Inequalities for the Finite Hilbert Transform of Convex Functions

Sever Silvestru Dragomir$^{1,2}$

February 19, 2019; accepted: October 3, 2019; published online: January 21, 2020.

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.

MSC. 26D15; 26D10.

Keywords. Finite Hilbert Transform, Convex functions

$^{1}$Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia, sever.dragomir@vu.edu.au http://rgmia.org/dragomir

$^{2}$DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science & Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa

1 Introduction

Allover this paper, we consider the finite Hilbert transform on the open interval $\left( a,b\right) $ defined by

\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*}

for$\; t\in \left( a,b\right) $ and for various classes of functions $f$ for which the above Cauchy Principal Value integral exists, see [ 12 , Section 3.2 ] or [ 16 , Lemma II.1.1 ] .

For several recent papers devoted to inequalities for the finite Hilbert transform $\left( Tf\right) $, see [ 2 ] – [ 10 ] , [ 13 ] – [ 15 ] and [ 17 ] – [ 18 ] .

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) $.

The following result concerning upper and lower bounds for the finite Hilbert transform of a convex function holds.

Theorem 1 Dragomir et al., 2001 [ 1 ]

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

\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}

where $\varphi \left( t\right) \in \left[ f_{-}^{\prime }\left( t\right) ,f_{+}^{\prime }\left( t\right) \right] $, $t\in \left( a,b\right) $.

Corollary 2

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

\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}

We observe that if we take $t=\frac{a+b}{2},$ then we get from (2) that

\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}

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.

2 Inequalities for Convex Functions

We can prove the following slightly more general result than Theorem 1.

Theorem 3

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

\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}

In particular, we have

\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}

Proof â–¼

The proof is similar to the one from [ 1 ] . For the sake of completeness we provide a proof here.

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

\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*}

Then, obviously

\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*}

from where we get the equality

\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}

for all $t\in \left( a,b\right) $.

By the convexity of $f$ we can state that for all $a\leq c{\lt}d\leq b$ we have

\begin{equation} \tfrac {f\left( d\right) -f\left( c\right) }{d-c}\geq \varphi \left( c\right) , \label{e.2.2} \end{equation}

where $\varphi \left( c\right) \in \left[ f_{-}^{\prime }\left( c\right) ,f_{+}^{\prime }\left( c\right) \right] $.

Using (8), we have

\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}

and

\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}

and then, by adding (8) and (9), we get

\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*}

Consequently, we have

\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*}

and by the identity (6), we deduce the first inequality in (4).

Similarly, by the convexity of $f$ we have for $a\leq c{\lt}d\leq b$

\begin{equation} \psi \left( d\right) \geq \tfrac {f\left( d\right) -f\left( c\right) }{d-c}, \label{e.2.5} \end{equation}

where $\psi \left( c\right) \in \left[ f_{-}^{\prime }\left( c\right) ,f_{+}^{\prime }\left( c\right) \right] $.

Using (10) we may state

\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*}

and

\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*}

By adding these inequalities and taking the limit, we get

\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*}

namely

\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*}

and by the identity (6), we deduce the second inequality in (4).

Proof â–¼

Remark 4

We observe that for $\psi =\varphi \in $ $\partial f,$ the subdifferential of $f,$ we recapture the inequality (4). If $f$ is differentiable on $\left( a,b\right) $ then we also get (2).

Corollary 5

Let $f:\left( a,b\right) \rightarrow \mathbb {R}$ be a convex mapping on $\left( a,b\right) $. Then

\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}

Proof â–¼

If we take the integral mean in (4), we get

\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}

Observe that

\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*}

and

\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*}

and by (12) we get the desired result

Proof â–¼

We have:

Theorem 6

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

\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}

In particular,

\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}

Proof â–¼

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 $\Phi _{\alpha }:I\setminus \left\{ \alpha \right\} \rightarrow \mathbb {R}$,

\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*}

is monotonic nondecreasing on $I\setminus \left\{ \alpha \right\} .$

Using this property for the function $f:\left( a,b\right) \rightarrow \mathbb {R}$, we have for $t\in \left( a,b\right) $ that

\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*}

for any $\tau \in \left( a,b\right) ,$ $\tau \neq t.$

By the gradient inequality for the convex function $f$ we also have

\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*}

and

\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*}

Therefore we have the following inequality

\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}

for $t,$ $\tau \in \left( a,b\right) $ and $\tau \neq t.$

If we take the $PV$ in (15), then we get

\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}

for $t\in \left( a,b\right) .$

Using the equality (6) we deduce the desired result (13).

Proof â–¼

Corollary 7

With the assumptions in Theorem 6 we have

\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}

The proof follows by (13) on taking the integral mean over $t$ on $\left[ a,b\right] .$

Proposition 8

With the assumptions in Theorem 6, the inequality (11) is better than the inequality (17). In fact, we have the chain of inequalities

\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}

Proof â–¼

We use the following Čebyšev’s inequality which states that, if $g,$ $h$ have the same monotonicity (opposite monotonicity) then

\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}

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 (19) we have

\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*}

which is equivalent to

\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*}

which proves the fifth inequality in (18).

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 (19) we have

\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*}

which proves the second inequality in (18).

Proof â–¼

We also have:

Theorem 9

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) $

\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}

In particular,

\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}

Proof â–¼

We use Grüss’ inequality for integrable functions $g,$ $h$

\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}

provided $m\leq g\left( t\right) \leq M,$ $n\leq h\left( t\right) \leq N$ for almost every $t\in \left[ a,b\right] .$

Using Grüss’ inequality for increasing functions, we have

\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}

and

\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}

where $t\in \left( a,b\right) $ and for small $\varepsilon {\gt}0.$

We have

\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*}

and

\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*}

Also

\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*}

and

\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*}

From (24) we get

\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}

while from (26) we get

\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}

for $t\in \left( a,b\right) $ and small $\varepsilon {\gt}0$.

For $t-a{\gt}\varepsilon {\gt}0$ we get from (27) that

\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}

and from (28) for $b-t{\gt}\varepsilon {\gt}0$ that

\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*}

or, that

\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}

If we add (29) and (30) and use the triangle inequality, then we get

\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*}

for $t\in \left( a,b\right) $ and $\min \left\{ t-a,b-t\right\} {\gt}\varepsilon {\gt}0.$

Taking the limit over $\varepsilon \rightarrow 0+$ we get

\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}

for $t\in \left( a,b\right) .$

Using the identity (6) we get from (31) the desired result (20).

Proof â–¼

3 A Numerical Example

For a differentiable convex function $f:\left( a,b\right) \rightarrow \mathbb {R}$, define the following expressions

\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*}

and

\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*}

By inequality (2), 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.

Consider the convex function $f:\left[ 1,2\right] \rightarrow \mathbb {R}$, $f(x)=\exp x.\mathbb {\ }$ Figure 1 contains the plots of $E_{l}\left( f\right) $, $T\left( f\right) $, $E_{u}\left( f\right) ,$ while Figure 2 shows the magnitude of the closeness.

\includegraphics[scale=0.3]{Q2DAEE00.png} — Figure 1 Plots of $E_{l}\left( f\right) $, $T\left( f\right) $, $E_{u}\left( f\right) $.

\includegraphics[scale=0.3]{Q2DAEE01.png} — Figure 2 Plots of $E_{l}\left( f\right) $, $T\left( f\right) $, $E_{u}\left( f\right) $.

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