SOME MULTIPLE INTEGRAL INEQUALITIES VIA THE DIVERGENCE THEOREM
SILVESTRUSEVERDRAGOMIR1;2
Abstract. In this paper, by the use of the divergence theorem, we establish some inequalities for functions de…ned 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 de…ned 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 (1.1) f (x; y) dxdy [( y) f (x; y) dx + (x ) f (x; y) dy] 2 ZZD I @D
1 @f (x; y) @f (x; y) x + y dxdy =: M ( ; ; f) 2 j j @x j j @y ZZD for all ; C and 2 (1.2) M ( ; ; f)
@f @f @x x dxdy + @y y dxdy; D; D j j D; D j j 1 1 8 RR RR > 1 =q 1=q > @f x q dxdy + @f y q dxdy > @x D @y D > D;p j j D;p j j > where p; q > 1; 1 + 1 = 1; > RR p q RR < > sup x @f + sup y @f ; > (x;y) D @x (x;y) D @y > 2 j j D;1 2 j j B;1 > where D;p> are the usual Lebesgue norms, we recall that kk : p 1=p D g (x; y) dxdy ; p 1; g := j j k kD;p 8 RR < sup(x;y) D g (x; y) ; p = : 2 j j 1 Applications for rectangles: and disks were also provided in [4]. For some recent double integral inequalities see [1], [2] and [3].
1991 Mathematics Subject Classi…cation. 26D15. Key words and phrases. Multiple integral inequalities, Divergence theorem, Green identity, Gauss-Ostrogradsky identity. 1 2 S.S.DRAGOMIR
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. If f : B C is a continuously di¤erentiable function de…ned on a open set containing B, then! by making use of the Gauss-Ostrogradsky identity, we have obtained the following inequality
1 (1.3) f (x; y; z) dxdydz (x ) f (x; y; z) dy dz 3 ^ ZZZB ZZ@B
+ (y ) f (x; y; z) dz dx + (z ) f (x; y; z) dx dy ^ ^ ZZ@B ZZ@B 1 @f (x; y; z) @f (x; y; z) x + y 3 j j @x j j @y ZZZB @f (x; y; z) + z dxdydz =: M ( ; ; ; f) j j @z
for all ; ; complex numbers. Moreover, we have the bounds (1.4) M ( ; ; ; f)
@f @f @x x dxdydz + @y y dxdydz B; B j j B; B j j 8 @f 1 1 + @z RRR z dxdydz; RRR > B; B j j > 1 > RRR > 1=q 1=q > @f x q dxdydz + @f y q dxdydz 1 > @x B @y B > B;p j j B;p j j 3 > @f 1=q 1 1 > + @z RRR z dxdydz ; p; q > 1; RRRp + q = 1; <> B;p B j j RRR > > sup x @f + sup y @f > (x;y;z) B @x (x;y;z) B @y > 2 j j B;1 2 j j B;1 > @f > + sup(x;y;z) B z @z : > 2 j j B;1 > > Applications:> for 3-dimensional balls were also given in [5]. For some triple inte- gral inequalities see [6] and [9]. Motivated by the above results, in this paper we establish several similar inequal- ities for multiple integrals for functions de…ned 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.
2. Some Preliminary Facts Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) boundary @B. Let F = (F1; :::; Fn) be a smooth vector …eld de…ned in Rn, or at least in B @B. Let n be the unit outward-pointing normal of @B. Then the Divergence Theorem[ states, see for instance [8]:
(2.1) div F dV = F ndA; ZB Z@B SOMEMULTIPLEINTEGRALINEQUALITIES 3 where n @F div F = F = i ; r @xi k=1 X 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 2 n n @Fk (x) (2.2) dx = Fk (x) nk (x) dA: @xk k=1 B k=1 @B X Z X Z By taking the real and imaginary part, we can extend the above equality for complex valued functions Fk; k 1; :::; n de…ned on B: 2 f g 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) we get for B R2 that @F (x ; x ) @F (x ; x ) (2.3) 1 1 2 dx dx + 2 1 2 dx dx @x 1 2 @x 1 2 ZB 1 ZB 2 = F1 (x1; x2) dx2 F2 (x1; x2) dx1; Z@B Z@B which is Green’s theorem in plane. If n = 3 and if @B is described as a level-set of a function of 3 variables i.e. @B = 3 x1; x2; x3 R G(x1; x2; x3) = 0 , then a vector pointing in the direction of n is 2 j grad G. We shall use the case where G (x1; x2; x3) = x3 g(x1; x2); (x1; x2) D; 2 a domain in R2 for some di¤erentiable 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) : 2 j Then ( gx1 ; gx2 ; 1) 2 2 1=2 n = ; dA = 1 + gx + gx dx1dx2 2 2 1=2 1 2 1 + gx1 + gx2 and ndA = ( gx ; gx ; 1) dx1dx2: 1 2 From (2.2) we get @F (x ; x ; x ) @F (x ; x ; x ) @F (x ; x ; x ) (2.4) 1 1 2 3 + 2 1 2 3 + 3 1 2 3 dx dx dx @x @x @x 1 2 3 ZB 1 2 3
= F1 (x1; x2; g(x1; x2)) gx (x1; x2) dx1dx2 1 ZD F1 (x1; x2; g(x1; x2)) gx (x1; x2)dx1dx2 2 ZD + F3 (x1; x2; g(x1; x2)) dx1dx2 ZD 4 S.S.DRAGOMIR 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 di¤erentiable function n de…ned in R , or at least in B @B and with complex values. If k; k C for n [ 2 k 1; :::; n with k = 1; then 2 f g k=1 P n @f (x) (3.1) f (x) dx = ( k kxk) dx @xk ZB k=1 ZB X n
+ ( kxk ) f (x) nk (x) dA: k k=1 @B X Z We also have
1 n @f (x) (3.2) f (x) dx = ( k xk) dx n @xk B k=1 B Z X Z 1 n + (xk ) f (x) nk (x) dA n k k=1 @B X Z for all k C, where k 1; :::; n : 2 2 f g Proof. Let x = (x1; :::; xn) B: We consider 2 Fk (x) = ( kxk ) f (x) ; k 1; :::; n k 2 f g and take the partial derivatives @Fk(x) to get @xk
@Fk (x) @f (x) = kf (x) + ( kxk k) ; k 1; :::; n : @xk @xk 2 f g If we sum this equality over k from 1 to n we get n n n @Fk (x) @f (x) (3.3) = kf (x) + ( kxk k) @xk @xk k=1 k=1 k=1 X X X n @f (x) = f (x) + ( kxk k) @xk k=1 X for all x = (x1; :::; xn) B: 2 Now, if we take the integral in the equality (3.3) over (x1; :::; xn) B we get 2 n n @Fk (x) @f (x) (3.4) dx = f (x) dx + ( kxk k) dx: @xk @xk B k=1 ! B k=1 B Z X Z X Z By the Divergence Theorem (2.2) we also have
n n @Fk (x) (3.5) dx = ( kxk k) f (x) nk (x) dA @xk B k=1 ! k=1 @B Z X X Z SOMEMULTIPLEINTEGRALINEQUALITIES 5 and by making use of (3.4) and (3.5) we get
n @f (x) f (x) dx + ( kxk k) dx @xk ZB k=1 ZB X n
= ( kxk ) f (x) nk (x) dA k k=1 @B X Z which gives the desired representation (3.1). 1 1 The identity (3.2) follows by (3.1) for k = and = ; k 1; :::; n : n k n k 2 f g For the body B we consider the coordinates for the centre of gravity
G (xB;1; :::; xB;n) de…ned by 1 xB;k := xkdx; k 1; :::; n ; V (B) 2 f g ZB where V (B) := xdx ZB is the volume of B: Corollary 1. With the assumptions of Theorem 1 we have n @f (x) (3.6) f (x) dx = k (xB;k xk) dx @xk ZB k=1 ZB X n
+ k (xk xB;k) f (x) nk (x) dA k=1 @B X Z and, in particular, 1 n @f (x) (3.7) f (x) dx = (xB;k xk) dx n @xk B k=1 B Z X Z 1 n + (xk xB;k) f (x) nk (x) dA: n k=1 @B X Z The proof follows by (3.1) on taking k = kxB;k; k 1; :::; n : For a function f as in Theorem 1 above, we de…ne the2 fpoints g @f(x) xk dx B @xk xB;@f;k := ; k 1; :::; n ; @f(x) dx 2 f g R B @xk provided that all denominators areR not zero. Corollary 2. With the assumptions of Theorem 1 we have n
(3.8) f (x) dx = k (xk xB;@f;k) f (x) nk (x) dA B k=1 @B Z X Z and, in particular, 1 n (3.9) f (x) dx = (xk xB;@f;k) f (x) nk (x) dA: n B k=1 @B Z X Z 6 S.S.DRAGOMIR
The proof follows by (3.1) on taking k = kxB;@f;k; k 1; :::; n and observing that 2 f g n @f (x) n @f (x) ( k kxk) dx = k (xB;@f;k xk) dx = 0: @xk @xk k=1 B k=1 B X Z X Z For a function f as in Theorem 1 above, we de…ne the points
xkf (x) nk (x) dA x ; := @B ; k 1; :::; n @B;f k f (x) n (x) dA R @B k 2 f g provided that all denominatorsR are not zero. Corollary 3. With the assumptions of Theorem 1 we have
n @f (x) (3.10) f (x) dx = k (x@B;f ;k xk) dx @xk B k=1 B Z X Z and, in particular,
1 n @f (x) (3.11) f (x) dx = (x@B;f ;k xk) dx: n @xk B k=1 B Z X Z The proof follows by (3.1) on taking k = kx@B;f ;k ; k 1; :::; n and observing that 2 f g n
( kxk ) f (x) nk (x) dA = 0: k k=1 @B X Z
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 di¤erentiable function n de…ned in R , or at least in B @B and with complex values. If k; k C for n [ 2 k 1; :::; n with k = 1; then 2 f g k=1 P n (4.1) f (x) dx ( kxk k) f (x) nk (x) dA B @B Z k=1 Z X n @f (x) k kxk dx j j @xk k=1 ZB X n @f(x) k=1 k kxk dx @x B j j k B; 8 1 P R > n q 1=q @f(x) > k kxk dx > k=1 B @xk > j j B;p > where p; q > 1; 1 + 1 = 1; > P R p q < > n @f(x) > k=1 supx B k kxk @x > 2 j j k B;1 > > P :> SOMEMULTIPLEINTEGRALINEQUALITIES 7
We also have
1 n (4.2) f (x) dx (xk k) f (x) nk (x) dA B n @B Z k=1 Z X n 1 @f (x) k xk dx n j j @xk k=1 ZB X n @f(x) k=1 k xk dx @x B j j k B; 8 1 P R > n q q 1=q @f(x) 1 > k xk dx > k=1 B @xk > j j B;p n > where p; q > 1; 1 + 1 = 1; > P R p q < > n @f(x) > k=1 supx B k xk @x > 2 j j k B;1 > > P for all k C, where k 1; :::; n : > 2 2 f g : Proof. By the identity (3.1) we have
n
f (x) dx ( kxk ) f (x) nk (x) dA k B k=1 @B Z X Z n n @f (x) @f (x) = ( k kxk) dx ( k kxk) dx @xk @xk k=1 ZB k=1 ZB X X n @f (x ) ( k kxk) dx; @xk k=1 ZB X which proves the …rst inequality in (4.1). By Hölder’sintegral inequality for multiple integrals we have
@f(x) supx B @x B k kxk dx 2 k j j 8 R p 1 =p @f (x) > @f( x) q 1=q > B @x B k kxk dx ( k kxk) dx > k j j B @xk > 1 1 Z > where p; q >1; p + q = 1; < R R
> @f(x) > supx B k kxk B @x dx > 2 j j k > > R@f : k kxk dx @x B j j k B; 8 1 R > q 1=q @f > k kxk dx > B @xk = > j j B;p ; > where p; q > 1; 1 + 1 = 1; > R p q < > @f > supx B k kxk @x > 2 j j k B;1 > > which proves the last part of (4.1). :> 8 S.S.DRAGOMIR
Corollary 4. With the assumptions of Theorem 2 we have
1 n (4.3) f (x) dx (xk xB;k) f (x) nk (x) dA B n @B Z k=1 Z X n 1 @f (x) xB;k xk dx n j j @xk k=1 ZB X n @f(x) k=1 xB;k xk dx @x B j j k B; 8 1 P R > n q q 1=q @f(x) 1 > xB;k xk dx > k=1 B @xk > j j B;p n > where p; q > 1; 1 + 1 = 1; > P R p q < > n @f(x) > k=1 supx B xB;k xk @x > 2 j j k B;1 > and > P :> 1 n @f (x) (4.4) f (x) dx x@B;f ;k xk dx n j j @xk ZB k=1 ZB X n @f(x) k=1 x@B;f ;k xk dx @x B j j k B; 8 1 P R > n q q 1=q @f(x) 1 > x@B;f ;k xk dx > k=1 B @xk > j j B;p n > where p; q > 1; 1 + 1 = 1; > P R p q < > n @f(x) > k=1 supx B x@B;f ;k xk @x : > 2 j j k B;1 > > P We also have the dual result: :> Theorem 3. With the assumption of Theorem 2 we have
n @f (x) (4.5) f (x) dx ( k kxk) dx B B @xk Z k=1 Z Xn
kxk nk (x) f (x) dA j kj j j j j k=1 Z@B X n f @B; k=1 @B kxk k nk (x) dA; k k 1 j j j j
8 Pn R q q 1=q > f @B;p k=1 @B kxk k nk (x) dA > k k 1 j 1 j j j > where p; q > 1; p + q = 1; <> P R f n sup x n (x) ; > @B;1 k=1 x @B k k k k > k k 2 jj j j jj > where :> P p 1=p @B f (x) dA ; p 1; f := j j k k@B;p 8 R < supx @B f (x) ; p = : 2 j j 1 : SOMEMULTIPLEINTEGRALINEQUALITIES 9
In particular,
1 n @f (x) (4.6) f (x) dx ( k xk) dx B n B @xk Z k=1 Z X n 1 xk nk (x) f (x) dA n j k j j j j j k=1 Z@B X n f @B; k=1 @B k xk nk (x) dA; k k 1 j j j j 8 P R 1 n q q 1=q > f @B;p k=1 @B k xk nk (x) dA > k k 1 j 1 j j j n > where p; q > 1; p + q = 1; <> P R f n sup [ x n (x) ] : > @B;1 k=1 x @B k k k > k k 2 j j j j > Proof. From the identity (3.1):> we have P
n @f (x) f (x) dx ( k kxk) dx @xk B k=1 B Z X Z n
= ( kxk k) f (x) nk (x) dA @B k=1 Z n X n
( kxk ) f (x) nk (x) dA ( kxk ) f (x) nk (x) dA; k j k j k=1 Z@B k=1 Z@B X X which proves the …rst inequality in (4.5). By Hölder’sinequality for functions de…ned on @B we have
@B kxk k nk (x) dA f @B; ; j j j j k k 1
8 R q q 1=q > @B kxk k nk (x) dA f @B;p kxk k nk (x) f (x) dA > j j1 j 1 j k k @B j j j j j j > where p; q > 1; p + q = 1; Z <> R
> supx @B kxk k nk (x) f @B;1 ; > 2 jj j j jj k k > > which proves the second part of the: inequality (4.5).
We also have:
Corollary 5. With the assumptions of Theorem 2 we have
1 n @f (x) (4.7) f (x) dx (xB;k xk) dx B n B @xk Z k=1 Z X n 1 xB;k xk nk (x) f (x) dA n j j j j j j k=1 @B X Z 10 S.S.DRAGOMIR
n f @B; k=1 @B xB;k xk nk (x) dA; k k 1 j j j j 8 P R 1 n q q 1=q > f @B;p k=1 @B xB;k xk nk (x) dA > k k 1 j 1 j j j n > where p; q > 1; p + q = 1; <> P R f n sup [ x x n (x) ] > @B;1 k=1 x @B B;k k k > k k 2 j j j j and > :> P 1 n (4.8) f (x) dx xB;@f;k xk nk (x) f (x) dA n j j j j j j ZB k=1 Z@B X n f @B; k=1 @B xB;@f;k xk nk (x) dA; k k 1 j j j j 8 P R 1 n q q 1=q > f @B;p k=1 @B xB;@f;k xk nk (x) dA > k k 1 j 1 j j j n > where p; q > 1; p + q = 1; <> P R f n sup [ x x n (x) ] : > @B;1 k=1 x @B B;@f;k k k > k k 2 j j j j > If we take n = 2 in Theorem:> 3, thenP we get other results from [4], while for n = 3 we recapture some results from [5].
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1Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia. E-mail address: [email protected] URL: http://rgmia.org/dragomir
2DST-NRF Centre of Excellence in the Mathematical, and Statistical Sciences, School of Computer Science, & Applied Mathematics, University of the Witwater- srand,, Private Bag 3, Johannesburg 2050, South Africa