Some new estimates of the ‘Jensen gap’

Abstract

Let $( \mu,\Omega )$ be a probability measure space. We consider the so-called ‘Jensen gap’

$$ J ( \varphi,\mu,f ) = \int_{\Omega}\varphi \bigl( f ( s ) \bigr)\,d\mu ( s ) -\varphi \biggl( \int_{\Omega }f ( s )\,d\mu ( s ) \biggr) $$

for some classes of functions φ. Several new estimates and equalities are derived and compared with other results of this type. Especially the case when φ has a Taylor expansion is treated and the corresponding discrete results are pointed out.

1 Introduction

Let $( \Omega,\mu )$ be a probability measure space i.e. $\mu ( \Omega ) =1$ and let f be a μ-measurable function on Ω. If φ is convex, then Jensen’s inequality

$$ \varphi \biggl( \int_{\Omega}f ( s )\,d\mu ( s ) \biggr) \leq \int_{\Omega}\varphi \bigl( f ( s ) \bigr)\,d\mu ( s ) $$

(1.1)

holds. This inequality can be traced back to Jensen’s original papers [1, 2] and is one of the most fundamental mathematical inequalities. One reason for that is that in fact a great number of classical inequalities can be derived from (1.1), see e.g. [3] and the references given therein. The inequality (1.1) cannot in general be improved since we have equality in (1.1) when $\varphi ( x ) \equiv x$. However, for special cases of functions (1.1) can be given in a more specific form e.g. by giving lower estimates of the so-called ‘Jensen gap’

$$ J ( \varphi,\mu,f ) = \int_{\Omega}\varphi \bigl( f ( s ) \bigr)\,d\mu ( s ) -\varphi \biggl( \int_{\Omega }f ( s )\,d\mu ( s ) \biggr) , $$

thus obtaining refined versions of (1.1).

We give a few examples of such results.

Example 1

(see [4])

Let φ be a superquadratic function i.e. $\varphi: [ 0,\infty ) \rightarrow \mathbb{R}$ is such that there exists a constant $C ( x )$, $x\geq0$, such that

$$ \varphi ( y ) \geq\varphi ( x ) +C ( x ) ( y-x ) +\varphi \bigl( \vert y-x \vert \bigr) $$

for $y\geq0$. For such functions we have the following estimate of the Jensen gap:

$$ J ( \varphi,\mu,f ) \geq \int_{\Omega}\varphi \biggl( \biggl\vert f ( s ) - \int_{\Omega}f ( s )\,d\mu ( s ) \biggr\vert \biggr)\,d\mu ( s ) . $$

Example 2

(see [5] and [6])

We say that a function $K ( x )$ in γ-superconvex if $\varphi ( x ) :=x^{-\gamma }K ( x )$ is convex. If φ is a differentiable convex, increasing function and $\varphi ( 0 ) = \lim_{z\rightarrow0+} z\varphi^{\prime} ( z ) =0$, then we have the following estimate of the Jensen gap:

$$ J ( K,\mu,f ) \geq\varphi ( z ) \int_{\Omega} \bigl( \bigl( f ( s ) \bigr) ^{\gamma}-z^{\gamma} \bigr)\,d\mu ( s ) +\varphi^{\prime} ( z ) \int_{\Omega} \bigl( f ( s ) \bigr) ^{\gamma} \bigl( f ( s ) -z \bigr)\,d\mu ( s ) \geq0, $$

for $z= \int_{\Omega}f ( s )\,d\mu ( s ) >0$ and $f\geq 0$, $f^{\gamma}$ when $\gamma\geq0$ are integrable functions on the probability measure space $( \Omega,\mu )$.

Remark 1

By using the results in Examples 1 and 2 it is possible to derive Hardy-type inequalities with other ‘breaking points’ (the point where the inequality reverses) than the usual breaking point $p=1$. See [5, 7, 8] and [9].

Remark 2

In the recent paper [6] it was proved that the notion of γ-superconvexity has sense also for the case $-1\leq\gamma \leq0$ and in fact this was used even to derive there some new two-sided Jensen type inequalities.

Example 3

(see [10])

In his paper Walker studied the Jensen gap for the special case $f\equiv1$ i.e. for $J ( \varphi,\mu) :=J ( \varphi,\mu,1 )$ and found an estimate of the type

$$ J ( \varphi,\mu ) \geq\frac{1}{2}C ( \varphi,\mu ) \biggl( \int_{\Omega}s^{2}\,d\mu ( s ) - \biggl( \int_{\Omega }s\,d\mu ( s ) \biggr) ^{2} \biggr) , $$

where the positive constant $C=C ( \varphi,\mu )$ is easily computed.

In his paper it was assumed that φ admits a Taylor power series representation $\varphi ( x ) =\sum_{n=1}^{\infty }a_{n}x^{n}$, $a_{n}\geq0$, $n=0,1,2,\ldots$ , $0< x\leq A<\infty$. In another recent paper Dragomir [11] derived some other Jensen integral inequalities for this power series case. A comparison between these two results and our results is given in our concluding remarks.

Inspired by these results, we derive some new results of the same type. In Theorem 1 we get an estimate like that of Walker in [10] but for the general case of $J ( \varphi,\mu,f )$. In Theorem 2 we prove another complement of the Walker result by considering the Jensen functional

$$ J_{\alpha} \bigl( t^{\alpha},\mu \bigr) = \int_{\Omega}y^{\alpha}\,d\mu ( y ) - \biggl( \int_{\Omega}y\,d\mu ( y ) \biggr) ^{\alpha }, \quad\alpha \geq2, $$

and get an estimate for this Jensen gap which even reduces to equality for $\alpha=N$, $N=2,3,\ldots$ . By using this result it is possible to derive a similar equality for the Jensen gap whenever it can be represented by a Taylor power series (see Theorem 3).

In Section 3 we show that our lower bound of the Jensen gap is better than the lower bound in [11] when the function that we deal with has a Taylor series expansion with non-negative coefficients. Moreover, we prove that by our technique we can in such cases derive also upper bounds and not only lower bounds as in [10].

2 The main results

Our first main result reads as follows.

Theorem 1

Let $\phi: [ 0,A ) \rightarrow \mathbb{R}$ have a Taylor power series representation on $[ 0,A )$, $0< A\leq\infty:\phi ( x ) =\sum_{n=0}^{\infty }a_{n}x^{n}$.

Let φ be a convex increasing function on $[ 0,A )$ that is related to ϕ by

$$ \varphi ( x ) =\frac{\phi ( x ) -\phi(0)}{x}=\sum_{n=0}^{\infty}a_{n+1}x^{n}. $$

(a) If $f\geq0$ and f, $f^{2}$, and $\phi\circ f$ are integrable functions on Ω, $z=\int_{\Omega}f\,d\mu>0$, where μ is a probability measure on Ω, then

$$ \int_{\Omega}\phi ( f )\,d\mu-\phi ( z ) \geq \biggl( \frac{\phi ( z ) -\phi(0)}{z} \biggr) ^{\prime} \biggl( \int_{\Omega}f^{2}\,d\mu-z^{2} \biggr) \geq0. $$

In other words,

$$\begin{aligned} J ( \phi,\mu,f ) =& \int_{\Omega}\phi ( f )\,d\mu -\phi ( z ) \\ =&\sum_{n=0}^{\infty}a_{n+1} \int_{\Omega}f^{n+1}\,d\mu-\sum _{n=0}^{\infty }a_{n+1}z^{n+1} \\ \geq&\sum_{n=0}^{\infty} ( n+1 ) a_{n+2}z^{n} \biggl( \int _{\Omega }f^{2}\,d\mu-z^{2} \biggr) \geq0. \end{aligned}$$

(b) For $\overline{x}=\sum_{i=1}^{m}\alpha_{i}x_{i}$, $\ \sum_{i=1}^{m}\alpha_{i}=1$, $0\leq\alpha_{i}\leq1$, $0\leq x_{i}< A$, $i=1,\ldots,m$, it yields

$$ \sum_{i=1}^{m}\alpha_{i}\phi ( x_{i} ) -\phi ( \overline {x} ) \geq \biggl( \frac{\phi ( \overline{x} ) -\phi(0)}{\overline{x}} \biggr) ^{\prime} \Biggl( \sum _{i=1}^{m}\alpha _{i}x_{i}^{2}-\overline{x}^{2} \Biggr) \geq0. $$

In other words,

$$ \sum_{i=1}^{m}\sum _{n=0}^{\infty}\alpha _{i}a_{n+1}x_{i}^{n+1}- \sum_{n=0}^{\infty}a_{n+1} \overline{x}^{n+1}\geq \sum_{n=0}^{\infty} ( n+1 ) a_{n+2}\overline{x}^{n} \Biggl( \sum _{i=1}^{m}\alpha_{i}x_{i}^{2}- \overline{x}^{2} \Biggr) \geq0. $$

Proof

For $\phi ( x ) =\sum_{n=0}^{\infty}a_{n}x^{n}$, $0\leq x< A$, by denoting the function $\psi: [ 0,A ) \rightarrow \mathbb{R} _{+}$ $\psi ( x ) =\phi ( x ) -\phi ( 0 ) =\sum_{n=0}^{\infty}a_{n+1}x^{n+1}$, $0\leq x< A$, and $\varphi ( x ) =\frac{\psi ( x ) }{x}\Leftrightarrow x\varphi ( x ) =\psi ( x )$, $0\leq x< A$, we see that $\psi ( x )$ is 1-quasiconvex function (see [6]), $\varphi ( x ) =\sum_{n=0}^{\infty}a_{n+1}x^{n}$, $0\leq x< A$, and $\varphi ^{\prime} ( x ) =\sum_{n=0}^{\infty} ( n+1 ) a_{n+2}x^{n}$.

The functions ϕ, ψ, φ, and $\varphi^{\prime}$ are differentiable functions on $[ 0,A )$. From the convexity of $\varphi ( x )$ we have

$$ \varphi ( y ) -\varphi ( x ) >\varphi^{\prime } ( x ) ( y-x ) ,\quad x,y\in [ 0,A ), $$

and, therefore,

$$ \psi ( y ) -\psi ( x ) =y\varphi ( y ) -x\varphi ( x ) \geq\varphi ( x ) ( y-x ) +\varphi^{\prime} ( x ) y ( y-x ) ,\quad x,y\geq0. $$

Since $\psi ( x ) =\phi ( x ) -\phi ( 0 )$ we get

$$ \phi ( y ) -\phi ( x ) =\psi ( y ) -\psi ( x ) \geq\varphi ( x ) ( y-x ) + \varphi ^{\prime} ( x ) y ( y-x ) . $$

Now using this inequality with $x=z$, $y=f$, and integrating, we find that

$$\begin{aligned} &\int_{\Omega}\phi ( f )\,d\mu-\phi ( z )\\ &\quad\geq\varphi ( z ) \biggl( \int_{\Omega}f\,d\mu- \int_{\Omega }z\,d\mu \biggr) +\varphi^{\prime} ( z ) \biggl( \int_{\Omega }f^{2}\,d\mu-z^{2} \biggr)\\ &\quad=0+ \biggl( \frac{\phi ( z ) -\phi ( 0 ) }{z} \biggr) ^{\prime} \biggl( \int_{\Omega}f^{2}\,d\mu-z^{2} \biggr) \geq0. \end{aligned}$$

In the last inequality we have used $z=\int_{\Omega}f\,d\mu>0$ and φbeing convex increasing, where $\varphi ( z ) =\frac{\phi ( z ) -\phi ( 0 ) }{z}$.

Hence (a) is proved and since (b) is just a special case of (a), the proof is complete. □

For the proof of our next main result we need the following lemma, which is also of independent interest.

Lemma 1

Let φ be a differentiable function on $I\subset \mathbb{R}$, and let $x,y\subseteq I$. Then, for $N=2,3,\ldots$ ,

$$\begin{aligned} &\varphi ( x ) \bigl( y^{N-1}-x^{N-1} \bigr) +\varphi ^{\prime } ( x ) y^{N-1} ( y-x ) \\ &\quad= \bigl( x^{N-1}\varphi ( x ) \bigr) ^{\prime} ( y-x ) + ( y-x ) ^{2}\sum_{k=1}^{N-1}y^{k-1} \bigl( x^{N-k-1}\varphi ( x ) \bigr) ^{\prime}. \end{aligned}$$

(2.1)

In particular, for $N=2$ we have

$$ \varphi ( x ) ( y-x ) +\varphi^{\prime} ( x ) y ( y-x ) = \bigl( x\varphi ( x ) \bigr) ^{\prime } ( y-x ) +\varphi^{\prime} ( x ) ( y-x ) ^{2}. $$

(2.2)

Proof

A simple calculation shows that (2.2) holds, i.e., that (2.1) holds for $N=2$. For $N=3$ (2.1) reads

$$\begin{aligned} &\varphi ( x ) \bigl( y^{2}-x^{2} \bigr) + \varphi^{\prime } ( x ) y^{2} ( y-x ) = \bigl( x^{2}\varphi ( x ) \bigr) ^{\prime} ( y-x ) + ( y-x ) ^{2} \bigl( \bigl( x\varphi ( x ) \bigr) ^{\prime}+y \varphi^{\prime} ( x ) \bigr) . \end{aligned}$$

(2.3)

Moreover, it is easy to verify the identity

$$\begin{aligned} &\varphi ( x ) \bigl( y^{2}-x^{2} \bigr) + \varphi^{\prime } ( x ) y^{2} ( y-x ) =\varphi^{\prime} ( x ) y ( y-x ) ^{2}+x\varphi ( x ) ( y-x ) + \bigl( x\varphi ( x ) \bigr) ^{\prime }y ( y-x ). \end{aligned}$$

(2.4)

By using (2.4) together with (2.2) and making some straightforward calculations we obtain (2.3). The general proof follows in the same way using induction and the more general (than (2.4)) identity

$$\begin{aligned}[b] &\varphi ( x ) \bigl( y^{N-1}-x^{N-1} \bigr) +\varphi ^{\prime } ( x ) y^{N-1} ( y-x )\\ &\qquad{}- \bigl[ \bigl( x\varphi ( x ) \bigr) \bigl( y^{N-2}-x^{N-2} \bigr) + \bigl( x\varphi ( x ) \bigr) ^{\prime }y^{N-2} ( y-x ) \bigr]\\ &\quad=\varphi^{\prime} ( x ) y^{N-2} ( y-x ) ^{2},\quad N=2,3,4, \ldots. \end{aligned} $$

 □

Now we are ready to state our next main result.

Theorem 2

Let μ be a probability measure on $\Omega= (0,\infty)$, $z=\int_{\Omega}y\,d\mu ( y ) >0$, and $N=2,3,\ldots$ . Then the refined Jensen-type inequality

$$ \int_{\Omega}y^{\alpha}\,d\mu ( y ) -z^{\alpha}\geq \int _{\Omega } ( y-z ) ^{2}\sum _{k=1}^{N-1} ( \alpha-k ) x^{k-1}z^{\alpha-k-1}\,d\mu,\quad y\geq0, $$

(2.5)

holds for any $\alpha\geq N$. Moreover, for $N-1<\alpha\leq N$ (2.5) holds in the reversed direction. In particular, for $\alpha=N$ we have equality in (2.5).

Proof

A convex differentiable function on $\varphi ( x )$ is characterized by

$$ \varphi ( y ) -\varphi ( x ) \geq\varphi^{\prime } ( x ) ( y-x ) $$

and this inequality holds in the reversed direction if $\varphi ( x )$ is concave. For $\varphi ( x ) =x$ we have equality. Therefore, when $\varphi ( x )$ is convex it yields

$$ \varphi ( y ) y^{N-1}-\varphi ( x ) x^{N-1}\geq \varphi ( x ) \bigl( y^{N-1}-x^{N-1} \bigr) +\varphi^{\prime} ( x ) y^{N-1} ( y-x ) ,\quad x,y\geq0. $$

Hence in view of Lemma 1 we find that

$$ \varphi ( y ) y^{N-1}-\varphi ( x ) x^{N-1}\geq \bigl( x^{N-1}\varphi ( x ) \bigr) ^{\prime} ( y-x ) + ( y-x ) ^{2}\sum_{k=1}^{N-1}y^{k-1} \bigl( x^{N-k-1}\varphi ( x ) \bigr) ^{\prime}. $$

By using this inequality with the convex function $\varphi ( x ) =x^{\alpha-N+1}$, $x\geq0$, $\alpha\geq N$, we obtain

$$ y^{\alpha}-x^{\alpha}\geq\alpha x^{\alpha-1} ( y-x ) + ( y-x ) ^{2}\sum_{k=1}^{N-1} ( \alpha-k ) y^{k-1}x^{\alpha -k-1}. $$

By now choosing $x=z$, integrating over Ω, and using the fact that $\int_{\Omega} ( y-z )\,d\mu ( y ) =0$ we obtain (2.5). For the reversed inequality we use the concave function $\varphi ( x ) =x^{\alpha-N+1}$, $( N-1 ) <\alpha\leq N$, and all inequalities above reverse. For $\alpha=N$ we get an equality, so the proof is complete. □

Corollary 1

Let $x_{i}\geq0$, $\alpha_{i}\geq0$, $i=1,2,\ldots,m$, $\sum_{i=1}^{m}\alpha_{i}=1$, and $\overline{x}=\sum_{i=1}^{m}\alpha _{i}x_{i}$. Then, for $N=2,3,\ldots$ ,

$$ \sum_{i=1}^{m}\alpha_{i}x_{i}^{\alpha}- \overline{x}^{\alpha}\geq \sum_{i=1}^{m} \alpha_{i} ( x_{i}-\overline{x} ) ^{2}\sum _{k=1}^{N-1} ( \alpha-k ) x_{i}^{k-1} \overline {x}^{\alpha -k-1} $$

(2.6)

holds for any $\alpha\geq N$. Moreover, for $N-1<\alpha\leq1$ (2.6) holds in the reversed direction. In particular, for $\alpha =N$, (2.6) reduces to an equality.

Our final main result reads as follows.

Theorem 3

Let $0< A\leq\infty$ and let $\phi: ( 0,A ] \rightarrow \mathbb{R}$ have a Taylor expansion $\phi ( x ) =\sum_{n=0}^{\infty }a_{n}x^{n}$, on $( 0,A ]$. If μ is a probability measure on $( 0,A ]$ and $z=\int_{0}^{A}x\,d\mu ( x ) >0$, then

$$ \int_{\Omega}\phi ( x )\,d\mu-\phi ( z ) =\sum _{n=2}^{\infty}a_{n} \int_{0}^{A} ( x-z ) ^{2}\sum _{k=1}^{n-1} ( n-k ) x^{k-1}z^{n-k-1}\,d\mu. $$

(2.7)

Proof

We note that

$$ \int_{0}^{A}\phi ( x )\,d\mu-\phi ( z ) = \int_{0}^{A}\sum_{n=0}^{\infty}a_{n} \bigl( x^{n}-z^{n} \bigr)\,d\mu =\sum _{n=0}^{\infty}a_{n} \int_{0}^{A} \bigl( x^{n}-z^{n} \bigr)\,d\mu. $$

Obviously, $\int_{0}^{A} ( x^{n}-z^{n} )\,d\mu=0$, for $n=0,1$, and hence (2.7) follows from the equality cases in (2.5) in Theorem 2, i.e. when $\alpha=N=2,3,\ldots$ .

The proof is complete. □

Corollary 2

Let $0< A\leq\infty$ and let $\phi: [ 0,A )$ have a Taylor expansion $\phi ( x ) =\sum_{n=0}^{\infty}a_{n}x^{n}$, on $[ 0,A )$. If $\overline{x}=\sum_{i=1}^{m}\alpha_{i}x_{i}$, $\sum_{i=1}^{m}\alpha_{i}=1$, $0\leq \alpha_{i}\leq1$, $0\leq x_{i}\leq A$, $i=1,2,\ldots,m$, then

$$ J=\sum_{i=1}^{m}\alpha_{i}\phi ( x_{i} ) -\phi ( \overline{x} ) =\sum _{n=2}^{\infty}a_{n} \Biggl( \sum _{i=1}^{m}\alpha _{i}x_{i}^{2}-\overline{x}^{2} \Biggr) \sum_{k=1}^{n-1} ( n-k ) x^{k-1}\overline{x}^{n-k-1}. $$

Corollary 3

Let $0< a< b<\infty$, and μ be a probability measure on $( a,b )$. Then we have the following estimate of the Jensen gap $J_{N}:=\int_{a}^{b}x^{N}\,d\mu- ( \int_{a}^{b}x\,d\mu ) ^{N}$, $N=2,3,\ldots$ :

$$ \frac{N ( N-1 ) }{2}a^{N-2}J_{2}\leq J_{N}\leq \frac{N ( N-1 ) }{2}b^{N-2}J_{2}. $$

(2.8)

Proof

We use Theorem 2 with $\alpha=N$ and find that

$$ J_{N}= \int_{a}^{b} ( x-z ) ^{2}\sum _{k=1}^{N-1} ( N-k ) x^{k-1}z^{N-k-1}\,d\mu. $$

We note that if $a< x< b$, then $a< z< b$ so that $a^{N-2}\leq x^{k-1}z^{N-k-1}\leq b^{N-2}$. Moreover, $\sum_{k=1}^{N-1} ( N-k ) =\frac{N ( N-1 ) }{2}$ and

$$ \int_{a}^{b} ( x-z ) ^{2}\,d\mu= \int_{a}^{b}x^{2}\,d\mu- \biggl( \int_{a}^{b}x\,d\mu \biggr) ^{2}=J_{2}, $$

so (2.8) is proved. □

Remark 3

For the case $N=2$ both inequalities in (2.8) reduce to equalities. Moreover, for the discrete case we have: If $0< a< x_{i}< b$, $\alpha_{i}\geq0$, $i=1,2,\ldots,m$, $\sum_{i=1}^{m}\alpha_{i}=1$, $\overline{x}=\sum_{i=1}^{m}\alpha_{i}x_{i}$, then, for $N=2,3,\ldots$ ,

$$\begin{aligned} &\frac{N ( N-1 ) }{2}a^{N-2} \Biggl( \sum_{i=1}^{m}a_{i}x_{i}^{2}-\overline{x}^{2} \Biggr) \\ &\quad\leq\sum_{i=1}^{m}a_{i}x_{i}^{N}- \overline{x}^{N}\leq\frac{N ( N-1 ) }{2}b^{N-2} \Biggl( \sum _{i=1}^{m}a_{i}x_{i}^{2}- \overline{x}^{2} \Biggr) . \end{aligned}$$

(2.9)

3 Final remarks and examples

In this section we present some recent interesting results of Dragomir [11] and Walker [10]. Moreover, we point out the corresponding special cases of our results and compare these results with those of [11] and [10].

Example 4

In Dragomir’s paper [11], Theorem 2, it was proved that for

$$ \phi ( x ) =\sum_{n=0}^{\infty}a_{n}x^{n},\quad a_{n}\geq0, $$

(3.1)

which converges on $0< x< R\leq\infty$, the following lower bound of the Jensen gap holds:

$$\begin{aligned} &\int_{\Omega}\phi\circ f\,d\mu-\phi \biggl( \int_{\Omega}f\,d\mu \biggr) \\ &\quad\geq\frac{1}{2} \biggl[ \int_{\Omega}f^{2}\,d\mu- \biggl( \int_{\Omega }f\,d\mu \biggr) ^{2} \biggr] \frac{\phi^{\prime} ( \int_{\Omega}f\,d\mu ) -\phi^{\prime} ( 0 ) }{\int_{\Omega}f\,d\mu}, \end{aligned}$$

(3.2)

when $( \Omega,\mu )$ is a probability measure space, $f\geq 0$, and f, $f^{2}$, and $\phi\circ f$ are integrable on Ω and $\int_{\Omega}f\,d\mu>0$.

Example 5

In Theorem 1 we proved that for convex increasing functions we get the inequalities

$$\begin{aligned} &\int_{\Omega}\phi\circ f\,d\mu-\phi \biggl( \int_{\Omega}f\,d\mu \biggr) \\ &\quad\geq \biggl[ \int_{\Omega}f^{2}\,d\mu- \biggl( \int_{\Omega}f\,d\mu \biggr) ^{2} \biggr] \biggl( \frac{\phi ( \int_{\Omega}f\,d\mu ) -\phi ( 0 ) }{\int_{\Omega}f\,d\mu} \biggr) ^{\prime}\geq0. \end{aligned}$$

(3.3)

A function that satisfies (3.1) is convex increasing and therefore Theorem 1 holds, which means that we get the inequalities in (3.3).

Remark 4

It is easily computed that when ϕ is of the form (3.1), then

$$ \frac{1}{2}\frac{\phi^{\prime} ( \int_{\Omega}f\,d\mu ) -\phi ^{\prime} ( 0 ) }{\int_{\Omega}f\,d\mu}\leq \biggl( \frac{\phi ( \int_{\Omega}f\,d\mu ) -\phi ( 0 ) }{\int_{\Omega }f\,d\mu} \biggr) ^{\prime} $$

(3.4)

holds, and from this we conclude that our bound in (3.3), when (3.1) is satisfied, is stronger than Dragomir’s (3.2). Indeed,

$$ \frac{1}{2}\frac{\phi^{\prime} ( z ) -\phi^{\prime} ( 0 ) }{z}=\sum_{n=0}^{\infty} \frac{1}{2} ( n+2 ) a_{n+2}z^{n} $$

and

$$ \biggl( \frac{\phi ( \int_{\Omega}f\,d\mu ) -\phi ( 0 ) }{\int_{\Omega}f\,d\mu} \biggr) ^{\prime}=\sum _{n=0}^{\infty} ( n+1 ) a_{n+2}z^{n}, $$

and our claim is obvious.

Example 6

In Theorem 3.1 in Walker’s paper [10], a lower bound for the Jensen gap is given for a function ϕ that satisfies (3.1):

$$ \int_{\Omega}\phi ( s )\,d\mu ( s ) -\phi \biggl( \int_{\Omega}\,d\mu ( s ) \biggr) \geq\mu ( 1,R ) \tau \frac{1}{2}\sum_{n=2}^{\infty}a_{n}n ( n-1 ) $$

where

$$ \tau= \int_{\Omega}s^{2}\,d\mu_{2} ( s ) - \biggl( \int_{\Omega }s\,d\mu_{2} ( s ) \biggr) ^{2} $$

when μ is a probability measure defined on $\Omega= ( 0,R )$ and $\mu_{2}$ is μ restricted and normalized to $( 1,R )$.

More generally, in Section 4 in [10], $\mu ( 1,R )$ was replaced by $\mu ( a,R )$ and we have

$$ \int_{\Omega}\phi ( s )\,d\mu ( s ) -\phi \biggl( \int_{\Omega}\,d\mu ( s ) \biggr) \geq\mu ( a,R ) \tau \frac{1}{2}\sum_{n=2}^{\infty}a^{n}a_{n}n ( n-1 ) , $$

(3.5)

where

$$ \tau= \int_{\Omega}s^{2}\,d\mu_{a} ( s ) - \biggl( \int_{\Omega }s\,d\mu_{a} ( s ) \biggr) ^{2}, $$

when $\mu_{a}$ is μ restricted and normalized to $\Omega= ( a,R )$.

From Corollary 3 and Remark 3 we easily get the following.

Example 7

Let $0< A\leq\infty$ and let $\phi: ( 0,A ] \rightarrow \mathbb{R}$ have Taylor expansion $\phi ( x ) =\sum_{n=0}^{\infty }a_{n}x^{n}$, $a_{n}\geq0$, $n=2,3,\ldots$ , on $( 0,A ]$. If μ is a probability measure on $( 0,A ]$, $0\leq a< b\leq A$, and $z=\int_{0}^{A}x\,d\mu ( x ) >0$, then

$$ \sum_{n=2}^{\infty}a_{n} \frac{n ( n-1 ) }{2}a^{n-2}J_{2}\leq J ( \phi,\mu ) \leq\sum _{n=2}^{\infty}a_{n} \frac{n ( n-1 ) }{2}b^{n-2}J_{2}. $$

(3.6)

Moreover, for the discrete case we have: If $0< a< x_{i}< b$, $\alpha _{i}\geq 0$, $i=1,2,\ldots,m$, $\sum_{i=1}^{m}a_{i}=1$, $\overline{x}=\sum_{i=1}^{m}\alpha_{i}x_{i}$, then, for $n=2,3,\ldots$ ,

$$\begin{aligned} &\sum_{n=2}^{\infty}a_{n} \frac{n ( n-1 ) }{2}a^{n-2} \Biggl( \sum_{i=1}^{m} \alpha_{i}x_{i}^{2}-\overline{x}^{2} \Biggr) \\ &\quad\leq\sum_{i=1}^{m}\alpha_{i} \bigl( \phi ( x_{i} ) -\phi ( \overline{x} ) \bigr) \leq\sum _{n=2}^{\infty}a_{n}\frac{n ( n-1 ) }{2}b^{n-2} \Biggl( \sum_{i=1}^{m}\alpha _{i}x_{i}^{2}-\overline{x}^{2} \Biggr) . \end{aligned}$$

Remark 5

The lower bound in (3.5) coincides with that in (3.6) when $a=1$. The lower bound in (3.6) is better than that in (3.5) when $a<1$, but Walker’s bound (3.5) is better than (3.6) for $a>1$. It seems not to be possible to derive an upper bound like that in (3.5) by using the method in [10].