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Horváth Journal of Inequalities and Applications 2011, 2011:26 http://www.journalofinequalitiesandapplications.com/content/2011/1/26

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A parameter-dependent refinement of the discrete Jensen’s inequality for convex and mid-convex functions László Horváth Correspondence: lhorvath@almos. vein.hu Department of Mathematics, University of Pannonia, 8200 Veszprém, Egyetem u. 10., Hungary

Abstract In this paper, a new parameter-dependent refinement of the discrete Jensen’s inequality is given for convex and mid-convex functions. The convergence of the introduced sequences is also studied. One of the proofs requires an interesting convergence theorem with probability theoretical background. We apply the results to define some new quasi-arithmetic and mixed symmetric means and study their monotonicity and convergence.

1 Introduction and the main results The considerations of this paper concern (A1) an arbitrarily given real vector space X, a convex subset C of X, and a finite subset {x1,..., xn} of C, where n ≥ 1 is fixed; (A2) a convex function f : C ® ℝ, and a discrete distribution p1,..., pn, which means that pj ≥ 0 with

n 

pj = 1;

j=1

(A 3 ) a mid-convex function f : C ® ℝ, and a discrete distribution p 1 ,..., p n with rational pj (1 ≤ j ≤ n). The function f : C ® ℝ is called convex if f (αx + (1 − α)y) ≤ αf (x) + (1 − α)f (y),

and mid-convex if x + y 1 1 f ≤ f (x) + f (y), 2 2 2

x, y ∈ C,

0 ≤ α ≤ 1,

x, y ∈ C.

For a variety of applications, the discrete Jensen’s inequalities are important: Theorem A. (see [1]) (a) If (A1) and (A2) are satisfied, then ⎛ ⎞ n n   ⎝ ⎠ f pj xj ≤ pj f (xj ). j=1

(1)

(2)

j=1

(b) If (A1) and (A3) are satisfied, then (2) also holds. Let N := {0, 1, 2,...} and let N+ := {1, 2,...}.

© 2011 Horváth; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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Various attempts have been made to refine inequality (2) in the following ways: Assume either (A 1 ) and (A 2 ) or (A 1 ) and (A 3 ). Let m ≥ 2 be an integer, and let I denote either the set {1,..., m} or the set N+. (B) Create a decreasing real sequence (Bk)kÎI such that Bk = Bk(f, xi, pi) (k Î I) is a sum whose index set is a subset of {1,..., n}k and ⎛ ⎞ n n   ⎝ ⎠ (3) f pj xj ≤ · · · ≤ Bk ≤ · · · ≤ B1 = pj f (xj ), k ∈ I. j=1

j=1

(C) Create an increasing real sequence (Ck)kÎI such that Ck = Ck(f, xi, pi) (k Î I) is a sum whose index set is a subset of {1,..., k}n and ⎛ ⎞ n n   ⎝ ⎠ (4) f pj xj = C1 ≤ · · · ≤ Ck ≤ · · · ≤ pj f (xj ), k ∈ I. j=1

j=1

The next two typical results belong to the group of refinements of type (B). These examples use p1 = · · · = pn = 1n. In [2], Pečarić and Volenec have constructed the sequence fk :=

1 (nk )



f

1≤i1 0 (1 ≤ j ≤ n). ⎧  1 ⎪ ⎪ r n ⎪  ⎪ ⎪ pj xrj , r = 0 ⎨ Mr = Mr (x1 , . . . , xn ; p1 , . . . , pn ) :=  j=1  . ⎪ n p ⎪  ⎪ j ⎪ xj , r = 0 ⎪ ⎩ j=1

If r ≠ 0, then the power means of order r belong to the means (12) (z : ]0, ∞[® ℝ, z (x) := xr), while we get the power means of order 0 by taking limit. Supported by the power means, we can introduce mixed symmetric means corresponding to (8):

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Definition 12 Let xj Î]0, ∞[ (1 ≤ j ≤ n), and let p1,..., pn be a discrete distribution with pj >0 (1 ≤ j ≤ n). Let l ≥ 1, and k Î N. We define the mixed symmetric means with respect to (8) by Ms,t (k, λ)

⎛ := ⎝

1



(n + λ − 1)k

(i1 ,...,in )∈Sk

⎞ ⎛ n  k! ⎝ λij pj ⎠ i1 ! . . . in ! j=1

⎞⎞ 1 s ⎜ i1 p in p ⎟⎟ λ λ ⎜ ⎟ ⎟ 1 n ·Mst ⎜x1 , ..., xn ; n , ..., n ⎟⎟ ,  i  i ⎠⎠ ⎝ λ j pj λ j pj ⎛

j=1

j=1

if s, t Î ℝ and s ≠ 0, and



M0,t (k, λ) :=

(i1 ,...,in )∈Sk

⎛

⎛

⎞⎞

⎜ ⎜ ⎜Mt ⎝

⎜ ⎟ λi1 p1 λin pn ⎟ ⎜ ⎟⎟ , ..., n ⎜x1 , ..., xn ; n ⎟⎟  i  i ⎠⎠ ⎝ λ j pj λ j pj j=1

1

k! k i !...i ! n (n + λ − 1) 1

 n 

 λij pj

j=1

,

j=1

where t Î ℝ. The monotonicity and the convergence of the previous means are studied in the next result. Proposition 13 Let xj Î]0, ∞[ (1 ≤ j ≤ n), and let p1,..., pn be a discrete distribution with pj >0 (1 ≤ j ≤ n). Let l ≥ 1, and k Î N. Suppose s, t Î ℝ such that s ≤ t. Then, (a) Mt = Ms,t (0, λ) ≥ · · · ≥ Ms,t (k, λ) ≥ · · · ≥ Ms ,

k ∈ N.

(b) In case of s, t ≠ 0 lim Ms,t (k, λ) = Ms

k→∞

for each fixed l >1, and lim Ms,t (k, λ) = Ms

λ→∞

for each fixed k Î N+. Proof. Assume s, t ≠ 0. Then, Proposition 9 (b) can be applied with g, h :]0, ∞[® ℝ, g(x) := xt, and h(x) := xs. If s = 0 or t = 0, the result follows by taking limit. ■

3 Some lemmas and the proofs of the main results Lemma 14 Let k Î N and (i1,..., in) Î Sk+1 be fixed. If we set z(i1 , . . . , in ) := {j ∈ {1, . . . , n}|ij = 0},

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then  j∈z(i1 ,...,in )

(k + 1)! k! . = i1 !...ij−1 !(ij − 1)!ij+1 !...in ! i1 !...in !

Proof. The lowest common denominator is i1!... in!. Combined with

n 

ij = k + 1, the

j=1

result follows. ■ The proof of Theorem 1. Proof. (a) We separate the proof of this part of the theorem into three steps. Let l ≥ 1 be fixed. I. Since S0 = {(0,..., 0)} ⎞ ⎞ ⎛ 0 λ p x j j n n ⎟ ⎜ j=1   ⎟ ⎜ C0 (λ) = ⎝ λ0 pj ⎠ f ⎜ n pj xj ⎠ . ⎟=f⎝ ⎝  0 ⎠ j=1 j=1 λ pj ⎞

⎛

⎛

n 

j=1

II. Next, we prove that Ck(l) ≤ Ck+1(l) (k Î N). It is easy to check that for every (i1,..., in) Î Sk n 

λij pj xj

j=1 n 

⎛ .

= λij pj

1 n+λ−1

j=1 n 

λij pj xj +(λ

n ⎜  ⎜ j=1 ⎜ n ⎝  l=1

−

n 

1)λil pl xl .

λij pj + (λ − 1)λil pl

j=1

λij pj +(λ

−

j=1 n 

λij pj

⎞ 1)λil pl ⎟

⎟ ⎟. ⎠

j=1

With the help of Theorem A, this yields that ⎞ ⎛ n  ij ij λ p x λ pj + (λ − 1)λil pl j j⎟ n ⎜ ⎜ j=1  1 ⎟ ⎜ j=1 ⎜ f⎜ n ⎟≤ ⎜ n  ⎝ ⎝  ij ⎠ n + λ − 1 l=1 λ pj λij pj ⎛

n 

j=1

⎛

j=1

n 

⎞⎞

ij il ⎜ j=1 λ pj xj + (λ − 1)λ pl xl ⎟⎟ ⎜ ⎟⎟ .f ⎜ n ⎟⎟ . ⎝  ij ⎠⎠ λ pj + (λ − 1)λil pl j=1

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Consequently, 

1

Ck (λ) ≤

k+1

k! i1 ! . . . in !

(n + λ − 1) (i1 ,...,in )∈Sk ⎛ ⎞⎞ ⎛ n  i ⎛ ⎞ j p x + (λ − 1)λil p x λ j j l l n ⎜  n ⎟⎟ ⎜ j=1  ⎜⎝ ⎟⎟ ⎜ . λij pj + (λ − 1)λil pl ⎠ f ⎜ n ⎜ ⎟⎟. ⎝ ⎝  ij ⎠⎠ j=1 l=1 λ pj + (λ − 1)λil pl

(13)

j=1

By Lemma 14, it is easy to see that the right-hand side of (13) can be written in the form ⎛ n ⎞  ij ⎞ ⎛ λ p x j j n ⎟  1 (k + 1)! ⎝ ij ⎠ ⎜ ⎟ ⎜ j=1 f λ p ⎟ ⎜ j n ⎠ ⎝  (n + λ − 1)k+1 (i1 ,...,in )∈Sk+1 i1 !...in ! j=1 i λ j pj j=1

which is just Ck+1(l). III. Finally, we prove that Ck (λ) ≤

n 

pj f (xj ),

k ∈ N+·

(14)

j=1

It follows from Theorem A that ⎞ n  k! ⎝ Ck (λ) ≤ λij pj f (xj )⎠ (n + λ − 1)k (i1 ,...,in )∈Sk i1 ! . . . in ! j=1 ⎛ ⎞ n   k! 1 ⎝ λij ⎠ pj f (xj ), k ∈ N+· = i1 ! . . . in ! (n + λ − 1)k 1

j=1



⎛

(15)

(i1 ,...,in )∈Sk

The multinomial theorem shows that  (i1 ,...,in )∈Sk

k! λij = (n + λ − 1)k , i1 ! . . . in !

1 ≤ j ≤ n,

hence (15) implies (14). ■ The proof of Theorem 3 (a) is based on the following interesting result. The s-algebra of Borel subsets of ℝn is denoted by B n. Lemma 15 Let p1,..., pn be a discrete distribution with n ≥ 2, and let l >1. Let l Î {1,..., n} be fixed. el denotes the vector in ℝn that has 0s in all coordinate positions except the lth, where it has a 1. Let q1,..., qn be also a discrete distribution such that qj >0 (1 ≤ j ≤ n) and ql > max(q1 , . . . ql−1 , ql+1 , . . . , qn ).

(16)

If ⎧ ⎨ g :

⎩

(t1 , . . . , tn ) ∈ Rn |tj > 0 (1 ≤ j ≤ n),

n  j=1

⎫ ⎬ tj = 1 → R ⎭

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is a bounded function for which τl := lim g el

exists in ℝ and pl >0, then 

lim

k→∞

(i1 ,...,in )∈Sk

⎛

⎞

⎜ λi1 p k! λin pn ⎟ ⎜ ⎟ 1 , ..., n qi11 . . . qinn g ⎜ n ⎟ = τl .  i ⎠ ⎝  ij i1 ! . . . in ! λ pj λ j pj j=1

(17)

j=1

Proof. To prove the result, we can obviously suppose that l = 1. For the sake of clarity, we shall denote the element (i1 ,..., in) of Sk by (i1k,..., ink) (k Î N+). Let ξk := (ξ1k,..., ξnk) (k Î N+) be a (ℝn, B n)-random variable on a probability space (, A, P) such that ξk has multinomial distribution of order k and with parameters q1,..., qn. A fundamental theorem of the statistics (see [9], Theorem 5.4.13), which is based on the multidimensional central limit theorem and the Cochran-Fisher theorem, implies that ⎛ ⎞ n  (ξjk − kqj )2 (18) lim P ⎝ < t⎠ = Fn−1 (t), t ∈ R, k→∞ kqj j=1

where Fn-1 means the distribution function of the Chi-squared distribution (c2-distribution) with n - 1 degrees of freedom. Choose 0 < ε < 1. Since Fn-1 is continuous and strictly increasing on ]0, ∞[, there exists a unique tε >0 such that Fn−1 (tε ) = 1 − ε.

Define

S1k :=

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

(i1k , . . . , ink ) ∈ Sk |

n  j=1

⎫ 2 ijk ⎪ ⎪ ⎬ − qj ) k k < tε . ⎪ qj ⎪ ⎭ (

The definition of the set S1k shows that  (i1k ,...,ink )∈S1k

⎛ ⎜ = P⎜ ⎝

n  j=1

  k! qi11k . . . qinnk = P (ξ1k , . . . , ξnk ) ∈ S1k i1k ! . . . ink !

(19)

⎞ 2 ⎞ ⎛ ξjk n 2 ( − qj )  ⎟ (ξ − kq ) jk j k ⎝ k < tε ⎟ < tε ⎠ ⎠=P qj kqj j=1

⎞ ⎛ ⎛ ⎞ n  (ξjk − kqj )2 = Fn−1 (tε ) + ⎝P ⎝ < tε ⎠ − Fn−1 (tε )⎠ kqj j=1

= 1 − ε + δε (k),

k ∈ N+ ,

(20)

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where by (18) lim δε (k) = 0.

(21)

k→∞

For j = 1,..., n construct the sequences (Ikj )k≥1 by  ∗     i    ijk  jk  j ∗ 1   Ik := ijk , if  − qj  = max  − qj  |(i1k , . . . , ink ) ∈ Sk , k  k

k ∈ N+·

(22)

We claim that j

Ik = qj , k→∞ k lim

(23)

1 ≤ j ≤ n.

Fix 1 ≤ j ≤ n. If (23) is false, then (22) yields that we can find a positive number r, a strictly increasing sequence (ku)u≥1 and points (i1ku , . . . , inku ) ∈ S1ku ,

such that    ijku    − q j  ≥ ρ, k u

u ∈ N+

(24)

u ∈ N+ ,

and therefore,

2 ijku − qj ρ2 ku ku ≥ ku → ∞ as u → ∞, qj qj contrary to (24). Let q := max (q2 , . . . , qn ).

It follows from (16) that γ :=

1 (q1 − q) > 0. 3

(25)

By (22) and (23), we can find an integer kg such that for each k > kg     j   ijk   Ik  − qj  ≤  − qj  < γ , (i1k , . . . , ink ) ∈ S1 , 1 ≤ j ≤ n. k k  k  Thus, for every k > kg ijk i1k > q1 − γ and < qj + γ , k k

2 ≤ j ≤ n,

(i1k , . . . , ink ) ∈ S1k ,

and hence, we get from (25) that i1k − ijk > kγ

2 ≤ j ≤ n,

(i1k , . . . , ink ) ∈ S1k ,

k > kγ .

(26)

We can see that i1k − ijk → ∞ as k → ∞,

2 ≤ j ≤ n,

(i1k , . . . , ink ) ∈ S1k .

(27)

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Now, set S2k := Sk \S1k (k ∈ N+ ), and consider the sequences ⎛ 

a1k :=

(i1k ,...,ink )∈S1k

⎞

⎜ λi1k p k! λink pn ⎟ ⎜ ⎟ 1 , ..., n qi11k . . . qinnk g ⎜ n ⎟,   ⎝ ⎠ i1k ! . . . ink ! λijk pj λijk pj j=1

j=1

and ⎞

⎛ 

a2k :=

(i1k ,...,ink )∈S2k

⎜ λi1k p k! λink pn ⎟ ⎟ ⎜ 1 , ..., n qi11k . . . qinnk g ⎜ n ⎟,  i ⎠ ⎝  ijk i1k ! . . . ink ! λ pj λ jk pj j=1

j=1

where k Î N+. The sum of these sequences is just the studied sequence in (17). Since p1 >0, we obtain from (27) that λi1k p1 = 1, n k→∞  λijk pj lim

(i1k , . . . , ink ) ∈ S1k ,

(28)

j=1

and λilk p1 = 0, n k→∞  λijk pj lim

2 ≤ l ≤ n,

(i1k , . . . , ink ) ∈ S1k .

(29)

j=1

According to (26), the convergence is uniform for all the possible sequences in (28) and (29); hence, for every ε1 >0, we can find an integer kε1 > kγ that for all k > kε1 ⎛ ⎞ ⎜ λi1k p λink pn ⎟ ⎜ ⎟ 1 τ1 − ε1 < g ⎜ n , ..., n ⎟ < τ1 + ε1 ,  i ⎠ ⎝  ijk λ pj λ jk pj j=1

(i1k , . . . , ink ) ∈ S1k .

j=1

Bringing in (19-20), we find that   P (ξ1k , . . . , ξnk ) ∈ S2k = ε − δε (k),

k ∈ N+ ,

and therefore, thanks to (19-20), (30) and the boundedness of g (|g| ≤ m) (1 − ε + δε (k))(τ1 − ε1 ) − (ε − δε (k))m ≤ a1k + a2k ≤ (1 − ε + δε (k))(τ1 + ε1 ) + (ε − δε (k))m,

k > kε1 .

Consequently, by (21) (1 − ε)(τ1 − ε1 ) − εm ≤ lim inf(a1k + a2k ) ≤ lim sup(a1k + a2k ) k→∞

≤ (1 − ε)(τ1 + ε1 ) + εm,

and this proves the convergence claim (17). The proof is now complete. ■ The proof of Theorem 3.

k→∞

(30)

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Proof. (a) We have only to observe that for every fixed 1 ≤ l ≤ n ⎛

lim

k→∞

1



(n + λ − 1)k

(i1 ,...,in )∈Sk

⎞

n 

⎜ j=1 λ pj xj ⎟ k! ⎜ ⎟ λil pl f ⎜ n ⎟ = pl f (xl ). ⎝  ij ⎠ i1 ! . . . in ! λ pj ij

j=1

The case pl = 0 is trivial. To prove the case pl >0, define the function ⎧ ⎨ g :

⎩

n 

(t1 , . . . , tn ) ∈ Rn |tj > 0 (1 ≤ j ≤ n),

j=1

⎫ ⎬ tj = 1 → R ⎭

by ⎛ g(t1 , . . . , tn ) := f ⎝

n 

⎞ tj x j ⎠ .

j=1

Consequently, the limit in (31) can be written in the form 

lim pl

k→∞

(i1 ,...,in )∈Sk

.



i1 il−1

il 1 λ k! 1 ··· i1 ! . . . in ! n + λ − 1 n+λ−1 n+λ−1 ⎛ ⎞

1 n+λ−1

il+1

···

1 n+λ−1

in ⎜ i1 λin pn ⎟ ⎜ λ p1 ⎟ g⎜ n , ..., n ⎟.  i ⎠ ⎝  ij j λ pj λ pj j=1

j=1

Now, we can apply Lemma 15 with qj =

1 , n+λ−1

1 ≤ j ≤ n,

j = l, and ql =

λ n+λ−1

and lim g = f (xl ), el

1 ≤ l ≤ n.

(b) Elementary considerations show this part of the theorem. The proof is complete. ■ The proof of Theorem 5. Proof. Theorem A confirms that f is bounded on the set ⎧ ⎫ n n ⎨ ⎬  G := tj xj ∈ C|tj ≥ 0 (1 ≤ j ≤ n), tj = 1 , ⎩ ⎭ j=1

j=1

where tj (1 ≤ j ≤ n) is also rational if f is mid-convex. It is elementary that for every (i1,..., in) Î Sk  λil 1, if il = k , 1 ≤ l ≤ n. lim = 0, if il < k λ→∞ (n + λ − 1)k

(31)

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By the definition of the set Sk, (0,..., 0, k, 0,..., 0) (the vector has 0s in all coordinate positions except the lth) is the only element of Sk for which il = k (1 ≤ l ≤ n). By using the boundedness of f on G, the previous assumptions imply the result, bringing the proof to an end. ■ Acknowledgements This study was supported by the Hungarian National Foundations for Scientific Research Grant No. K73274. Competing interests The author declares that he has no competing interests. Received: 8 March 2011 Accepted: 25 July 2011 Published: 25 July 2011 References 1. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge Mathematical Library Series. Cambridge University Press (1967) 2. Pečarić, JE, Volenec, V: Interpolation of the Jensen inequality with some applications, Österreích. Akad Wiss Math-Natur KI Sitzungsber II. 197, 463–467 (1988) 3. Pečarić, JE, Svrtan, D: Unified approach to refinements of Jensen’s inequalities. Math Inequal Appl. 5, 45–47 (2002) 4. Horváth, L, Pečarić, JE: A refinement of the discrete Jensen’s inequality. Math Inequal Appl. (to appear) 5. Horváth, L: A method to refine the discrete Jensen’s inequality for convex and mid-convex functions. doi:10.1016/j. mcm.2011.05.060 6. Horváth, L: Inequalities corresponding to the classical Jensen’s inequality. J Math Inequal. 3, 189–200 (2009) 7. Xiao, Z-G, Srivastava, HM, Zhang, Z-H: Further refinements of the Jensen inequalities based upon samples with repetitions. Math Comput Mod. 51, 592–600 (2010) 8. Mitrinović, DS, Pečarić, JE, Fink, AM: Classical and New Inequalities in Analysis, vol. 61 of Mathematics and Its Applications. Kluwer Academic Publishers, Dordrecht (1993) 9. Dacunha-Castello, D, Duflo, M: Probability and Statistics I. Springer-Verlag, New York (1986) doi:10.1186/1029-242X-2011-26 Cite this article as: Horváth: A parameter-dependent refinement of the discrete Jensen’s inequality for convex and mid-convex functions. Journal of Inequalities and Applications 2011 2011:26.

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