Some Multiple Integral Inequalities Via the Divergence Theorem

Journal of Mathematical Inequalities Volume 14, Number 3 (2020), 661–671 doi:10.7153/jmi-2020-14-42

SOME MULTIPLE INTEGRAL INEQUALITIES VIA THE DIVERGENCE THEOREM

SILVESTRU SEVER DRAGOMIR

(Communicated by J. Pecari´ˇ c)

Abstract. In this paper, by the use of the divergence theorem, we establish some inequalities for functions deﬁned on closed and bounded subsets of the Euclidean space Rn, n 2.

1. Introduction

Let ∂D be a simple, closed counterclockwise curve bounding a region D and f deﬁned on an open set containing D and having continuous partial derivatives on D. In the recent paper [4], by the use of Green’s identity, we have shown among others that 1 f (x,y)dxdy− [(β − y) f (x,y)dx+(x − α) f (x,y)dy] D 2 ∂D 1 ∂ f (x,y) ∂ f (x,y) |α − x| + |β − y| dxdy =: M (α,β; f ) (1.1) 2 D ∂x ∂y for all α, β ∈ C and ⎧ ⎪ ∂ f |α − | + ∂ f |β − | ⎪ ∂ D x dxdy ∂ D y dxdy; ⎪ x D,∞ y D,∞ ⎪ ⎪ ⎨⎪ ∂ / ∂ / f ( |α − x|q dxdy)1 q + f ( |β − y|q dxdy)1 q (α,β ) ∂x , D ∂y , D M ; f ⎪ D p D p ⎪ where p, q > 1, 1 + 1 = 1; ⎪ p q ⎪ ⎪ ⎩⎪ |α − | ∂ f + |β − | ∂ f , sup(x,y)∈D x ∂ sup(x,y)∈D y ∂ x D,1 y B,1 (1.2)

Mathematics subject classiﬁcation (2010): 26D15. Keywords and phrases: Multiple integral inequalities, divergence theorem, Green identity, Gauss- Ostrogradsky identity.

Ð c ,Zagreb 661 Paper JMI-14-42 662 S. S. DRAGOMIR

· where D,p are the usual Lebesgue norms, we recall that ⎧ ⎨ ( | ( , )|p )1/p , D g x y dxdy p 1; = g D,p : ⎩ | ( , )|, = ∞. sup(x,y)∈D g x y p

Applications for rectangles and disks were also provided in [4]. For some recent double integral inequalities see [1], [2]and[3]. We also considered similar inequalities for 3-dimensional bodies as follows, see [5]. Let B be a solid in the three dimensional space R3 bounded by an orientable closed surface ∂B.Iff : B → C is a continuously differentiable function deﬁned on a open set containing B, then by making use of the Gauss-Ostrogradsky identity,we have obtained the following inequality 1 f (x,y,z)dxdydz− (x − α) f (x,y,z)dy∧ dz B 3 ∂B + (y − β) f (x,y,z)dz∧ dx+ (z − γ) f (x,y,z)dx∧ dy ∂B ∂B 1 ∂ f (x,y,z) ∂ f (x,y,z) ∂ f (x,y,z) |α − x| + |β − y| + |γ − z| dxdydz 3 B ∂x ∂y ∂z =:M (α,β,γ; f ) (1.3) for all α, β, γ complex numbers. Moreover, we have the bounds

M (α,β,γ; f ) ⎧ ⎪ ∂ f |α − | + ∂ f |β − | ⎪ ∂ B x dxdydz ∂ B y dxdydz ⎪ x B,∞ y B,∞ ⎪ ⎪ + ∂ f |γ − | ⎪ ∂ B z dxdydz; ⎪ z B,∞ ⎪ ⎪ ⎪ ∂ / ∂ / ⎨ f ( |α − x|q dxdydz)1 q + f ( |β − y|q dxdydz)1 q 1 ∂x , B ∂y , B Bp B p (1.4) 3 ⎪ + ∂ f ( |γ − | )1/q , , > , 1 + 1 = ⎪ ∂z B z dxdydz p q 1 p q 1; ⎪ B,p ⎪ ⎪ ⎪ ∂ f ∂ f ⎪ sup( , , )∈ |α − x| + sup( , , )∈ |β − y| ⎪ x y z B ∂x , x y z B ∂y , ⎪ B1 B 1 ⎩ + |γ − | ∂ f . sup(x,y,z)∈B z ∂ z B,1

Applications for 3-dimensional balls were also given in [5]. For some triple integral inequalities see [6]and[9]. Motivated by the above results, in this paper we establish several similar inequalities for multiple integrals for functions deﬁned on bonded subsets of Rn (n 2) with smooth (or piecewise smooth) boundary ∂B. To achieve this goal we make use of the well known divergence theorem for multiple integrals as summarized below. SOME MULTIPLE INTEGRAL INEQUALITIES 663

2. Some preliminary facts

Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) n boundary ∂B.LetF =(F1,...,Fn) be a smooth vector ﬁeld deﬁned in R , or at least in B∪ ∂B.Letn be the unit outward-pointing normal of ∂B. Then the Divergence Theorem states, see for instance [8]: divFdV = F · ndA, (2.1) B ∂B where n ∂ = ∇ · = Fk , divF F ∑ ∂ k=1 xk dV is the element of volume in Rn and dA is the element of surface area on ∂B. If n =(n1,...,nn), x =(x1,...,xn) ∈ B and use the notation dx for dV we can write (2.1) more explicitly as

n n ∂Fk (x) ∑ dx = ∑ Fk (x)nk (x)dA. (2.2) ∂ ∂ k=1 B xk k=1 B By taking the real and imaginary part, we can extend the above equality for complex valued functions Fk, k ∈{1,...,n} deﬁned on B. If n = 2, the normal is obtained by rotating the tangent vector through 90◦ (in the correct direction so that it points out). The quantity tds can be written (dx1,dx2) along the surface, so that ndA := nds =(dx2,−dx1) Here t is the tangent vector along the boundary curve and ds is the element of arc-length. From (2.2)wegetforB ⊂ R2 that ∂F1 (x1,x2) ∂F2 (x1,x2) dx1dx2+ dx1dx2 = F1 (x1,x2)dx2− F2 (x1,x2)dx1, B ∂x1 B ∂x2 ∂B ∂B (2.3) which is Green’s theorem in plane. Ifn = 3andif∂B is described as a level-set of a function of 3 variables i.e. 3 ∂B = x1,x2,x3 ∈ R | G(x1,x2,x3)=0 , then a vector pointing in the direction of n is gradG. We shall use the case where G(x1,x2,x3)=x3 − g(x1,x2), (x1,x2) ∈ D, a domain in R2 for some differentiable function g on D and B corresponds to the inequality x3 < g(x1,x2), namely 3 B = (x1,x2,x3) ∈ R | x3 < g(x1,x2) .

Then (−g ,−g ,1) / = x1 x2 , = + 2 + 2 1 2 n / dA 1 gx gx dx1dx2 1 + g2 + g2 1 2 1 2 x1 x2 664 S. S. DRAGOMIR and =(− ,− , ) . ndA gx1 gx2 1 dx1dx2 From (2.2)weget ∂F (x ,x ,x ) ∂F (x ,x ,x ) ∂F (x ,x ,x ) 1 1 2 3 + 2 1 2 3 + 3 1 2 3 dx dx dx ∂x ∂x ∂x 1 2 3 B 1 2 3 = − ( , , ( , )) ( , ) − ( , , ( , )) ( , ) F1 x1 x2 g x1 x2 gx1 x1 x2 dx1dx2 F1 x1 x2 g x1 x2 gx2 x1 x2 dx1dx2 D D

+ F3 (x1,x2,g(x1,x2))dx1dx2 (2.4) D which is the Gauss-Ostrogradsky theorem in space.

3. Identities of interest

We have the following identity of interest:

THEOREM 1. Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) boundary ∂B. Let f be a continuously differentiable function de- n ﬁned in R ,oratleastinB∪ ∂B and with complex values. If αk, βk ∈ C for ∈{ ,..., } ∑n α = , k 1 n with k=1 k 1 then n ∂ f (x) n f (x)dx= ∑ (βk − αkxk) dx+ ∑ (αkxk − βk) f (x)nk (x)dA. (3.1) ∂ ∂ B k=1 B xk k=1 B We also have 1 n ∂ f (x) 1 n f (x)dx = ∑ (γk − xk) dx + ∑ (xk − γk) f (x)nk (x)dA (3.2) ∂ ∂ B n k=1 B xk n k=1 B for all γk ∈ C,wherek∈{1,...,n}.

Proof. Let x =(x1,...,xn) ∈ B. We consider

Fk (x)=(αkxk − βk) f (x), k ∈{1,...,n}

∂ ( ) and take the partial derivatives Fk x to get ∂xk

∂Fk (x) ∂ f (x) = αk f (x)+(αkxk − βk) , k ∈{1,...,n}. ∂xk ∂xk If we sum this equality over k from 1 to n we get n ∂ ( ) n n ∂ ( ) n ∂ ( ) Fk x = α ( )+ (α − β ) f x = ( )+ (α − β ) f x ∑ ∂ ∑ k f x ∑ kxk k ∂ f x ∑ kxk k ∂ k=1 xk k=1 k=1 xk k=1 xk (3.3) SOME MULTIPLE INTEGRAL INEQUALITIES 665 for all x =(x1,...,xn) ∈ B. Now, if we take the integral in the equality (3.3) over (x1,...,xn) ∈ B we get n ∂ ( ) n ∂ ( ) Fk x = ( ) + (α − β ) f x . ∑ ∂ dx f x dx ∑ kxk k ∂ dx (3.4) B k=1 xk B k=1 B xk By the Divergence Theorem (2.2)wealsohave n n ∂Fk (x) ∑ dx = ∑ (αkxk − βk) f (x)nk (x)dA (3.5) ∂ ∂ B k=1 xk k=1 B and by making use of (3.4)and(3.5)weget n ∂ f (x) n f (x)dx + ∑ (αkxk − βk) dx = ∑ (αkxk − βk) f (x)nk (x)dA ∂ ∂ B k=1 B xk k=1 B which gives the desired representation (3.1). α = 1 β = 1 γ , ∈{ ,..., }. The identity (3.2) follows by (3.1)for k n and k n k k 1 n For the body B we consider the coordinates for the centre of gravity

G(xB,1,...,xB,n) deﬁned by 1 xB,k := xkdx, k ∈{1,...,n}, V (B) B where V (B) := xdx B is the volume of B.

COROLLARY 1. With the assumptions of Theorem 1 we have n ∂ f (x) n f (x)dx = ∑ αk xB,k − xk dx + ∑ αk xk − xB,k f (x)nk (x)dA ∂ ∂ B k=1 B xk k=1 B (3.6) and, in particular, 1 n ∂ f (x) 1 n f (x)dx = ∑ xB,k − xk dx + ∑ xk − xB,k f (x)nk (x)dA. ∂ ∂ B n k=1 B xk n k=1 B (3.7)

The proof follows by (3.1)ontakingβk = αkxB,k, k ∈{1,...,n}. For a function f as in Theorem 1 above, we deﬁne the points ∂ f (x) xk ∂ dx = B xk , ∈{ ,..., }, xB,∂ f ,k : ∂ ( ) k 1 n f x dx B ∂xk provided that all denominators are not zero. 666 S. S. DRAGOMIR

COROLLARY 2. With the assumptions of Theorem 1 we have n f (x)dx = ∑ αk xk − xB,∂ f ,k f (x)nk (x)dA (3.8) ∂ B k=1 B and, in particular, 1 n f (x)dx = ∑ xk − xB,∂ f ,k f (x)nk (x)dA. (3.9) ∂ B n k=1 B

The proof follows by (3.1)ontakingβk = αkxB,∂ f ,k, k ∈{1,...,n} and observing that n ∂ ( ) n ∂ ( ) (β − α ) f x = α − f x = . ∑ k kxk ∂ dx ∑ k xB,∂ f ,k xk ∂ dx 0 k=1 B xk k=1 B xk For a function f as in Theorem 1 above, we deﬁne the points ( ) ( ) , = ∂B xk f x nk x dA, ∈{ ,..., } x∂B, f k : ( ) ( ) k 1 n ∂B f x nk x dA provided that all denominators are not zero.

COROLLARY 3. With the assumptions of Theorem 1 we have n ∂ ( ) ( ) = α , − f x f x dx ∑ k x∂B, f k xk ∂ dx (3.10) B k=1 B xk and, in particular, n ∂ ( ) ( ) = 1 , − f x . f x dx ∑ x∂B, f k xk ∂ dx (3.11) B n k=1 B xk

The proof follows by (3.1)ontakingβk = αkx∂B, f ,k , k ∈{1,...,n} and observing that n ∑ (αkxk − βk) f (x)nk (x)dA = 0. ∂ k=1 B

4. Some integral inequalities

We have the following result generalizing the inequalities from the introduction:

THEOREM 2. Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) boundary ∂B. Let f be a continuously differentiable function de- n ﬁned in R ,oratleastinB∪ ∂B and with complex values. If αk, βk ∈ C for SOME MULTIPLE INTEGRAL INEQUALITIES 667

∈{ ,..., } ∑n α = , k 1 n with k=1 k 1 then n ( ) − (α − β ) ( ) ( ) f x dx ∑ kxk k f x nk x dA B = ∂B k 1 ⎧ ⎪ ∑n |β − α | ∂ f (x) ⎪ k=1 B k kxk dx ∂ ⎪ xk B,∞ ⎪ ⎪ ⎪ / ∂ ( ) n ⎨ n q 1 q f x ∂ f (x) ∑ = ( |βk − αkxk| dx) ∂ k 1 B xk , ∑ |βk − αkxk| dx B p (4.1) ∂ ⎪ 1 1 k=1 B xk ⎪ where p, q > 1, + = 1; ⎪ p q ⎪ ⎪ ⎩⎪ n |β − α | ∂ f (x) ∑k=1 supx∈B k kxk ∂ xk B,1

We also have n ( ) − 1 ( − γ ) ( ) ( ) f x dx ∑ xk k f x nk x dA B n = ∂B k 1 ⎧ ⎪ ∑n |γ − | ∂ f (x) ⎪ k=1 B k xk dx ∂ ⎪ xk B,∞ ⎪ ⎪ ⎪ / ∂ ( ) n ⎨ n q q 1 q f x 1 ∂ f (x) 1 ∑ = |γk − xk| dx ∂ k 1 B xk , ∑ |γk − xk| dx B p (4.2) ∂ ⎪ 1 1 n k=1 B xk n ⎪ where p, q > 1, + = 1; ⎪ p q ⎪ ⎪ ⎩⎪ n |γ − | ∂ f (x) ∑k=1 supx∈B k xk ∂ xk B,1 for all γk ∈ C,wherek∈{1,...,n}.

Proof. By the identity (3.1)wehave n f (x)dx− ∑ (αkxk − βk) f (x)nk (x)dA B ∂B k=1 n ∂ ( ) n ∂ ( ) = (β − α ) f x (β − α ) f x ∑ k kxk ∂ dx ∑ k kxk ∂ dx k=1 B xk k=1 B xk n ∂ ( ) (β − α ) f x , ∑ k kxk ∂ dx k=1 B xk which proves the ﬁrst inequality in (4.1). 668 S. S. DRAGOMIR

By H¨older’s integral inequality for multiple integrals we have ∂ f (x) (βk − αkxk) dx ∂x B k ⎧ ⎧ ∂ ⎪ ∂ f (x) |β − α | ⎪ |β − α | f ⎪ supx∈B ∂ k kxk dx ⎪ B k kxk dx ∂x ⎪ xk B ⎪ k B,∞ ⎪ ⎪ ⎪ ⎪ ⎪ p 1/p ⎪ / ∂ ⎨ ∂ f (x) q 1/q ⎨ q 1 q f ( |β − α | ) ( |βk − αkxk| dx) ∂ B ∂x B k kxk dx = B xk , , ⎪ k ⎪ B p ⎪ , > , 1 + 1 = ⎪ where p, q > 1, 1 + 1 = 1; ⎪ where p q 1 p q 1; ⎪ p q ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ |β − α | ∂ f (x) ⎩⎪ |β − α | ∂ f supx∈B k kxk ∂ dx supx∈B k kxk ∂ B xk xk B,1 which proves the last part of (4.1).

COROLLARY 4. With the assumptions of Theorem 2 we have n ( ) − 1 − ( ) ( ) f x dx ∑ xk xB,k f x nk x dA B n = ∂B k 1 ⎧ ⎪ ∑n − ∂ f (x) ⎪ k=1 B xB,k xk dx ∂ ⎪ xk B,∞ ⎪ ⎪ ⎪ / ∂ ( ) n ⎨ n q q 1 q f x 1 ∂ f (x) 1 ∑ = xB,k − xk dx ∂ k 1 B xk , ∑ xB,k − xk dx B p (4.3) ∂ ⎪ 1 1 n k=1 B xk n ⎪ where p, q > 1, + = 1; ⎪ p q ⎪ ⎪ ⎩⎪ n − ∂ f (x) ∑k=1 supx∈B xB,k xk ∂ xk B,1 and n ∂ ( ) ( ) 1 , − f x f x dx ∑ x∂B, f k xk dx B n B ∂x ⎧k=1 k ⎪ ∑n , − ∂ f (x) ⎪ k=1 B x∂B, f k xk dx ∂ ⎪ xk B,∞ ⎪ ⎪ ⎪ / ⎨ ∑n q , − q 1 q ∂ f (x) 1 k=1 B x∂B, f k xk dx ∂ xk B,p (4.4) n ⎪ , > , 1 + 1 = ⎪ where p q 1 p q 1; ⎪ ⎪ ⎪ ∂ ( ) ⎩ ∑n , − f x . k=1 supx∈B x∂B, f k xk ∂ xk B,1

We also have the dual result: SOME MULTIPLE INTEGRAL INEQUALITIES 669

THEOREM 3. With the assumption of Theorem 2 we have n ∂ ( ) ( ) − (β − α ) f x f x dx ∑ k kxk ∂ dx B k=1 B xk ⎧ n ⎪ f ∂ ,∞ ∑ = ∂ |αkxk − βk||nk (x)|dA; ⎪ B k 1 B ⎪ ⎨⎪ n ∑n ( |α −β |q| ( )|q )1/q f ∂B,p k=1 ∂B kxk k nk x dA ∑ |αkxk − βk||nk (x)||f (x)|dA ∂ ⎪ , > , 1 + 1 = k=1 B ⎪ where p q 1 p q 1; ⎪ ⎪ ⎩ n ||α − β || ( )||, f ∂B,1 ∑k=1 supx∈∂B kxk k nk x (4.5) where ⎧ ⎨ ( | ( )|p )1/p , ∂B f x dA p 1; = f ∂B,p : ⎩ | ( )|, = ∞. supx∈∂B f x p

In particular, n ∂ ( ) ( ) − 1 (γ − ) f x f x dx ∑ k xk dx B n = B ∂xk k 1 ⎧ n ⎪ f ∂ ,∞ ∑ = ∂ |γk − xk||nk (x)|dA; ⎪ B k 1 B ⎪ ⎨⎪ n ∑n ( |γ − |q | ( )|q )1/q 1 1 f ∂B,p k=1 ∂B k xk nk x dA ∑ |γk − xk||nk (x)||f (x)|dA n ∂ n ⎪ where p, q > 1, 1 + 1 = 1; k=1 B ⎪ p q ⎪ ⎩⎪ ∑n [|γ − || ( )|]. f ∂B,1 k=1 supx∈∂B k xk nk x (4.6)

Proof. From the identity (3.1)wehave n ∂ f (x) f (x)dx− ∑ (βk − αkxk) dx B B ∂xk k=1 n n =∑ (αkxk − βk) f (x)nk (x)dA ∑ (αkxk − βk) f (x)nk (x)dA ∂ ∂ k=1 B k=1 B n ∑ |(αkxk − βk) f (x)nk (x)|dA, ∂ k=1 B which proves the ﬁrst inequality in (4.5). 670 S. S. DRAGOMIR

By H¨older’s inequality for functions deﬁned on ∂B we have ⎧ ⎪ ∂ |αkxk − βk||nk (x)|dA f ∂ ,∞ ; ⎪ B B ⎪ ⎨⎪ ( |α − β |q | ( )|q )1/q ∂B kxk k nk x dA f ∂B,p |αkxk − βk||nk (x)||f (x)|dA ∂ ⎪ , > , 1 + 1 = B ⎪ where p q 1 p q 1; ⎪ ⎪ ⎩ ||α − β || ( )|| , supx∈∂B kxk k nk x f ∂B,1 which proves the second part of the inequality (4.5). We also have:

COROLLARY 5. With the assumptions of Theorem 2 we have n ∂ ( ) ( ) − 1 − f x f x dx ∑ xB,k xk ∂ dx B n k=1 B xk n 1 ∑ xB,k − xk |nk (x)||f (x)|dA n = ∂B ⎧k 1 ⎪ ∑n − | ( )| ⎪ f ∂B,∞ k=1 ∂B xB,k xk nk x dA; ⎪ ⎪ ⎨ n q q 1/q 1 f ∑ x , − x |n (x)| dA ∂B,p k=1 ∂B B k k k (4.7) n ⎪ where p, q > 1, 1 + 1 = 1; ⎪ p q ⎪ ⎩⎪ ∑n − | ( )| f ∂B,1 k=1 supx∈∂B xB,k xk nk x and n 1 f (x)dx ∑ xB,∂ f ,k − xk |nk (x)||f (x)|dA B n = ∂B ⎧k 1 n − | ( )| ⎪ f ∂ ,∞ ∑ = ∂ xB,∂ f ,k xk nk x dA; ⎪ B k 1 B ⎪ ⎪ / ⎨ n q q 1 q 1 f ∑ x ,∂ , − x |n (x)| dA ∂B,p k=1 ∂B B f k k k (4.8) n ⎪ where p, q > 1, 1 + 1 = 1; ⎪ p q ⎪ ⎩⎪ n − | ( )| . f ∂B,1 ∑k=1 supx∈∂B xB,∂ f ,k xk nk x

If we take n = 2 in Theorem 3, then we get other results from [4], while for n = 3 we recapture some results from [5].

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(Received June 26, 2019) Silvestru Sever Dragomir Mathematics, College of Engineering & Science Victoria University PO Box 14428, Melbourne City, MC 8001, Australia DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences School of Computer Science & Applied Mathematics, University of the Witwatersrand Private Bag 3, Johannesburg 2050, South Africa e-mail: [email protected]

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