplier operator and is widely used in the theory of Fourier transforms. The Hilbert transform is the main part of the singular integral equations on the real line (see <span class="cite">
	[
	<a href="index.html#24" >24</a>
	]
</span>). Therefore, approximations of the Hilbert transform are of great interest. </p>
<p>Many papers have dealt with the numerical approximation of the Hilbert Transform in the case of bounded intervals and the reader can refer to <span class="cite">
	[
	<a href="index.html#1" >1</a>
	, 
	<a href="index.html#3" >3</a>
	, 
	<a href="index.html#6" >6</a>
	, 
	<a href="index.html#7" >7</a>
	, 
	<a href="index.html#9" >9</a>
	, 
	<a href="index.html#10" >10</a>
	, 
	<a href="index.html#12" >12</a>
	, 
	<a href="index.html#13" >13</a>
	, 
	<a href="index.html#15" >15</a>
	, 
	<a href="index.html#16" >16</a>
	, 
	<a href="index.html#17" >17</a>
	, 
	<a href="index.html#20" >20</a>
	, 
	<a href="index.html#21" >21</a>
	, 
	<a href="index.html#22" >22</a>
	, 
	<a href="index.html#25" >25</a>
	, 
	<a href="index.html#28" >28</a>
	, 
	<a href="index.html#30" >30</a>
	, 
	<a href="index.html#31" >31</a>
	, 
	<a href="index.html#32" >32</a>
	, 
	<a href="index.html#37" >37</a>
	, 
	<a href="index.html#38" >38</a>
	]
</span> and the references given there. On the other hand, the literature concerning the numerical integration on unbounded intervals is by far poorer than the one on bounded intervals. The case of the Hilbert Transform has been considered very little and the reader can consult <span class="cite">
	[
	<a href="index.html#2" >2</a>
	, 
	<a href="index.html#8" >8</a>
	, 
	<a href="index.html#11" >11</a>
	, 
	<a href="index.html#12" >12</a>
	, 
	<a href="index.html#19" >19</a>
	, 
	<a href="index.html#20" >20</a>
	, 
	<a href="index.html#23" >23</a>
	, 
	<a href="index.html#26" >26</a>
	, 
	<a href="index.html#27" >27</a>
	, 
	<a href="index.html#34" >34</a>
	, 
	<a href="index.html#35" >35</a>
	, 
	<a href="index.html#36" >36</a>
	, 
	<a href="index.html#39" >39</a>
	]
</span>. In particular, in <span class="cite">
	[
	<a href="index.html#19" >19</a>
	]
</span> the authors assume that the function \(u\) is analytic in the strip \(\{ z \in {\mathbb C}: |\Im z|{\lt}d\} \), in which case they show that the series \(\frac{2}{\pi } \sum _{k \in {\mathbb Z}, k \ne even} \frac{u(t+k \delta )}{-k}\) uniformly converges to \((Hu)(t)\) as \(\delta \to 0\). In <span class="cite">
	[
	<a href="index.html#5" >5</a>
	]
</span> the author replaces the above series with the following one \(\frac{1}{\pi } \sum _{k \in {\mathbb Z}} \frac{u(t+(k+1/2) \delta )}{-k-1/2}\) for a suitable choice of the step \(\delta \to 0\). </p>
<p>This article is devoted to the approximation of the Hilbert transform of functions from \(L_{p}({\mathbb R})\) by operators of the form </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  (H_{\delta }u)(t)=\tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {u(t+(k+1/2) \delta )}{-k-1/2}, \  \   \delta {\gt}0 \end{equation*}
</div>
<p> which were introduced in <span class="cite">
	[
	<a href="index.html#19" >19</a>
	]
</span>. </p>
<p>In <a href="index.html#sec.2">section 2</a> we present the properties of the approximating operators \(H_{\delta }\). We show that the operators \(H_{\delta }\) are bounded maps in the space \(L_{p}({\mathbb R})\), \(1{\lt}p{\lt} \infty \) and </p>
<div class="displaymath" id="a0000000006">
  \[  H_{\delta }^{2}=-I  \]
</div>
<p> in \(L_{p}({\mathbb R})\) (??). </p>
<p>In <a href="index.html#sec.3">section 3</a> we give an approximation of the singular integral with Hilbert kernel </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  (S\varphi )(t)=\tfrac {1}{2\pi } \int _{-\pi }^{\pi }\cot \tfrac {t-\tau }{2}\varphi (\tau )d \tau , \  \  t \in T=[-\pi ,\pi ) \end{equation*}
</div>
<p> by a sequence of operators </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  (S_{n}\varphi )(t)=\tfrac {1}{n} \sum _{k=0}^{n-1}\cot \left(- \tfrac {\pi (2k+1)}{2n} \right) \varphi \left(t+\tfrac {\pi (2k+1)}{n} \right), \  \  n \in {\mathbb N}. \end{equation*}
</div>
<p> in \(L_{p}(T)\). We show that the operators \(S_{n}\) are uniformly bounded in \(L_{p}(T)\) and strongly converges to the operator \(S\) in \(L_{p}(T)\), \(1{\lt}p{\lt}\infty \) (??). </p>
<p>In <a href="index.html#sec.4">section 4</a> we give an approximation of the Hilbert transform \(H\) by the operators \(H_{\delta }\). We show that for any \(\delta {\gt}0\) the sequence of operators \(\{ H_{\delta /n} \} _{n \in {\mathbb N}}\) strongly converges to the operator \(H\) in \(L_{p}({\mathbb R})\), \(1{\lt}p{\lt}\infty \) (<a href="index.html#th.9">theorem 9</a>). </p>
<p>Note that in this paper the singular integral with Hilbert kernel and the Hilbert transform is approximated by operators preserving the main properties of these operators (see: <a href="index.html#th.2">theorem 2</a> and <a href="index.html#3.1" class="eqref">6</a>, <a href="index.html#3.2" class="eqref">7</a>). This leads to give an approximation of the singular integral and the Hilbert transform of the functions from \(L_{p}\), \(1{\lt}p{\lt} \infty \), but other approximate methods can only be applied to continuous or piecewise continuous functions. </p>
<h1 id="sec.2">2 Properties of the approximating operators \(H_{\delta }\)</h1>

<p>Let \(l_{p}\), \(1 \le p {\lt} \infty \), the space of all sequences \(b=\{ b_{n} \} _{n \in {\mathbb Z}}\) with finite norm \(\| b\| _{l_{p}}=\left(\sum _{n \in {\mathbb Z}}|b_{n}|^{p} \right)^{1/p}\). The sequence \(h(b)=\{ (h(b))_{n} \} _{n \in {\mathbb Z}}\) is called the discrete Hilbert transform of the sequence \(b=\{ b_{n} \} _{n \in {\mathbb Z}}\), where \((h(b))_{n}=\sum _{m \ne n } \frac{b_{m}}{n-m}\), \(n \in {\mathbb Z}\). </p>
<p>M. Riesz (see <span class="cite">
	[
	<a href="index.html#29" >29</a>
	]
</span>) proved that if \(b \in l_{p}\), \(1 {\lt} p {\lt} \infty \), then \(h(b) \in l_{p}\) and the inequality </p>
<div class="equation" id="2.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.1} \| h(b)\| _{l_{p}} \le C_{p} \| b\| _{l_{p}} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> holds, where \(C_{p}\) is constant depending only on \(p\). </p>
<p>We will use a modified version of the discrete Hilbert transform: \((\tilde{h}(b))_{n}=\sum _{m \in {\mathbb Z} } \frac{b_{m}}{n-m-1/2}\), \(n \in {\mathbb Z}\). K. Andersen <span class="cite">
	[
	<a href="index.html#4" >4</a>
	]
</span> proved that the inequality <a href="index.html#2.1" class="eqref">1</a> is also valid for the transform \(\tilde{h}\), that is, there exist \(\tilde{C} _{p}{\gt}0\) such that the inequality </p>
<div class="equation" id="2.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.2} \| \tilde{h}(b)\| _{l_{p}} \le \tilde{C}_{p} \| b\| _{l_{p}} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> holds for any \(b \in l_{p}\), \(1 {\lt} p {\lt} \infty \). </p>
<p>In the following theorems we prove that the operators \(H_{\delta }\) are bounded maps in the space \(L_{p}({\mathbb R})\) and \(H_{\delta }^{2}=-I \) in \(L_{p}({\mathbb R})\), \(1{\lt}p{\lt} \infty \). </p>
<p><div class="theorem_thmwrapper " id="th.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For any \(\delta {\gt}0\) the operator \(H_{\delta }\) is bounded in the space \(L_{p}({\mathbb R})\), \(1{\lt}p{\lt}\infty \), and the inequality </p>
<div class="equation" id="2.3">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.3} \| H_{\delta }\| _{L_{p}({\mathbb R}) \to L_{p}({\mathbb R})} \le \| \tilde{h}\| _{l_{p} \to l_{p}} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> holds. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let \(u \in L_{p}({\mathbb R})\), \(1{\lt}p{\lt} \infty \). For any \(t \in {\mathbb R}\) </p>
<div class="displaymath" id="a0000000010">
  \begin{align*}  \tilde{h}(\{ u(t+\delta /2 +n \delta ) \} _{n \in {\mathbb Z}})& =\left\{  \tfrac {1}{\pi } \sum _{m \in {\mathbb Z}} \tfrac {u(t +\delta /2 +m\delta )}{n-m-1/2}\right\} _{n \in {\mathbb Z}}= \\ & =\left\{  \tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {u(t+\delta /2 +k \delta +n \delta )}{-k-1/2}\right\} _{n \in {\mathbb Z}}= \{ (H_{\delta }u)(t+n\delta ) \} _{n \in {\mathbb Z}}. \end{align*}
</div>
<p> Then, by inequality <a href="index.html#2.2" class="eqref">2</a>, for almost all \(t \in {\mathbb R}\) </p>
<div class="displaymath" id="a0000000011">
  \begin{align*}  \| \{ (H_{\delta }u)(t+n\delta ) \} _{n \in {\mathbb Z}}\| _{l_{p}}& =\| \tilde{h}(\{ u(t+\delta /2 +n \delta ) \} _{n \in {\mathbb Z}}) \| _{l_{p}} \\ & \le \| \tilde{h}\| _{l_{p} \to l_{p}} \cdot \Big\| \{ u(t+\delta /2 +n \delta ) \} _{n \in {\mathbb Z}} \Big\| _{l_{p}}. \end{align*}
</div>
<p> It follows that </p>
<div class="displaymath" id="a0000000012">
  \begin{align*}  \| H_{\delta }u \| _{L_{p}({\mathbb R})}^{p}& =\int _{R}|(H_{\delta }u)(t)|^{p}dt = \sum _{n \in {\mathbb Z}} \int _{(n-1/2)\delta }^{(n+1/2)\delta }|(H_{\delta }u)(t)|^{p}dt \\ & =\sum _{n \in {\mathbb Z}} \int _{-\delta /2}^{\delta /2}|(H_{\delta }u)(t+n\delta )|^{p}dt = \int _{-\delta /2}^{\delta /2} \sum _{n \in {\mathbb Z}} |(H_{\delta }u)(t+n\delta )|^{p}dt \\ & = \int _{-\delta /2}^{\delta /2} \| \{ (H_{\delta }u)(t+n\delta )\} _{n \in {\mathbb Z}} \| _{l_{p}}^{p}dt \\ &  \le \| \tilde{h}\| _{l_{p} \to l_{p}}^{p} \cdot \int _{-\delta /2}^{\delta /2} \| \{ u(t+\delta /2+n\delta )\} _{n \in {\mathbb Z}} \| _{l_{p}}^{p}dt \\ & = \| \tilde{h}\| _{l_{p} \to l_{p}}^{p} \cdot \int _{-\delta /2}^{\delta /2} \sum _{n \in {\mathbb Z}}|u(t+\delta /2+n\delta )|^{p}dt \\ & =\| \tilde{h}\| _{l_{p} \to l_{p}}^{p} \cdot \sum _{n \in {\mathbb Z}} \int _{-\delta /2}^{\delta /2} |u(t+\delta /2+n\delta )|^{p}dt \\ & = \| \tilde{h}\| _{l_{p} \to l_{p}}^{p} \cdot \sum _{n \in {\mathbb Z}} \int _{n\delta }^{(n+1)\delta } |u(t)|^{p}dt = \| \tilde{h}\| _{l_{p} \to l_{p}}^{p} \cdot \| u\| _{L_{p}({\mathbb R})}^{p}. \end{align*}
</div>

<p><div class="theorem_thmwrapper " id="th.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For any \(\delta {\gt}0\) and \(u \in L_{p}({\mathbb R})\), \(1{\lt}p{\lt} \infty \) the following inequality holds: </p>
<div class="equation" id="2.4">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.4} H_{\delta } (H_{\delta }u)(t)=-u(t). \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>For any \(u \in L_{p}({\mathbb R})\) we have </p>
<div class="displaymath" id="a0000000014">
  \begin{align}  H_{\delta } (H_{\delta }u)(t)& =-\tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {(H_{\delta }u)(t+(k+1/2) \delta )}{k+1/2} =\tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {1}{k+1/2} \cdot \tfrac {1}{\pi } \sum _{m \in {\mathbb Z}} \tfrac {u(t+(k+m+1) \delta )}{m+1/2} \nonumber \\ & =\tfrac {1}{\pi ^{2}} \sum _{k \in {\mathbb Z}} \sum _{m \in {\mathbb Z}} \tfrac {u(t+(k+m+1) \delta )}{(k+1/2)(m+1/2)}=\tfrac {1}{\pi ^{2}} \sum _{k \in {\mathbb Z}} \sum _{n \in {\mathbb Z}} \tfrac {u(t+n \delta )}{(k+1/2)(n-k-1/2)} \nonumber \\ \label{2.5} & =\tfrac {1}{\pi ^{2}} \sum _{n \in {\mathbb Z}} \left( \sum _{k \in {\mathbb Z}} \tfrac {1}{(k+1/2)(n-k-1/2)} \right) u(t+n \delta ). \end{align}
</div>
<p> Since for \(n=0\) </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  \sum _{k \in {\mathbb Z}} \tfrac {1}{(k+1/2)(n-k-1/2)}=-4\sum _{k \in {\mathbb Z}} \tfrac {1}{(2k+1)^{2}}=-\pi ^{2}, \end{equation*}
</div>
<p> and for \(n \ne 0\) </p>
<div class="displaymath" id="a0000000016">
  \begin{align*}  \sum _{k \in {\mathbb Z}} \tfrac {1}{(k+1/2)(n-k-1/2)}& =\sum _{k \in {\mathbb Z}} \tfrac {1}{n} \left[\tfrac {1}{k+1/2}+\tfrac {1}{n-k-1/2} \right] \\ & = \tfrac {1}{n} \lim _{N \to \infty }\sum _{|k| \le N} \left[\tfrac {1}{k+1/2}+\tfrac {1}{n-k-1/2} \right]=0, \end{align*}
</div>
<p> then equality <a href="index.html#2.4" class="eqref">4</a> follows from <a href="index.html#2.5" class="eqref">5</a>. </p>
<h1 id="sec.3">3 Approximation of the singular integral with Hilbert kernel</h1>

<p>Denote by \(L_{p}(T)\), \(1 \le p{\lt}\infty \), the space of all measurable, \(2\pi \)-periodic functions with finite norm \(\|  \varphi \| _{L_{p}(T)}=\left(\int _{T}|\varphi (t)|^{p}dt \right)^{1/p}\), where \(T=[-\pi ,\pi )\), and by \(L_{p}([a,b])\) the space of all measurable functions on the interval \([a,b] \subset {\mathbb R}\) with finite norm \(\|  \varphi \| _{L_{p}([a,b])}=\left(\int _{a}^{b}|\varphi (t)|^{p}dt \right)^{1/p}\). </p>
<p>It is well known that (see <span class="cite">
	[
	<a href="index.html#40" >40</a>
	]
</span>) the singular integral with Hilbert kernel </p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  (S\varphi )(t)=\tfrac {1}{2\pi } \int _{-\pi }^{\pi }\cot \tfrac {t-\tau }{2}\varphi (\tau )d \tau , \  \  t \in T, \end{equation*}
</div>
<p> is a bounded map in the space \(L_{p}(T)\), \(1 {\lt}p{\lt}\infty \) and for any \(\varphi \in L_{p}(T)\) </p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  (S^{2} \varphi )(t)=-\varphi (t)+ \tfrac {1}{2\pi } \int _{-\pi }^{\pi }\varphi (\tau )d \tau , \  \  t \in T. \end{equation*}
</div>
<p>Consider in \(L_{p}(T)\), \(1{\lt} p{\lt}\infty \) the sequence of operators </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  (S_{n}\varphi )(t)=\tfrac {1}{n} \sum _{k=0}^{n-1}\cot \left(- \tfrac {\pi (2k+1)}{2n} \right) \varphi \left(t+\tfrac {\pi (2k+1)}{n} \right), \  \  n \in {\mathbb N}. \end{equation*}
</div>
<p>It is easy to verify that if </p>
<div class="displaymath" id="a0000000020">
  \[  \varphi (t)=\tfrac {a_{0}}{2}+\sum _{m=1}^{\infty }(a_{m}\cos mt+b_{m}\sin mt),  \]
</div>
<p> then </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  (S_{n}\varphi ) (t)=\sum _{m=1}^{\infty }\lambda _{m}^{(n)}(a_{m}\cos mt+b_{m}\sin mt), \end{equation*}
</div>
<p> where \(\lambda _{m}^{(n)}=1\) for \(m=\overline{1,n-1}\), \(\lambda _{n}^{(n)}=\lambda _{2n}^{(n)}=0\), \(\lambda _{m}^{(n)}=-1\) for \(m=\overline{n+1,2n-1}\) and \(\lambda _{m+2n}^{(n)}=\lambda _{m}^{(n)}\) for \(m \in {\mathbb Z}\). It follows from here that for any trigonometric polynomial \(P(t )\) of order at most \(n-1\) </p>
<div class="equation" id="3.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3.1} (S_{n}P)(t)=(SP)(t), \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> and for any \(\varphi \in L_{p}(T)\) </p>
<div class="equation" id="3.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3.2} (S^{2}_{n}\varphi )(t)=-\varphi (t)+\tfrac {1}{n} \sum _{k=0}^{n-1} \varphi (t+\tfrac {2 \pi k}{n}), \  \  n \in {\mathbb N}. \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>In the following theorems we prove that the sequence of operators \(S_{n}\) are uniformly bounded in \(L_{p}(T)\) and strongly converges to the operator \(S\) in \(L_{p}(T)\), \(1{\lt}p{\lt}\infty \). </p>
<p><div class="theorem_thmwrapper " id="th.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Operators \(S_{n}\) are uniformly bounded in \(L_{p}(T)\), \(1{\lt}p{\lt}\infty \), and for any \(n\in {\mathbb N}\) the inequality </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  \| S_{n} \| _{L_{p}(T) \to L_{p}(T)} \le 4+2 \| \tilde{h} \| _{l_{p} \to l_{}p} \end{equation*}
</div>
<p> holds. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let \(\varphi \in L_{p}(T)\). Define the function \(u(t)=\varphi (t)\) for \(t \in [-2\pi , 2\pi ]\) and \(u(t)=0\) for \(t \in {\mathbb R} \setminus [-2\pi , 2\pi ]\). Then \(u \in L_{p}({\mathbb R})\), and therefore, it follows from <a href="index.html#th.1">theorem 1</a> that for any \(\delta {\gt}0\) </p>
<div class="equation" id="3.3">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3.3} \| H_{\delta }u\| _{L_{p}({\mathbb R})} \le \| \tilde{h}\| _{l_{p} \to l_{p}} \cdot \| u\| _{L_{p}({\mathbb R})}= 2\| \tilde{h}\| _{l_{p} \to l_{p}} \cdot \| \varphi \| _{L_{p}(T)}. \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> Since for any \(t \in [-\pi , \pi ]\) </p>
<div class="displaymath" id="a0000000024">
  \begin{align*}  (S_{n}\varphi )(t)& =\tfrac {1}{n} \sum _{k=0}^{n-1}\cot \left(- \tfrac {\pi (2k+1)}{2n} \right) \varphi \left(t+\tfrac {\pi (2k+1)}{n} \right) \\ & =\tfrac {1}{n} \sum _{k \in \Delta _{n}}\cot \left(- \tfrac {\pi (2k+1)}{2n} \right) \varphi \left(t+\tfrac {\pi (2k+1)}{n} \right), \end{align*}
</div>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  (H_{2\pi /n}u)(t)=\tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {u(t+\frac{\pi (2k+1)}{n}) }{-k-1/2} =\tfrac {1}{\pi } \sum _{k \in \Delta _{n}} \tfrac {u(t+\frac{\pi (2k+1)}{n}) }{-k-1/2}+\tfrac {1}{\pi } \sum _{k \in \tilde{\Delta }_{n}} \tfrac {u(t+\frac{\pi (2k+1)}{n}) }{-k-1/2}, \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  \Delta _{n}& =\left\{ k \in {\mathbb Z}: \left[\tfrac {-n+1}{2} \right] \le k \le \left[\tfrac {n-1}{2} \right] \right\} , \\ \tilde{\Delta }_{n}& =\left\{ k \in {\mathbb Z}: |k| \le 2n, \  k{\gt} \left[\tfrac {n-1}{2} \right] \   \mbox{ or } \  k{\lt} \left[\tfrac {-n+1}{2} \right] \right\} , \end{align*}
</div>
<p> then for any \(t \in [-\pi , \pi ]\) we have </p>
<div class="displaymath" id="a0000000027">
  \begin{align} & (H_{2\pi /n}u)(t)-(S_{n} \varphi )(t) = \nonumber \\ & =\tfrac {1}{n} \sum _{k \in \Delta _{n}}\left[\cot \tfrac {\pi (2k+1)}{2n} - \tfrac {2n}{\pi (2k+1)} \right] \varphi \left(t+\tfrac {\pi (2k+1)}{n} \right)+\tfrac {1}{\pi } \sum _{k \in \tilde{\Delta }_{n}} \tfrac {u(t+\frac{\pi (2k+1)}{n}) }{-k-1/2}.\label{3.4} \end{align}
</div>
<p> It follows from <a href="index.html#3.4" class="eqref">9</a> and from inequality \(| \cot x -1/x | \le 2/ \pi \) for \(0{\lt}|x| \le \pi /2\) that </p>
<div class="displaymath" id="a0000000028">
  \begin{align} & \|  H_{2\pi /n}u-S_{n} \varphi \| _{L_{p}([-\pi ,\pi ])} \le \nonumber \\ & \le \tfrac {1}{n} \sum _{k \in \Delta _{n}}\tfrac {2}{\pi }\|  \varphi \| _{L_{p}(T)}+ \tfrac {1}{\pi } \sum _{k \in \tilde{\Delta }_{n}}\tfrac {2}{n }\|  \varphi \| _{L_{p}(T)} \le 4 \|  \varphi \| _{L_{p}(T)}.\label{3.5} \end{align}
</div>
<p> From <a href="index.html#3.3" class="eqref">8</a> and <a href="index.html#3.5" class="eqref">10</a> we have </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  \|  S_{n} \varphi \| _{L_{p}(T)} & \le \|  H_{2\pi /n}u-S_{n} \varphi \| _{L_{p}([-\pi ,\pi ])} + \|  H_{2\pi /n}u \| _{L_{p}({\mathbb R})} \\ & \le \left(4+2 \| \tilde{h} \| _{l_{p} \to l_{}p} \right) \cdot \|  \varphi \| _{L_{p}(T)}. \end{align*}
</div>

<p><div class="theorem_thmwrapper " id="th.4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p> The sequence of operators \(S_{n}\) strongly converges to the operator \(S\) in \(L_{p}(T)\), \(1{\lt}p{\lt}\infty \), and for any \(\varphi \in L_{p}(T)\) the following estimate holds: </p>
<div class="equation" id="3.6">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3.6} \|  S \varphi - S_{n} \varphi \| _{L_{p}(T)} \le \left(4+\|  S \| _{L_{p}(T) \to L_{p}(T)}+2 \| \tilde{h} \| _{l_{p} \to l_{}p} \right) \cdot E_{n-1}^{p}(\varphi ), \   n \in {\mathbb N}, \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> where \(E_{n-1}^{p}(\varphi )\) – is the best approximation of the function \(\varphi \) in the metric \(L_{p}(T)\) by trigonometric polynomials of order at most \(n-1\), \(n \in {\mathbb N}\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000030">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Suppose that </p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  q_{n-1} (t)=\tfrac {a_{0}}{2}+\sum _{m=1}^{n-1}(a_{m}\cos mt+b_{m}\sin mt) \end{equation*}
</div>
<p> is the best approximation of the function \(\varphi \) in the metric \(L_{p}(T)\) by trigonometric polynomials of order at most \(n-1\), \(n \in {\mathbb N}\). Then it follows from the equality </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  (S_{n}q_{n-1})(t)=(Sq_{n-1})(t) \end{equation*}
</div>
<p> that </p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  (S \varphi - S_{n} \varphi )(t)=S (\varphi - q_{n-1})(t)- S_{n} (\varphi - q_{n-1})(t). \end{equation*}
</div>
<p> Then </p>
<div class="displaymath" id="a0000000034">
  \begin{align*}  \|  S \varphi - S_{n} \varphi \| _{L_{p}(T)} & \le \left(\|  S \| _{L_{p}(T) \to L_{p}(T)}+\|  S_{n} \| _{L_{p}(T) \to L_{p}(T)} \right) \cdot \| \varphi - q_{n-1}\| _{L_{p}(T)} \\ & \le \left(4+\|  S \| _{L_{p}(T) \to L_{p}(T)}+2 \| \tilde{h} \| _{l_{p} \to l_{}p} \right) \cdot E_{n-1}^{p}(\varphi ). \end{align*}
</div>

<h1 id="sec.4">4 Approximation of the Hilbert transform</h1>

<p>Consider the regular integral operator </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  (\mathrm{K}\varphi )(t)=\tfrac {1}{2 \pi }\int _{-\pi }^{\pi }K(t,\tau )\varphi (\tau )d\tau , \  \  t \in T, \end{equation*}
</div>
<p> where \(K(t,\tau )\) is a continuous function on \([-\pi , \pi ]^{2}\) , and the sequence of operators </p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  (\mathrm{K}_{n}\varphi )(t)=\tfrac {1}{n}\sum _{k=0}^{n-1}K\left(t,t+\tfrac {\pi (2k+1)}{n}\right)\varphi \left(t+\tfrac {\pi (2k+1)}{n}\right), \  \  t \in T, \  \   n \in {\mathbb N}, \end{equation*}
</div>
<p> where \(K(t,\tau )=K(t,\tau -2\pi )\) for \((t,\tau ) \in [-\pi ,\pi ]\times (\pi ,3\pi )\). </p>
<p><div class="lemma_thmwrapper " id="lm.5">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The sequence of operators \(\{  \mathrm{K}_{n} \} \) strongly converges to the operator \(\mathrm{K}\) in \(L_{p}(T)\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000037">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>First assume that \(K(t,\tau )\) is a \(2\pi \)-periodic function by \(\tau \) . Denote </p>
<div class="displaymath" id="a0000000038">
  \begin{equation*}  \| K\| _{\infty }=\max _{t,\tau \in [-\pi ,\pi ] } |K(t,\tau )|, \  \  E_{n}(K)=\inf \| K-\Phi _{n}\| _{\infty }, \end{equation*}
</div>
<p> where \(\Phi _{n}(t,\tau )=\tfrac {\alpha _{0}(t)}{2}+\sum _{m=1}^{n}(\alpha _{m}(t)\cos m\tau +\beta _{m}(t)\sin m\tau )\), and infimum is taken over all trigonometric polynomials \(\alpha _{m}(t)\), \(m=\overline{0,n}\), \(\beta _{m}(t)\), \(m=\overline{1,n}\) of order at most \(n\). </p>
<p>Denote \(n_{0}=\left[\tfrac {n-1}{2} \right]\). Suppose that </p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  q_{n_{0}} (t)=\tfrac {a_{0}}{2}+\sum _{m=1}^{n_{0}}(a_{m}\cos mt+b_{m}\sin mt) \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  \Phi _{n_{0}}^{(0)}(t,\tau )=\tfrac {\alpha _{0}^{(0)}(t)}{2}+\sum _{m=1}^{n_{0}}(\alpha _{m}^{(0)}(t)\cos m\tau +\beta _{m}^{(0)}(t)\sin m\tau ) \end{equation*}
</div>
<p> are the best approximations of the functions \(\varphi \) and \(K\) by trigonometric polynomials of order at most \(n_{0}\). </p>
<p>For any trigonometric polynomial \(r_{n-1}(t)\) of order at most \(n-1\), the equality </p>
<div class="displaymath" id="a0000000041">
  \begin{equation*}  \tfrac {1}{2 \pi }\int _{-\pi }^{\pi }r_{n-1}(\tau )d\tau =\tfrac {1}{n}\sum _{k=0}^{n-1}r_{n-1}\left(t+\tfrac {\pi (2k+1)}{n}\right) \end{equation*}
</div>
<p> holds. Therefore </p>
<div class="displaymath" id="a0000000042">
  \begin{align*} & (\mathrm{K}\varphi )(t)- (\mathrm{K}_{n}\varphi )(t)= \\ & = (\mathrm{K}-\mathrm{K}_{n})(\varphi -q_{n_{0}})(t)+ \tfrac {1}{2 \pi }\int _{-\pi }^{\pi } \left[K(t,\tau )-\Phi _{n_{0}}^{(0)}(t,\tau ) \right]q_{n_{0}}(\tau ) d\tau \\ & \quad +\tfrac {1}{n}\sum _{k=0}^{n-1} \left[ K(t,t+\tau _{k}^{(n)} )-\Phi _{n_{0}}^{(0)} (t,t+\tau _{k}^{(n)} ) \right] q_{n_{o}} (t+\tau _{k}^{(n)} ), \end{align*}
</div>
<p> where \(\tau _{k}^{(n)}=\frac{\pi (2k+1)}{n} \), \(k \in {\mathbb Z}\). It follows from here and from inequalities </p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  \| \mathrm{K}\| _{L_{p}(T) \to L_{p}(T)} \le \| K\| _{\infty }, \  \   \| \mathrm{K}_{n}\| _{L_{p}(T) \to L_{p}(T)} \le \| K\| _{\infty } \end{equation*}
</div>
<p> that </p>
<div class="displaymath" id="a0000000044">
  \begin{equation*}  \| \mathrm{K}\varphi - \mathrm{K}_{n}\varphi \| _{L_{p}(T)} \le 2\| K\| _{\infty } E_{n_{0}}^{p}(\varphi )+2E_{n_{0}}(K) \left[\|  \varphi \| _{L_{p}(T)}+ E_{n_{0}}^{p}(\varphi ) \right]. \end{equation*}
</div>
<p> This completes the proof of the lemma in this case. Now consider the general case. </p>
<p>Let \(\varphi \in L_{p}(T)\) and \(\varepsilon {\gt}0\). Denote </p>
<div class="displaymath" id="a0000000045">
  \begin{align*}  K^{\ast }(t,\tau )& =K(t,\tau ) \   \mbox{ for } \   (t,\tau ) \in [-\pi , \pi ] \times [-\pi , \pi -\delta _{\varepsilon }], \\ K^{\ast }(t,\tau )& =K(t,\pi -\delta _{\varepsilon })+\tfrac {\tau -\pi +\delta _{\varepsilon }}{\delta _{\varepsilon }}[K(t,-\pi )-K(t,\pi -\delta _{\varepsilon })] \\ & \mbox{ for } \  \   (t,\tau ) \in [-\pi , \pi ] \times [\pi -\delta _{\varepsilon }, \pi ], \\ K^{\ast }(t,\tau +2\pi )& =K^{\ast }(t,\tau ) \   \mbox{ for any} \   (t,\tau ) \in [-\pi , \pi ] \times {\mathbb R}, \end{align*}
</div>
<p> where \(\delta _{\varepsilon }=\min \left\{  2\pi \cdot \left(\frac{\varepsilon }{8\| K\| _{\infty } \|  \varphi \| _{L_{p}(T)}} \right)^{\frac{p}{p-1}}, \frac{\pi \varepsilon }{8\| K\| _{\infty } \|  \varphi \| _{L_{p}(T)}}, 1 \right\} \). </p>
<p>Since the function \(K^{\ast }(t,\tau )\) is continuous and \(2\pi \)-periodic by \(\tau \), the sequence of operators </p>
<div class="displaymath" id="a0000000046">
  \begin{equation*}  (\mathrm{K}_{n}^{\ast }\varphi )(t)=\tfrac {1}{n}\sum _{k=0}^{n-1}K^{\ast }(t,t+\tau _{k}^{(n)})\varphi (t+\tau _{k}^{(n)} ), \  \  t \in T, \  \   n \in {\mathbb N} \end{equation*}
</div>
<p> strongly converges to the operator </p>
<div class="displaymath" id="a0000000047">
  \begin{equation*}  (\mathrm{K}^{\ast }\varphi )(t)=\tfrac {1}{2 \pi }\int _{-\pi }^{\pi }K^{\ast }(t,\tau )\varphi (\tau )d\tau \end{equation*}
</div>
<p> in \(L_{p}(T)\). Therefore, the inequality </p>
<div class="displaymath" id="a0000000048">
  \begin{equation*}  \| \mathrm{K}_{n}^{\ast }\varphi -\mathrm{K}^{\ast }\varphi \| _{L_{p}(T)} {\lt} \varepsilon /2 \end{equation*}
</div>
<p> is satisfied for large values of \(n\). Moreover, since </p>
<div class="displaymath" id="a0000000049">
  \begin{align*}  \| \mathrm{K}\varphi -\mathrm{K}^{\ast }\varphi \| _{L_{p}(T)}&  \le \tfrac {1}{2 \pi } \left[\int _{-\pi }^{\pi } \left( \int _{\pi -\delta _{\varepsilon }}^{\pi } |K(t,\tau )-K^{\ast }(t,\tau )| |\varphi (\tau )| d\tau \right)^{p} dt \right]^{1/p} \\ & \le \tfrac {\| K\| _{\infty }}{\pi } \left[\int _{-\pi }^{\pi } \left( \int _{\pi -\delta _{\varepsilon }}^{\pi } |\varphi (\tau )| d\tau \right)^{p} dt \right]^{1/p} \\ & \le \tfrac {2\| K\| _{\infty }}{(2\pi )^{1-1/p}}(\delta _{\varepsilon })^{1-1/p} \|  \varphi \| _{L_{p}([\pi -\delta _{\varepsilon }, \pi ])} \le \tfrac {\varepsilon }{4}, \end{align*}
</div>
<p> and for \(n \ge \tfrac {16\| K\| _{\infty }\|  \varphi \| _{L_{p}(T)}}{ \varepsilon } \) </p>
<div class="displaymath" id="a0000000050">
  \begin{equation*}  \| \mathrm{K}_{n}\varphi -\mathrm{K}_{n}^{\ast }\varphi \| _{L_{p}(T)} \le \tfrac {1}{n} \cdot \left(\tfrac {n}{2\pi }\cdot \delta _{\varepsilon } +1 \right) \cdot 2 \| K\| _{\infty } \|  \varphi \| _{L_{p}(T)} \le \tfrac {\varepsilon }{4}, \end{equation*}
</div>
<p> then for sufficiently large values \(n\) we have </p>
<div class="displaymath" id="a0000000051">
  \begin{align*} & \| \mathrm{K}_{n}\varphi -\mathrm{K}\varphi \| _{L_{p}(T)} \le \\ & \le \| \mathrm{K}_{n}\varphi -\mathrm{K}_{n}^{\ast }\varphi \| _{L_{p}(T)}+\| \mathrm{K}_{n}^{\ast }\varphi -\mathrm{K}^{\ast }\varphi \| _{L_{p}(T)}+\| \mathrm{K}^{\ast }\varphi -\mathrm{K}\varphi \| _{L_{p}(T)} {\lt} \varepsilon . \end{align*}
</div>

<p><div class="corollary_thmwrapper " id="cor.6">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">6</span>
  </div>
  <div class="corollary_thmcontent">
  <p> The sequence of operators </p>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  (\widetilde{\mathrm{K}}_{n}\varphi )(t)=\tfrac {1}{n}\sum _{\{ k \in {\mathbb Z}:t+\tau _{k}^{(n)} \in [-\pi , \pi ] \} }K(t,t+\tau _{k}^{(n)})\varphi (t+\tau _{k}^{(n)}), \   t \in [-\pi ,\pi ], \   n \in {\mathbb N} \end{equation*}
</div>
<p> strongly converges to the operator \(\mathrm{K}\) in \(L_{p}([-\pi ,\pi ])\). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="cor.7">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">7</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If the function \(K(t, \tau )\) is continuous on \([\pi m, \pi m+ 2\pi q] \times [-\pi , \pi ]\), then for any \(\varphi \in L_{p}(T)\) the sequence of functions </p>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  (\widetilde{\mathrm{K}}_{n}\varphi )(t)=\tfrac {1}{n}\sum _{\big\{ k \in {\mathbb Z}:t+\tau _{k}^{(n)} \in [-\pi , \pi ] \big\} }K(t,t+\tau _{k}^{(n)})\varphi (t+\tau _{k}^{(n)}), \   t \in [\pi m, \pi m+2\pi q], \end{equation*}
</div>
<p> converges to the function </p>
<div class="displaymath" id="a0000000054">
  \begin{equation*}  (\mathrm{K}\varphi )(t)=\tfrac {1}{2 \pi }\int _{-\pi }^{\pi }K(t,\tau )\varphi (\tau )d\tau , \  \  t \in [\pi m, \pi m+2\pi q] \end{equation*}
</div>
<p> in \(L_{p}( [\pi m, \pi m+2\pi q])\), where \(m \in {\mathbb Z}\), \(q \in {\mathbb N}\). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="cor.8">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">8</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If the function \(K_{0}(t)\) is continuous on \( [-\pi , \pi ] \), then the sequence of operators </p>
<div class="displaymath" id="a0000000055">
  \begin{equation*}  (\mathrm{K}_{n}^{0}\varphi )(t)=\tfrac {1}{2n}\sum _{k=-n}^{n-1}K_{0}\left(\tfrac {\pi (2k+1)}{2n}\right)\varphi \left(t+\tfrac {\pi (2k+1)}{2n}\right), \   t \in T, \   n \in {\mathbb N} \end{equation*}
</div>
<p> strongly converges to the operator </p>
<div class="displaymath" id="a0000000056">
  \begin{equation*}  (\mathrm{K}^{0}\varphi )(t)=\tfrac {1}{2 \pi }\int _{-\pi }^{\pi }K_{0}(\tau )\varphi (t+\tau )d\tau , \  \  t \in T \end{equation*}
</div>
<p> in \(L_{p}(T)\). </p>

  </div>
</div> </p>
<p>In the following theorem we prove that for any \(\delta {\gt}0\) the sequence of operators \(\{ H_{\delta /n} \} _{n \in {\mathbb N}}\) strongly converges to the operator \(H\) in \(L_{p}({\mathbb R})\), \(1{\lt}p{\lt}\infty \). </p>
<p><div class="theorem_thmwrapper " id="th.9">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">9</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For any \(\delta {\gt}0\) the sequence of the operators \(\{ H_{\delta /n}\} _{n \in {\mathbb N}}\) strongly converges to the operator \(H\) in \(L_{p}({\mathbb R})\), that is for any \(u \in L_{p}({\mathbb R})\) the following inequality holds: </p>
<div class="displaymath" id="a0000000057">
  \begin{equation*}  \lim _{n \to \infty } \| H_{\delta /n}u-Hu\| _{L_{p}({\mathbb R})}=0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000058">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>For simplicity of presentation we have divided the proof into three steps. </p>
<p><b class="bfseries">Step 1.</b> Let us first prove that the operator </p>
<div class="displaymath" id="a0000000059">
  \begin{equation*}  (H^{\ast } \varphi )(t)= \tfrac {1}{ \pi }\int _{t-\pi }^{t+\pi }\tfrac {\varphi (\tau )}{t-\tau }d\tau \end{equation*}
</div>
<p> is a bounded operator in \(L_{p}(T)\). Indeed, for any \(\varphi \in L_{p}(T)\) we have </p>
<div class="displaymath" id="a0000000060">
  \begin{align}  (H^{\ast } \varphi )(t)& = \tfrac {1}{ \pi }\int _{t-\pi }^{t+\pi }\tfrac {\varphi (\tau )}{t-\tau }d\tau = \tfrac {1}{ \pi }\int _{t-\pi }^{t+\pi }\left[ \tfrac {1}{t-\tau }-\tfrac {1}{2}\cot \tfrac {t-\tau }{2} \right] \varphi (\tau )d\tau +(S \varphi )(t) \nonumber \\ \label{4.1} & = \tfrac {1}{ 2\pi }\int _{-\pi }^{\pi }\left[ \cot \tfrac {\tau }{2}-\tfrac {2}{\tau } \right] \varphi (t+\tau )d\tau +(S \varphi )(t). \end{align}
</div>
<p> Since the function </p>
<div class="displaymath" id="a0000000061">
  \begin{equation*}  K_{0}(\tau )= \cot \tfrac {\tau }{2}-\tfrac {2}{\tau } \  \  \mbox{for} \  \   \tau \ne 0, \  \   K_{0}=0 \end{equation*}
</div>
<p> is continuous on \([-\pi , \pi ]\), then it follows from <a href="index.html#4.1" class="eqref">12</a> and from <a href="index.html#cor.8">corollary 8</a> that the operator \(H^{\ast }\) is bounded in \(L_{p}(T)\). </p>
<p>Consider the sequence of operators </p>
<div class="displaymath" id="a0000000062">
  \begin{equation*}  (H_{n}^{\ast }\varphi )(t)=\tfrac {1}{\pi }\sum _{k=-n}^{n-1} \tfrac {1}{-k-1/2} \varphi \left(t+\tfrac {\pi (2k+1)}{2n}\right), \   t \in T, \  n \in {\mathbb N}. \end{equation*}
</div>
<p> Since for any \(\varphi \in L_{p}(T)\) </p>
<div class="displaymath" id="a0000000063">
  \begin{align*}  (H_{n}^{\ast }\varphi )(t)& =\tfrac {1}{2n}\sum _{k=-n}^{n-1} \left[ \cot \left( \tfrac {\pi (2k+1)}{4n} \right)- \tfrac {4n}{\pi (2k+1)} \right] \varphi \left(t+\tfrac {\pi (2k+1)}{2n}\right) + (S_{2n}\varphi )(t)= \\ & =\tfrac {1}{2n}\sum _{k=-n}^{n-1} K_{0} \left( \tfrac {\pi (2k+1)}{2n} \right) \varphi \left(t+\tfrac {\pi (2k+1)}{2n}\right) + (S_{2n}\varphi )(t), \end{align*}
</div>
<p> then it follows from <a href="index.html#th.4">theorem 4</a> and from <a href="index.html#cor.8">corollary 8</a> that the sequence of operators \(H_{n}^{\ast }\) strongly converges to the operator \(H^{\ast }\) in \(L_{p}(T)\). </p>
<p><b class="bfseries">Step 2.</b> Let us first prove that the sequence of operators </p>
<div class="displaymath" id="a0000000064">
  \begin{equation*}  (H_{\pi /(4n)}u)(t)=\tfrac {1}{\pi }\sum _{k \in {\mathbb Z}} \tfrac {1}{-k-1/2} u \left(t+\tfrac {\pi (k+1/2)}{4n}\right), \   t \in {\mathbb R}, \  n \in {\mathbb N} \end{equation*}
</div>
<p> strongly converges to the operator \(H\) in \(L_{p}({\mathbb R})\). At first assume that \(\operatorname {supp} \  u \subset [-\pi /4, \pi /4]\). Denote by \(\varphi \) \(2\pi \)-periodic function, coinciding with the function \(u\) on \( [-\pi /4, \pi /4]\) and equal to zero in \(T \setminus [-\pi /4, \pi /4]\). Since for any \(t \in [-\pi /2, \pi /2]\) </p>
<div class="equation" id="4.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.2} (Hu) (t)=\tfrac {1}{\pi }\int _{-\pi /4}^{\pi /4}\tfrac {u(\tau )}{t-\tau }d\tau =(H^{\ast }\varphi )(t), \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<div class="displaymath" id="a0000000065">
  \begin{align}  (H_{\pi /n}u)(t)& =\tfrac {1}{\pi }\sum _{k=-n}^{n-1} \tfrac {1}{-k-1/2} u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \nonumber \\ \label{4.3} & =\tfrac {1}{\pi }\sum _{k=-n}^{n-1} \tfrac {1}{-k-1/2} \varphi \left(t+\tfrac {\pi (k+1/2)}{n}\right)=(H_{n}^{\ast }\varphi )(t), \end{align}
</div>
<p> and the sequence of operators \(H_{n}^{\ast }\) strongly converges to the operator \(H^{\ast }\) in \(L_{p}(T)\), then it follows from <a href="index.html#4.2" class="eqref">13</a> and <a href="index.html#4.3" class="eqref">14</a> that for any \(\varepsilon {\gt}0\) for large values of \(n\) </p>
<div class="displaymath" id="a0000000066">
  \begin{align}  \| H_{\pi /n}u-Hu\| _{L_{p}([-\pi /2,\pi /2])}& =\| H_{n}^{\ast }\varphi -H^{\ast }\varphi \| _{L_{p}([-\pi /2,\pi /2])} \nonumber \\ \label{4.4} & \le \| H_{n}^{\ast }\varphi -H^{\ast }\varphi \| _{L_{p}(T)} {\lt} \varepsilon . \end{align}
</div>
<p> Due to the inequalities </p>
<div class="displaymath" id="a0000000067">
  \begin{align*}  |(Hu) (t)| & \le \tfrac {1}{\pi }\int _{-\pi /4}^{\pi /4}\left| \tfrac {u(\tau )}{t-\tau } \right|d\tau \le \tfrac {\| u\| _{L_{1}([-\pi /4, \pi /4])}}{\pi (|t|-\pi /4)}, \  |t|{\gt}\pi /4, \\ |(H_{\pi /n}u)(t)| & \le \tfrac {1}{\pi }\sum _{k \in Z_{(n)}^{(t)}}\tfrac {1}{|k+1/2|} \left|u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \right| \\ & \le \tfrac {1}{n(|t|-\pi /4)}\sum _{k \in Z_{(n)}^{(t)}} \left|u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \right|, \  \   |t|{\gt}\pi /4, \end{align*}
</div>
<p> where \(Z_{(n)}^{(t)}=\{ k \in {\mathbb Z}:t+\tfrac {\pi (k+1/2)}{n} \in [-\pi /4, \pi /4] \} \), we get that for any \(M{\gt}2\pi \) </p>
<div class="displaymath" id="a0000000068">
  \begin{align*} & \| Hu\| _{L_{p}([M,\infty ])} \le \tfrac {\| u\| _{L_{1}([-\pi /4, \pi /4])}}{\pi }\cdot \left(\int _{M}^{\infty }\tfrac {dt}{(|t|-\pi /4)^{p}} \right)^{1/p} = \tfrac {\| u\| _{L_{1}([-\pi /4, \pi /4])}}{\pi (p-1)^{1/p}(M-\pi /4)^{1-1/p} }, \\ & \| H_{\pi /n}u\| _{L_{p}([M,\infty ])} \le \tfrac {1}{n} \left[\int _{M}^{\infty }\tfrac {1}{(|t|-\frac{\pi }{4})^{p}} \left( \sum _{k \in Z_{(n)}^{(t)}} \left|u (t+\tfrac {\pi (k+1/2)}{n}) \right| \right)^{p}dt \right]^{1/p} \\ & \le \tfrac {1}{n^{1/p}} \left[\int _{M}^{\infty }\tfrac {1}{(|t|-\frac{\pi }{4})^{p}} \sum _{k \in Z_{(n)}^{(t)}} \left|u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \right|^{p}dt \right]^{1/p} \\ & = \tfrac {1}{n^{1/p}} \left[\sum _{m=0}^{\infty }\int _{M+\frac{\pi m}{n}}^{M+\frac{\pi (m+1)}{n}}\tfrac {1}{(|t|-\frac{\pi }{4})^{p}} \sum _{k \in Z_{(n)}^{(t)}} \left|u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \right|^{p}dt \right]^{1/p} \\ & \le \tfrac {1}{n^{1/p}} \left[\sum _{m=0}^{\infty }\tfrac {1}{(M+\frac{\pi m}{n}-\frac{\pi }{4})^{p}}\int _{M+\frac{\pi m}{n}}^{M+\frac{\pi (m+1)}{n}} \sum _{k \in Z_{(n)}^{(t)}} \left|u \left(t+\tfrac {\pi (k+1/2)}{n}\right) \right|^{p}dt \right]^{1/p} \\ & = \tfrac {1}{n^{1/p}} \left[\sum _{m=0}^{\infty }\tfrac {1}{(M+\frac{\pi m}{n}-\frac{\pi }{4})^{p}} \| u\| _{L_{p}([-\pi /4,\pi /4])} \right]^{1/p} \\ & \le \tfrac {\| u\| _{L_{p}([-\pi /4,\pi /4])}}{n^{1/p}} \left[\tfrac {n/ \pi }{(p-1)(M-\frac{\pi }{4}-\frac{\pi }{n})^{p-1}} \right]^{1/p} \\ & = \tfrac {\| u\| _{L_{p}([-\pi /4,\pi /4])}}{\pi ^{1/p}(p-1)^{1/p}(M-\frac{\pi }{4}-\frac{\pi }{n})^{1-1/p}}. \end{align*}
</div>
<p>Similar inequalities holds for \(\| Hu\| _{L_{p}([-\infty , -M])}\) and for \(\| H_{\pi /n}u\| _{L_{p}([-\infty , -M])}\). Therefore, for any \(\varepsilon {\gt}0\) there exist \(m_{0} \ge 4\) such that </p>
<div class="equation" id="4.5">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.5} \| Hu\| _{L_{p}(R \setminus [-\frac{\pi m_{0}}{2}, \frac{\pi m_{0}}{2}] )} < \varepsilon , \  \   \| H_{\pi /n}u\| _{L_{p}(R \setminus [-\frac{\pi m_{0}}{2}, \frac{\pi m_{0}}{2}] )} < \varepsilon . \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p>Since the function \(\frac{1}{t-\tau }\) is continuous on a rectangle \([2 \pi , 2 \pi m_{0}] \times [-\pi , \pi ]\), then it follows from <a href="index.html#cor.7">corollary 7</a> that the sequence of functions </p>
<div class="displaymath" id="a0000000069">
  \begin{align*}  (\mathrm{W}_{n}\varphi )(t)& =\tfrac {2}{n}\sum _{\{ k \in {\mathbb Z}:t+\frac{\pi (2k+1)}{n} \in [-\pi , \pi ] \} }\tfrac {\varphi (t+\pi (2k+1)/n)}{-\pi (2k+1)/n} = \\ & =\tfrac {1}{\pi }\sum _{\{ k \in {\mathbb Z}:t+\frac{\pi (2k+1)}{n} \in [-\pi , \pi ] \} }\tfrac {\varphi (t+\pi (2k+1)/n)}{-k-1/2}, \   \   n \in {\mathbb N} \end{align*}
</div>
<p> converges to the function </p>
<div class="displaymath" id="a0000000070">
  \begin{equation*}  (\mathrm{W} \varphi )(t)= \int _{-\pi }^{\pi }\tfrac {\varphi (\tau )}{t-\tau }d \tau \end{equation*}
</div>
<p> in \(L_{p}([2 \pi , 2\pi m_{0}])\). Denote by \(\psi \) the function, defined on \([-\pi , \pi ]\) by the equality \(\psi (\tau )=u(\tau /4)\). Then it follows from the equations </p>
<div class="displaymath" id="a0000000071">
  \begin{align*}  (Hu) (t)& =\tfrac {1}{\pi }\int _{-\pi /4}^{\pi /4}\tfrac {u(\tau )}{t-\tau }d\tau =\tfrac {1}{\pi }\int _{-\pi }^{\pi }\tfrac {u(\tau /4)}{4t-\tau }d\tau =(\mathrm{W}\psi )(4t), \   t \in [\pi /2, \pi m_{0}/2], \\ (H_{\pi /(4n)}u)(t)& =\tfrac {1}{\pi }\sum _{k \in Z_{(4n)}^{(t)}}\tfrac {u (t+\pi (k+1/2)/4n)}{-k-1/2} = \tfrac {1}{\pi }\sum _{k \in Z_{(4n)}^{(t)}}\tfrac {\psi (4t+\pi (k+1/2)/n)}{-k-1/2} \\ & = (\mathrm{W}_{n}\psi )(4t), \quad t \in [\pi /2, \pi m_{0}/2], \end{align*}
</div>
<p> that the sequence of functions \(H_{\pi /(4n)}u\) converges to the function \(Hu\) in \(L_{p}([\pi /2, \pi m_{0}/2])\). Therefore, for large values of \(n\) </p>
<div class="equation" id="4.6">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.6} \|  H_{\pi /(4n)}u-Hu\| _{L_{p}([\pi /2, \pi m_{0}/2])} < \varepsilon . \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> Similarly, for large values on \(n\) </p>
<div class="equation" id="4.7">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.7} \|  H_{\pi /(4n)}u-Hu\| _{L_{p}([-\pi m_{0}/2, -\pi /2])} < \varepsilon . \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> It follows from <a href="index.html#4.4" class="eqref">15</a>–<a href="index.html#4.7" class="eqref">18</a> that in the case \(\operatorname {supp} u \subset [-\pi /4, \pi /4]\) </p>
<div class="equation" id="4.8">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.8} \lim _{n \to \infty } \|  H_{\pi /(4n)}u-Hu\| _{L_{p}({\mathbb R})} =0 . \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p>Now suppose that \(\operatorname {supp} u \subset [-\pi m/4, \pi m/4]\) for some \(m \in {\mathbb N}\). Denote by \(u_{0}\) the function, defined on \( [-\pi /4, \pi /4]\) by the equation \(u_{0}(t)=u(m t)\). Then for any \(t \in R\) </p>
<div class="displaymath" id="a0000000072">
  \begin{align*}  (Hu) (t)& =\tfrac {1}{\pi }\int _{-\pi m/4}^{\pi m/4}\tfrac {u(\tau )}{t-\tau }d\tau =\tfrac {1}{\pi }\int _{-\pi /4}^{\pi /4}\tfrac {u(m \tau )}{t-\tau }m d\tau =(H u_{0})(t/m), \\ (H_{\pi /(4n)}u)(t)& =\tfrac {1}{\pi }\sum _{\{ k \in {\mathbb Z}:t+\frac{\pi (k+1/2)}{4n} \in [-\frac{\pi m}{4}, \tfrac {\pi m}{4}] \} }\tfrac {u (t+\pi (k+1/2)/4n)}{-k-1/2} \\ & = \tfrac {1}{\pi }\sum _{k \in Z_{(4mn)}^{(t/m)}}\tfrac {u_{0} (t/m+\pi (k+1/2)/(4mn))}{-k-1/2}= (H_{\pi /(4mn)}u_{0})(t/m). \end{align*}
</div>
<p> Since equation <a href="index.html#4.8" class="eqref">19</a> holds for \(u_{0}\), we obtain that </p>
<div class="displaymath" id="a0000000073">
  \begin{equation*}  \lim _{n \to \infty } \| H_{\pi /(4n)}u-Hu \| _{L_{p}({\mathbb R})} = m^{1/p} \lim _{n \to \infty } \| H_{\pi /(4mn)}u_{0}-Hu_{0} \| _{L_{p}({\mathbb R})}=0. \end{equation*}
</div>
<p>Now consider the general case. Let us prove that equation <a href="index.html#4.8" class="eqref">19</a> holds for any \(u \in L_{p}({\mathbb R})\). For any \(u \in L_{p}({\mathbb R})\) and \(\varepsilon {\gt}0\) there exist \(m \in {\mathbb N}\) such that </p>
<div class="equation" id="4.9">
<p>
  <div class="equation_content">
    \begin{equation}  \label{4.9} \| u-u_{m}\| _{L_{p}({\mathbb R})}< \varepsilon , \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> where \(u_{m}(t)=u(t) \cdot \chi _{[-\pi m/4, \pi m/4](t)} \). Since equation <a href="index.html#4.8" class="eqref">19</a> holds for \(u_{m}\), and it follows from <a href="index.html#2.3" class="eqref">3</a>, <a href="index.html#4.9" class="eqref">20</a> that </p>
<div class="displaymath" id="a0000000074">
  \begin{align*} & \| H_{\pi /(4n)}(u-u_{m})-H(u-u_{m}) \| _{L_{p}({\mathbb R})} \le \\ & \le \left[ \| H_{\pi /(4n)} \| _{L_{p}({\mathbb R}) \to L_{p}({\mathbb R})} + \| H \| _{L_{p}({\mathbb R}) \to L_{p}({\mathbb R})} \right] \cdot \| u-u_{m} \| _{L_{p}({\mathbb R})} \\ & \le \varepsilon \cdot \left[ \| \tilde{h}\| _{l_{p} \to l_{p}} + \| H \| _{L_{p}({\mathbb R}) \to L_{p}({\mathbb R})} \right], \end{align*}
</div>
<p> then we get that the equation <a href="index.html#4.8" class="eqref">19</a> also holds for the function \(u\). </p>
<p><b class="bfseries">Step 3.</b> Let us prove that for any \(\delta {\gt}0\) the secuence of the operators \(\{  H_{\delta /n} \} _{n \in {\mathbb N}}\) strongly converges to the operator \(H\) in \(L_{p}({\mathbb R})\). Let \(u \in L_{p}({\mathbb R})\). Denote \(w(t)=u(4\delta t/ \pi )\), \(t \in {\mathbb R}\). Then for any \(t \in {\mathbb R}\) </p>
<div class="displaymath" id="a0000000075">
  \begin{align}  (Hu) (t)& =\tfrac {1}{\pi }\int _{R}\tfrac {u(\tau )}{t-\tau }d\tau =\tfrac {1}{\pi }\int _{R}\tfrac {w(\pi \tau /(4 \delta ))}{t-\tau }d\tau =\nonumber \\ \label{4.10} & = \tfrac {1}{\pi }\int _{R}\tfrac {w(\tau )}{\pi t /(4 \delta )-\tau }d\tau = (Hw)(\pi t /(4 \delta )), \end{align}
</div>
<div class="displaymath" id="a0000000076">
  \begin{align}  (H_{\delta /n }u)(t)& =\tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {u(t+(k+1/2) \delta /n)}{-k-1/2}=\nonumber \\ \label{4.11} & = \tfrac {1}{\pi } \sum _{k \in {\mathbb Z}} \tfrac {w(\pi t /(4 \delta )+\pi (k+1/2) /(4n))}{-k-1/2} = (H_{\pi /(4n)}w)(\pi t /(4 \delta )). \end{align}
</div>
<p> Since \(\lim _{n \to \infty } \|  H_{\pi /(4n)}w-Hw\| _{L_{p}({\mathbb R})} =0\), then it follows from <a href="index.html#4.10" class="eqref">21</a>, <a href="index.html#4.11" class="eqref">22</a> that </p>
<div class="displaymath" id="a0000000077">
  \begin{equation*}  \lim _{n \to \infty } \|  H_{\delta /n}u-Hu\| _{L_{p}({\mathbb R})} =0. \end{equation*}
</div>

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