# On Some Metric Inequalities and Applications

Authors: Hassen Aydi; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

We derive a new inequality in metric spaces and provide its geometric interpretation. Some applications of our result are given, including metric inequalities in Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities. Our main result is a generalization of that obtained by Dragomir and Gosa.


---

## Body

## 1. Introduction

In [1], Dragomir and Gosa established the following interesting inequality in metric spaces.Theorem 1.
(Dragomir-Gosa inequality).
LetV be a nonempty set equipped with a metric δ. Let N≥2 be a natural number, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(1)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑj≤infϑ∈V∑i=1Nξiδϑi,ϑ.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficientC=1 on the right-hand side of (1) (in front of inf) cannot be replaced by a smaller real number.In the special case whenξi=1/N, for all i∈1,2,⋯,N, (1) reduces to
(2)∑i=1N−1∑j=i+1Nδϑi,ϑj≤Ninfϑ∈V∑i=1Nδϑi,ϑ.Inequality (2) can be considered a polygonal-type inequality. Namely, it has the following geometric interpretation: let Ω be a polygon in a metric space with N vertices and ϑ be an arbitrary point in the space. Then, the sum of all edges and diagonals of Ω is less than N-times the sum of the distances from ϑ to the vertices of Ω.In the same paper [1], the authors presented some interesting applications of inequality (1) to normed linear spaces and pre-Hilbert spaces.In this paper, motivated by the above-mentioned work, a generalization of inequality (1) is obtained and its geometric interpretation is provided. Moreover, some applications of our result are given, including metric inequalities in Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities.


## 2. Main Result and Some Consequences

We first recall briefly the notion of metric spaces (see, e.g., [2]). Let V be a nonempty set and δ:V×V⟶0,∞ be a given function. We say that δ is a metric on V if the following conditions hold: for all ϑ,ρ,κ∈V,
(i)
δϑ,ρ=0, if and only if ϑ=ρ(ii)
δϑ,ρ=δρ,ϑ(iii)
δϑ,ρ≤δϑ,κ+δκ,ρIn this case, we say thatV,δ is a metric space.Letℕ be the set of positive natural numbers. Our main result is the following:Theorem 2.
LetV be a nonempty set equipped with a metric δ. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(3)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm≤12infϑ∈V2∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficient1/2 on the right-hand side of (3) cannot be replaced by a smaller real number.Proof.
By the triangle inequality, for allϑ∈V, one has
(4)δϑi,ϑj≤δϑi,ϑ+δϑ,ϑj,i,j∈1,2,⋯,N,which yields
(5)δϑi,ϑjm≤δϑi,ϑ+δϑ,ϑjm.
On the other hand, by the binomial theorem, one has(6)δϑi,ϑ+δϑ,ϑjm=∑k=0mmkδϑi,ϑkδϑ,ϑjm−k,where
(7)mk=m!m−k!k!.
Hence, combining (5) with (6), one obtains
(8)δϑi,ϑjm≤∑k=0mmkδϑi,ϑkδϑ,ϑjm−k.
Multiplying the above inequality byξiξj and taking the sum over i and j, it holds that
(9)∑i,j=1Nξiξjδϑi,ϑjm≤∑i,j=1Nξiξj∑k=0mmkδϑi,ϑkδϑ,ϑjm−k.
Notice that due to the symmetry ofδ and the fact that δu,u=0, u∈V, one has
(10)∑i,j=1Nξiξjδϑi,ϑjm=2∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm.
Furthermore, using that∑i=1Nξi=1, one obtains
(11)∑i,j=1Nξiξj∑k=0mmkδϑi,ϑkδϑ,ϑjm−k=∑k=0mmk∑i,j=1Nξiξjδϑi,ϑkδϑ,ϑjm−k=∑k=0mmk∑i=1Nξiδϑi,ϑk∑j=1Nξjδϑ,ϑjm−k=∑k=0mmk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k=2∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k,i.e.,
(12)∑i,j=1Nξiξj∑k=0mmkδϑi,ϑkδϑ,ϑjm−k=2∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k.
Hence, it follows from (9), (10), and (12) that
(13)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm≤122∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k.
Notice that the above inequality holds for allϑ∈V. So, taking the infimum over ϑ∈V, (3) follows.
Suppose now that there exists a certain constantM>0 such that
(14)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm≤M2∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k.
TakingN=2, ϑ1≠ϑ2, ξ1=λ, and ξ2=1−λ, where λ∈0,1, one obtains
(15)λ1−λδϑ1,ϑ2m≤2Mλδϑ1,ϑm+1−λδϑ2,ϑm+M∑k=1m−1mkλδϑ1,ϑk+1−λδϑ2,ϑkλδϑ1,ϑm−k+1−λδϑ2,ϑm−k,for all ϑ∈V. In particular, for ϑ=ϑ1, one deduces that
(16)λ1−λδϑ1,ϑ2m≤2M1−λδϑ1,ϑ2m+M1−λ2δϑ1,ϑ2m∑k=1m−1mk,which yields (since ϑ1≠ϑ2)
(17)λ≤2M+M1−λ∑k=1m−1mk.
Taking the limit asλ⟶1− in the above inequality, it holds that M≥1/2. The proof is then complete.Remark 3.
Takingm=1 in Theorem 2, (3) reduces to (1).Corollary 4.
LetV be a nonempty set equipped with a metric δ. Let m,N∈ℕ, N≥2, and ϑii=1N⊂V. Then,
(18)∑i=1N−1∑j=i+1Nδϑi,ϑjm≤infϑ∈VN∑i=1Nδϑi,ϑm+12∑k=1m−1mk∑i=1Nδϑi,ϑk∑i=1Nδϑ,ϑim−k.Proof.
Using (3) with
(19)ξi=1N,i∈1,2,⋯,N,(18) follows.Remark 5.
Takingm=1 in Corollary 4, (18) reduces to (2).
In the special casem=2, one deduces form Corollary 4 the following result.Corollary 6.
LetV be a nonempty set equipped with a metric δ. Let N∈ℕ, N≥2, and ϑii=1N⊂V. Then,
(20)∑i=1N−1∑j=i+1Nδϑi,ϑj2≤infϑ∈VN∑i=1Nδϑi,ϑ2+∑i=1Nδϑi,ϑ2.Inequality (20) has the following geometric interpretation.Corollary 7.
LetΩ be a polygon in a metric space with N vertices, and ϑ be an arbitrary point in the space. Then, the sum of the squares of all edges and diagonals of Ω is less than N-times the sum of the squares of the distances from ϑ to the vertices of Ω plus the square of the sum of the distances from ϑ to the vertices of Ω.
Givenρ>0 and ϑ∈V, where V,δ is a metric space, we denote by Bϑρ¯ the closed ball in V with center ϑ and radius ρ, namely,
(21)Bϑρ¯=u∈V:δϑ,u≤ρ.Corollary 8.
LetV be a nonempty set equipped with a metric δ. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Suppose that there exist ρ>0 and ϑ∈V such that ϑii=1N⊂Bϑρ¯. Then,
(22)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm≤2m−1ρm.Proof.
Sinceϑii=1N⊂Bϑρ¯, one has
(23)δϑi,ϑ≤ρ,i∈1,2,⋯,N.
Hence, using (3), (23), and the fact that ∑i=1Nξi=1, one deduces that
(24)∑i=1N−1∑j=i+1Nξiξjδϑi,ϑjm≤122∑i=1Nξiδϑi,ϑm+∑k=1m−1mk∑i=1Nξiδϑi,ϑk∑i=1Nξiδϑ,ϑim−k≤122ρm+ρm∑k=1m−1mk=ρm22+∑k=0mmk−2=2m−1ρm.
The proof is complete.Remark 9.
Takingm=1 in Corollary 8, one obtains Corollary 2 in [1].


## 3. Applications

### 3.1. A Metric Inequality in Lebesgue Spaces

Consider a measure spaceχ,M,μ and a real number r∈1,∞. We denote by Lrχ,M,μ the space of measurable functions f such that
(25)∫χfrdμ<∞.Proposition 10.
Letm,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and fii=1N⊂Lrχ,M,μ. Then,
(26)∑i=1N−1∑j=i+1Nξiξj∫χfi−fjrdμm/r≤12inff∈Lrχ,M,μΛf,where
(27)Λf=2∑i=1Nξi∫χfi−frdμm/r+∑k=1m−1mk∑i=1Nξi∫χfi−frdμk/r∑i=1Nξi∫χfi−frdμm−k/r,for all f∈Lrχ,M,μ. Moreover, this inequality is optimal, in the sense that the multiplicative coefficient 1/2 on the right-hand side of (26) cannot be replaced by a smaller real number.Proof.
Consider the distance function(28)δf,g=∫χf−grdμ1/r,f,g∈Lrχ,M,μ.
Then,Lrχ,M,μ,δ is a metric space (see, e.g., [3]). Hence, using (3) with δ as defined above, (26) follows. The optimality of (26) follows from Theorem 2.


### 3.2. A Matrix Inequality

We denote byMnℝ the set of square matrices of size n∈ℕ, n≥2, with real number coefficients. Let M∈Mnℝ. We denote by ρM the spectral radius of M, namely,
(29)ρM=maxλi:i=1,2,⋯,n,where λi,i∈1,2,⋯,n, are the eigenvalues of M. We denote by σmaxM the largest singular value of M, namely,
(30)σmaxM=ρMMt,where Mt is the transpose of M. For more details on matrix analysis, see, for example, [4, 5].Proposition 11.
Letm,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and Mii=1N⊂Mnℝ. Then,
(31)∑i=1N−1∑j=i+1NξiξjσmaxMi−Mjm≤12infM∈MnℝμM,where
(32)μM=2∑i=1NξiσmaxMi−Mm+∑k=1m−1mk∑i=1NξiσmaxMi−Mk∑i=1NξiσmaxM−Mim−k.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficient1/2 on the right-hand side of (31) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:Mnℝ×Mnℝ⟶0,∞ defined by
(33)δA,B=σmaxA−B,A,B∈Mnℝ.
Then,δ is a metric on Mnℝ (see, e.g., [5]). Hence, using (3) with δ as defined above, (31) follows. The optimality of (31) follows from Theorem 2.


### 3.3. A Multiplicative Metric Inequality

We first recall the notion of multiplicative metric spaces (see [6]). A multiplicative metric on a nonempty set V is a function σ:V×V⟶1,∞ satisfying the following properties: for all ϑ,ρ,κ∈V,
(i)
σϑ,ρ=1, if and only if ϑ=ρ(ii)
σϑ,ρ=σρ,ϑ(iii)
σϑ,ρ≤σϑ,κσκ,ρIn this case, we say thatV,σ is a multiplicative metric space.Example 12.
Letσ:0,∞×0,∞⟶1,∞ be the function defined by
(34)σϑ,ρ=ϑρ−1,ifϑ≥ρ,ρϑ−1,ifϑ<ρ.
Thenσ is a multiplicative metric on 0,∞.Proposition 13.
LetV be a nonempty set equipped with a multiplicative metric σ. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(35)∑i=1N−1∑j=i+1Nξiξjlnσϑi,ϑjm≤12infϑ∈V2∑i=1Nξilnσϑi,ϑm+∑k=1m−1mk∑i=1Nξilnσϑi,ϑk∑i=1Nξilnσϑ,ϑim−k.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficient1/2 on the right-hand side of (35) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:V×V⟶0,∞ defined by
(36)δϑ,ρ=lnσϑ,ρ,ϑ,ρ∈V.
It can be easily seen thatδ is a metric on V. Then, using (3) with δ as defined above, (35) follows. The optimality of (35) follows from Theorem 2.


### 3.4. A Partial Metric Inequality

We first recall briefly some notions on partial metric spaces (see, e.g., [7–10]).A partial metric on a nonempty setV is a function η:V×V⟶0,∞ satisfying the following properties: for all ϑ,ρ,κ∈V,
(37)ηϑ,ϑ=ηρ,ρ=ηϑ,ρ⇔ϑ=ρ,ηϑ,ϑ≤ηϑ,ρ,ηϑ,ρ=ηρ,ϑ,ηϑ,ρ≤ηϑ,κ+ηκ,ρ−ηκ,κ.In this case, we say thatV,η is a partial metric space.Example 14.
Letη:0,∞×0,∞⟶0,∞ be the function defined by
(38)ηϑ,ρ=maxϑ,ρ,ϑ,ρ≥0.
Then,η is a partial metric on 0,∞.Proposition 15.
LetV be a nonempty set equipped with a partial metric η. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(39)∑i=1N−1∑j=i+1Nξiξj2ηϑi,ϑj−ηϑi,ϑi−ηϑj,ϑjm≤12infϑ∈VΞϑ,where
(40)Ξϑ=2∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑm+∑k=1m−1mk∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑk×∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑm−k,for all ϑ∈V. Moreover, the inequality is optimal, in the sense that the multiplicative coefficient 1/2 on the right-hand side of (39) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:V×V⟶0,∞ defined by
(41)δϑ,ρ=2ηϑ,ρ−ηϑ,ϑ−ηρ,ρ,ϑ,ρ∈V.
It can be easily seen thatδ is a metric on V. Then, using (3) with δ as defined above, (39) follows. The optimality of (39) follows from Theorem 2.


## 3.1. A Metric Inequality in Lebesgue Spaces

Consider a measure spaceχ,M,μ and a real number r∈1,∞. We denote by Lrχ,M,μ the space of measurable functions f such that
(25)∫χfrdμ<∞.Proposition 10.
Letm,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and fii=1N⊂Lrχ,M,μ. Then,
(26)∑i=1N−1∑j=i+1Nξiξj∫χfi−fjrdμm/r≤12inff∈Lrχ,M,μΛf,where
(27)Λf=2∑i=1Nξi∫χfi−frdμm/r+∑k=1m−1mk∑i=1Nξi∫χfi−frdμk/r∑i=1Nξi∫χfi−frdμm−k/r,for all f∈Lrχ,M,μ. Moreover, this inequality is optimal, in the sense that the multiplicative coefficient 1/2 on the right-hand side of (26) cannot be replaced by a smaller real number.Proof.
Consider the distance function(28)δf,g=∫χf−grdμ1/r,f,g∈Lrχ,M,μ.
Then,Lrχ,M,μ,δ is a metric space (see, e.g., [3]). Hence, using (3) with δ as defined above, (26) follows. The optimality of (26) follows from Theorem 2.


## 3.2. A Matrix Inequality

We denote byMnℝ the set of square matrices of size n∈ℕ, n≥2, with real number coefficients. Let M∈Mnℝ. We denote by ρM the spectral radius of M, namely,
(29)ρM=maxλi:i=1,2,⋯,n,where λi,i∈1,2,⋯,n, are the eigenvalues of M. We denote by σmaxM the largest singular value of M, namely,
(30)σmaxM=ρMMt,where Mt is the transpose of M. For more details on matrix analysis, see, for example, [4, 5].Proposition 11.
Letm,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and Mii=1N⊂Mnℝ. Then,
(31)∑i=1N−1∑j=i+1NξiξjσmaxMi−Mjm≤12infM∈MnℝμM,where
(32)μM=2∑i=1NξiσmaxMi−Mm+∑k=1m−1mk∑i=1NξiσmaxMi−Mk∑i=1NξiσmaxM−Mim−k.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficient1/2 on the right-hand side of (31) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:Mnℝ×Mnℝ⟶0,∞ defined by
(33)δA,B=σmaxA−B,A,B∈Mnℝ.
Then,δ is a metric on Mnℝ (see, e.g., [5]). Hence, using (3) with δ as defined above, (31) follows. The optimality of (31) follows from Theorem 2.


## 3.3. A Multiplicative Metric Inequality

We first recall the notion of multiplicative metric spaces (see [6]). A multiplicative metric on a nonempty set V is a function σ:V×V⟶1,∞ satisfying the following properties: for all ϑ,ρ,κ∈V,
(i)
σϑ,ρ=1, if and only if ϑ=ρ(ii)
σϑ,ρ=σρ,ϑ(iii)
σϑ,ρ≤σϑ,κσκ,ρIn this case, we say thatV,σ is a multiplicative metric space.Example 12.
Letσ:0,∞×0,∞⟶1,∞ be the function defined by
(34)σϑ,ρ=ϑρ−1,ifϑ≥ρ,ρϑ−1,ifϑ<ρ.
Thenσ is a multiplicative metric on 0,∞.Proposition 13.
LetV be a nonempty set equipped with a multiplicative metric σ. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(35)∑i=1N−1∑j=i+1Nξiξjlnσϑi,ϑjm≤12infϑ∈V2∑i=1Nξilnσϑi,ϑm+∑k=1m−1mk∑i=1Nξilnσϑi,ϑk∑i=1Nξilnσϑ,ϑim−k.
Moreover, the inequality is optimal, in the sense that the multiplicative coefficient1/2 on the right-hand side of (35) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:V×V⟶0,∞ defined by
(36)δϑ,ρ=lnσϑ,ρ,ϑ,ρ∈V.
It can be easily seen thatδ is a metric on V. Then, using (3) with δ as defined above, (35) follows. The optimality of (35) follows from Theorem 2.


## 3.4. A Partial Metric Inequality

We first recall briefly some notions on partial metric spaces (see, e.g., [7–10]).A partial metric on a nonempty setV is a function η:V×V⟶0,∞ satisfying the following properties: for all ϑ,ρ,κ∈V,
(37)ηϑ,ϑ=ηρ,ρ=ηϑ,ρ⇔ϑ=ρ,ηϑ,ϑ≤ηϑ,ρ,ηϑ,ρ=ηρ,ϑ,ηϑ,ρ≤ηϑ,κ+ηκ,ρ−ηκ,κ.In this case, we say thatV,η is a partial metric space.Example 14.
Letη:0,∞×0,∞⟶0,∞ be the function defined by
(38)ηϑ,ρ=maxϑ,ρ,ϑ,ρ≥0.
Then,η is a partial metric on 0,∞.Proposition 15.
LetV be a nonempty set equipped with a partial metric η. Let m,N∈ℕ, N≥2, ξii=1N⊂0,∞, ∑i=1Nξi=1, and ϑii=1N⊂V. Then,
(39)∑i=1N−1∑j=i+1Nξiξj2ηϑi,ϑj−ηϑi,ϑi−ηϑj,ϑjm≤12infϑ∈VΞϑ,where
(40)Ξϑ=2∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑm+∑k=1m−1mk∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑk×∑i=1Nξi2ηϑi,ϑ−ηϑi,ϑi−ηϑ,ϑm−k,for all ϑ∈V. Moreover, the inequality is optimal, in the sense that the multiplicative coefficient 1/2 on the right-hand side of (39) cannot be replaced by a smaller real number.Proof.
Consider the functionδ:V×V⟶0,∞ defined by
(41)δϑ,ρ=2ηϑ,ρ−ηϑ,ϑ−ηρ,ρ,ϑ,ρ∈V.
It can be easily seen thatδ is a metric on V. Then, using (3) with δ as defined above, (39) follows. The optimality of (39) follows from Theorem 2.


## 4. Conclusion

A new inequality in metric spaces is proved. This inequality is a generalization of that derived by Dragomir and Gosa [1]. Moreover, we provided a geometric interpretation of our main result (see Corollary 7) and discussed some special cases including Lebesgue spaces, matrices inequalities, multiplicative metric inequalities, and partial metric inequalities.



---

*Source: 3842879-2020-10-05.xml*