On an Extension of Variant of Jensen’s Inequality for Convex Function

Liang-Cheng Wang^{∗}

Abstract

In this paper, we extend a variant of Jensen’s inequality for convex function.

2000 Mathematics Subject Classification. Primary 26D015; Secondary 26D07.

Key Words. variant, Jensen’s inequality, convex function

1 Introduction and Main results

Let f be a given convex function defined on a non-empty interval lI ⊂ lR. And let x_{i} ∈ lI, ω_{i} > 0(i = 1, 2, ⋯, n; n ≥ 2), ∑_{i=1}^{n} ω_{i} = 1. It is well-known that the following Jensen’s inequality [1,2] holds

f (∑_{i=1}^{n} ω_{i}x_{i}) ≤ ∑_{i=1}^{n} ω_{i}f (x_{i}) .  (1.1)

Under the above hypotheses and x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n}, A. M. Mercer [3] gave a variant of inequality (1.1):

f (x_{1} + x_{n} − ∑_{i=1}^{n} ω_{i}x_{i}) ≤ f (x_{1}) + f (x_{n}) − ∑_{i=1}^{n} ω_{i}f (x_{i}) .  (1.2)

In this note, our purpose is to prove the following extension of inequality (1.2):

Theorem 1.1. Let f be a convex function on lI containing x_{i}, and let t_{i} ∈ lR/{0}(i = 1, 2, ⋯, n; n ≥ 2). If {x_{1}, x_{2} ⋯, x_{n}} be monotonic sequence, and

0 ≤ ∑_{i=k}^{n} t_{i} ≤ ∑_{i=1}^{n} t_{i} = T_{n}, T_{n} > 0, (k = 2, 3, ⋯, n)  (1.3)

∗Department of Mathematics, Daxian Teacher’s College, Dazhou 635000, Sichuan Province, China. This author is partially supported by the Key Research Foundation of the Daxian Teacher’s College under Grant 2003-81. E-mail: wangliangcheng@163.com.

or equivalent form of (1.3)

0 ≤ ∑_{i=1}^{m} t_{i} ≤ ∑_{i=1}^{n} t_{i} = T_{n}, T_{n} > 0(m = 1, 2, ⋯, n − 1),  (1.4)

then we have

f (\frac{1}{T_{n}} ∑_{i=1}^{n} t_{i}x_{i}) ≤ \frac{1}{T_{n}} ∑_{i=1}^{n} t_{i}f (x_{i}) .  (1.5)

Corollary 1.2. Let f be a convex function on lI containing x_{i}, t_{i} ∈ lR/{0}(i = 1, 2, ⋯, n; n ≥ 2). And let x_{1} = min{x_{i}|1 ≤ i ≤ n}, x_{n} = max{x_{i}|1 ≤ i ≤ n}. If

∑_{i=2}^{n−1} |t_{i}| ≤ min{t_{1}, t_{n}}, ∑_{i=1}^{n} t_{i} = T_{n} > 0, (∑_{i=2}^{2−1} |t_{i}| = 0) ,  (1.6)

then we also obtain (1.5).

Remark If we choose \frac{t_{i}}{T_{n}} = 1 − ω_{i}(i = 1, n) and \frac{t_{i}}{T_{n}} = −ω_{i}(i = 2, 3, ⋯, n − 1) and x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n} in Corollary 1.2, then inequality (1.5) reduces to inequality (1.2).

Towards proving Theorem 1.1 we shall need the following lemma [4]:

Lemma 1.3. Let f be a convex function on a closed interval [a, b](a < b). Then f−^{′} and f+^{′} exist in (a, b). For any x_{1}, x_{2} ∈ (a, b), x_{1} < x_{2}, we have

f−^{′}(x_{1}) ≤ f+^{′}(x_{1}) ≤ f−^{′}(x_{2}) ≤ f+^{′}(x_{2});  (1.7)

For any x_{1}, x_{2} ∈ [a, b], x_{1} < x_{2}, then ∃ξ ∈ (x_{1}, x_{2}) such that

f−^{′}(ξ) ≤ \frac{f(x_{2}) − f(x_{1})}{x_{2} − x_{1}} ≤ f+^{′}(ξ).  (1.8)

2 The Proof of Theorem and Corollary

Proof of Theorem 1.1. Let x_{0} = T_{n}^{−1} ∑_{i=1}^{n} t_{i}x_{i}. We write

T^{(k)} = T_{n}^{−1} ∑_{i=k}^{n} t_{i} (k = 2, 3, ⋯, n)

and

T_{(m)} = T_{n}^{−1} ∑_{i=1}^{m} t_{i} (m = 1, 2, ⋯, n − 1).

Using (1.3) and (1.4), we get 0 ≤ T^{(k)} ≤ 1(k = 2, 3, ⋯, n) and 0 ≤ T_{(m)} ≤ 1(m = 1, 2, ⋯, n − 1).

1. When x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n}, we have

x_{1} ≤ x_{1}T^{−1} ∑_{i=1}^{n} t_{i} + ∑_{k=2}^{n} T^{(k)} (x_{k} − x_{k−1}) = x_{0} = ∑_{m=1}^{n−1} T_{(m)} (x_{m} − x_{m+1}) + x_{n}T^{−1} ∑_{i=1}^{n} t_{i} ≤ x_{n}.

(2.1)

Case 1. x_{1} = x_{0}. From (2.1), T^{(k)} ≥ 0 and x_{k} − x_{k−1} ≥ 0 we have T^{(k)} (x_{k} − x_{k−1}) = 0 i. e. T^{(k)} = 0 or x_{k} = x_{k−1}, (k = 2, 3, ⋯, n). Then we get

T^{(k)} (f(x_{k}) − f(x_{k−1})) = 0,  (k = 2, 3, ⋯, n).  (2.2)

Using (2.2), we have

T_{n}^{−1} ∑_{i=1}^{n} t_{i}f(x_{i}) − f (T_{n}^{−1} ∑_{i=1}^{n} t_{i}x_{i})

= f(x_{1})T^{−1} ∑_{i=1}^{n} t_{i} + ∑_{k=2}^{n} T^{(k)} (f(x_{k}) − f(x_{k−1})) − f(x_{0})
= f(x_{1}) − f(x_{0}) = 0,

which is equality in (1.5).

Case 2. x_{n} = x_{0}. By same arguments of proof for case 1, we can also get equality in (1.5).

Case 3. x_{p} = x_{0}(2 ≤ p ≤ n − 1). Using lemma 1.3 we have

T_{n}^{−1} ∑_{i=1}^{n} t_{i}f(x_{i}) − f (T^{−1} ∑_{i=1}^{n} t_{i}x_{i})

= f(x_{1}) − f(x_{p}) + ∑_{k=2}^{n} T^{(k)} (f(x_{k}) − f(x_{k−1}))
= ∑_{k=2}^{p} (T^{(k)} − 1) (f(x_{k}) − f(x_{k−1})) + ∑_{k=p+1}^{n} T^{(k)} (f(x_{k}) − f(x_{k−1}))
≥ ∑_{k=2}^{p} (T^{(k)} − 1) (x_{k} − x_{k−1}) f+^{′} (ξ_{k})

+ ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k}) , where ξ_{k} ∈ (x_{k−1}, x_{k}), by (1.8),

≥ ∑_{k=2}^{p} (T^{(k)} − 1) (x_{k} − x_{k−1}) f+^{′} (ξ_{p}) ,  by (1.7),

+ ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k})

= [∑_{k=2}^{p} T^{(k)} (x_{k} − x_{k−1}) − (x_{0} − x_{1})] f+^{′} (ξ_{p}) + ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k})
= [∑_{k=2}^{p} T^{(k)} (x_{k} − x_{k−1}) − ∑_{k=2}^{n} T^{(k)} (x_{k} − x_{k−1})] f+^{′} (ξ_{p}) ,  by (2.1),

+ ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k})

= ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) (f−^{′} (ξ_{k}) − f+^{′} (ξ_{p})) ≥ 0,  by (1.7),

which is (1.5). Case 4. x_{p−1} < x_{0} < x_{p}(2 ≤ p ≤ n). When 2 ≤ p ≤ n − 1, using lemma 1.3 we have

\begin{matrix}T_{n}^{−1} ∑_{i=1}^{n} t_{i}f(x_{i}) − f (T^{−1} ∑_{i=1}^{n} t_{i}x_{i}) \\ = f(x_{1}) − f(x_{0}) + ∑_{k=2}^{n} T^{(k)} (f(x_{k}) − f(x_{k−1})) \\ = ∑_{k=2}^{p−1} (T^{(k)} − 1) (f(x_{k}) − f(x_{k−1})) + (T^{(p)} − 1) (f(x_{0}) − f(x_{p−1})) \\ +T^{(p)} (f(x_{p}) − f(x_{0})) + ∑_{k=p+1}^{n} T^{(k)} (f(x_{k}) − f(x_{k−1})) \\ ≥ ∑_{k=2}^{p−1} (T^{(k)} − 1) (x_{k} − x_{k−1}) f+^{′} (ξ_{k}) ,  ξ_{k} ∈ (x_{k} − x_{k−1}) , \\ + (T^{(p)} − 1) (x_{0} − x_{p−1}) f+^{′} (ξ_{p}) ,  ξ_{p} ∈ (x_{0}, x_{p−1}) , \\ +T^{(p)} (x_{p} − x_{0}) f−^{′} (ξp^{′}) ,  ξp^{′} ∈ (x_{p}, x_{0}) , \\ + ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k}) ,  ξ_{k} ∈ (x_{k} − x_{k−1}) ,  by (1.8), \\ ≥ ∑_{k=2}^{p−1} (T^{(k)} − 1) (x_{k} − x_{k−1}) f+^{′} (ξ_{p}) + (T^{(p)} − 1) (x_{0} − x_{p−1}) f+^{′} (ξ_{p}) \\ +T^{(p)} (x_{p} − x_{0}) f+^{′} (ξ_{p}) + ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k}) ,  by (1.7), \\ = ∑_{k=2}^{p−1} T^{(k)} (x_{k} − x_{k−1}) f+^{′} (ξ_{p}) − (x_{0} − x_{1}) f+^{′} (ξ_{p}) + T^{(p)} (x_{0} − x_{p−1}) f+^{′} (ξ_{p}) \\ +T^{(p)} (x_{p} − x_{0}) f+^{′} (ξ_{p}) + ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k}) \\ = [∑_{k=2}^{p} T^{(k)} (x_{k} − x_{k−1}) − ∑_{k=2}^{n} T^{(k)} (x_{k} − x_{k−1})] f+^{′} (ξ_{p}) ,  by (2.1), \\ + ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) f−^{′} (ξ_{k}) \\ = ∑_{k=p+1}^{n} T^{(k)} (x_{k} − x_{k−1}) (f−^{′} (ξ_{k}) − f+^{′} (ξ_{p})) ≥ 0,  by (1.7), \end{matrix} (2.3)

which is (1.5). When p = n i. e. x_{n−1} < x_{0} < x_{n}, We agree

∑_{k=n+1}^{n} ⋯ = 0.  (2.4)

In (2.3), taking p = n and using (2.4), we also get (1.5).

2. When x_{1} ≥ x_{2} ≥ ⋯ ≥ x_{n}, we have

x_{1} ≥ x_{1}T^{−1} ∑_{i=1}^{n} t_{i} + ∑_{k=2}^{n} T^{(k)} (x_{k} − x_{k−1}) = x_{0} = ∑_{m=1}^{n−1} T_{(m)} (x_{m} − x_{m+1}) + x_{n}T^{−1} ∑_{i=1}^{n} t_{i} ≥ x_{n}

(2.5)

and

\begin{matrix}T_{n}^{−1} ∑_{i=1}^{n} t_{i}f (x_{i}) − f (T_{n}^{−1} ∑_{i=1}^{n} t_{i}x_{i}) \\ = ∑_{m=1}^{n−1} T_{(m)} (f(x_{m}) − f(x_{m+1})) + f(x_{n}) − f(x_{0}). \end{matrix}  (2.6)

Using (2.5), (2.6) and the same arguments of proof for x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n}, we also obtain (1.5).

The proof of Theorem 1.1 is completed.

Proof of Corollary 1.2. Case 1. n = 2. From (1.6), we have t_{1} > 0 and t_{n} = t_{2} > 0. By (1.1), we get (1.5).

Case 2. n > 2. Let {m_{2}, m_{3}, ⋯, m_{n−1}} be a permutation of {2, 3, ⋯, n − 1} so that x_{1} ≤ x_{m_{2}} ≤ ⋯ ≤ x_{m_{k}} ≤ x_{m_{k+1}} ≤ ⋯ ≤ x_{m_{n−1}} ≤ x_{n}. Using (1.6), we get

0 ≤ ∑_{i=k}^{n−1} t_{m_{i}} + t_{n} ≤ t_{1} + ∑_{i=2}^{n−1} t_{m_{i}} + t_{n} = T_{n}, T_{n} > 0, (k = 2, 3, ⋯, n − 1).  (2.7)

Using (2.7) and Theorem 1.1, we obtain Corollary 1.2.

The proof of Corollary 1.2 is completed.

References

[1] D.S. Mitrinović, J.E. pečarić and A.M. Fink, Classical and New Inequalities in Analysis,Kluwer Academic Publishers, Dordrecht, 1993.

[2] L.-C. Wang and X. Zhang, Inequalities Generated by a Chain of Jensen Inequality on Convex Function, J. Kodia Math. 2004. (In the press).

[3] A.M. Mercer, A Variant of Jensen’s Inequality, J. Inequalities in Pure and Appl. Math. 4(4), (2003).

[4] L.-C. Wang, Convex Functions and Their Inequalities, Sichuan University Press, Chengdu, China, 2001. (In Chinese).