SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE

RGMIA Reserch Report Collection, Vol., No., 998 http://sci.vut.edu.u/ rgmi SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE S.S. DRAGOMIR Astrct. Some clssicl nd new integrl inequlities of Grüss type re presented. Grüss Integrl Inequlity In 935, G. Grüss, proved the following integrl inequlity which gives n estimtion for the integrl of product in terms of the product of integrls (see for emple [, p. 96]) f () g () d f () d g () d (Φ ϕ) (Γ γ) ; 4 provided tht f nd g re two integrle functions on [, ] nd stisfying the condition (.) ϕ f () Φ, γ g () Γ for ll [, ]. The constnt 4 is the est possile nd is chieved for f () g () sgn ( ) +. We give here weighted version of Grüss inequlity Theorem.. Let f nd g e two functions defined nd integrle on [, ]. If (.) holds for ech [, ], where ϕ, Φ, γ, Γ re given rel constnts, nd h : [, ] [0, ) is integrle nd h()d > 0, then (.) h()d f () g () h()d f () h()d g () h()d (Φ ϕ) (Γ γ) 4 h()d nd the constnt 4 is the est possile. For the ske of completeness we give here simple proof of this fct which is similr with the clssicl one for unweighted cse (compre with [, p. 96]). Dte. Decemer, 998 99 Mthemtics Suject Clssifiction. Primry 6D5. Key words nd phrses. Grüss Inequlity, Cey sev Inequlity, Hölder Inequlity

96 Drgomir Let us note tht the following equlity is vlid: (.3) h()d f () g () h()d (.4) ( f () h()d h()d h()d ) h()d g () h()d (f () f (y)) (g () g (y)) h()h(y)ddy. Applying Cuchy-Bunikowski-Schwrz s integrl inequlity for doule integrls we hve ( h()d ) (f () f (y)) (g () g (y)) h()h(y)ddy ( h()d ) (f () f (y)) h()h(y)ddy ( h()d ) f () h()d h()d g () h()d h()d (g () g (y)) h()h(y)ddy h()d h()d f () h()d The following equlity lso holds f () h()d h()d h()d g () h()d. f () h()d

Grüss Integrl Inequlities 97 Φ h()d f () h()d h()d h()d (Φ f ()) (f () ϕ) h()d. As, (Φ f ()) (f () ϕ) 0 for ech [, ], then (.5) Φ h()d f () h()d h()d f () h()d Similrly, we hve (.6) Γ h()d g () h()d h()d g () h()d h()d f () h()d ϕ f () h()d h()d h()d f () h()d ϕ. g () h()d h()d Now, y (.3), (.4), (.5) nd (.6) we get (.7) f () g () h()d h()d Φ h()d f () h()d h()d f () h()d h()d g () h()d γ. g () h()d h()d f () h()d ϕ

98 Drgomir Γ h()d g () h()d h()d g () h()d γ. Using the elementry inequlity for rel numers: 4pq (p + q), p, q R we cn stte 4 Φ (.8) h()d f () h()d h()d f () h()d ϕ (Φ ϕ) nd (.9) 4 Γ h()d g () h()d h()d g () h()d γ (Γ γ). Now, comining (.7) with (.8) nd (.9) we deduce the desired inequlity (.). To prove the shrpness of (.), let choose h(), f () g () sgn ( ) + for ll [, ]. Then f () d, f () d g () d 0, Φ ϕ Γ γ nd the equlity in (.) is relized. For other inequlities of Grüss type see the ook [], where mny other references re given. We omit the detils.

Grüss Integrl Inequlities 99 The Cse When Both Mppings Are Lipschitzin The following inequlity of Grüss type for lipschitzin mppings holds : Theorem.. Let f, g : [, ] R e two lipschitzin mppings with the constnts L > 0 nd L > 0, i.e., (.) f () f (y) L y, g () g (y) L y for ll, y [, ]. If p : [, ] [0, ) is integrle, then (.) p () d p () f () g () d p () f () d p () g () d L L nd the inequlity is shrp. Proof. By (.) we hve tht p () d p () d p () d (f () f (y)) (g () g (y)) L L ( y) for ll, y [, ]. Multiplying y p () p (y) 0 nd integrting on [, ], we get p () p (y) (f () f (y)) (g () g (y)) ddy nd L L As it is esy to see tht p () p (y) (f () f (y)) (g () g (y)) ddy p () p (y) ( y) ddy. (f () f (y)) (g () g (y)) p () p (y) ddy p () d p () f () g () d p () p (y) ( y) ddy p () d p () f () d p () d p () g () d p () d the inequlity (.) is thus otined. Now, if we chose f () L, g () L, then f is L lipschitzin, g is L lipschitzin nd the equlity in (.) is relized for ny p s ove.

00 Drgomir Corollry.. Under the ove ssumptions, we hve f () g () d (.3) f () d g () d The constnt is the est possile. L L ( ). We note tht the ove corollry is nturl generliztion of well-known result y Cey sev (see for emple [, p. 97]) : Corollry.3. Let f, g : [, ] R e two differentile mppings whose derivtives re ounded on (, ). Denote f f (t) <. Then we hve the inequlity: (.4) sup t (,) f () g () d f () d g () d The constnt is the est possile. f g ( ). 3 The Cse When f Is Lipschitzin We re le now to prove nother inequlity of Grüss type ssuming tht only one mpping is lipschitzin s follows: Theorem 3.. Let f : [, ] R e M lipschitzin mpping on [, ]. Then we hve the inequlity: f () g () d (3.) f () d g () d M g provided tht g L [, ] [ M (p + ) (p + ) ] p ( ) p g q provided tht g L q [, ] p > nd p + q ; Proof. We hve tht ( )3 M g 3 provided tht g L [, ]. f () g (y) f (y) g (y) M y g (y) for ll, y [, ], from where, y integrtion on [, ], we get tht (f () g (y) f (y) g (y)) ddy M y g (y) ddy.

Grüss Integrl Inequlities 0 But (f () g (y) f (y) g (y)) ddy Now, if g L [, ], then f () d g () d ( ) y g (y) ddy ( ) m y g (y) dy ( ) g. (,y) [,] f () g () d. Now, ssume tht p > nd p + q, g L q [, ]. Then y Hölder s integrl inequlity we hve: where nd then we get p y p ddy K : y g (y) ddy y p ddy g (y) q ddy y p dy + q y p dy d y p dy d [ ] ( ) p+ + ( ) p+ d p + [ y g (y) ddy (p + ) (p + ) Finlly, ssuming tht g L [, ], we hve tht The theorem is thus proved. y g (y) ddy g The following corollry is importnt in pplictions. y ddy K p ( ) q g q ( )p+ (p + ) (p + ) ] p ( ) + p g q. ( )3 3 g. Corollry 3.. Let f : [, ] R e differentile mpping whose derivtive is ounded on (, ). Then we hve the inequlity: f () g () d (3.) f () d g () d

0 Drgomir f g provided tht g L [, ] [ ] p ( ) p f (p + ) (p + ) g q provided tht g L q [, ] p >, p + q ; ( ) 3 f g provided tht g L [, ]. 4 The Cse When f Is M g-lipschitzin Another generliztion of Grüss integrl inequlity is emodied in the following theorem: Theorem 4.. Let f, g : [, ] R e two integrle mppings on [, ] such tht (4.) f () f (y) M g () g (y) for ll, y [, ]. Then we hve the inequlity: (4.) p () d M p () f () g () d p () d p () g () d p () f () d p () g () d p () g () d where p : [, ] [0, ) is n ritrry integrle function on [, ]. The inequlity (4.) is shrp. Proof. By condition (4.) we hve (f () f (y)) (g () g (y)) M (g () g (y)) for ll, y [, ]. Multiplying y p () p (y) 0 nd integrting on [, ] we get p () p (y) (f () f (y)) (g () g (y)) ddy M p () p (y) (f () f (y)) (g () g (y)) ddy p () p (y) (g () g (y)) ddy which is clerly equivlent to (4.). Now, if we choose f () M, g (), then the equlity in the ove inequlity is relized for ny p s ove. The following corollry is importnt for pplictions.

Grüss Integrl Inequlities 03 Corollry 4.. Let f, g : [, ] R e two differentile mppings with g () 0 on (, ) nd there eists constnt M > 0 so tht: f (4.3) () g () M for ll (, ). Then we hve the inequlity (4.).The inequlity is shrp. Proof. Use the Cuchy s men vlue theorem, i.e., for every, y [, ] with y, there eists c etween nd y so tht Consequently, for ech, y [, ] we hve f () f (y) g () g (y) f (c) g (c). f () f (y) M g () g (y) i.e., (4.) holds. Applying Theorem 4., we get (4.3). Remrk 4.. Under the ssumption of Corollry 4. we cn choose M ssuming tht the norm is finite. sup (,) f () g () f g, Remrk 4.. If f, g re s in the ove theorem, then we hve the inequlity (4.4) M f () g () d f () d g () d g () d nd the inequlity is shrp.. If f, g re s in Corollry 4., then we hve the inequlity f g nd the inequlity is shrp. f () g () d g () d f () d g () d g () d g () d 5 The Cse When Both Mppings Are of Hölder Type In this section we point out Grüss type inequlity for mppings stisfying the condition of Hölder s follows :

04 Drgomir Theorem 5.. Suppose tht f is of r Hölder type nd g is of s Hölder, i.e., (5.) f () f (y) H y r nd g () g (y) H y s for ll, y [, ], where H, H > 0 nd r, s (0, ] re fied. Then we hve the inequlity: f () g () d (5.) f () d g () d Proof. By the ssumption (5.) we hve for ll, y [, ]. Integrting on [, ] we get H H ( ) r+s (r + s + ) (r + s + ). (f () f (y)) (g () g (y)) H H y r+s (f () f (y)) (g () g (y)) ddy Now, we oserve tht : nd s (f () f (y)) (g () g (y)) ddy H H y r+s ddy y r+s dy d ( y) r+s dy + y r+s ddy. (y ) r+s dy d [ ] ( ) r+s+ + ( ) r+s+ d r + s + ( ) r+s+ (r + s + ) (r + s + ) (f () f (y)) (g () g (y)) ddy ( ) we get the desired inequlity (5.). f () g () d f () d g () d

Grüss Integrl Inequlities 05 6 The Cse When f nd g Belong to Some L p -Spces In this section we point out some inequlities of Grüss type for differentile mppings whose derivtives elong firstly to L (, ), then to L p (, ) (p > ) nd finlly to L (, ). Theorem 6.. Let f, g : [, ] R e two differentile mppings on (, ) nd p : [, ] [0, ) is integrle on [, ]. If f, g L (, ), then we hve the inequlity (6.) p () d p () f () g () d f g p () p (y) Moreover, the inequlity (6.)is shrp. p () d f (t) dt p () f () d p () d Proof. Let oserve tht for ny, y [, ] we hve tht (f () f (y)) (g () g (y)) As f, g L (, ), then we hve f (t) dt g (z) dz ddy p () g () d p () d. f (t) g (z) dtdz. p () p (y) (f () f (y)) (g () g (y)) g (z) dz p () p (y) f g ( y) p () p (y) for ll, y [, ]. By the properties of the modulus, we hve (6.) p () p (y) (f () f (y)) (g () g (y)) ddy p () p (y) f (t) dt g (z) dz ddy f g ( y) p () p (y) ddy, from where we get the desired inequlity (6.).

06 Drgomir To prove the shrpness of (6.), let consider the mppings f () α +, g () γ + δ (α, γ > 0,, δ R) on [, ]. A simple clcultion gives p () d p () f () g () d p () p (y) f g αγ f (t) dt p () d ( y) p () p (y) ddy p () f () d g (z) dz ddy p () d which proves tht we cn hve equlity in ll inequlities in (6.). p () g () d p () d The following corollry holds. Corollry 6.. With the ove ssumptions on the mppings f, g, we hve : f () g () d (6.3) f () d g () d f (t) dt The constnts nd, respectively, re the est possile. g (z) dz ddy f g ( ). Remrk 6.. We shll show tht some time the estimtion given y clssicl Grüss inequlity for the difference f () g () d f () d g () d is etter thn the estimtion (6.3) nd some other time the other wy round. Let f, g : [0, ] [0, ) given y f () p, g () q, p, q >. Then ϕ inf f () 0, Φ sup f () ; [0,] [0,] Also we hve γ inf g () 0, Γ sup g (). [0,] [0,] f () p p, g () q q, [0, ]

Grüss Integrl Inequlities 07 nd oviously f p, g q. Now, we oserve tht nd 4 (Φ ϕ) (Γ γ) 4 f g ( ) pq. Consequently, if pq > 3, then the ound provided y Grüss inequlity is etter thn the ound provided y (6.3). If pq < 3 (p, q > ) then (6.3) is etter thn (.). Remrk 6.. The inequlity (6.3) is lso refinement of Čey sev s inequlity emodied in Corollry.. The following theorem lso holds Theorem 6.3. Let f, g : [, ] R e two differentile mppings on (, ) nd p : [, ] [0, ) is integrle on [, ]. If f L α (, ), g L (, ) with α > nd α +, then we hve the inequlity (6.4) p () d p () f () g () d p () f () d p () g () d p () p (y) y p () p (y) y f α g Note tht, the first inequlity in (6.4) is shrp. f (t) α dt ddy g (t) α dt ddy y p () p (y) ddy. Proof. Using Hölder s inequlity for doule integrls, we hve f (t) g (z) dtdz y α y α f (t) α dtdz f (t) α dt α α f (t) α dt g (z) dtdz y g (t) dt α g (z) α dz.

08 Drgomir Now, s in the proof of Theorem 6., we hve : p () p (y) (f () f (y)) (g () g (y)) ddy p () p (y) p () p (y) y f (t) g (z) dtdz ddy α f (t) α dt Using gin Hölder s inequlity for doule integrls, we hve (6.5) nd, s p () p (y) y α f (t) α dt p () p (y) y p () p (y) y g (z) dz g (z) dz f (t) α dt ddy g (z) dz ddy α ddy. ddy (6.6) p () p (y) (f () f (y)) (g () g (y)) ddy p () d p () f () g () d p () f () d the inequlity (6.5) nd (6.6) provide the first inequlity in (6.4). Now, let oserve tht f (t) α dt f α α, y g (z) dz g for ll, y [, ], nd then p () p (y) y f (t) α dt ddy p () g () d α

Grüss Integrl Inequlities 09 f α f α g p () p (y) y p () p (y) y ddy α p () p (y) y ddy g (z) dt ddy g p () p (y) y ddy nd the second inequlity in (6.4) is lso proved. For the shrpness of the first inequlity in (6.4), let consider the mppings f, g : [, ] R, f () m + n, g () s + z with m, t > 0. Then, oviously p () d p () f () g () d p () f () d p () g () d ms p () p (y) ( y) ddy nd then f (t) α dt mα y, p () p (y) y g (z) dz s y f (t) α dt ddy α ms ms p () p (y) y p () p (y) y ddy p () p (y) ( y) ddy α g (z) dz ddy nd the equlity is relized in the first inequlity in (6.4). The following corollry holds. p () p (y) y ddy

0 Drgomir Corollry 6.4. Let f, g e s ove. Then we hve the inequlity f () g () d (6.7) f () d g () d ( ) ( ) y y f (t) α dt ddy g (t) dt ddy α The first inequlity in (6.7) is shrp. 6 f α g ( ). In similr wy we cn prove the following theorem: Theorem 6.5. Let f, g : [, ] R e two differentile mppings on (, ). If f L (, ) nd g L (, ) then we hve the inequlities: (6.8) p () d p () f () g () d p () f () d p () g () d p () p (y) y sup f (t) t [,y] g (z) dz ddy f g The first inequlity in (6.8) is shrp. The following corollry lso holds. p () p (y) y ddy. Corollry 6.6. Under the ove ssumptions for the mppings f nd g, we hve f () g () d (6.9) f () d g () d ( ) p () p (y) y sup f (t) t [,y] g (z) dz ddy The first inequlity in (6.9) is shrp. 6 f g ( ).

Grüss Integrl Inequlities Remrk 6.3. We note tht some time the upper ound provided y (6.4) is etter thn the upper ound given y (6.8) nd other time, the other wy round. Indeed, choosing f, g : [0, ] R, f () p, g () q (p, q > ) we hve f () p p, g () q q, f p, g, f p α [α (p ) + ] α nd g q q [ (q ) + ] where α, > nd α +. Also, let A : 6 f g ( ) p 6 nd B : 6 f α g pq ( ). 6 [α (p ) + ] α [ (q ) + ] If we choose α, we get A B [(p + ) (q + )] q which cn e greter or less thn for different vlues of p, q >. [] References MITRINOVIĆ, D.S. ; PEČARIĆ, J.E. ; FINK, A.M. ; Clssicl nd New Inequlities in Anlysis, Kluwer Acdemic Pulishers, Dordrecht, 993. School of Communictions nd Informtics, Victori University of Technology, PO Bo 448, Melourne City MC, Victori 800, Austrli. E-mil ddress: sever@mtild.vut.edu.u