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)