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A monotonic refinement of Levinson’s inequality
Journal of Inequalities and Applications volume 2015, Article number: 162 (2015)
Abstract
In this paper we give a monotonic refinement of the probabilistic version of Levinson’s inequality based on the monotonic refinement of Jensen’s inequality obtained by Cho et al. (Panam. Math. J. 12:4350, 2002).
Introduction
Levinson’s inequality and its converse are summarized in the following result taken from Bullen [1].
Theorem 1.1
(a) If \(f:[a,b]\rightarrow\mathbb{R}\) is 3convex and \(p_{i}\), \(x_{i}\), \(y_{i}\), \(i=1,2,\ldots, n\), are such that \(p_{i}>0\), \(\sum_{i=1}^{n}p_{i}=1\), \(a\leq x_{i},y_{i}\leq b\),
and
for some \(c\in[a,b]\), then
where \(\overline{x}=\sum_{i=1}^{n}p_{i}x_{i}\) and \(\overline{y}=\sum_{i=1}^{n}p_{i}y_{i}\) denote the weighted arithmetic means.
(b) If for a continuous function f inequality (3) holds for all n, all \(c\in[a,b]\), all 2n distinct points \(x_{i}, y_{i} \in [a,b]\) satisfying (1) and (2) and all weights \(p_{i}>0\) such that \(\sum_{i=1}^{n}p_{i}=1\), then f is 3convex.
Levinson [2] originally proved the inequality for functions \(f: (0,2c) \to\mathbb{R}\) such that \(f''' \geq0\). Popoviciu [3] showed that the assumption of nonnegativity of the third derivative can be weakened to 3convexity of f. Bullen [1] gave another proof of the inequality (rescaled to a general interval \([a,b]\)) as well as its converse given in part (b) of Theorem 1.1. Pečarić and Raşa [4] extended the inequality by using the method of index set functions; in the process they weakened assumption (1) and obtained a monotonic refinement of the inequality.
The above version of the inequality assumes that the sequences \(x_{i}\)’s and \(y_{i}\)’s are symmetrically distributed around the point c. Mercer [5] made a significant improvement by replacing this condition of symmetric distribution with the weaker one that the variances of the two sequences are equal.
Theorem 1.2
If \(f:[a,b]\rightarrow\mathbb{R}\) satisfies \(f''' \geq0\) and \(p_{i}\), \(x_{i}\), \(y_{i}\), \(i = 1,2,\ldots,n\), are such that \(p_{i}>0\), \(\sum_{i=1}^{n}p_{i}=1\), \(a\leqslant x_{i},y_{i}\leqslant b\), (1) holds and
then (3) holds.
Witkowski [6] extended Mercer’s result to 3convex functions and a more general probabilistic setting. Baloch et al. [7] showed that inequality (3) holds for a larger class of functions they introduced and called 3convex functions at a point.
Definition 1.3
Let I be an interval in \(\mathbb{R}\) and \(c\in I\). A function \(f: I\to\mathbb{R} \) is said to be 3convex at point c if there exists a constant A such that the function \(F(s) = f(s)  \frac{A}{2} s^{2} \) is concave on \(I\cap(\infty,c]\) and convex on \(I \cap[c ,\infty)\).
Baloch et al. [7] also proved the converse of the inequality, i.e., 3convex functions at a point are the largest class of functions for which Levinson’s inequality holds under the equal variances assumption. Probabilistic version of Levinson’s inequality and its converse are summarized in the following result taken from Pečarić et al. [8].
Theorem 1.4
(a) Let \(f: [a,b] \to\mathbb{R}\) be 3convex at point c and \(X:\Omega\to[a,c]\) and \(Y:\Omega\to[c,b]\) be two random variables such that \(\operatorname{Var}(X) = \operatorname{Var}(Y)\). Then
(b) Let \(f:[a,b] \to\mathbb{R}\) be continuous and \(c\in(a,b)\) fixed. Suppose that inequality (4) holds for all discrete random variables X and Y taking two values \(x_{1}, x_{2} \in[a,c]\) and \(y_{1}, y_{2} \in[c,b]\), respectively, each with probability \(\frac{1}{2}\) and such that \(\operatorname{Var}(X) = \operatorname{Var}(Y)\) (i.e. \(x_{2} x_{1} = y_{2}  y_{1}\)). Then f is 3convex at c.
Remark 1.5
Results in [8] were stated for f defined on an arbitrary interval I. In that case, the finiteness of \(\operatorname{Var}(X) = \operatorname{Var}(Y)\), \(\mathbb{E}[f(X)]\) and \(\mathbb{E}[f(Y)]\) needs to be assumed. For simplicity, in this paper we will work with the closed interval \([a,b]\) since in this case the function f and all random variables are bounded and the aforementioned finiteness assumptions are satisfied.
If X and Y are discrete random variables taking values \(x_{i}\) and \(y_{i}\), respectively, with probabilities \(p_{i}\), then Theorem 1.4(a) gives Theorem 1.2. In [8] it was proven that a function defined on an interval is 3convex if and only if it is 3convex at every point of the interval. Therefore, the converse stated in Theorem 1.4(b) strengthens the converse stated in Theorem 1.1(b).
Theorem 1.4 shows that 3convex functions at a point are characterized by Levinson’s inequality in a similar way that convex functions are characterized by Jensen’s inequality. Cho et al. [9] constructed two mappings connected to Jensen’s inequality and proved their monotonicity and convexity properties. Throughout the rest of the paper Ω denotes a measurable space with a finite measure μ, and we assume all mappings to be measurable. Further, \(\mathbb{E}[\cdot]\) and \(\operatorname{Var}(\cdot)\) denote the expectation and variance operators with respect to the probability measure \(\frac{1}{\mu(\Omega)} \mu\), i.e., for \(z:\Omega\to \mathbb{R}\),
The following is a result from [9].
Theorem 1.6
Let \(f:[a,b]\to\mathbb{R}\) be convex, \(x: \Omega\to[a,b]\) and \(H,V: [0,1]\to\mathbb{R}\) the mappings
and
Then:

(a)
the mappings H and V are convex on \([0,1]\),

(b)
the mapping H is nondecreasing on \([0,1]\), while the mapping V is nonincreasing on \([0,\frac{1}{2} ]\) and nondecreasing on \([\frac{1}{2} ,1]\),

(c)
the following equalities hold:
$$\begin{aligned} & \inf_{t\in[0,1]} H(t) = H(0) = f\bigl(\mathbb{E}[x]\bigr), \\ & \sup_{t\in[0,1]} H(t) = H(1) = \mathbb{E}\bigl[f(x)\bigr], \\ & \inf_{t\in[0,1]} V(t) = V\biggl(\frac{1}{2} \biggr) = \frac{1}{\mu(\Omega)^{2}} \int_{\Omega} \int_{\Omega} f \biggl(\frac{ x(s) + x(u)}{2} \biggr)\,d\mu(s)\,d\mu(u), \\ & \sup_{t\in[0,1]} V(t) = V(0)=V(1) = \mathbb{E}\bigl[f(x)\bigr], \end{aligned}$$ 
(d)
the following inequality holds for all \(t\in[0,1]\):
$$V(t) \geq\max\bigl\{ H(t), H(1t)\bigr\} . $$
Remark 1.7
Theorem 1.6 was proven in [9] for the case when Ω is an interval in \(\mathbb{R}\) and μ is a measure with density, i.e., \(d\mu(s) = p(s)\,ds\). But, from the proofs given there, it is obvious that the statements hold under the more general setting given here.
If we denote \(x_{(t)} (s) = tx(s) + (1t) \mathbb{E}[x]\), then \(H(t) = \mathbb{E}[f(x_{(t)})]\). As t ranges from 0 to 1, the function (i.e., random variable) \(x_{(t)}\) ranges from the constant \(\mathbb{E}[x]\) to the function x itself. In the process the expectation \(\mathbb{E}[f(x_{(t)})]\) increases by the monotonicity property from Theorem 1.6(b). Therefore, for \(0\leq s\leq t \leq1\), the following monotonic refinement of Jensen’s inequality holds:
Furthermore, if \(x'\) and \(x''\) are two independent identically distributed ‘copies’ of x on the product space \(\Omega\times\Omega \), then \(V(t) = \mathbb{E}[f(\tilde{x}_{(t)})]\), where \(\tilde{x}_{(t)} = tx' + (1t) x''\), and Theorem 1.6(d) can be interpreted as \(\mathbb{E}[f(x_{(t)})] \leq \mathbb{E}[f(\tilde{x}_{(t)})]\).
In this paper we will construct the corresponding two mappings in connection with Levinson’s inequality and show their monotonicity and convexity properties.
Main results
The following is our main result.
Theorem 2.1
Let \(f: [a,b] \to\mathbb{R}\) be 3convex at point c, \(x:\Omega\to [a,c]\) and \(y:\Omega\to[c,b]\) such that \(\operatorname{Var}(x)= \operatorname{Var}(y)\) and \(H,V: [0,1] \to\mathbb{R}\) the mappings
and
Then:

(a)
the mappings H and V are convex on \([0,1]\),

(b)
the mapping H is nondecreasing on \([0,1]\), while the mapping V is nonincreasing on \([0,\frac{1}{2} ]\) and nondecreasing on \([\frac{1}{2} ,1]\),

(c)
the following equalities hold:
$$\begin{aligned}& \inf_{t\in[0,1]} H(t) = H(0) = f\bigl(\mathbb{E}[y]\bigr)  f\bigl(\mathbb{E}[x] \bigr), \\& \sup_{t\in[0,1]} H(t) = H(1) = \mathbb{E}\bigl[f(y)\bigr]  \mathbb{E}\bigl[f(x) \bigr], \\& \begin{aligned}[b] \inf_{t\in[0,1]} V(t) ={}& V\biggl( \frac{1}{2} \biggr) = \frac{1}{\mu(\Omega)^{2}} \int_{\Omega} \int _{\Omega} \biggl[ f \biggl(\frac{ y(s) + y(u)}{2} \biggr) \\ & {}  f \biggl(\frac{ x(s) + x(u)}{2} \biggr) \biggr]\,d\mu(s)\,d\mu(u), \end{aligned} \\& \sup_{t\in[0,1]} V(t) = V(0) = V(1) = \mathbb{E}\bigl[f(y)\bigr]  \mathbb{E}\bigl[f(x)\bigr], \end{aligned}$$ 
(d)
the following inequality holds for all \(t\in[0,1]\):
$$V(t) \geq\max\bigl\{ H(t), H(1t)\bigr\} . $$
Proof
Let the constant A be as in Definition 1.3, i.e., such that the function \(F(s) = f(s)  \frac{A}{2} s^{2}\) is concave on \([a,c]\) and convex on \([c,b]\).
Since the function y takes values in \([c,b]\), so does the function \(y_{(t)} = ty + (1t) \mathbb{E}[y]\) for every \(t\in[0,1]\). Furthermore, since the function F is convex on \([c,b]\), by Theorem 1.6 the mapping
is convex and nondecreasing on \([0,1]\). We have
Similarly, the function \(x_{(t)} = tx + (1t)\mathbb{E}[x]\) takes values in \([a,c]\) for every \(t\in[0,1]\) and −F is convex on \([a,c]\), so by Theorem 1.6 the mapping
is convex and nondecreasing on \([0,1]\), and we have
Let us also denote the (constant) mapping \(H_{3} (t) = \frac{A}{2} ( \mathbb{E}^{2}[y]  \mathbb{E}^{2}[x] )\). All three of the mappings \(H_{i}\), \(i=1,2,3\), are convex and nondecreasing and, therefore, so is their sum. Since \(\operatorname{Var}(x)=\operatorname{Var}(y)\), we have \(H= H_{1} + H_{2} + H_{3}\), and this proves the convexity and monotonicity properties of H from parts (a) and (b), while the first two equalities in (c) follow by simple calculation.
As for the mapping V, first of all, it is easy to see that \(V(t) = V(1t)\) for all \(t\in[0,1]\), that is, V is symmetric with respect to \(t=\frac{1}{2}\). Next, since y takes values in \([c,b]\) and F is convex on that interval, by Theorem 1.6 the mapping
is convex on \([0,1]\) and nondecreasing on \([\frac{1}{2}, 1]\). We have
Similarly, since x takes values in \([a,c]\) and −F is convex on that interval, by Theorem 1.6 the mapping
is convex on \([0,1]\) and nondecreasing on \([\frac{1}{2}, 1]\) and we have
Let us also denote the (constant) mapping \(V_{3}(t) = \frac{A}{2} (\mathbb{E}[y^{2}]  \mathbb{E}[x^{2}] )\). All three of the mappings \(V_{i}\), \(i=1,2,3\), are convex and nondecreasing on \([\frac{1}{2},1]\) and, therefore, so is their sum. Since \(\operatorname{Var}(x)=\operatorname{Var}(y)\), we have \(V= V_{1} + V_{2} + V_{3}\). Furthermore, since V is symmetric around \(t=\frac{1}{2}\), it follows that it is nonincreasing on \([0,\frac{1}{2}]\), its minimum is attained at \(t=\frac{1}{2}\) and its maximum is attained at \(t=0\) and \(t=1\). This proves the convexity and monotonicity properties of V.
Finally, as for part (d), since V is symmetric around \(t=\frac{1}{2}\) and H is nondecreasing, it is enough to prove that \(V(t) \geq H(t)\) for \(t\in[\frac{1}{2},1]\). This inequality holds since \(V_{1} (t) \geq H_{1} (t)\) and \(V_{2} (t) \geq H_{2} (t)\) by Theorem 1.6(d) and \(V_{3}(t) = H_{3}(t)\) since \(\operatorname{Var}(x) =\operatorname{Var}(y)\) and this finishes the proof. □
A monotonic refinement of Levinson’s inequality (4) based on Theorem 2.1 is the following: if \(x_{(t)}\) and \(y_{(t)}\) for \(t\in[0,1]\) are as in the proof of Theorem 2.1, then \(H(t) =\mathbb{E}[f(y_{(t)})]  \mathbb{E}[f(x_{(t)})]\) and for \(0\leq s \leq t \leq1\) it holds
Remark 2.2
The convexity and monotonicity property of the mapping H in the case when x and y are two discrete random variables taking values \(x_{i}\) and \(y_{i}\), respectively, with probabilities \(p_{i}\), \(i=1,\ldots,n\), was proven in [7].
Remark 2.3
The assumption of equal variances in Theorem 2.1 can be weakened. If we denote \(B = A ( \operatorname{Var}(y)  \operatorname{Var}(x) )\), then the assumption \(\operatorname{Var}(x)=\operatorname{Var}(y)\) can be relaxed to \(B\geq0\). Indeed, what we have shown in the proof of Theorem 2.1 is that
where \(H_{4} (t) = \frac{1}{2} B t^{2}\) and \(V_{4}(t) = Bt(t1)\). For \(B\geq0\) the mapping \(H_{4}\) is convex and nondecreasing, while the mapping \(V_{4}\) is convex, symmetric around \(t=\frac{1}{2}\) and nondecreasing on \([\frac{1}{2},1]\). Therefore, the convexity and monotonicity properties of H and V are preserved.
Furthermore, \(V_{3} (t)  H_{3} (t) = \frac{1}{2} B\), so from \(V_{1}(t) \geq H_{1}(t)\), \(V_{2} (t) \geq H_{2} (t)\) and \(V_{3}(t)+V_{4}(t)  H_{3}(t)  H_{4}(t) = \frac{1}{2} B (1t)^{2} \geq0 \) it follows that \(V(t) \geq H(t)\), i.e., part (d) also holds.
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Acknowledgements
This work has been fully supported by Croatian Science Foundation under the project 5435.
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Jakšetić, J., Pečarić, J. & Praljak, M. A monotonic refinement of Levinson’s inequality. J Inequal Appl 2015, 162 (2015). https://doi.org/10.1186/s1366001506828
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MSC
 26D15
Keywords
 Levinson’s inequality
 Jensen’s inequality
 3convexity at a point