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<h1>Inequalities for the Finite Hilbert Transform of Convex Functions</h1>
<p class="authors">
<span class="author">Sever Silvestru Dragomir\(^{1,2}\)</span>
</p>
<p class="date">February 19, 2019; accepted: October 3, 2019; published online: January 21, 2020.</p>
</div>
<div class="abstract"><p> In this paper we obtain some new inequalities for the finite Hilbert transform of convex functions. Applications for some particular functions of interest are provided as well. </p>
<p><b class="bf">MSC.</b> 26D15; 26D10. </p>
<p><b class="bf">Keywords.</b> Finite Hilbert Transform, Convex functions </p>
</div>
<p>\(^{1}\)Mathematics, College of Engineering &amp; Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia, <span class="tt">sever.dragomir@vu.edu.au</span> <span class="tt">http://rgmia.org/dragomir</span> </p>
<p>\(^{2}\)DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science &amp; Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa </p>
<h1 id="a0000000002">1 Introduction</h1>
<p>Allover this paper, we consider the finite Hilbert transform on the open interval \(\left( a,b\right) \) defined by</p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  \left( Tf\right) \left( a,b;t\right) :=\tfrac {1}{\pi }PV\int _{a}^{b}\tfrac {f\left( \tau \right) }{\tau -t}d\tau :=\lim \limits _{\varepsilon \rightarrow 0+}\Big[ \int _{a}^{t-\varepsilon }+\int _{t+\varepsilon }^{b}\Big] \tfrac {f\left( \tau \right) }{\pi \left( \tau -t\right) }d\tau \end{equation*}
</div>
<p>for\(\; t\in \left( a,b\right) \) and for various classes of functions \(f\) for which the above Cauchy Principal Value integral exists, see <span class="cite">
	[
	<a href="#G" >12</a>
	, 
	Section 3.2
	]
</span> or <span class="cite">
	[
	<a href="#MP" >16</a>
	, 
	Lemma II.1.1
	]
</span>. </p>
<p>For several recent papers devoted to inequalities for the finite Hilbert transform \(\left( Tf\right) \), see <span class="cite">
	[
	<a href="#DDF2" >2</a>
	]
</span>–<span class="cite">
	[
	<a href="#DSS4" >10</a>
	]
</span>, <span class="cite">
	[
	<a href="#LG" >13</a>
	]
</span>–<span class="cite">
	[
	<a href="#LL" >15</a>
	]
</span> and <span class="cite">
	[
	<a href="#WGL" >17</a>
	]
</span>–<span class="cite">
	[
	<a href="#WLX" >18</a>
	]
</span>. </p>
<p>Now, if we assume that the mapping \(f:\left( a,b\right) \rightarrow \mathbb {R}\) is convex on \(\left( a,b\right) \), then it is locally Lipschitzian on \(\left( a,b\right) \) and then the finite Hilbert transform of \(f\) exists in every point \(t\in \left( a,b\right) \). </p>
<p>The following result concerning upper and lower bounds for the finite Hilbert transform of a convex function holds. </p>
<p><div class="theorem_thmwrapper " id="t.A">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
    <span class="theorem_thmtitle">Dragomir <i class="it">et al.</i>, 2001 <span class="cite">
	[
	<a href="#DDF" >1</a>
	]
</span></span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a convex function on \(\left( a,b\right) \) and \(t\in \left( a,b\right) .\) Then we have </p>
<div class="displaymath" id="e.1.1">
  \begin{align} &  \tfrac {1}{\pi }\left[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) \right] \leq \label{e.1.1} \\ &  \leq \left( Tf\right) \left( a,b;t\right) \notag \\ &  \leq \tfrac {1}{\pi }\left[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( b\right) -f\left( t\right) +\varphi \left( t\right) \left( t-a\right) \right] , \notag \end{align}
</div>
<p>where \(\varphi \left( t\right) \in \left[ f_{-}^{\prime }\left( t\right) ,f_{+}^{\prime }\left( t\right) \right] \), \(t\in \left( a,b\right) \). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="c.A">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a differentiable convex function on \(\left( a,b\right) \) and \(t\in \left( a,b\right) .\) Then we have</p>
<div class="displaymath" id="e.1.2">
  \begin{align} &  \tfrac {1}{\pi }\big[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( t\right) -f\left( a\right) +f^{\prime }\left( t\right) \left( b-t\right) \big]\leq \label{e.1.2} \\ &  \leq \left( Tf\right) \left( a,b;t\right) \notag \\ &  \leq \tfrac {1}{\pi }\big[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( b\right) -f\left( t\right) +f^{\prime }\left( t\right) \left( t-a\right) \big] . \notag \end{align}
</div>

  </div>
</div> </p>
<p>We observe that if we take \(t=\frac{a+b}{2},\) then we get from (<a href="#e.1.2">2</a>) that</p>
<div class="displaymath" id="e.1.3">
  \begin{align} &  \tfrac {1}{\pi }\big[ f\big( \tfrac {a+b}{2}\big) -f\left( a\right) +\tfrac {1}{2}f^{\prime }\big( \tfrac {a+b}{2}\big) \left( b-a\right) \big] \label{e.1.3} \\ &  \leq \left( Tf\right) \big( a,b;\tfrac {a+b}{2}\big) \notag \\ &  \leq \tfrac {1}{\pi }\big[ f\left( b\right) -f\big( \tfrac {a+b}{2}\big) +\tfrac {1}{2}f^{\prime }\big( \tfrac {a+b}{2}\big) \left( b-a\right) \big] . \notag \end{align}
</div>
<p>In this paper we obtain some new inequalities for the finite Hilbert transform of convex functions. Applications for some particular functions of interest are provided as well. </p>
<h1 id="a0000000004">2 Inequalities for Convex Functions</h1>
<p>We can prove the following slightly more general result than Theorem <a href="#t.A">1</a>. </p>
<p><div class="theorem_thmwrapper " id="t.2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a convex mapping on \(\left( a,b\right) \). Then for \(t\in \left( a,b\right) \) and \(\varphi \left( t\right) ,\) \(\psi \left( t\right) \in \left[ f_{-}^{\prime }\left( t\right) ,f_{+}^{\prime }\left( t\right) \right] \) we have</p>
<div class="displaymath" id="e.2.1">
  \begin{align} &  \tfrac {1}{\pi }\big[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) \big] \leq \label{e.2.1} \\ &  \leq \left( Tf\right) \left( a,b;t\right) \notag \\ &  \leq \tfrac {1}{\pi }\big[ f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) +f\left( b\right) -f\left( t\right) +\psi \left( t\right) \left( t-a\right) \big] . \notag \end{align}
</div>
<p>In particular, we have</p>
<div class="displaymath" id="e.2.1.a">
  \begin{align} &  \tfrac {1}{\pi }\big[ f\big( \tfrac {a+b}{2}\big) -f\left( a\right) +\tfrac {1}{2}\varphi \big( \tfrac {a+b}{2}\big) \left( b-a\right) \big]\leq \label{e.2.1.a} \\ &  \leq \left( Tf\right) \left( a,b;\tfrac {a+b}{2}\right) \notag \\ &  \leq \tfrac {1}{\pi }\big[ f\left( b\right) -f\big( \tfrac {a+b}{2}\big) +\tfrac {1}{2}\psi \big( \tfrac {a+b}{2}\big) \left( b-a\right) \big] . \notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000005">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The proof is similar to the one from <span class="cite">
	[
	<a href="#DDF" >1</a>
	]
</span>. For the sake of completeness we provide a proof here. </p>
<p>As for the mapping \(f:\left( a,b\right) \rightarrow \mathbb {R}\), \(f\left( t\right) =1\), \(t\in \left( a,b\right) \), we have </p>
<div class="displaymath" id="a0000000006">
  \begin{align*}  \left( Tf\right) \left( a,b;t\right) &  =\tfrac {1}{\pi }PV\int _{a}^{b}\tfrac {1}{\tau -t}d\tau \\ &  =\tfrac {1}{\pi }\lim \limits _{\varepsilon \rightarrow 0+}\left[ \int _{a}^{t-\varepsilon }\tfrac {1}{\tau -t}d\tau +\int _{t+\varepsilon }^{b}\tfrac {1}{\tau -t}d\tau \right] \\ &  =\tfrac {1}{\pi }\ln \big( \tfrac {b-t}{t-a}\big) ,\; \; t\in \left( a,b\right) . \end{align*}
</div>
<p>Then, obviously </p>
<div class="displaymath" id="a0000000007">
  \begin{align*}  \left( Tf\right) \left( a,b;t\right) &  =\tfrac {1}{\pi }PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) +f\left( t\right) }{\tau -t}d\tau \\ &  =\tfrac {1}{\pi }PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau +\tfrac {f\left( t\right) }{\pi }PV\int _{a}^{b}\tfrac {1}{\tau -t}d\tau \end{align*}
</div>
<p>from where we get the equality </p>
<div class="equation" id="Equ">
<p>
  <div class="equation_content">
    \begin{equation}  \left( Tf\right) \left( a,b;t\right) -\tfrac {f\left( t\right) }{\pi }\ln \big( \tfrac {b-t}{t-a}\big) =\tfrac {1}{\pi }PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \label{Equ} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>for all \(t\in \left( a,b\right) \). </p>
<p>By the convexity of \(f\) we can state that for all \(a\leq c{\lt}d\leq b\) we have </p>
<div class="equation" id="e.2.2">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {f\left( d\right) -f\left( c\right) }{d-c}\geq \varphi \left( c\right) , \label{e.2.2} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>where \(\varphi \left( c\right) \in \left[ f_{-}^{\prime }\left( c\right) ,f_{+}^{\prime }\left( c\right) \right] \). </p>
<p>Using (<a href="#e.2.3">8</a>), we have </p>
<div class="equation" id="e.2.3">
<p>
  <div class="equation_content">
    \begin{equation}  \int _{a}^{t-\varepsilon }\tfrac {f\left( t\right) -f\left( \tau \right) }{t-\tau }d\tau \geq \int _{a}^{t-\varepsilon }\varphi \left( \tau \right) d\tau \label{e.2.3} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>and </p>
<div class="equation" id="e.2.4">
<p>
  <div class="equation_content">
    \begin{equation}  \int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \geq \int _{t+\varepsilon }^{b}l\left( t\right) d\tau =\varphi \left( t\right) \left( b-t-\varepsilon \right) \label{e.2.4} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>and then, by adding (<a href="#e.2.3">8</a>) and (<a href="#e.2.4">9</a>), we get </p>
<div class="displaymath" id="a0000000008">
  \begin{align*} &  \lim \limits _{\varepsilon \rightarrow 0+}\left[ \int _{a}^{t-\varepsilon }\tfrac {f\left( t\right) -f\left( \tau \right) }{t-\tau }d\tau +\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right]\geq \\ &  \geq \lim \limits _{\varepsilon \rightarrow 0+}\left[ \int _{a}^{t-\varepsilon }\varphi \left( \tau \right) d\tau +\varphi \left( t\right) \left( b-t-\varepsilon \right) \right] \\ &  =\int _{a}^{t}\varphi \left( \tau \right) d\tau +\varphi \left( t\right) \left( b-t\right) =f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) . \end{align*}
</div>
<p>Consequently, we have </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \geq f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) \end{equation*}
</div>
<p>and by the identity (<a href="#Equ">6</a>), we deduce the first inequality in (<a href="#e.2.1">4</a>). </p>
<p>Similarly, by the convexity of \(f\) we have for \(a\leq c{\lt}d\leq b\) </p>
<div class="equation" id="e.2.5">
<p>
  <div class="equation_content">
    \begin{equation}  \psi \left( d\right) \geq \tfrac {f\left( d\right) -f\left( c\right) }{d-c}, \label{e.2.5} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p>where \(\psi \left( c\right) \in \left[ f_{-}^{\prime }\left( c\right) ,f_{+}^{\prime }\left( c\right) \right] \). </p>
<p>Using (<a href="#e.2.5">10</a>) we may state </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  \int _{a}^{t-\varepsilon }\tfrac {f\left( t\right) -f\left( \tau \right) }{t-\tau }d\tau \leq \int _{a}^{t-\varepsilon }\psi \left( t\right) d\tau =\psi \left( t\right) \left( t-\varepsilon -a\right) \end{equation*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  \int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \leq \int _{t+\varepsilon }^{b}\psi \left( \tau \right) d\tau =f\left( b\right) -f\left( t+\varepsilon \right) . \end{equation*}
</div>
<p>By adding these inequalities and taking the limit, we get</p>
<div class="displaymath" id="a0000000012">
  \begin{align*} &  \lim \limits _{\varepsilon \rightarrow 0+}\left[ \int _{a}^{t-\varepsilon }\tfrac {f\left( t\right) -f\left( \tau \right) }{t-\tau }d\tau +\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right]\leq \\ &  \leq \lim \limits _{\varepsilon \rightarrow 0+}\left[ \psi \left( t\right) \left( t-\varepsilon -a\right) +f\left( b\right) -f\left( t+\varepsilon \right) \right] \\ &  =\psi \left( t\right) \left( t-a\right) +f\left( b\right) -f\left( t\right) , \end{align*}
</div>
<p>namely</p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \leq \psi \left( t\right) \left( t-a\right) +f\left( b\right) -f\left( t\right) \end{equation*}
</div>
<p>and by the identity (<a href="#Equ">6</a>), we deduce the second inequality in (<a href="#e.2.1">4</a>). <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="r.2.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p>We observe that for \(\psi =\varphi \in \) \(\partial f,\) the subdifferential of \(f,\) we recapture the inequality (<a href="#e.2.1">4</a>). If \(f\) is differentiable on \(\left( a,b\right) \) then we also get (<a href="#e.1.2">2</a>). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="c.2.1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">5</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a convex mapping on \(\left( a,b\right) \). Then</p>
<div class="displaymath" id="e.2.6">
  \begin{align}  \tfrac {2}{\pi }\left( \tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) dt-f\left( a\right) \right) &  \leq \tfrac {1}{b-a}\int _{a}^{b}\left( Tf\right) \left( a,b;t\right) dt-\tfrac {1}{\pi }\tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) \ln \big( \tfrac {b-t}{t-a}\big) dt \label{e.2.6} \\ &  \leq \tfrac {2}{\pi }\left[ f\left( b\right) -\tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) dt\right] .\nonumber \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000015">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> If we take the integral mean in (<a href="#e.2.1">4</a>), we get </p>
<div class="displaymath" id="e.2.7">
  \begin{align} &  \tfrac {1}{\pi }\left[ \tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) \ln \left( \tfrac {b-t}{t-a}\right) +\tfrac {1}{b-a}\int _{a}^{b}\left[ f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) \right] dt\right] \leq \label{e.2.7} \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}\left( Tf\right) \left( a,b;t\right) dt \notag \\ &  \leq \tfrac {1}{\pi }\left[ \tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) \ln \left( \tfrac {b-t}{t-a}\right) +\tfrac {1}{b-a}\int _{a}^{b}\left[ f\left( b\right) -f\left( t\right) +\psi \left( t\right) \left( t-a\right) \right] \right] . \notag \end{align}
</div>
<p>Observe that </p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  \int _{a}^{b}\left[ f\left( t\right) -f\left( a\right) +\varphi \left( t\right) \left( b-t\right) \right] dt=2\left( \int _{a}^{b}f\left( t\right) dt-f\left( a\right) \left( b-a\right) \right) \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  \int _{a}^{b}\left[ f\left( b\right) -f\left( t\right) +\psi \left( t\right) \left( t-a\right) \right] =2\left( f\left( b\right) \left( b-a\right) -\int _{a}^{b}f\left( t\right) dt\right) \end{equation*}
</div>
<p>and by (<a href="#e.2.7">12</a>) we get the desired result <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We have: </p>
<p><div class="theorem_thmwrapper " id="t.2.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a convex mapping on \(\left( a,b\right) \) with finite lateral derivatives \(f_{+}^{\prime }\left( a\right) \) and \(f_{-}\left( b\right) .\) Then for \(t\in \left( a,b\right) \) we have </p>
<div class="displaymath" id="e.2.8">
  \begin{align}  \tfrac {1}{\pi }\left( b-a\right) f_{+}^{\prime }\left( a\right) &  \leq \tfrac {1}{\pi }\left( b-a\right) \tfrac {f\left( t\right) -f\left( a\right) }{t-a} \label{e.2.8} \\ &  \leq \left( Tf\right) \left( a,b;t\right) -\tfrac {f\left( t\right) }{\pi }\ln \big( \tfrac {b-t}{t-a}\big) \notag \\ &  \leq \tfrac {1}{\pi }\left( b-a\right) \tfrac {f\left( b\right) -f\left( t\right) }{b-t}\leq \tfrac {1}{\pi }\left( b-a\right) f_{-}\left( b\right) . \notag \end{align}
</div>
<p>In particular, </p>
<div class="displaymath" id="e.2.8.a">
  \begin{align}  \tfrac {1}{\pi }\left( b-a\right) f_{+}^{\prime }\left( a\right) &  \leq \tfrac {2}{\pi }\left( b-a\right) \tfrac {f\big( \frac{a+b}{2}\big) -f\left( a\right) }{b-a} \label{e.2.8.a} \\ &  \leq \left( Tf\right) \big( a,b;\tfrac {a+b}{2}\big) \notag \\ &  \leq \tfrac {2}{\pi }\left( b-a\right) \tfrac {f\left( b\right) -f\big( \frac{a+b}{2}\big) }{b-a}\leq \tfrac {1}{\pi }\left( b-a\right) f_{-}\left( b\right) . \notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We recall that if \(\Phi :I\rightarrow \mathbb {R}\) is a continuous convex function on the interval of real numbers \(I\) and \(\alpha \in I\) then the divided difference function \(\Phi _{\alpha }:I\setminus \left\{  \alpha \right\}  \rightarrow \mathbb {R}\), </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  \Phi _{\alpha }\left( t\right) :=\left[ \alpha ,t;\Phi \right] :=\tfrac {\Phi \left( t\right) -\Phi \left( \alpha \right) }{t-\alpha } \end{equation*}
</div>
<p>is monotonic nondecreasing on \(I\setminus \left\{  \alpha \right\}  .\) </p>
<p>Using this property for the function \(f:\left( a,b\right) \rightarrow \mathbb {R}\), we have for \(t\in \left( a,b\right) \) that</p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \tfrac {f\left( a\right) -f\left( t\right) }{a-t}\leq \tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\leq \tfrac {f\left( b\right) -f\left( t\right) }{b-t} \end{equation*}
</div>
<p>for any \(\tau \in \left( a,b\right) ,\) \(\tau \neq t.\) </p>
<p>By the gradient inequality for the convex function \(f\) we also have </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  \tfrac {f\left( t\right) -f\left( a\right) }{t-a}\geq f_{+}^{\prime }\left( a\right) \text{ for }t\in \left( a,b\right) \end{equation*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  \tfrac {f\left( b\right) -f\left( t\right) }{b-t}\leq f_{-}\left( b\right) \text{ for }t\in \left( a,b\right) . \end{equation*}
</div>
<p>Therefore we have the following inequality</p>
<div class="equation" id="e.2.9">
<p>
  <div class="equation_content">
    \begin{equation}  f_{+}^{\prime }\left( a\right) \leq \tfrac {f\left( t\right) -f\left( a\right) }{t-a}\leq \tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\leq \tfrac {f\left( b\right) -f\left( t\right) }{b-t}\leq f_{-}\left( b\right) \label{e.2.9} \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p>for \(t,\) \(\tau \in \left( a,b\right) \) and \(\tau \neq t.\) </p>
<p>If we take the \(PV\) in (<a href="#e.2.9">15</a>), then we get </p>
<div class="displaymath" id="e.2.10">
  \begin{align}  f_{+}^{\prime }\left( a\right) \left( b-a\right) &  \leq \tfrac {f\left( t\right) -f\left( a\right) }{t-a}\left( b-a\right) \label{e.2.10} \\ &  \leq PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \notag \\ &  \leq \tfrac {f\left( b\right) -f\left( t\right) }{b-t}\left( b-a\right) \leq f_{-}\left( b\right) \left( b-a\right) \notag \end{align}
</div>
<p>for \(t\in \left( a,b\right) .\) </p>
<p>Using the equality (<a href="#Equ">6</a>) we deduce the desired result (<a href="#e.2.8">13</a>). <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="c.2.2">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">7</span>
  </div>
  <div class="corollary_thmcontent">
  <p>With the assumptions in Theorem <a href="#t.2.2">6</a> we have</p>
<div class="displaymath" id="e.2.11">
  \begin{align}  \tfrac {1}{\pi }\left( b-a\right) f_{+}^{\prime }\left( a\right) &  \leq \tfrac {1}{\pi }\int _{a}^{b}\tfrac {f\left( t\right) -f\left( a\right) }{t-a}dt \label{e.2.11} \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}\left( Tf\right) \left( a,b;t\right) dt-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( t\right) }{\pi }\ln \big( \tfrac {b-t}{t-a}\big) dt \notag \\ &  \leq \tfrac {1}{\pi }\int _{a}^{b}\tfrac {f\left( b\right) -f\left( t\right) }{b-t}dt\leq \tfrac {1}{\pi }\left( b-a\right) f_{-}\left( b\right) . \notag \end{align}
</div>

  </div>
</div> </p>
<p>The proof follows by (<a href="#e.2.8">13</a>) on taking the integral mean over \(t\) on \(\left[ a,b\right] .\) </p>
<p><div class="proposition_thmwrapper " id="p.2.1">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">8</span>
  </div>
  <div class="proposition_thmcontent">
  <p>With the assumptions in Theorem <a href="#t.2.2">6</a>, the inequality (<a href="#e.2.6">11</a>) is better than the inequality (<a href="#e.2.11">17</a>). In fact, we have the chain of inequalities</p>
<div class="displaymath" id="e.2.12">
  \begin{align}  \tfrac {1}{\pi }\left( b-a\right) f_{+}^{\prime }\left( a\right) &  \leq \tfrac {1}{\pi }\int _{a}^{b}\tfrac {f\left( t\right) -f\left( a\right) }{t-a}dt \label{e.2.12} \\ &  \leq \tfrac {2}{\pi }\left( \tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) dt-f\left( a\right) \right) \notag \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}\left( Tf\right) \left( a,b;t\right) dt-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( t\right) }{\pi }\ln \big( \tfrac {b-t}{t-a}\big) dt \notag \\ &  \leq \tfrac {2}{\pi }\left[ f\left( b\right) -\tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) dt\right] \notag \\ &  \leq \tfrac {1}{\pi }\int _{a}^{b}\tfrac {f\left( b\right) -f\left( t\right) }{b-t}\leq \tfrac {1}{\pi }\left( b-a\right) f_{-}\left( b\right) . \notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000025">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We use the following Čebyšev’s inequality which states that, if \(g,\) \(h\) have the same monotonicity (opposite monotonicity) then </p>
<div class="equation" id="Ceb">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {1}{b-a}\int _{a}^{b}g\left( t\right) h\left( t\right) dt\geq \left( \leq \right) \tfrac {1}{b-a}\int _{a}^{b}g\left( t\right) dt\tfrac {1}{b-a}\int _{a}^{b}h\left( t\right) dt. \label{Ceb} \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p>Now, since \(g\left( t\right) =\frac{f\left( b\right) -f\left( t\right) }{b-t} \) is nondecreasing on \(\left( a,b\right) \) and \(h\left( t\right) =b-t\) is decreasing on \(\left[ a,b\right] ,\) then by (<a href="#Ceb">19</a>) we have</p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  \tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( b\right) -f\left( t\right) }{b-t}\left( b-t\right) dt\leq \tfrac {1}{2}\int _{a}^{b}\tfrac {f\left( b\right) -f\left( t\right) }{b-t}dt, \end{equation*}
</div>
<p>which is equivalent to</p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  2\left[ f\left( b\right) -\tfrac {1}{b-a}\int _{a}^{b}f\left( t\right) dt\right] \leq \int _{a}^{b}\tfrac {f\left( b\right) -f\left( t\right) }{b-t}dt, \end{equation*}
</div>
<p>which proves the fifth inequality in (<a href="#e.2.12">18</a>). </p>
<p>Also, since \(g\left( t\right) =\frac{f\left( t\right) -f\left( a\right) }{t-a}\) is nondecreasing on \(\left( a,b\right) \) and \(h\left( t\right) =t-a\) is increasing on \(\left[ a,b\right] ,\) then by (<a href="#Ceb">19</a>) we have</p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  \tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( t\right) -f\left( a\right) }{t-a}\left( t-a\right) dt\leq \tfrac {1}{2}\int _{a}^{b}\tfrac {f\left( t\right) -f\left( a\right) }{t-a}dt, \end{equation*}
</div>
<p>which proves the second inequality in (<a href="#e.2.12">18</a>). <div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We also have: </p>
<p><div class="theorem_thmwrapper " id="t.2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">9</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left( a,b\right) \rightarrow \mathbb {R}\) be a convex mapping on \(\left( a,b\right) \). Then for \(t\in \left( a,b\right) \)</p>
<div class="displaymath" id="e.2.13">
  \begin{multline}  \bigg\vert \left( Tf\right) \left( a,b;t\right) -\tfrac {f\left( t\right) }{\pi }\ln \left( \tfrac {b-t}{t-a}\right) \label{e.2.13} -\tfrac {2}{\pi }\bigg( \tfrac {1}{b-t}\int _{t}^{b}f\left( \tau \right) d\tau -\tfrac {1}{t-a}\int _{a}^{t}f\left( \tau \right) d\tau \bigg) \bigg\vert \leq \\ \leq \tfrac {1}{2\pi }\left( t-a\right) \left[ f_{-}^{\prime }\left( t\right) -\tfrac {f\left( t\right) -f\left( a\right) }{t-a}\right] +\tfrac {1}{2\pi }\left( b-t\right) \left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-f_{+}^{\prime }\left( t\right) \right] . \end{multline}
</div>
<p>In particular, </p>
<div class="displaymath" id="e.2.13.a">
  \begin{align} &  \bigg\vert \left( Tf\right) \left( a,b;\tfrac {a+b}{2}\right) -\tfrac {4}{\pi }\bigg( \tfrac {1}{b-a}\int _{\frac{a+b}{2}}^{b}f\left( \tau \right) d\tau -\tfrac {1}{b-a}\int _{a}^{\frac{a+b}{2}}f\left( \tau \right) d\tau \bigg) \bigg\vert \leq \label{e.2.13.a} \\ &  \leq \tfrac {1}{4\pi }\left( b-a\right) \left[ 4\tfrac {\frac{f\left( b\right) +f\left( a\right) }{2}-f\left( \frac{a+b}{2}\right) }{b-a}-\left( f_{+}^{\prime }\left( \tfrac {a+b}{2}\right) -f_{-}^{\prime }\left( \tfrac {a+b}{2}\right) \right) \right] \notag \\ &  \leq \tfrac {1}{\pi }\left[ \tfrac {f\left( b\right) +f\left( a\right) }{2}-f\left( \tfrac {a+b}{2}\right) \right] . \notag \end{align}
</div>

  </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> We use Grüss’ inequality for integrable functions \(g,\) \(h\) </p>
<div class="displaymath" id="Gru">
  \begin{align} &  \left\vert \tfrac {1}{b-a}\int _{a}^{b}g\left( t\right) h\left( t\right) dt-\tfrac {1}{b-a}\int _{a}^{b}g\left( t\right) dt\tfrac {1}{b-a}\int _{a}^{b}h\left( t\right) dt\right\vert \leq \label{Gru} \\ &  \leq \tfrac {1}{4}\left( M-m\right) \left( N-n\right) , \notag \end{align}
</div>
<p>provided \(m\leq g\left( t\right) \leq M,\) \(n\leq h\left( t\right) \leq N\) for almost every \(t\in \left[ a,b\right] .\) </p>
<p><br /></p>
<p>Using Grüss’ inequality for increasing functions, we have</p>
<div class="displaymath" id="e.2.14">
  \begin{multline}  \bigg\vert \int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\left( \tau -t\right) d\tau \label{e.2.14} -\int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \tfrac {1}{t-\varepsilon -a}\int _{a}^{t-\varepsilon }\left( \tau -t\right) d\tau \bigg\vert \leq \\ \leq \tfrac {1}{4}\left( t-\varepsilon -a\right) \left( t-\varepsilon -a\right) \left[ \tfrac {f\left( t-\varepsilon \right) -f\left( t\right) }{t-\varepsilon -t}-\tfrac {f\left( a\right) -f\left( t\right) }{a-t}\right] \end{multline}
</div>
<p>and</p>
<div class="displaymath" id="e.2.15">
  \begin{align} & \left\vert \int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\left( \tau -t\right) d\tau \right. \left. -\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \tfrac {1}{b-t-\varepsilon }\int _{t+\varepsilon }^{b}\left( \tau -t\right) d\tau \right\vert \leq \label{e.2.15} \\ & \leq \tfrac {1}{4}\left( b-t-\varepsilon \right) \left( b-t-\varepsilon \right) \left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-\tfrac {f\left( t+\varepsilon \right) -f\left( t\right) }{t+\varepsilon -t}\right] \nonumber \end{align}
</div>
<p>where \(t\in \left( a,b\right) \) and for small \(\varepsilon {\gt}0.\) </p>
<p>We have</p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  \int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\left( \tau -t\right) d\tau =\int _{a}^{t-\varepsilon }f\left( \tau \right) d\tau -f\left( t\right) \left( t-\varepsilon -a\right) \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  \int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}\left( \tau -t\right) d\tau =\int _{t+\varepsilon }^{b}f\left( \tau \right) d\tau -f\left( t\right) \left( b-t-\varepsilon \right) . \end{equation*}
</div>
<p>Also</p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  \tfrac {1}{t-\varepsilon -a}\int _{a}^{t-\varepsilon }\left( \tau -t\right) d\tau =\tfrac {\varepsilon ^{2}-\left( a-t\right) ^{2}}{2\left( t-\varepsilon -a\right) }=-\tfrac {\left( t-a+\varepsilon \right) }{2} \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \tfrac {1}{b-t-\varepsilon }\int _{t+\varepsilon }^{b}\left( \tau -t\right) d\tau =\tfrac {\left( b-t\right) ^{2}-\varepsilon ^{2}}{2\left( b-t-\varepsilon \right) }=\tfrac {\left( b-t+\varepsilon \right) }{2}. \end{equation*}
</div>
<p>From (<a href="#e.2.14">24</a>) we get</p>
<div class="displaymath" id="e.2.16">
  \begin{align} &  \left\vert \int _{a}^{t-\varepsilon }f\left( \tau \right) d\tau -f\left( t\right) \left( t-\varepsilon -a\right) +\tfrac {t-a+\varepsilon }{2}\int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right\vert \leq \label{e.2.16} \\ &  \leq \tfrac {1}{4}\left( t-\varepsilon -a\right) \left( t-\varepsilon -a\right) \left[ \tfrac {f\left( t\right) -f\left( t-\varepsilon \right) }{\varepsilon }-\tfrac {f\left( a\right) -f\left( t\right) }{a-t}\right] \notag \end{align}
</div>
<p>while from (<a href="#e.2.15">26</a>) we get</p>
<div class="displaymath" id="e.2.17">
  \begin{align} &  \left\vert \int _{t+\varepsilon }^{b}f\left( \tau \right) d\tau -f\left( t\right) \left( b-t-\varepsilon \right) -\tfrac {b-t+\varepsilon }{2}\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right\vert \leq \label{e.2.17} \\ &  \leq \tfrac {1}{4}\left( b-t-\varepsilon \right) \left( b-t-\varepsilon \right) \left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-\tfrac {f\left( t+\varepsilon \right) -f\left( t\right) }{\varepsilon }\right] \notag \end{align}
</div>
<p>for \(t\in \left( a,b\right) \) and small \(\varepsilon {\gt}0\). </p>
<p>For \(t-a{\gt}\varepsilon {\gt}0\) we get from (<a href="#e.2.16">27</a>) that</p>
<div class="displaymath" id="e.2.18">
  \begin{align} &  \left\vert \tfrac {1}{2}\int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau +\tfrac {1}{t-a+\varepsilon }\int _{a}^{t-\varepsilon }f\left( \tau \right) d\tau -f\left( t\right) \right\vert \leq \label{e.2.18} \\ &  \leq \tfrac {1}{4}\left( t-\varepsilon -a\right) \left[ \tfrac {f\left( t\right) -f\left( t-\varepsilon \right) }{\varepsilon }-\tfrac {f\left( a\right) -f\left( t\right) }{a-t}\right] \notag \end{align}
</div>
<p><br />and from (<a href="#e.2.17">28</a>) for \(b-t{\gt}\varepsilon {\gt}0\) that </p>
<div class="displaymath" id="a0000000035">
  \begin{align*} &  \left\vert \tfrac {1}{b-t+\varepsilon }\int _{t+\varepsilon }^{b}f\left( \tau \right) d\tau -f\left( t\right) \tfrac {b-t-\varepsilon }{b-t+\varepsilon }-\tfrac {1}{2}\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right\vert \leq \\ &  \leq \tfrac {1}{4}\tfrac {\left( b-t-\varepsilon \right) \left( b-t-\varepsilon \right) }{b-t+\varepsilon }\left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-\tfrac {f\left( t+\varepsilon \right) -f\left( t\right) }{\varepsilon }\right] \end{align*}
</div>
<p>or, that </p>
<div class="displaymath" id="e.2.19">
  \begin{align} &  \left\vert \tfrac {1}{2}\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau -\tfrac {1}{b-t+\varepsilon }\int _{t+\varepsilon }^{b}f\left( \tau \right) d\tau +f\left( t\right) \tfrac {b-t-\varepsilon }{b-t+\varepsilon }\right\vert \leq \label{e.2.19} \\ &  \leq \tfrac {1}{4}\tfrac {\left( b-t-\varepsilon \right) \left( b-t-\varepsilon \right) }{b-t+\varepsilon }\left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-\tfrac {f\left( t+\varepsilon \right) -f\left( t\right) }{\varepsilon }\right] . \notag \end{align}
</div>
<p>If we add (<a href="#e.2.18">29</a>) and (<a href="#e.2.19">30</a>) and use the triangle inequality, then we get </p>
<div class="displaymath" id="a0000000036">
  \begin{align*} &  \left\vert \tfrac {1}{2}\int _{a}^{t-\varepsilon }\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau +\tfrac {1}{2}\int _{t+\varepsilon }^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau \right.+ \\ &  \quad \left. +\tfrac {1}{t-a+\varepsilon }\int _{a}^{t-\varepsilon }f\left( \tau \right) d\tau -f\left( t\right) -\tfrac {1}{b-t+\varepsilon }\int _{t+\varepsilon }^{b}f\left( \tau \right) d\tau +f\left( t\right) \tfrac {b-t-\varepsilon }{b-t+\varepsilon }\right\vert \leq \\ &  \leq \tfrac {1}{4}\left( t-\varepsilon -a\right) \left[ \tfrac {f\left( t\right) -f\left( t-\varepsilon \right) }{\varepsilon }-\tfrac {f\left( a\right) -f\left( t\right) }{a-t}\right] \\ &  \quad +\tfrac {1}{4}\tfrac {\left( b-t-\varepsilon \right) \left( b-t-\varepsilon \right) }{b-t+\varepsilon }\left[ \tfrac {f\left( b\right) -f\left( t\right) }{b-t}-\tfrac {f\left( t+\varepsilon \right) -f\left( t\right) }{\varepsilon }\right] \end{align*}
</div>
<p>for \(t\in \left( a,b\right) \) and \(\min \left\{  t-a,b-t\right\}  {\gt}\varepsilon {\gt}0.\) </p>
<p>Taking the limit over \(\varepsilon \rightarrow 0+\) we get </p>
<div class="displaymath" id="e.2.20">
  \begin{align} &  \left\vert \tfrac {1}{2}PV\int _{a}^{b}\tfrac {f\left( \tau \right) -f\left( t\right) }{\tau -t}d\tau +\tfrac {1}{t-a}\int _{a}^{t}f\left( \tau \right) d\tau -\tfrac {1}{b-t}\int _{t}^{b}f\left( \tau \right) d\tau \right\vert \leq \label{e.2.20} \\ &  \leq \tfrac {1}{4}\left( t-a\right) \left[ f_{-}^{\prime }\left( t\right) -\tfrac {f\left( t\right) -f\left( a\right) }{t-a}\right] +\tfrac {1}{4}\left( b-t\right) \left[ \frac{f\left( b\right) -f\left( t\right) }{b-t}-f_{+}^{\prime }\left( t\right) \right] \notag \end{align}
</div>
<p>for \(t\in \left( a,b\right) .\) </p>
<p>Using the identity (<a href="#Equ">6</a>) we get from (<a href="#e.2.20">31</a>) the desired result (<a href="#e.2.13">20</a>). <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> </p>
<h1 id="a0000000038">3 A Numerical Example</h1>
<p>For a differentiable convex function \(f:\left( a,b\right) \rightarrow \mathbb {R}\), define the following expressions </p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  E_{l}\left( f;a,b,t\right) =\tfrac {f\left( t\right) -f\left( a\right) +f^{\prime }\left( t\right) \left( b-t\right) }{\pi }+\tfrac {f\left( t\right) }{\pi }\ln \left( \tfrac {b-t}{t-a}\right) ,\; t\in \left( a,b\right) \end{equation*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  E_{u}\left( f;a,b,t\right) =\tfrac {f\left( t\right) }{\pi }\ln \left( \tfrac {b-t}{t-a}\right) +\tfrac {f\left( b\right) -f\left( t\right) +f^{\prime }\left( t\right) \left( t-a\right) }{\pi }. \end{equation*}
</div>
<p>By inequality (<a href="#e.1.2">2</a>), we know that \(E_{l}\left( f;a,b,t\right) \) is a lower bound for the finite Hilbert transform \(\left( Tf\right) \left( a,b,t\right) \) while \(E_{u}\left( f;a,b,t\right) \) is an upper bound. </p>
<p>Consider the convex function \(f:\left[ 1,2\right] \rightarrow \mathbb {R}\), \(f(x)=\exp x.\mathbb {\  }\) Figure <a href="#fig1">1</a> contains the plots of \(E_{l}\left( f\right) \), \(T\left( f\right) \), \(E_{u}\left( f\right) ,\) while Figure <a href="#fig2">2</a> shows the magnitude of the closeness. </p>
<figure id="fig1">
  <div class="centered"> <img src="img-0001.png" alt="\includegraphics[scale=0.3]{Q2DAEE00.png}" style="width:252.89999999999998px; height:178.79999999999998px" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Plots of \(E_{l}\left( f\right) \), \(T\left( f\right) \), \(E_{u}\left( f\right) \).</span> 
</figcaption>


</figure>
<figure id="fig2">
  <div class="centered"> <img src="img-0002.png" alt="\includegraphics[scale=0.3]{Q2DAEE01.png}" style="width:252.89999999999998px; height:178.79999999999998px" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Plots of \(E_{l}\left( f\right) \), \(T\left( f\right) \), \(E_{u}\left( f\right) \).</span> 
</figcaption>


</figure>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</a> </p>
</dd>
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</a> </p>
</dd>
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</dd>
</dl>


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