R E S E A R C H
Open Access
Notions of generalized
s
-convex functions
on fractal sets
Adem Kılıçman
1*and Wedad Saleh
2*Correspondence:
akilic@upm.edu.my
1Department of Mathematics and
Institute for Mathematical Research, University Putra Malaysia, Serdang, 43400, Malaysia
Full list of author information is available at the end of the article
Abstract
The purpose of this article is to present some new inequalities for products of generalized convex and generalizeds-convex functions on fractal sets. Furthermore, some applications are given.
MSC: 26A51; 26D07; 26D15; 53C22
Keywords: s-convex functions; fractal space; local fractional derivative
1 Introduction
Letf: I⊂R−→Rα. For anyx
,x∈Iandγ ∈[, ] if the inequality
fγx+ ( –γ)x
≤γαf(x) + ( –γ)αf(x)
holds, thenf is called a generalized convex function onI []. Inα= , we have convex function, convexity is defined only in geometrical terms as being the property of a function whose graph bears tangents only under it [].
The convexity of functions plays a significant role in many fields, for example, in biolog-ical system, economy, optimization, and so on [–].
In recent years, the fractal theory has received significantly remarkable attention from scientists and engineers. In the sense of Mandelbrot, a fractal set is one whose Hausdorff dimension strictly exceeds the topological dimension [, ]. Many researchers studied the properties of functions on fractal space and constructed many kinds of fractional calcu-lus by using different approaches [–]. Particularly, in [], Yang gave the analysis of local fractional functions on fractal space systematically, which includes local fractional calculus and the monotonicity of function.
LetRαbe the real line numbers on fractal space. Then by using Gao-Yang-Kang’s
con-cept one can explain the definitions of the local fractional derivative and local fractional integral as in [–]. Now ifrα
,rαandrα∈Rα( <α≤), then
() rα+rα∈Rα,rαrα∈Rα,
() rα
+rα=rα+rα = (r+r)α= (r+r)α,
() rα
+ (rα+rα) = (rα+rα) +rα,
() rα
rα=rαrα = (rr)α= (rr)α,
() rα
(rαrα) = (rαrα)rα,
() rα
(rα+rα) = (rαrα) + (rαrα),
() rα
+ α= α+rα =rα andrα·α= α·rα =rα.
Let us start with some definitions as regards the local fractional calculus onRα.
Definition .[] Letybe a local fractional continuous function on the interval [a,a].
The local fractional integral of the functiony(m) of orderαis defined by
aI
(α)
ay(m) =
( +α)– a
a
y(μ)(dμ)α
=( +α)– lim
μ−→
n
i=
y(μi)(μi)α
withμi=μi+–μiandμ=max{μi:i= , , . . . ,n– }where [μi,μi+],i= , , . . . ,n–
,μ=a <μ<· · ·<μn–<μn=a is a partition of the interval [a,a] and is the
well-known Gamma function(m) =∞tm–e–tdt.
In [], Mo and Sui introduced the definitions of two kinds of generalizeds-convex func-tions on fractal sets as follows.
Definition .
(i) A functionf:R+−→Rα, is called a generalizeds-convex ( <s< ) in the first sense
if
f(γa+γa)≤γsαf(a) +γsαf(a) ()
for alla,a∈R+and allγ,γ≥withγs+γs= . This class of functions is
denoted byGK
s.
(ii) A functionf:R+−→Rα, is called a generalizeds-convex ( <s< ) in the second
sense if () holds for alla,a∈R+and allγ,γ≥withγ+γ= . This class of
functions is denoted byGK
s.
In the same paper, [], Mo and Sui proved that all functions fromGKs, fors∈(, ), are nonnegative.
In this article, recall thatgsandgs are the generic classes consistent for those
func-tions that are generalizeds-convex in the first sense, and in the second sense, respectively. It is well known that there are many important established inequalities for the class of generalized convex functions, however, one of the most famous is known as the general-ized Hermit-Hadamard inequality, or the ‘generalgeneral-ized Hadamard inequality’ and stated as follows (see []): letf be a generalized convex function on [a,a]⊆R,a<a, then
f
a+a
≤ ( +α) (a–a)αa
Ia(α)f(x)≤f(a) +f(a) α .
2 Basic results for generalizeds-convex functions Lemma . If f ∈GK
s or f ∈GKs,then f(γx+γx)≤γαsf(x) +γαsf(x)
So, we might re-write the definitions of generalizeds-convexity in the first sense and the second sense as follows.
Definition . A functionf:I⊂R+−→Rα is called a generalizeds-convex in the first
sense if
fγx+
–γs
sx
≤γαsf(x) +
–γsαf(x)
for allx,x∈Iand for all ≤γ ≤.
Definition . A functionf:I⊂R+−→Rαis called a generalizeds-convex in the second
sense if
fγx+ ( –γ)x
≤γαsf(x) + ( –γ)αsf(x)
for allx,x∈Iand for all ≤γ ≤.
Theorem . The classes GK, GK and the class of generalized convex functions are equivalent when the domain is restricted toR+.
Proof Simply an issue of applying the definitions.
Now, some results as regards generalizeds-convex functions are given.
Remark .
(i) Iff∈GK
s, thenf( x+x
s )≤
f(x)+f(x)
α .
(ii) Iff∈GK
s, thenf(x+x)≤
f(x)+f(x)
αs .
(iii) For a function that is both generalizedgsandgs, there is a one-to-one
correspondence between the set of all pairs of the sort(γ,η)with respect togsand
the set of all pairs of the sort(γ,η)with respect togs. (So we may write eachγ as a
γsand eachηas aηsandvice versa. This happens in the light of the fact that
≤γ,η≤,≤s≤.)
Theorem . If a function f ∈GK
s and f ∈GKs,then f(γx+ηx)≤γαsf(x) +ηαsf(x)≤γαsf(x) +ηαsf(x)
for some{γi,ηi,i= , } ⊂[, ].
Proof It follows from the one-to-one correspondence proved before. To eachγ,ηsuch
that γ+η = , there correspondsγ, η whereγs+ηs= andγ ≥γ, η≥η since
{γi,ηi,i= , } ⊂[, ].
Theorem . If a function f ∈GK
sand f ∈GKsand its domain concurs with its counter-domain,then fof ∈GK
Proof f(γx+ ( –γs)
sx)≤γαs
f(x) + ( –γs)αf(x), then
fγαsf(x) +
–γsαf(x)
≤γαsαsff(x)
+ –γsααsff(x)
=γαsfof(x) +βαsfof(x).
3 Inequalities for generalizeds-convex functions Theorem . Let g: [, ]−→Rαbe a function such that
g(γ) =
(a–a)α(( +α))
a
a
a
a
fγx+ ( –γ)x
(dx)α(dx)α,
where f: [a,a]−→Rαand f ∈GKs.Then:
(i) g∈GK
s in[, ].Iff ∈GKs,theng∈GKs.
(ii) αg() = αg() = α
(a–a)α(+α)aI
(α)
af(x)is an upper bound forg(γ).
(iii) α(–s)g(γ)≥g( ) =
(a–a)α((+α))
a
a
a
a f(
x+x
)(dx)
α(dx
)α,γ ∈[, ].
Proof (i) Take{γ,γ} ⊂[, ],γ+γ= ,t,t∈Dandf ∈GKs, then we have g(γt+γt)
=
(a–a)α(( +α))
a
a
a
a
f(γt+γt)x
+ – (γt+γt)
x
(dx)α(dx)α
≤
(a–a)α(( +α))
a
a
a
a
γαsf(tx+x–tx)
+γαsf(tx+x–tx)
(dx)α(dx)α
=γαs
(a–a)α(( +α))
a
a
a
a
ftx+ ( –t)x
(dx)α(dx)α
+γαs
(a–a)α(( +α))
a
a
a
a
ftx+ ( –t)x
(dx)α(dx)α
=γαsg(t) +γαsg(t),
which implies thatg∈GK
s in [, ].
(ii) Sincef(γx+ ( –γ)x)≤γαsf(x) + ( –γ)αsf(x) we have
A= (( +α))
a
a
a
a
fγx+ ( –γ)x
(dx)α(dx)α
≤
(( +α))
a
a
a
a
γαsf(x)(dx)α(dx)α
+
(( +α))
a
a
a
a
( –γ)αsf(x)(dx)α(dx)α
=γαs
( +α)(a–a)
α
aI
(α)
af(x) + ( –γ)
αs
( +α)(a–a)
α
aI
(α)
af(x)
= (γ
αs+ ( –γ)αs)
( +α) (a–a)
α
aI
(α)
sinceγαs≤α. Because ( –γ)αs≤αas well, we have
A≤
α
( +α)(a–a)
α
aI
(α)
af(x).
(iii) Sincef is generalizeds-convex in the second sense,
f(γx+ ( –γ)x) +f(( –γ)x+γx)
αs ≥f
x+x
for allγ ∈[, ] andx,x∈[a,b].
If we integrate on [a,a]×[a,a], we obtain
αs(( +α))
a
a
a
a
fγx+ ( –γ)x
+f( –γ)x+γx
(dx)α(dx)α
≥
(( +α))
a
a
a
a
f
x+x
(dx)α(dx)α
⇒
α(s–)(a
–a)α(( +α))
a
a
a
a
fγx+ ( –γ)x
(dx)α(dx)α
≥
(a–a)α(( +α))
a
a
a
a
f
x+x
(dx)α(dx)α,
which means that
α(–s)g(γ)≥g
.
The proof is complete.
Assume the following functions:
(i) gF: [, ]−→R
α, defined by
gF(γ) =
(( +α))
a
a
a
a
fγx+ ( –γ)x
F(x)F(x)(dx)α(dx)α,
wheref: [a,a]⊂R+−→Rαandf ∈GKs.
(ii) gF: [, ]−→R
α, defined by
gF(γ) =
(( +α))
a
a
a
a
fγx+ ( –γ)x
F(x)F(x)(dx)α(dx)α,
wheref: [a,a]⊂R+−→Rαandf ∈GKs. Theorem . The following holds:
(i) gFandgF are both symmetric aboutγ =
.
(iii) We have the upper bound
αg
F() =
α
(( +α))
a
a
a
a
f(x)F(x)F(x)(dx)α(dx)α
for the functiongF(γ).
Proof (i) gF and gF are both symmetric aboutγ =
because gF(γ) =gF( –γ) and
gF(γ) =gF( –γ).
(ii) We get
gF
γt+ ( –γ)t
=
(( +α))
a
a
a
a
fγt+ ( –γ)t
x
+ –γt+ ( –γ)t
x
F(x)F(x)(dx)α(dx)α.
But
γt+ ( –γ)t
x+ ( –
γt+ ( –γ)t
x
=γtx+ ( –γ)tx+γx+ ( –γ)x–γtx– ( –γ)tx
=γ(tx+x–tx) + ( –γ)(tx+x–tx).
Sincef∈GK
s,
gF
γt+ ( –γ)t
=
(( +α))
a
a
a
a
fγt+ ( –γ)t
x
+ –γt+ ( –γ)t
x
F(x)F(x)(dx)α(dx)α
≤γαs (( +α))
a
a
a
a
ftx+ ( –t)x
F(x)F(x)(dx)α(dx)α
+ ( –γ)αs (( +α))
a
a
a
a
ftx+ ( –t)x
F(x)F(x)(dx)α(dx)α
=γαsgF(t) + ( –γ)
αsg F(t),
which proves thatgF∈GK
s.
(iii) From the definition and the assumptions, we get
gF(γ) =
(( +α))
a
a
a
a
fγx+ ( –γ)x
F(x)F(x)(dx)α(dx)α
≤γαs (( +α))
a
a
a
a
f(x)F(x)F(x)(dx)α(dx)α
+ ( –γ)αs (( +α))
a
a
a
a
f(x)F(x)F(x)(dx)α(dx)α
=γαs+ ( –γ)αs (( +α))
a
a
a
a
≤ α (( +α))
a
a
a
a
f(x)F(x)F(x)(dx)α(dx)α
= αgF().
Theorem . Let f,f: [a,a]⊂R+−→Rα,a<a,be nonnegative,and generalized
s-convex functions in the second sense.If f∈GKsand f∈GK
son[a,a]for someγ∈[, ]
and s,s∈(, ],then
(( +α))
a
a
a
a
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)α(dx)α(dx)α
≤ α( + (γ+γ)α)
( + (γ+γ+ )α)
(( +α))(a–a)
α
a
a
f(x)f(x)(dx)α
+
α(( +α))βα(a–a) α T
(a,b) +T(a,b)
,
where
βα=
γαs( –γ)αs(dγ)α,
T(a,a) =f(a)f(a) +f(a)f(a),
and
T(a,a) =f(a)f(a) +f(a)f(a).
Proof Sincef∈GKsandf∈GK
son [a,a],
f
γx+ ( –γ)x
≤γαsf
(x) + ( –γ)αsf(x),
f
γx+ ( –γ)x
≤γαsf
(x) + ( –γ)αsf(x),
for allγ ∈[, ]. Sincefandfare nonnegative,
f
γx+ ( –γ)x
f
γx+ ( –γ)x
≤γα(s+s)f
(x)f(x) + ( –γ)α(s+s)f(x)f(x)
+γαs( –γ)αsf
(x)f(x) +γαs( –γ)αsf(x)f(x).
Integrating both sides of the above inequality over [, ], we obtain
( +α)
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)α
≤
( +α)f(x)f(x)
γα(s+s)(dγ)α
+
( +α)f(x)f(x)
( –γ)α(s+s)(dγ)α
+
( –α)f(x)f(x)
+
( +α)f(x)f(x)
γαs( –γ)αs(dγ)α
= ( + (s+s)α) ( + (s+s+ )α)
f(x)f(x) +
( + (s+s)α)
( + (s+s+ )α)
f(x)f(x)
+
( +α)βα f(x)f(x) +f(x)f(x)
.
If we integrate both sides of the above inequalities on [a,a]×[a,a], we obtain
(( +α))
a
a
a
a
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)α(dx)α(dx)α
≤ ( + (s+s)α)
( + (s+s+ )α)
(( +α))
a
a
a
a
f(x)f(x)(dx)α(dx)α
+ a
a
a
a
f(x)f(x)(dx)α(dx)α
+
(( +α))βα
a
a
a
a
f(x)f(x)(dx)α(dx)α
+ a
a
a
a
f(x)f(x)(dx)α(dx)α
= ( + (s+s)α) ( + (s+s+ )α)
(( +α))(a–a)
α
× a
a
f(x)f(x)(dx)α+
a
a
f(x)f(x)(dx)α
+
(( +α))βα
a
a
f(x)(dx)α
a
a
f(x)(dx)α
+ a
a
f(x)(dx)α
a
a
f(x)(dx)α
.
By using the right half of the generalized Hadamard inequality on the right side of the above inequality, we have
(( +α))
a
a
a
a
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)α(dx)α(dx)α
≤ α( + (s+s)α)
( + (s+s+ )α)
(( +α))(a–a)
α
a
a
f(x)f(x)(dx)α
+
α(( +α))βα(a–a) α
T(a,a) +T(a,a)
.
The proof is complete.
Remark .
(i) Ifα= , in Theorem ., then
β=
γs( –γ)s(dγ) =(s+ )(s+ )
and
a
a
a
a
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)(dx)(dx)
≤
s+s+
(a–a)
a
a
f(x)f(x)dx
+(s+ )(s+ ) (s+s+ )
(a–a) T(a,a) +T(a,a)
.
(ii) Ifα= , ands=s= , then(s(+)s+s(+)s+)=. Also,
β=
γs( –γ)s(dγ) =(s+ )(s+ )
(s+s+ )
,
which implies that
a
a
a
a
f
γx+ ( –γ)x
f
γx+ ( –γ)x
(dγ)(dx)(dx)
≤
(a–a) a
a
f(x)f(x)dx
+
(a–a)
T
(a,a) +T(a,a)
.
Theorem . Let f,f: [a,a]⊂R+−→Rα,a<a,be nonnegative and generalized
s-convex functions in the second sense.If f∈GKsand f∈GK
son[a,a]for someγ∈[, ]
and s,s∈(, ],then
(( +α))
a
a
f
γx+ ( –γ)
a+b
f
γx+ ( –γ)
a+b
(dγ)α(dx)α
≤ ( + (s+s)α)
( + (s+s+ )α)
( +α)
a
a
f(x)f(x)(dx)α
+
α( +α)
( + (s+s)α)
α( + (s
+s+ )α)
+
( +α)βα
×(a–a)α T(a,a) +T(a,a)
,
whereβα,T(a,a),and T(a,a)are defined in Theorem..
Proof Sincef∈GKsandf∈GK
son [a,a],
f
γx+ ( –γ)
a+a
≤γαsf
(x) + ( –γ)αsf
a+a
,
f
γx+ ( –γ)
a+a
≤γαsf
(x) + ( –γ)αsf
a+a
,
f
γx+ ( –γ)
a+a
f
γx+ ( –γ)
a+a
≤γα(s+s)f
(x)f(x) + ( –γ)α(s+s)f
a+a
f
a+a
+γαs( –γ)αsf
(x)f
a+a
+γαs( –γ)αsf
a+a
f(x).
Integrating both sides of the above inequality over [, ], we obtain
( +α)
f
γx+ ( –γ)
a+a
f
γx+ ( –γ)
a+a
(dγ)α
≤
( +α)f(x)f(x)
γα(s+s)(dγ)α
+
( +α)f
a+a
f
a+a
( –γ)α(s+s)(dγ)α
+
( +α)f(x)f
a+a
γαs( –γ)αs(dγ)α
+
( +α)f
a+a
f(x)
γαs( –γ)αs(dγ)α
= ( + (s+s)α) ( + (s+s+ )α)
f(x)f(x)
+ ( + (s+s)α) ( + (s+s+ )α)
f
a+a
f
a+a
+
( +α)βα
f(x)f
a+a
+f
a+a
f(x)
.
If we integrate both sides of the above inequalities on [a,a], we obtain
(( +α))
a a f
γx+ ( –γ)
a+a
f
γx+ ( –γ)
a+a
(dγ)α(dx
)α
≤ ( + (s+s)α)
( + (s+s+ )α)
( +α)
a
a
f(x)f(x)(dx)α
+ ( + (s+s)α) ( + (s+s+ )α)
( +α)(a–a)
αf
a+a
f
a+a
+
(( +α))βα
f
a+a
a
a
f(x)(dx)α+f
a+a
a
a
f(x)(dx)α
≤ ( + (s+s)α)
( + (s+s+ )α)
( +α)
a
a
f(x)f(x)(dx)α
+ ( + (s+s)α) ( + (s+s+ )α)
( +α)(a–a)
αf(a) +f(a)
α
f(a) +f(a)
α
+
(( +α))(a–a)
α
βα
αf(a) +f(a) α
f(a) +f(a)
α
= ( + (s+s)α) ( + (s+s+ )α)
( +α)
a
a
+ ( + (s+s)α) α( + (s
+s+ )α)
( +α)(a–a)
α
T(a,a) +T(a,a)
+
α(( +α))(a–a)
αβ
α T(a,a) +T(a,a)
.
Then
(( +α))
a
a
f
γx+ ( –γ)
a+a
f
γx+ ( –γ)
a+a
(dγ)α(dx
)α
≤ ( + (s+s)α)
( + (s+s+ ))α
( +α)
a
a
f(x)f(x)(dx)α
+
α( +α)
( + (s+s)α)
α( + (s
+s+ ))α
+
( +α)βα
×(a–a)α T(a,a) +T(a,a)
.
The proof is complete.
Remark . Ifα= , then
a
a
f
γx+ ( –γ)
a+a
f
γx+ ( –γ)
a+a
(dγ)(dx)
≤
s+s+
a
a
f(x)f(x)(dx)
+
(s+s+ )
+(s+ )(s+ ) (s+s+ )
(a–a)T(a,a) +T(a,a)
.
Theorem . Let f,f: [a,a]⊂R+−→Rα,a<a,be nonnegative and generalized
s-convex functions in the second sense.If f∈GKsand f∈GK
son[a,a]for someγ∈[, ]
and s,s∈(, ],then
(( +α))
a
a
f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
(dγ)α(dx)α
≤ ( + (s+s)α)
( + (s+s+ )α)
( +α)
a
a
f(x)f(x)(dx)α
+
α( +α)
( + (s+s)α)
α( + (s
+s+ )α)
+
( +α)βα
×(a–a)α T(a,a) +T(a,a)
,
whereβα,T(a,a)and T(a,a)are defined in Theorem..
Proof The proof of this theorem can easily be given like in Theorem ..
Theorem . Let f,f: [a,a]⊂R+−→Rα,a<a,be nonnegative,and generalized
(( +α))
a a f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
(dγ)α(dx
)α
≤( + α) ( + α)
( +α)
a
a
f(x)f(x)(dx)α
+
α( +α)
( + α) α( + α)+
( +α)βα
(a–a)α T(a,a) +T(a,a)
,
where T(a,a)and T(a,a)are defined in Theorem.,but
βα=
γα( –γ)α(dγ)α= ( +α) ( + α)–
( + α) ( + α).
Proof Sincefandfare generalized convex on [a,a],
f
γa+a
+ ( –γ)x
≤γαf
a+a
+ ( –γ)αf(x),
f
γa+a
+ ( –γ)x
≤γαf
a+a
+ ( –γ)αf
(x),
for allx∈[a,a],γ ∈[, ]. Sincefandfare nonnegative,
f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
≤γαf
a+a
f
a+a
+ ( –γ)αf
(x)f(x)
+γα( –γ)αf
a+a
f(x) +γα( –γ)αf(x)f
a+a
.
Integrating both sides of the above inequality over [, ], we obtain
( +α)
f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
(dγ)α
≤
( +α)f
a+a
f
a+a
γα(dγ)α
+
( +α)f(x)f(x)
( –γ)α(dγ)α
+
( +α)f
a+a
f(x)
γα( –γ)α(dγ)α
+
( +α)f(x)f
a+a
γα( –γ)α(dγ)α
=( + α) ( + α)f
a+a
f
a+a
+( + α)
( + α)f(x)f(x)
+
( +α)βα
f
a+a
f(x) +f(x)f
a+a
If we integrate both sides of the above inequalities on [a,a], we obtain
(( +α))
a a f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
(dγ)α(dx)α
≤( + α) ( + α)
( +α)(a–a)
α
f
a+a
f
a+a
+( + α) ( + α)
( +α)
a
a
f(x)f(x)(dx)α
+
(( +α))βα
f
a+a
a
a
f(x)(dx)α+f
a+a
a
a
f(x)(dx)α
≤( + α) ( + α)
( +α)(a–a)
αf(a) +f(a)
α
f(a) +f(a)
α
+( + α) ( + α)
( +α)
a
a
f(x)f(x)(dx)α
+
(( +α))βα
αf(a) +f(a)
α
f(a) +f(a)
α
(a–a)α
=( + α) ( + α)
( +α)
a
a
f(x)f(x)(dx)α+
α( +α)
( + α) α( + α)
+
( +α)βα
(a–a)α T(a,a) +T(a,a)
.
The proof is complete.
Remark . Ifα= , then
a a f
γa+a
+ ( –γ)x
f
γa+a
+ ( –γ)x
(dγ)(dx)
≤
a
a
f(x)f(x)(dx) +
(a–a) T(a,a) +T(a,a)
.
4 Applications
Example . Leta,a∈R+,a<a, anda–a≤, then
α
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
K(a,a)
+
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
G(a,a)
≤α
( + α) ( +α)( + α)
(aα–aα) (a–a)α
+
α
(( +α))βαA (a
where
βα=
γα( –γ)α(dγ)α,
A(a,a) =
aα
+aα
α , a,a≥,
G(a,a) =
aα
aα
, a
,a≥,
and
K(a,a) =
aα
+aα
α
, a,a≥.
Proof Iff∈GKsandf∈GK
s on [a,a] for someγ ∈[, ] ands,s∈(, ], then, by
Theorem ., iff,f: [, ]−→[α, α],f(x) =xαs,f(x) =xαs, wherex∈[a,a],s=s=
anda–a≤, then
(( +α))
a
a
a
a
γx+ ( –γ)x
α
(dγ)α(dx)α(dx)α
≤α( + α) ( + α)
(( +α))(a–a)
α
a
a
xα(dx)α
+
α(( +α))βα a α
+a
α
+
α
aαaα(a–a)α
⇒
(a–a)α
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
aα+aα
+
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
aα
aα
≤α
( + α) ( +α)( + α)
(a–a)α
aα–aα
+
α(( +α))βα(a–a) α
aα+aα ⇒
(a–a)α
α
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
K(a,a)
+
α
( +α)
( + α) ( + α)
+ α
( +α) ( + α)–
( + α) ( + α)
( +α) ( + α)
G(a,a)
≤α
( + α) ( +α)( + α)
(a–a)α
aα–aα
+
α
(( +α))βα(a–a) αA(a
,a).
The proof is complete.
Remark . Ifα= , then
K
(a ,a) +
G
(a ,a)≤
a
–a
a–a
+
A
(a ,a)
⇒
K(a,a) + G(a,a)≤
a+aa+a
+ A(a,a),
where
A(a,a) =
a+a
, a,a≥, is the arithmetic mean, G(a,a) = (aa)
, a,a≥, is the geometric mean, and
K(a,a) =
a+a
, a,a≥, is the quadratic mean.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally in the design and progress of the study. Both authors read and approved the final manuscript.
Author details
1Department of Mathematics and Institute for Mathematical Research, University Putra Malaysia, Serdang, 43400,
Malaysia.2Department of Mathematics, University Putra Malaysia, Serdang, 43400, Malaysia.
Acknowledgements
The authors would like to thank the referees for valuable suggestions and comments, which helped the authors to improve this article substantially.
Received: 24 June 2015 Accepted: 15 September 2015
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