In this present investigation, the authors obtain Fekete-Szegö inequality for certain normalized analytic function f(ζ) defined on the open unit disk for which
(f′(ζ)ϑ(ζf′(ζ)f(ζ))1−ϑ≺1+sinζ;(0≤ϑ≤1)
lies in a region starlike with respect to 1 and symmetric with respect to the real axis. As a special case of this result, the Fekete-Szegö inequality for a class of functions defined through Poisson distribution series is obtained. Further, we discuss the second Hankel inequality for functions in this new class.
Citation: Huo Tang, Gangadharan Murugusundaramoorthy, Shu-Hai Li, Li-Na Ma. Fekete-Szegö and Hankel inequalities for certain class of analytic functions related to the sine function[J]. AIMS Mathematics, 2022, 7(4): 6365-6380. doi: 10.3934/math.2022354
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In this present investigation, the authors obtain Fekete-Szegö inequality for certain normalized analytic function f(ζ) defined on the open unit disk for which
(f′(ζ)ϑ(ζf′(ζ)f(ζ))1−ϑ≺1+sinζ;(0≤ϑ≤1)
lies in a region starlike with respect to 1 and symmetric with respect to the real axis. As a special case of this result, the Fekete-Szegö inequality for a class of functions defined through Poisson distribution series is obtained. Further, we discuss the second Hankel inequality for functions in this new class.
The famous Young's inequality, as a classical result, state that: if a,b>0 and t∈[0,1], then
atb1−t≤ta+(1−t)b | (1.1) |
with equality if and only if a=b. Let p,q>1 such that 1/p+1/q=1. The inequality (1.1) can be written as
ab≤app+bqq | (1.2) |
for any a,b≥0. In this form, the inequality (1.2) was used to prove the celebrated Hölder inequality. One of the most important inequalities of analysis is Hölder's inequality. It contributes wide area of pure and applied mathematics and plays a key role in resolving many problems in social science and cultural science as well as in natural science.
Theorem 1 (Hölder inequality for integrals [11]). Let p>1 and 1/p+1/q=1. If f and g are real functions defined on [a,b] and if |f|p,|g|q are integrable functions on [a,b] then
∫ba|f(x)g(x)|dx≤(∫ba|f(x)|pdx)1/p(∫ba|g(x)|qdx)1/q, | (1.3) |
with equality holding if and only if A|f(x)|p=B|g(x)|q almost everywhere, where A and B are constants.
Theorem 2 (Hölder inequality for sums [11]). Let a=(a1,...,an) and b=(b1,...,bn) be two positive n-tuples and p,q>1 such that 1/p+1/q=1. Then we have
n∑k=1akbk≤(n∑k=1apk)1/p(n∑k=1bqk)1/q. | (1.4) |
Equality hold in (1.4) if and only if ap and bq are proportional.
In [10], İşcan gave new improvements for integral ans sum forms of the Hölder inequality as follow:
Theorem 3. Let p>1 and 1p+1q=1. If f and g are real functions defined on interval [a,b] and if |f|p, |g|q are integrable functions on [a,b] then
∫ba|f(x)g(x)|dx≤1b−a{(∫ba(b−x)|f(x)|pdx)1p(∫ba(b−x)|g(x)|qdx)1q+(∫ba(x−a)|f(x)|pdx)1p(∫ba(x−a)|g(x)|qdx)1q} | (1.5) |
Theorem 4. Let a=(a1,...,an) and b=(b1,...,bn) be two positive n-tuples and p,q>1 such that 1/p+1/q=1. Then
n∑k=1akbk≤1n{(n∑k=1kapk)1/p(n∑k=1kbqk)1/q+(n∑k=1(n−k)apk)1/p(n∑k=1(n−k)bqk)1/q}. | (1.6) |
Let E be a nonempty set and L be a linear class of real valued functions on E having the following properties
L1: If f,g∈L then (αf+βg)∈L for all α,β∈R;
L2: 1∈L, that is if f(t)=1,t∈E, then f∈L;
We also consider positive isotonic linear functionals A:L→R is a functional satisfying the following properties:
A1: A(αf+βg)=αA(f)+β A(g) for f,g∈L and α,β∈R;
A2: If f∈L, f(t)≥0 on E then A(f)≥0.
Isotonic, that is, order-preserving, linear functionals are natural objects in analysis which enjoy a number of convenient properties. Functional versions of well-known inequalities and related results could be found in [1,2,3,4,5,6,7,8,9,11,12].
Example 1. i.) If E=[a,b]⊆R and L=L[a,b], then
A(f)=∫baf(t)dt |
is an isotonic linear functional.
ii.)If E=[a,b]×[c,d]⊆R2 and L=L([a,b]×[c,d]), then
A(f)=∫ba∫dcf(x,y)dxdy |
is an isotonic linear functional.
iii.)If (E,Σ,μ) is a measure space with μ positive measure on E and L=L(μ) then
A(f)=∫Efdμ |
is an isotonic linear functional.
iv.)If E is a subset of the natural numbers N with all pk≥0, then A(f)=∑k∈Epkfk is an isotonic linear functional. For example; If E={1,2,...,n} and f:E→R,f(k)=ak, then A(f)=∑nk=1ak is an isotonic linear functional. If E={1,2,...,n}×{1,2,...,m} and f:E→R,f(k,l)=ak,l, then A(f)=∑nk=1∑ml=1ak,l is an isotonic linear functional.
Theorem 5 (Hölder's inequality for isotonic functionals [13]). Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p−1+q−1=1. If w,f,g≥0 on E and wfp,wgq,wfg∈L then we have
A(wfg)≤A1/p(wfp)A1/q(wgq). | (2.1) |
In the case 0<p<1 and A(wgq)>0 (or p<0 and A(wfp)>0), the inequality in (2.1) is reversed.
Remark 1. i.) If we choose E=[a,b]⊆R, L=L[a,b], w=1 on E and A(f)=∫ba|f(t)|dt in the Theorem 5, then the inequality (2.1) reduce the inequality (1.3).
ii.) If we choose E={1,2,...,n}, w=1 on E, f:E→[0,∞),f(k)=ak, and A(f)=∑nk=1ak in the Theorem 5, then the inequality (2.1) reduce the inequality (1.4).
iii.) If we choose E=[a,b]×[c,d],L=L(E), w=1 on E and A(f)=∫ba∫dc|f(x,y)|dxdy in the Theorem 5, then the inequality (2.1) reduce the following inequality for double integrals:
∫ba∫dc|f(x,y)||g(x,y)|dxdy≤(∫ba∫dc|f(x,y)|pdx)1/p(∫ba∫dc|g(x,y)|qdx)1/q. |
The aim of this paper is to give a new general improvement of Hölder inequality for isotonic linear functional. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals.
Theorem 6. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p−1+q−1=1. If α,β,w,f,g≥0 on E, αwfg,βwfg,αwfp,αwgq,βwfp,βwgq,wfg∈L and α+β=1 on E, then we have
i.)
A(wfg)≤A1/p(αwfp)A1/q(αwgq)+A1/p(βwfq)A1/q(βwgq) | (3.1) |
ii.)
A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)≤A1/p(wfp)A1/q(wgq). | (3.2) |
Proof. ⅰ.) By using of Hölder inequality for isotonic functionals in (2.1) and linearity of A, it is easily seen that
A(wfg)=A(αwfg+βwfg)=A(αwfg)+A(βwfg)≤A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq). |
ⅱ.) Firstly, we assume that A1/p(wfp)A1/q(wgq)≠0. then
A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)=(A(αwfp)A(wfp))1/p(A(αwgq)A(wgq))1/q+(A(βwfp)A(wfp))1/p(A(βwgq)A(wgq))1/q, |
By the inequality (1.1) and linearity of A, we have
A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)≤1p[A(αwfp)A(wfp)+A(βwfp)A(wfp)]+1q[A(αwgq)A(wgq)+A(βwgq)A(wgq)]=1. |
Finally, suppose that A1/p(wfp)A1/q(wgq)=0. Then A1/p(wfp)=0 or A1/q(wgq)=0, i.e. A(wfp)=0 or A(wgq)=0. We assume that A(wfp)=0. Then by using linearity of A we have,
0=A(wfp)=A(αwfp+βwfp)=A(αwfp)+A(βwfp). |
Since A(αwf),A(βwf)≥0, we get A(αwfp)=0 and A(βwfp)=0. From here, it follows that
A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)=0≤0=A1/p(wfp)A1/q(wgq). |
In case of A(wgq)=0, the proof is done similarly. This completes the proof.
Remark 2. The inequality (3.2) shows that the inequality (3.1) is better than the inequality (2.1).
If we take w=1 on E in the Theorem 6, then we can give the following corollary:
Corollary 1. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p−1+q−1=1. If α,β,f,g≥0 on E, αfg,βfg,αfp,αgq,βfp,βgq,fg∈L and α+β=1 on E, then we have
i.)
A(fg)≤A1/p(αfp)A1/q(αgq)+A1/p(βfq)A1/q(βgq) | (3.3) |
ii.)
A1/p(αfp)A1/q(αgq)+A1/p(βfp)A1/q(βgq)≤A1/p(fp)A1/q(gq). |
Remark 3. i.) If we choose E=[a,b]⊆R, L=L[a,b], α(t)=b−tb−a,β(t)=t−ab−a on E and A(f)=∫ba|f(t)|dt in the Corollary 1, then the inequality (3.3) reduce the inequality (1.5).
ii.) If we choose E={1,2,...,n}, α(k)=kn,β(k)=n−kn on E, f:E→[0,∞),f(k)=ak, and A(f)=∑nk=1ak in the Theorem1, then the inequality (3.3) reduce the inequality (1.6).
We can give more general form of the Theorem 6 as follows:
Theorem 7. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p−1+q−1=1. If αi,w,f,g≥0 on E, αiwfg,αiwfp,αiwgq,wfg∈L,i=1,2,...,m, and ∑mi=1αi=1 on E, then we have
i.)
A(wfg)≤m∑i=1A1/p(αiwfp)A1/q(αiwgq) |
ii.)
m∑i=1A1/p(αiwfp)A1/q(αiwgq)≤A1/p(wfp)A1/q(wgq). |
Proof. The proof can be easily done similarly to the proof of Theorem 6.
If we take w=1 on E in the Theorem 6, then we can give the following corollary:
Corollary 2. Let L satisfy conditions L1, L2, and A satisfy conditions A1, A2 on a base set E. Let p>1 and p−1+q−1=1. If αi,f,g≥0 on E, αifg,αifp,αigq,fg∈L,i=1,2,...,m, and ∑mi=1αi=1 on E, then we have
i.)
A(fg)≤m∑i=1A1/p(αifp)A1/q(αigq) | (3.4) |
ii.)
m∑i=1A1/p(αifp)A1/q(αigq)≤A1/p(fp)A1/q(gq). |
Corollary 3 (Improvement of Hölder inequality for double integrals). Let p,q>1 and 1/p+1/q=1. If f and g are real functions defined on E=[a,b]×[c,d] and if |f|p,|g|q∈L(E) then
∫ba∫dc|f(x,y)||g(x,y)|dxdy≤4∑i=1(∫ba∫dcαi(x,y)|f(x,y)|pdx)1/p(∫ba∫dcαi(x,y)|g(x,y)|qdx)1/q, | (3.5) |
where α1(x,y)=(b−x)(d−y)(b−a)(d−c),α2(x,y)=(b−x)(y−c)(b−a)(d−c),α3(x,y)=(x−a)(y−c)(b−a)(d−c),,α4(x,y)=(x−a)(d−y)(b−a)(d−c) on E
Proof. If we choose E=[a,b]×[c,d]⊆R2, L=L(E), α1(x,y)=(b−x)(d−y)(b−a)(d−c),α2(x,y)=(b−x)(y−c)(b−a)(d−c),α3(x,y)=(x−a)(y−c)(b−a)(d−c),α4(x,y)=(x−a)(d−y)(b−a)(d−c) on E and A(f)=∫ba∫dc|f(x,y)|dxdy in the Corollary 1, then we get the inequality (3.5).
Corollary 4. Let (ak,l) and (bk,l) be two tuples of positive numbers and p,q>1 such that 1/p+1/q=1. Then we have
n∑k=1m∑l=1ak,lbk,l≤4∑i=1(n∑k=1m∑l=1αi(k,l)apk,l)1/p(n∑k=1m∑l=1αi(k,l)bqk,l)1/q, | (3.6) |
where α1(k,l)=klnm,α2(k,l)=(n−k)lnm,α3(k,l)=(n−k)(m−l)nm,α4(k,l)=k(m−l)nm on E.
Proof. If we choose E={1,2,...,n}×{1,2,...,m}, α1(k,l)=klnm,α2(k,l)=(n−k)lnm,α3(k,l)=(n−k)(m−l)nm,α4(k,l)=k(m−l)nm on E, f:E→[0,∞),f(k,l)=ak,l, and A(f)=∑nk=1∑ml=1ak,l in the Theorem1, then we get the inequality (3.6).
In [14], Sarıkaya et al. gave the following lemma for obtain main results.
Lemma 1. Let f:Δ⊆R2→R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If ∂2f∂t∂s∈L(Δ), then the following equality holds:
f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy]=(b−a)(d−c)4∫10∫10(1−2t)(1−2s)∂2f∂t∂s(ta+(1−t)b,sc+(1−s)d)dtds. |
By using this equality and Hölder integral inequality for double integrals, Sar\i kaya et al. obtained the following inequality:
Theorem 8. Let f:Δ⊆R2→R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If |∂2f∂t∂s|q,q>1, is convex function on the co-ordinates on Δ, then one has the inequalities:
|f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)4(p+1)2/p[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q, | (4.1) |
where
A=12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy], |
1/p+1/q=1 and fst=∂2f∂t∂s.
If Theorem 8 are resulted again by using the inequality (3.5), then we get the following result:
Theorem 9. Let f:Δ⊆R2→R be a partial differentiable mapping on Δ=[a,b]×[c,d] in R2with a<b and c<d. If |∂2f∂t∂s|q,q>1, is convex function on the co-ordinates on Δ, then one has the inequalities:
|f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)41+1/p(p+1)2/p{[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}, | (4.2) |
where
A=12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy], |
1/p+1/q=1 and fst=∂2f∂t∂s.
Proof. Using Lemma 1 and the inequality (3.5), we find
|f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)4∫10∫10|1−2t||1−2s||fst(ta+(1−t)b,sc+(1−s))|dtds≤(b−a)(d−c)4{(∫10∫10ts|1−2t|p|1−2s|pdtds)1/p×(∫10∫10ts|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10t(1−s)|1−2t|p|1−2s|pdtds)1/p×(∫10∫10t(1−s)|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10(1−t)s|1−2t|p|1−2s|pdtds)1/p×(∫10∫10(1−t)s|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10(1−t)(1−s)|1−2t|p|1−2s|pdtds)1/p×(∫10∫10(1−t)(1−s)|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q}. | (4.3) |
Since |fst|q is convex function on the co-ordinates on Δ, we have for all t,s∈[0,1]
|fst(ta+(1−t)b,sc+(1−s))|q≤ts|fst(a,c)|q+t(1−s)|fst(a,d)|q+(1−t)s|fst(a,c)|q+(1−t)(1−s)|fst(a,c)|q | (4.4) |
for all t,s∈[0,1]. Further since
∫10∫10ts|1−2t|p|1−2s|pdtds=∫10∫10t(1−s)|1−2t|p|1−2s|pdtds=∫10∫10(1−t)s|1−2t|p|1−2s|pdtds | (4.5) |
=∫10∫10(1−t)(1−s)|1−2t|p|1−2s|pdtds=14(p+1)2, | (4.6) |
a combination of (4.3) - (4.5) immediately gives the required inequality (4.2).
Remark 4. Since η:[0,∞)→R,η(x)=xs,0<s≤1, is a concave function, for all u,v≥0 we have
η(u+v2)=(u+v2)s≥η(u)+η(v)2=us+vs2. |
From here, we get
I={[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}≤2{[6|fst(a,c)|q+3|fst(a,d)|q+6|fst(b,c)|q+3|fst(b,d)|q72]1/q+[3|fst(a,c)|q+6|fst(a,d)|q+3|fst(b,c)|q+6|fst(b,d)|q72]1/q} |
≤4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q |
Thus we obtain
(b−a)(d−c)41+1/p(p+1)2/pI≤(b−a)(d−c)41+1/p(p+1)2/p4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q}≤(b−a)(d−c)4(p+1)2/p{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q}. |
This shows that the inequality (4.2) is better than the inequality (4.1).
The aim of this paper is to give a new general improvement of Hölder inequality via isotonic linear functional. An important feature of the new inequality obtained here is that many existing inequalities related to the Hölder inequality can be improved. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals. Similar method can be applied to the different type of convex functions.
This research didn't receive any funding.
The author declares no conflicts of interest in this paper.
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