New Refinements for integral and sum forms of Hölder inequality
aa r X i v : . [ m a t h . G M ] J a n NEW REFINEMENTS FOR INTEGRAL AND SUM FORMS OFH ¨OLDER INEQUALITY ˙IMDAT ˙IS¸CAN
Abstract.
In this paper, new refinements for integral and sum forms ofH¨older inequality are established. We note that many existing inequalitiesrelated to the H¨older inequality can be improved via obtained new inequalitiesin here, we show this in an application Introduction
The famous Young inequality for two scalars is the t -weighted arithmetic-geometricmean inequality. This inequality says that if x, y > t ∈ [0 , x t y − t ≤ tx + (1 − t ) y with equality if and only if a = b. Let p, q > /p + 1 /q = 1. Theinequality (1.1) can be written as(1.2) xy ≤ x p p + y q q for any x, y ≥
0. In this form, the inequality (1.2) was used to prove the celebratedH¨older inequality. One of the most important inequalities of analysis is H¨older’sinequality. It contributes wide area of pure and applied mathematics and plays akey role in resolving many problems in social science and cultural science as wellas in natural science.
Theorem 1 (H¨older Inequality for Integrals [8]) . Let p > and /p + 1 /q = 1 . If f and g are real functions defined on [ a, b ] and if | f | p , | g | q are integrable functionson [ a, b ] then (1.3) Z ba | f ( x ) g ( x ) | dx ≤ Z ba | f ( x ) | p dx ! /p Z ba | g ( x ) | q dx ! /q , 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 (H¨older Inequality for Sums [8]) . Let a = ( a , ..., a n ) and b = ( b , ..., b n ) be two positive n-tuples and p, q > such that /p + 1 /q = 1 . Then we have (1.4) n X k =1 a k b k ≤ n X k =1 a pk ! /p n X k =1 b qk ! /q . Mathematics Subject Classification.
Primary 26D15; Secondary 26A51.
Key words and phrases.
H¨older Inequality, Young Inequality, Integral Inequalities, Hermite-Hadamard Type Inequality.
Equality hold in (1.4) if and only if a p and b q are proportional. Of course the H¨older’s inequality has been extensively explored and tested toa new situation by a number of scientists. Many generalizations and refinementsfor H¨older’s inequality have been obtained so far. See, for example, [1, 3, 4, 5, 6,7, 8, 9, 10, 11] and the references therein. In this paper, by using a simple proofmethod some new refinements for integral and sum forms of H¨older’s inequality areobtained. 2.
Main Results
Theorem 3.
Let p > and /p + 1 /q = 1 . If f and g are real functions definedon [ a, b ] and if | f | p , | g | q are integrable functions on [ a, b ] , then i.) Z ba | f ( x ) g ( x ) | dx (2.1) ≤ b − a Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q ii.) 1 b − a Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q ≤ Z ba | f ( x ) | p dx ! /p Z ba | g ( x ) | q dx ! /q . (2.2) ¨OLDER INTEGRAL INEQUALITY 3 Proof. i.)
First method for Proof (Short method):
By using of H¨older inequality in(1.3), it is easily seen that Z ba | f ( x ) g ( x ) | dx = 1 b − a (Z ba (cid:12)(cid:12)(cid:12) ( b − x ) /p f ( x )( b − x ) /q g ( x ) (cid:12)(cid:12)(cid:12) dx + Z ba (cid:12)(cid:12)(cid:12) ( x − a ) /p f ( x )( x − a ) /q g ( x ) (cid:12)(cid:12)(cid:12) dx ) ≤ b − a Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q . Second method for Proof (Long method):
Applying (1.3) on the subinterval [ a, λb +(1 − λ ) a ] and on the subinterval [ λb + (1 − λ ) a, b ], respectively, we get Z λb +(1 − λ ) aa | f ( x ) g ( x ) | dx ≤ Z λb +(1 − λ ) aa | f ( x ) | p dx ! /p Z λb +(1 − λ ) aa | g ( x ) | q dx ! /q and Z bλb +(1 − λ ) a | f ( x ) g ( x ) | dx ≤ Z bλb +(1 − λ ) a | f ( x ) | p dx ! /p Z bλb +(1 − λ ) a | g ( x ) | q dx ! /q . Adding the resulting inequalities, we get: Z ba | f ( x ) g ( x ) | dx ≤ Z λb +(1 − λ ) aa | f ( x ) | p dx ! /p Z λb +(1 − λ ) aa | g ( x ) | q dx ! /q + Z bλb +(1 − λ ) a | f ( x ) | p dx ! /p Z bλb +(1 − λ ) a | g ( x ) | q dx ! /q . (2.3)By the change of variable x = ub + (1 − u ) a ; on the right hand sides integrals in(2.3), we have Z ba | f ( x ) g ( x ) | dx ≤ ( b − a ) Z λ | f ( ub + (1 − u ) a ) | p du ! /p Z λ | g ( ub + (1 − u ) a ) | q du ! /q + (cid:18)Z λ | f ( ub + (1 − u ) a ) | p du (cid:19) /p (cid:18)Z λ | g ( ub + (1 − u ) a ) | q du (cid:19) /q ) . ˙IMDAT ˙IS¸CAN Integrating both sides of this inequality over [0 ,
1] with respect to λ we obtain that Z ba | f ( x ) g ( x ) | dx ≤ ( b − a ) Z Z λ | f ( ub + (1 − u ) a ) | p du ! /p Z λ | g ( ub + (1 − u ) a ) | q du ! /q dλ + Z (cid:18)Z λ | f ( ub + (1 − u ) a ) | p du (cid:19) /p (cid:18)Z λ | g ( ub + (1 − u ) a ) | q du (cid:19) /q dλ ) . Then, By applying of H¨older inequality for the right hand sides integrals in the lastinequality, we have Z ba | f ( x ) g ( x ) | dx ≤ ( b − a ) Z Z λ | f ( ub + (1 − u ) a ) | p dudλ ! /p Z Z λ | g ( ub + (1 − u ) a ) | q dudλ ! /q + (cid:18)Z Z λ | f ( ub + (1 − u ) a ) | p dudλ (cid:19) /p (cid:18)Z Z λ | g ( ub + (1 − u ) a ) | q dudλ (cid:19) /q ) . By Fubini theorem and the change of variable u = ( x − a ) / ( b − a ) we get Z ba | f ( x ) g ( x ) | dx ≤ ( b − a ) ((cid:18)Z Z u | f ( ub + (1 − u ) a ) | p dλdu (cid:19) /p (cid:18)Z Z u | g ( ub + (1 − u ) a ) | q dλdu (cid:19) /q + (cid:18)Z Z u | f ( ub + (1 − u ) a ) | p dλdu (cid:19) /p (cid:18)Z Z u | g ( ub + (1 − u ) a ) | q dλdu (cid:19) /q ) = ( b − a ) ((cid:18)Z (1 − u ) | f ( ub + (1 − u ) a ) | p du (cid:19) /p (cid:18)Z (1 − u ) | g ( ub + (1 − u ) a ) | q du (cid:19) /q + (cid:18)Z u | f ( ub + (1 − u ) a ) | p du (cid:19) /p (cid:18)Z u | g ( ub + (1 − u ) a ) | q du (cid:19) /q ) = 1 b − a Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q . b) First let us consider the case Z ba | f ( x ) | p dx ! /p Z ba | g ( x ) | q dx ! /q = 0 . ¨OLDER INTEGRAL INEQUALITY 5 Then, f ( x ) = 0 for almost everywhere x ∈ [ a, b ] or g ( x ) = 0 for almost everywhere x ∈ [ a, b ] . Thus, we have Z ba | f ( x ) g ( x ) | dx = 0 . Therefore the inequality (2.2) is trivial in this case.Finally, we consider the case I = Z ba | f ( x ) | p dx ! /p Z ba | g ( x ) | q dx ! /q = 0 . Then 1( b − a ) I Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q ≤ b − a R ba ( b − x ) | f ( x ) | p dx R ba | f ( x ) | p dx ! /p R ba ( b − x ) | g ( x ) | q dx R ba | g ( x ) | q dx ! /q + R ba ( x − a ) | f ( x ) | p dx R ba | f ( x ) | p dx ! /p R ba ( x − a ) | g ( x ) | q dx R ba | g ( x ) | q dx ! /q . Applying (1.1) on the right hand sides integrals of the last inequality1( b − a ) I Z ba ( b − x ) | f ( x ) | p dx ! /p Z ba ( b − x ) | g ( x ) | q dx ! /q + Z ba ( x − a ) | f ( x ) | p dx ! /p Z ba ( x − a ) | g ( x ) | q dx ! /q ≤ b − a ( R ba ( b − x ) | f ( x ) | p dxp R ba | f ( x ) | p dx + R ba ( b − x ) | g ( x ) | q dxq R ba | g ( x ) | q dx + R ba ( x − a ) | f ( x ) | p dxp R ba | f ( x ) | p dx + R ba ( x − a ) | g ( x ) | q dxq R ba | g ( x ) | q dx ) = 1 p + 1 q = 1 . This completes the proof. (cid:3)
Remark 1.
The inequality (2.2) show that the inequality (2.1) is better than theinequality (1.3).
The more general versions of Theorem 3 can be given as follow:
Theorem 4.
Let p > and /p + 1 /q = 1 . If f and g are real functions definedon [ a, b ] and if | f | p , | g | q are integrable functions on [ a, b ] , then ˙IMDAT ˙IS¸CAN i.) Z ba | f ( x ) g ( x ) | dx (2.4) ≤ Z ba α ( x ) | f ( x ) | p dx ! /p Z ba α ( x ) | g ( x ) | q dx ! /q + Z ba β ( x ) | f ( x ) | p dx ! /p Z ba β ( x ) | g ( x ) | q dx ! /q , where α, β : [ a, b ] → [0 , ∞ ) are continuous functions such that α ( x ) + β ( x ) = 1 , x ∈ [ a, b ] . ii.) Z ba | f ( x ) g ( x ) | dx ≤ n X i =1 Z ba α i ( x ) | f ( x ) | p dx ! /p Z ba α i ( x ) | g ( x ) | q dx ! /q where α i : [ a, b ] → [0 , ∞ ) , i = 1 , , ...n, are continuous functions such that P ni =1 α i ( x ) =1 , x ∈ [ a, b ] . Proof.
The proof of Theorem is easily seen by using similar method the proof ofTheorem 3. (cid:3)
Remark 2.
It is easily seen that the inequalities obtained in Theorem 4 are thebest than the inequality (1.3).
Remark 3. i.) In the inequality (2.4) of Theorem 4, if we take α ( x ) = sin x and β ( x ) = cos x , then we have Z ba | f ( x ) g ( x ) | dx ≤ Z ba sin x | f ( x ) | p dx ! /p Z ba sin x | g ( x ) | q dx ! /q + Z ba cos x | f ( x ) | p dx ! /p Z ba cos x | g ( x ) | q dx ! /q . ii.) In the inequality (2.4) of Theorem 4, if we take α ( x ) = b − xb − a and β ( x ) = x − ab − a ,then we have the inequality (2.1). Theorem 5.
Let a = ( a , ..., a n ) and b = ( b , ..., b n ) be two positive n-tuples and p, q > such that /p + 1 /q = 1 . Then ¨OLDER INTEGRAL INEQUALITY 7 i.) n X k =1 a k b k ≤ n n X k =1 ka pk ! /p n X k =1 kb qk ! /q (2.5) + n X k =1 ( n − k ) a pk ! /p n X k =1 ( n − k ) b qk ! /q . ii.)1 n n X k =1 ka pk ! /p n X k =1 kb qk ! /q + n X k =1 ( n − k ) a pk ! /p n X k =1 ( n − k ) b qk ! /q (2.6) ≤ n X k =1 a pk ! /p n X k =1 b qk ! /q . Proof. i.) By using of H¨older inequality in (1.4), it is easily seen that n X k =1 a k b k = n X k =1 (cid:18) kn + n − kn (cid:19) a k b k = 1 n ( n X k =1 k /p a k k /q b k + n X k =1 ( n − k ) /p a k ( n − k ) /q b k ) ≤ n n X k =1 ka pk ! /p n X k =1 kb qk ! /q + n X k =1 ( n − k ) a pk ! /p n X k =1 ( n − k ) b qk ! /q . ii.) First let us consider the case n X k =1 a pk ! /p n X k =1 b qk ! /q = 0 . Then a k = 0 for k = 1 , , .., n or b k = 0 for k = 1 , , .., n. Thus, we have n X k =1 a k b k = 0 . Therefore the inequality (2.6) is trivial in this case.Finally, we consider the case S = n X k =1 a pk ! /p n X k =1 b qk ! /q = 0 . ˙IMDAT ˙IS¸CAN Then 1 nS n X k =1 ka pk ! /p n X k =1 kb qk ! /q + n X k =1 ( n − k ) a pk ! /p n X k =1 ( n − k ) b qk ! /q = 1 n ((cid:18) P nk =1 ka pk P nk =1 a pk (cid:19) /p (cid:18) P nk =1 kb qk P nk =1 b qk (cid:19) /q + (cid:18) P nk =1 ( n − k ) a pk P nk =1 a pk (cid:19) /p (cid:18) P nk =1 ( n − k ) b qk P nk =1 b qk (cid:19) /q ) . Applying (1.1) on the right hand sides sums of the last inequality1 nS n X k =1 ka pk ! /p n X k =1 kb qk ! /q + n X k =1 ( n − k ) a pk ! /p n X k =1 ( n − k ) b qk ! /q ≤ n (cid:26) P nk =1 ka pk p P nk =1 a pk + P nk =1 kb qk q P nk =1 b qk + P nk =1 ( n − k ) a pk p P nk =1 a pk + P nk =1 ( n − k ) b qk q P nk =1 b qk (cid:27) = 1 p + 1 q = 1 . This completes the proof. (cid:3)
Remark 4.
The inequality (2.6) show that the inequality (2.5) is better than theinequality (1.4).
The more general versions of Theorem 3 can be given as follow:
Theorem 6.
Let a = ( a , ..., a n ) and b = ( b , ..., b n ) be two positive n-tuples and p, q > such that /p + 1 /q = 1 . i.) If c = ( c , ..., c n ) and d = ( d , ..., d n ) be two positive n-tuples such that c k + d k = 1 , k = 1 , , ..., n. Then n X k =1 a k b k ≤ n X k =1 c k a pk ! /p n X k =1 c k b qk ! /q (2.7) + n X k =1 d k a pk ! /p n X k =1 d k b qk ! /q . ii.) If c ( i ) = (cid:16) c ( i )1 , ..., c ( i ) n (cid:17) , i = 1 , , ..., m be positive n-tuples such that P mi =1 c ( i ) k =1 , k = 1 , , ..., n. Then n X k =1 a k b k ≤ m X i =1 n X k =1 c ( i ) k a pk ! /p n X k =1 c ( i ) k b qk ! /q Proof.
The proof of Theorem is easily seen by using similar method the proof ofTheorem 3. (cid:3)
Remark 5.
It is easily seen that the inequalities obtained in Theorem 6 are thebest than the inequality (1.4). ¨OLDER INTEGRAL INEQUALITY 9
Remark 6. i.) In the inequality (2.7) of Theorem 6, if we take c = (cid:0) sin , ..., sin n (cid:1) and d = (cid:0) cos , ..., cos n (cid:1) , then we have n X k =1 a k b k ≤ n X k =1 sin ka pk ! /p n X k =1 sin kb qk ! /q + n X k =1 cos ka pk ! /p n X k =1 cos kb qk ! /q . ii.) In the inequality (2.7) of Theorem 6, if we take c = (cid:0) n , n ..., (cid:1) and d = (cid:0) n − n , n − n ..., (cid:1) , then we have the inequality (2.5). An Application
In [2], Dragomir et al. gave the following lemma for obtain main results.
Lemma 1.
Let f : I ◦ ⊆ R → R be a differentiable mapping on I ◦ , a, b ∈ I ◦ with a < b and q > . If f ∈ L [ a, b ] , then the following equality holds: f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx = b − a Z (1 − t ) f ( ta + (1 − t ) b ) dt. By using this equality and H¨older integral inequality Dragomir et al. obtainedthe following inequality:
Theorem 7.
Let f : I ◦ ⊆ R → R be a differentiable mapping on I ◦ , a, b ∈ I ◦ with a < b . If the new mapping | f ′ | q is convex on [ a, b ] , then the following inequalityholds: (3.1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ b − a p + 1) /p (cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q , where /p + 1 /q = 1 . If Theorem 7 are resulted again by using the inequality (2.1) in Theorem 3, thenwe get the following result:
Theorem 8.
Let f : I ◦ ⊆ R → R be a differentiable mapping on I ◦ , a, b ∈ I ◦ with a < b . If the new mapping | f ′ | q convex on [ a, b ] , then the following inequality holds: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (3.2) ≤ b − a p + 1) /p ((cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q + (cid:20) | f ′ ( a ) | q + 2 | f ′ ( b ) | q (cid:21) /q ) , where /p + 1 /q = 1 . Proof.
Using Lemma 1 and the inequality (2.1), we find (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (3.3) ≤ b − a Z | − t | | f ( ta + (1 − t ) b ) | dt ≤ b − a ((cid:18)Z (1 − t ) | − t | p dt (cid:19) /p (cid:18)Z (1 − t ) | f ( ta + (1 − t ) b ) | q dt (cid:19) /q + (cid:18)Z t | − t | p dt (cid:19) /p (cid:18)Z t | f ( ta + (1 − t ) b ) | q dt (cid:19) /q ) . Using the convexity of | f ′ | q , we have Z t | f ( ta + (1 − t ) b ) | q dt ≤ Z t [ t | f ( a ) | q + (1 − t ) | f ( b ) | q ] dt = 2 | f ′ ( a ) | q + | f ′ ( b ) | q , (3.4)and Z (1 − t ) | f ( ta + (1 − t ) b ) | q dt = Z t | f ( tb + (1 − t ) a ) | q dt ≤ | f ′ ( a ) | q + 2 | f ′ ( b ) | q , (3.5)Further, since Z t | − t | p dt = Z (1 − t ) | − t | p dt = 12( p + 1) , a combination of (3.3)-(3.5) immediately gives the required inequality (3.2). (cid:3) Remark 7.
Since h : [0 , ∞ ) → R , h ( x ) = x s , < s ≤ , is a concave function, forall u, v ≥ we have h (cid:18) u + v (cid:19) = (cid:18) u + v (cid:19) s ≥ h ( u ) + h ( v )2 = u s + v s . From here, we get (cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q + 12 (cid:20) | f ′ ( a ) | q + 2 | f ′ ( b ) | q (cid:21) /q ≤ (cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q Thus we obtain b − a p + 1) /p ((cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q + (cid:20) | f ′ ( a ) | q + 2 | f ′ ( b ) | q (cid:21) /q ) ≤ b − a p + 1) /p (cid:20) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:21) /q . This show us that the inequality (3.2) is the best than the inequality (3.1). ¨OLDER INTEGRAL INEQUALITY 11
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Department of Mathematics, Faculty of Arts and Sciences,, Giresun University,28200, Giresun, Turkey.
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