We extended the Mercer inequlaity, Fejér-Hermite-Hadamard, and Jensen inequalities for strongly convex functions. Moreover, we obtained several results in information theory and mathematical analysis using obtained inequalities.
Citation: Yamin Sayyari, Mana Donganont, Mehdi Dehghanian, Morteza Afshar Jahanshahi. Strongly convex functions and extensions of related inequalities with applications to entropy[J]. AIMS Mathematics, 2024, 9(5): 10997-11006. doi: 10.3934/math.2024538
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We extended the Mercer inequlaity, Fejér-Hermite-Hadamard, and Jensen inequalities for strongly convex functions. Moreover, we obtained several results in information theory and mathematical analysis using obtained inequalities.
For a,b∈R,a<b, and I:=[a,b], the function φ:I→R is convex if
φ(tx1+(1−t)x2)≤tφ(x1)+(1−t)φ(x2) |
and φ is named strongly convex (S-C) with modulus k if
φ(tx1+(1−t)x2)≤tφ(x1)+(1−t)φ(x2)−kt(1−t)(x1−x2)2, |
for all x1,x2∈I and all t∈[0,1]
Mercer inequality [8], Hermite-Hadamard inequality (H-H inequality) [6], and Jensen's inequality [7] are some types of important inequalities in different fields of mathematical analysis and optimization.
In [4], Fejér investigated an extension of the H-H inequality. Azócar et al. [1] obtained the following Fejér inequality for the S-C function.
Because of the comprehensive results of the Jensen's, Mercer, and H-H inequalities, some researchers exteneded their studies via mappings of various types (see [2,3,10,17,18]).
The theory of convex function is significant in entropy estimation and optimization [11,12,13,14,15,16].
An equivalent condition for the convexity of a continuous function is given in [21]. Also, in [22], S. Zlobec generalized some of the basic integral properties of convex functions.
Throughout this article, suppose that x={xi}⊆[a,b] and t={ti},0≤ti with ∑ni=1ti=1.
Theorem 1.1. [9] Suppose that φ:I→R is an S-C function with modulus k, then
φ(n∑i=1tixi)≤n∑i=1tiφ(xi)−kV(x), |
for all x1,x2,…,xn∈[a,b],ti≥0(i=1,…,n) with ∑ni=1ti=1,ˉx=∑ni=1tixi, and V(x):=∑ni=1ti(xi−ˉx)2.
Definition 1.1. The Shannon entropy of a probability distribution t is defined by
H(t):=−n∑i=1tilog(ti). |
Proposition 1.1. [19] Let η:=min{ti:i=1,…,n} and ϑ:=max{ti:i=1,…,n}, then
m(η,ϑ):=ηlog(2ηη+ϑ)+ϑlog(2ϑη+ϑ)≤logn−H(t)≤log((η+ϑ)24ηϑ):=M(η,ϑ). |
Our main goal is to obtain some intersting Jensen, Mercer, and Fejér-Hermite-Hadamard inequalities for S-C functions. Furthermore, we use those inequalities in mathematical analysis and the Shannon entropy to obtain a strong bound for entropy of a probability distribution.
In this section, we generelize the Mercer, Jensen type, and Fejér-Hermite-Hadamard inequalities for S-C functions.
Lemma 2.1. Suppose that φ:I→R is an S-C function with modulus k. If w1,w2,w3∈I, and w1<w2<w3, then
(i) φ(w2)−φ(w1)2≤φ(w2+w32)−φ(w1+w32)−k4(w2−w1)(2w3−w2−w1),
(ii) φ(w3)−φ(w2)2≥φ(w1+w32)−φ(w1+w22)+k4(w3+w2−2w1)(w3−w2).
Proof. Since w1<w2<w2+w32<w3, there are s,t∈[0,1], s+t=1 such that w2=s(w2+w32)+tw1. Therefore,
φ(w1)−φ(w2)2+φ(w2+w32)=12[φ(w1)−φ(tw1+s(w2+w32))]+φ(w2+w32)≥12[φ(w1)−(tφ(w1)+sφ(w2+w32)−kts(w2+w32−w1)2)]+φ(w2+w32)=s2φ(w1)+2−s2φ(w2+w32)+k2ts(w2+w32−w1)2≥φ(s2w1+2−s2(w2+w32))+(ks(2−s)4+k2ts)(w2+w32−w1)2=φ(s2w1+(w2+w32)−12(w2−tw1))+k4(4s−3s2)(w2+w3−2w12)2=φ(w1+w32)+k4(4(w2−w1)(2w3−w2−w1)(w2+w3−2w1)2)(w2+w3−2w12)2=φ(w1+w32)+k4(w2−w1)(2w3−w2−w1). |
Similarly, we obtain (ⅱ) by putting w2=t(w1+w22)+sw3, where w1<w1+w22<w2<w3.
Theorem 2.1. Assume that φ:[a,b]→R is an S-C function with modulus k and
Δk(p,q)=φ(p)+φ(q)−2φ(p+q2)−k2(p−q)2, |
where a≤p,q≤b, then
maxp,qΔk(p,q)=Δk(a,b). | (2.1) |
Proof. Taking w1=a,w2=p, and w3=b in the part (ⅰ) of Lemma 2.1 and w1=p,w2=q, and w3=b in the part (ⅱ) of Lemma 2.1, we gain
φ(p)−φ(a)2≤φ(p+b2)−φ(a+b2)−k4(p−a)(2b−p−a), | (2.2) |
φ(b)−φ(q)2≥φ(p+b2)−φ(p+q2)+k4(b−q)(b+q−2p), | (2.3) |
respectively.
By (2.3), we get
φ(q)−φ(b)2≤φ(p+q2)−φ(p+b2)−k4(b−q)(b+q−2p). | (2.4) |
Now, from (2.2) and (2.4), we conclude (2.1).
Corollary 2.1. Assume that φ:[a,b]→R is an S-C function with modulus k and x∈[a,b], then
φ(a+b−x)−k2(a+b−2x)2≤φ(a)+φ(b)−φ(x)−k2(b−a)2. | (2.5) |
Proof. Replacing p by x and q by a+b−x in Theorem 2.1 gives the desired result.
Lemma 2.2. Let λ,μ≥0 and λ+μ=1 and let φ:[a,b]→R be an S-C function with modulus k, then
φ(λ(a+b)+(μ−λ)n∑i=1tixi)≤λφ(a)+λφ(b)+(μ−λ)n∑i=1tiφ(xi)−k[n∑i=1ti(xi−ˉx)2+λ2(b−a)2+λμ(a+b−2ˉx)2−λ2n∑i=1ti(a+b−2xi)2], | (2.6) |
for all x1,x2,…,xn∈[a,b],ti≥0(1≤i≤n) with ∑ni=1ti=1 and ˉx=∑ni=1tixi.
Proof. By applying Theorem 1.1 and (2.5), we obtain
φ(λ(a+b)+(μ−λ)n∑i=1tixi)=φ(λn∑i=1ti(a+b−xi)+μn∑i=1tixi)≤λφ(n∑i=1ti(a+b−xi))+μφ(n∑i=1tixi)−kλμ(a+b−2n∑i=1tixi)2≤λ(n∑i=1tiφ(a+b−xi)−kn∑i=1ti(xi−ˉx)2)+μφ(n∑i=1tixi)−kλμ(a+b−2n∑i=1tixi)2≤λ(φ(a)+φ(b)−n∑i=1tiφ(xi)−k2(b−a)2+k2n∑i=1ti(a+b−2xi)2)−kλn∑i=1ti(xi−ˉx)2+μn∑i=1tiφ(xi)−kμn∑i=1ti(xi−ˉx)2−kλμ(a+b−2ˉx)2=λφ(a)+λφ(b)+(μ−λ)n∑i=1tiφ(xi)+kλ2n∑i=1ti(a+b−2xi)2−k[n∑i=1ti(xi−ˉx)2+λ2(b−a)2+λμ(a+b−2ˉx)2]. |
Corollary 2.2. Let λ,μ∈[0,1] with λ+μ=1, and let φ be an S-C function with modulus k on [a,b] and x∈[a,b], then
φ(λ(a+b)+(μ−λ)x)−kλ2(a+b−2x)2≤λφ(a)+λφ(b)+(μ−λ)φ(x)−kλ[12(b−a)2+μ(a+b−2x)2]. |
Corollary 2.3. Let φ be an S-C function with modulus k on [a,b] and s,t,λ,μ∈[0,1] with s+t=1 and λ+μ=1, then
φ(λ(a+b)+(μ−λ)(sa+tb))+(λ−μ)φ(sa+tb)−kλ2(s−t)2(b−a)2≤λφ(a)+λφ(b)−kλ[12(b−a)2+λμ(a+b−2sa−2tb)2]. |
Proof. Substitute x by sa+tb in Corollary 2.2 to get the inequality.
Corollary 2.4. Let φ be an S-C function with modulus k on [a,b] and s,t∈[0,1],s+t=1, then
2φ(a+b2)≤φ(sa+tb)+φ(ta+sb)−k2(s−t)2(b−a)2≤φ(a)+φ(b)−k2(b−a)2. |
Proof. Set λ=1 in Corollary 2.3 to get the righthand side of the inequality. On the other hand, by definition, we have
φ(a+b2)=φ(sa+tb2+ta+sb2)≤12φ(sa+tb)+12φ(ta+sb)−k4(s−t)2(a−b)2. |
The Corollary 2.4 shows a kind of pre-H-H inequalities.
Next, we prove a generalization of the H-H type inequality for S-C functions.
Theorem 2.2. Assume that φ:[a,b]→R is an S-C function with modulus k and ω is a nonnegative function on [a,b], then
2φ(a+b2)∫baω(u)du≤∫ba(ω(u)+ω(a+b−u))φ(u)du−k2∫ba(a+b−2u)2ω(u)du≤(φ(a)+φ(b)−k2(b−a)2)∫baω(u)du. |
Proof. By multiplying the two-sided inequality in Corollary 2.4 with ω(a+t(b−a)), integrating with respect to variable t from 0 to 1, and by interchanging u=a+t(b−a), we get
2φ(a+b2)∫baω(u)du≤∫ba(φ(u)+φ(a+b−u))ω(u)du−k2∫ba(a+b−2u)2ω(u)du≤(φ(a)+φ(b)−k2(b−a)2)∫baω(u)du, |
since
∫ba(ω(u)+ω(a+b−u))φ(u)du=∫ba(φ(u)+φ(a+b−u))ω(u)du. |
Corollary 2.5. Let φ:[a,b]→R be an S-C function with modulus k, then
φ(a+b2)+k12(b−a)2≤1b−a∫baφ(u)du≤φ(a)+φ(b)2−k6(b−a)2. |
Proof. Putting ω≡1 in Theorem 2.2 and after some calculations, the desired inequality follows.
If we consider ω(u)=uα−1 in Theorem 2.2, then we have the following result.
Corollary 2.6. Let α>0, 0<a<b, and φ:[a,b]→R be an S-C function with modulus k, then
2φ(a+b2)≤αbα−aα∫ba(uα−1+(a+b−u)α−1)φ(u)du−kα2(bα−aα)∫ba(a+b−2u)2uα−1du≤φ(a)+φ(b)−k2(b−a)2. |
If α→0+ in Corollary 2.6, then we get the next corollary.
Corollary 2.7. Let 0<a<b and φ:[a,b]→R be an S-C function with modulus k, then
2φ(a+b2)log(ba)a+b≤∫baφ(u)u(a+b−u)du−k2((a+b)2log(ba)−2b2+2a2)≤(φ(a)+φ(b)−k2(b−a)2)log(ba)a+b. |
Let φ:[a,b]→R be an S-C function with modulus k and 0≤s,t≤1 such that s+t=1. We define
Δ∗k(s,t,x,y)=sφ(x)+tφ(y)−φ(sx+ty)−kst(x−y)2, |
for all x,y∈[a,b].
Theorem 2.3. Let φ:[a,b]→R be an S-C function with modulus k, then
maxs,t∈t;x,y∈[a,b]Δ∗k(s,t,x,y)≤Δk(a,b). |
Proof. First, we prove that
Δ∗k(s,t,x,y)≤Δk(x,y), |
for all s,t∈t and x,y∈[a,b].
Now,
Δk(x,y)−Δ∗k(s,t,x,y)=tφ(x)+sφ(y)+φ(sx+ty)+kst(x−y)2−2φ(x+y2)−k2(x−y)2≥φ(tx+sy)+φ(sx+ty)+2kst(x−y)2−k2(x−y)2−2φ(x+y2)=φ(tx+sy)+φ(sx+ty)−k2(s−t)2(x−y)2−2φ(x+y2)≥2φ((tx+sy)+(sx+ty)2)−2φ(x+y2)=0. |
The remains of the proof is an application of Theorem 2.1.
Theorem 2.4. Suppose that φ:[a,b]→R is an S-C function with modulus k, then
Jφ(t,x):=n∑i=1tiφ(xi)−φ(n∑i=1tixi)≤φ(a)+φ(b)−2φ(a+b2)−k4(b−a)2−k(b−xr)(xr−a), |
where (b−xr)(xr−a)=mini{(b−xi)(xi−a)}.
Proof. Since xi∈[a,b], there is a sequence {λi},λi∈[0,1] with xi=λia+(1−λi)b for i=1,2,…. Thus,
n∑i=1tiφ(xi)−φ(n∑i=1tixi)=n∑i=1tiφ(λia+(1−λi)b)−φ(n∑i=1ti(λia+(1−λi)b))≤n∑i=1ti(λiφ(a)+(1−λi)φ(b)−kλi(1−λi)(b−a)2)−φ(an∑i=1tiλi+bn∑i=1ti(1−λi)). |
Setting λ:=∑ni=1tiλi and μ:=1−∑ni=1tiλi, by the use of Theorem 2.3, we gain
Jφ(t,x)≤λφ(a)+μφ(b)−φ(λa+μb)−k(b−a)2n∑i=1tiλi(1−λi)≤φ(a)+φ(b)−2φ(a+b2)−k2(b−a)2+kλμ(b−a)2−kλr(1−λr)(b−a)2, |
because
λr(1−λr)=(b−xr)(xr−a)(b−a)2≤(b−xi)(xi−a)(b−a)2=λi(1−λi), |
for all i=1,2,…,n. Hence,
Jφ(t,x)≤φ(a)+φ(b)−2φ(a+b2)−k4(b−a)2−k(b−xr)(xr−a). |
In the following, we obtain an application of Lemma 2.2, which improves the results from [8].
Proposition 3.1. Let 0<a≤xi≤b, ti≥0(i=1,2,…,n) with ∑ni=1ti=1,λ,μ≥0, and λ+μ=1, then
˜Gλ≤˜Aλe−12b2(V(x)+12(b−a)2−λ2ˉy)≤˜Aλ, | (3.1) |
where ˜Aλ:=λ(a+b)+(μ−λ)ˉx,˜Gλ:=(ab)λ∏ni=1xti(λ−μ)i,yi:=(a+b−2xi)2, and y:={yi}.
Proof. Letting φ(x):=−log(x) and k=12b2 in Lemma 2.2, the desired inequality follows.
Remark 3.1. Putting λ=1 in (3.1), we obtain
˜G≤˜Ae−12b2(V(x)+12(b−a)2−12ˉy)≤˜A, |
where ˜A=a+b−ˉx and ˜G=ab∏ni=1xtii (see [8]).
Example 3.1. Assume that β≥2 and 0<a<b, then φ(x)=xβ is an S-C function with modulus k:=β(β−1)2aβ−2 on [a,b]. Further, if β=2n(n=1,2,…), then φ(x)=xβ is an S-C function with modulus k on arbitrary interval [a,b].
Proof. Let x,y∈[a,b]. Define the following functions on [0,1]:
g(t):=(tx+(1−t)y)β+β(β−1)2aβ−2t(1−t)(x−y)2 |
and
h(t):=txβ+(1−t)yβ. |
Since g(0)=h(0), g(1)=h(1), h′′≡0, and
g′′(t)=β(β−1)(x−y)2((tx+(1−t)y)β−2−aβ−2)≥0, |
g(t)≤h(t) for every t∈[0,1]. Therefore, xβ is an S-C function with modulus β(β−1)2aβ−2 on [a,b].
In the next proposition, we give an extension of the pre-Grüss inequality (see [5,20]):
n∑i=1tix2i−(n∑i=1tixi)2≤14(b−a)2. |
Proposition 3.2. Assume that xi∈[a,b] and {ti}∈t, then
n∑i=1tix2i−(n∑i=1tixi)2≤14(b−a)2−(b−xr)(xr−a), |
where (b−xr)(xr−a)=mini{(b−xi)(xi−a)}.
Proof. It follows from Example 3.1 and Theorem 2.4 with φ(x)=x2.
In this subsection, new Shannon entropy bounds are found, that improve the entropy bounds from [19].
Proposition 3.3. Assume that η:=min{ti:i=1,…,n} and ϑ:=max{ti:i=1,…,n}, then
0≤logn−H(t)≤log((η+ϑ)24ηϑ)−(ϑ−η)28ϑ2:=Γ(η,ϑ)≤M(η,ϑ). |
Proof. Replace φ(x) by −log(x) in Theorem 2.4 and consider k=12b2,xi:=1ti for all i=1,…,n.
Proposition 3.4. Assume that φ:[a,b]→R is an S-C function with modulus k, then
1nn∑i=1φ(xi)−φ(n∑i=1xin)≤φ(a)+φ(b)−2φ(a+b2)−k4(b−a)2−k(b−xr)(xr−a), |
where (b−xr)(xr−a)=mini{(b−xi)(xi−a)}.
Proof. Set ti=1n for every i=1,…,n in Theorem 2.4.
Proposition 3.5. Assume that η:=min{ti:i=1,…,n} and ϑ:=max{ti:i=1,…,n}, then
0≤logn−H(t)≤n{ηlog(2ηη+ϑ)+ϑlog(2ϑη+ϑ)−18ϑ(ϑ−η)2}:=Λ(η,ϑ). |
Proof. Replace φ(x) by xlog(x) in Proposition 3.4 and modulus k=12b,xi:=ti for all i=1,…,n.
Example 3.2. Let j≥2 be an integer, n=10j, η=10−j−1, and ϑ=10−j+1, then
M(η,ϑ)−Γ(η,ϑ)≃0.1225 |
and
n.m(η,ϑ)−Λ(η,ϑ)≃12.25. |
In this work, we have stablished some new inequalities such as the Mercer, Fejér-Hermite-Hadamard, and Jensen inequalities for strongly convex functions. Next, using these inequalities, we get some applications in analysis and entropy of probability distributions.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors declare that they have no competing interests.
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