Comments on the article Opial inequality in q-Calculus
Tatjana Z. Mirkovic, Slobodan B. Trickovic, Miomir S. Stankovic
aa r X i v : . [ m a t h . G M ] N ov Comments on the article Opial inequality in q -Calculus Tatjana Z. Mirkovic , Slobodan B. Triˇckovi´c , Miomir S. Stankovi´c College of Applied Professional Studies, Filipa Filipovi´ca 20, 17000 Vranje, Serbia, Department of Mathematics, Faculty of Civil Engineering,Aleksandra Medvedeva 14, 18000 Niˇs, Serbia, Mathematical Institute of the Serbian Academy of Sciences and Arts,Kneza Mihaila 36, 11001 Belgrade, Serbia
Abstract
We give corrections concerned with the proofs of the theorems fromthe paper [9], where, by using quite a elementary method based on somesimple observations and applications of some fundamental inequalities, anew general Opial type integral inequality in q -Calculus was established,Opial inequalities in q -Calculus involving two functions and their firstorder derivatives were investigated, and several particular cases were dis-cussed. In the recent paper [9] a generalization of the Opial integral inequality Z h | f ( x ) f ′ ( x ) | dx ≤ h Z h ( f ′ ( x )) dx (1)in q -calculus was given. Here we eliminate some inaccuracies by simplifying andmodifying the proofs of the theorems.First of all, we present necessary definitions and facts from the q -calculus,where q is a real number satisfying 0 < q <
1, and q -natural number is definedby [ n ] q = 1 − q n − q = q n − + · · · + q + 1 , . . . , n ∈ N . Definition 1.1.
Let f be a function defined on an interval ( a, b ) ⊂ R , so that qx ∈ ( a, b ) for all x ∈ ( a, b ) . For < q < , we define the q -derivative as ( D q f )( x ) = f ( x ) − f ( qx ) x − qx , x = 0; D q f (0) = lim x → D q f ( x ) . (2)1n the paper [8], Jackson defined q -integral, which in the q -calculus bears hisname. Definition 1.2.
The q -integral on [0 , a ] is Z a f ( x ) d q x = a (1 − q ) ∞ X j =0 q j f ( aq j ) . On this basis, in the same paper, Jackson defined an integral on [ a, b ] Z ba f ( x ) d q x = Z b f ( x ) d q x − Z a f ( x ) d q x, (3)For a positive integer n and a = bq n , using the left-hand side integral of (3),in the paper [7], Gauchman introduced the q -restricted integral Z ba f ( x ) d q x = Z bbq n f ( x ) d q x = b (1 − q ) n − X j =0 q j f ( q j b ) . (4) Definition 1.3.
The real function f defined on [ a, b ] is called q -increasing ( q -decreasing ) on [ a, b ] if f ( qx ) ≤ f ( x ) ( f ( qx ) ≥ f ( x )) for x, qx ∈ [ a, b ] . It is easy to see that if the function f is increasing (decreasing), then it is q -increasing ( q -decreasing) too. Our main results are contained in three theorems.
Theorem 2.1.
Let f ( x ) be q -decreasing function on [ a, b ] with f ( bq ) = 0 .Then, for any p ≥ , there holds Z ba | D q f ( x ) || f ( x ) | p d q x ≤ ( b − a ) p Z ba | D q f ( x ) | p +1 d q x. (5) Proof.
Using Definition 1.1 and (4), we have Z ba | D q f ( x ) || f ( x ) | p d q x = Z bbq n (cid:12)(cid:12)(cid:12)(cid:12) f ( x ) − f ( qx ) x − qx (cid:12)(cid:12)(cid:12)(cid:12) | f ( x ) | p d q x = b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:12) | f ( bq j ) | p , whence, taking into account that f ( x ) is q -decreasing, we have n − X j =0 | f ( bq j ) − f ( bq j +1 ) || f ( bq j ) | p ≤ | f ( bq n ) | p n − X j =0 | f ( bq j ) − f ( bq j +1 ) | .
2n view of f ( bq n ) = n − P j =0 f ( bq j +1 ) − f ( bq j ), we obtain | f ( bq n ) | p = (cid:12)(cid:12)(cid:12) n − X j =0 f ( bq j +1 ) − f ( bq j ) (cid:12)(cid:12)(cid:12) p ≤ (cid:16) n − X j =0 (cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) (cid:12)(cid:12)(cid:17) p , so that | f ( bq n ) | p n − X j =0 (cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) (cid:12)(cid:12) ≤ (cid:16) n − X j =0 (cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) (cid:12)(cid:12)(cid:17) p +1 . Thus Z ba | D q f ( x ) || f ( x ) | p d q x ≤ (cid:16) n − X j =0 (cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) (cid:12)(cid:12)(cid:17) p +1 . (6)The right-hand side of this inequality we can write in the form of (cid:16) n − X j =0 (cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) (cid:12)(cid:12)(cid:17) p +1 = (cid:16) n − X j =0 (cid:12)(cid:12) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) p +1 = (cid:16) b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) p +1 . After rewriting q j = ( q j ) pp +1 ( q j ) p +1 , and applying H¨older’s inequality to thelast sum, we have n − X j =0 ( q j ) pp +1 ( q j ) p +1 (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) ≤ (cid:16) n − X j =0 (cid:0) ( q j ) pp +1 (cid:1) p +1 p (cid:17) pp +1 × (cid:16) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) p +1 (cid:17) p +1 . After raising both sides to the power p + 1, we find (cid:16) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) p +1 ≤ (cid:16) n − X j =0 q j (cid:17) p n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) p +1 . Multiplying this inequality by b p +1 (1 − q ) p +1 , and relying on the formula forthe sum of the first n terms of the geometric series, we arrive at the inequality (cid:16) b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) p +1 ≤ b p (1 − q n ) p × b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) p +1 . (7)3onsidering that b p (1 − q n ) p = ( b − bq n ) p = ( b − a ) p , taking into account (6),we have proved the inequality Z ba | D q f ( x ) || f ( x ) | p d q x ≤ ( b − a ) p (cid:16) b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) p +1 (cid:17) . Referring to (4), there holds b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) p +1 = Z ba | D q f ( x ) | p +1 d q x, whereby we prove the theorem. Remark 2.2.
In particular, by taking p = 1 , the inequality (5) in Theorem 2.1 reduces to the following Opial’s inequality in q -Calculus. Z ba | D q f ( x ) || f ( x ) | d q x ≤ ( b − a ) Z ba | D q f ( x ) | d q x. The following theorems are concerned with q -monotonic functions. Theorem 2.3. If f ( x ) and g ( x ) are q -decreasing functions on [ a, b ] satisfying f ( bq ) = 0 and g ( bq ) = 0 , then there holds the inequality Z ba (cid:0) f ( x ) D q g ( x ) + g ( qx ) D q f ( x ) (cid:1) d q x ≤ b − a Z ba (cid:0) ( D q f ( x )) + ( D q g ( x )) (cid:1) d q x. (8) Proof.
Replacing (2) in the integral Z ba (cid:0) f ( x ) D q g ( x ) + g ( qx ) D q f ( x ) (cid:1) d q x, we obtain Z bbq n (cid:18) f ( x ) g ( x ) − g ( qx ) x − qx + g ( qx ) f ( x ) − f ( qx ) x − qx (cid:19) d q x, whence, using Gauchman q -restricted integral, we have b (1 − q ) (cid:16) n − X j =0 q j f ( bq j ) g ( bq j ) − g ( bq j +1 ) bq j − bq j +1 + n − X j =0 q j g ( bq j +1 ) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:17) = n − X j =0 (cid:0) f ( bq j )( g ( bq j ) − g ( bq j +1 )) + g ( bq j +1 )( f ( bq j ) − f ( bq j +1 )) (cid:1) = n − X j =0 (cid:0) f ( bq j ) g ( bq j ) − g ( bq j +1 ) f ( bq j +1 ) (cid:1) = − f ( bq n ) g ( bq n ) . − xy ≤ ( x + y ) , x, y ∈ R , and consideringthat f ( bq n ) = n − X j =0 (cid:0) f ( bq j +1 ) − f ( bq j ) (cid:1) , g ( bq n ) = n − X j =0 (cid:0) g ( bq j +1 ) − g ( bq j ) (cid:1) , we find − f ( bq n ) g ( bq n ) ≤ (cid:16) n − X j =0 (cid:0) f ( bq j +1 ) − f ( bq j ) (cid:1)(cid:17) + (cid:16) n − X j =0 (cid:0) g ( bq j +1 ) − g ( bq j ) (cid:1)(cid:17) ! Applying (7) for p = 1, knowing that f ( x ) and g ( x ) are q -decreasing, we obtain (cid:16) b (1 − q ) n − X j =0 q j f ( bq j +1 ) − f ( bq j ) bq j − bq j +1 (cid:17) ≤ b (1 − q n ) b (1 − q ) n − X j =0 q j (cid:16) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:17) as well as (cid:16) b (1 − q ) n − X j =0 q j g ( bq j +1 ) − g ( bq j ) bq j − bq j +1 (cid:17) ≤ b (1 − q n ) b (1 − q ) n − X j =0 q j (cid:16) g ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:17) . Since b (1 − q n ) = b − a , making use of (4), we have Z ba (cid:0) f ( x ) D q g ( x ) + g ( qx ) D q f ( x ) (cid:1) d q x ≤ b − a Z ba (cid:0) ( D q f ( x )) + ( D q g ( x )) (cid:1) d q x, whereby (8) is proved. Theorem 2.4. If f ( x ) and g ( x ) are q -decreasing functions on [ a, b ] satisfying f ( bq ) = g ( bq ) = 0 , then there holds the inequality Z ba | f ( x ) | s | g ( x ) | t d q x ≤ ( b − a ) s + t Z ba (cid:16) ss + t | D q f ( x ) | s + t d q x + ts + t | D q g ( x ) | s + t d q x (cid:17) . (9) Proof.
First, we apply (4) to the left-hand side of (9), and have Z ba | f ( x ) | s | g ( x ) | t d q x = b (1 − q ) n − X i =0 q i | f ( bq i ) | s | g ( bq i ) | t . For real numbers z, w ≥ s, t >
0, we rely on the elementary inequality z s w t ≤ ss + t z s + t + ts + t w s + t . z = ( q i ) s + t | f ( bq i ) | , w = ( q i ) s + t | g ( bq i ) | , we find n − X i =0 q i | f ( bq i ) | s | g ( bq i ) | t = n − X i =0 (cid:0) ( q i ) s + t | f ( bq i ) | (cid:1) s (cid:0) ( q i ) s + t | g ( bq i ) | (cid:1) t ≤ ss + t n − X i =0 q i | f ( bq i ) | s + t + ts + t n − X i =0 q i | g ( bq i ) | s + t . Considering that f and g are q -decreasing functions, so | f ( bq i ) | s + t ≤ | f ( bq n ) | s + t and | g ( bq i ) | s + t ≤ | g ( bq n ) | s + t , the last inequality becomes n − X i =0 q i | f ( bq i ) | s | g ( bq i ) | t ≤ − q n − q (cid:16) ss + t | f ( bq n ) | s + t + ts + t | g ( bq n ) | s + t (cid:17) . However, there holds | f ( bq n ) | s + t = (cid:12)(cid:12)(cid:12) n − X i =0 f ( bq i +1 ) − f ( bq i ) (cid:12)(cid:12)(cid:12) s + t ≤ (cid:16) n − X i =0 | f ( bq i +1 ) − f ( bq i ) | (cid:17) s + t | g ( bq n ) | s + t = (cid:12)(cid:12)(cid:12) n − X i =0 g ( bq i +1 ) − g ( bq i ) (cid:12)(cid:12)(cid:12) s + t ≤ (cid:16) n − X i =0 | g ( bq i +1 ) − g ( bq i ) | (cid:17) s + t , so that we have Z ba | f ( x ) | s | g ( x ) | t d q x = b (1 − q ) n − X i =0 q i | f ( bq i ) | s | g ( bq i ) | t ≤ b (1 − q n ) ss + t (cid:16) n − X i =0 | f ( bq i +1 ) − f ( bq i ) | (cid:17) s + t + ts + t (cid:16) n − X i =0 | g ( bq i +1 ) − g ( bq i ) | (cid:17) s + t ! ≤ b (1 − q n )( s + t ) h s (cid:16) b (1 − q ) n − X i =0 q i (cid:12)(cid:12)(cid:12) f ( bq i ) − f ( bq i +1 ) bq i − bq i +1 (cid:12)(cid:12)(cid:12)(cid:17) s + t + t (cid:16) b (1 − q ) n − X i =0 q i (cid:12)(cid:12)(cid:12) g ( bq i ) − g ( bq i +1 ) bq i − bq i +1 (cid:12)(cid:12)(cid:12)(cid:17) s + t i , (10)Here we follow the same procedure as in the proof of Theorem 2.1. So, afterrewriting q i = ( q i ) s + t − s + t ( q j ) s + t , and applying H¨older’s inequality to both sumson the right side of the last inequality, for the first sum we have n − X j =0 ( q j ) s + t − s + t ( q j ) s + t (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) ≤ (cid:16) n − X j =0 (cid:0) ( q j ) s + t − s + t (cid:1) s + ts + t − (cid:17) s + t − s + t × (cid:16) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) s + t (cid:17) s + t , n − X j =0 ( q j ) s + t − s + t ( q j ) s + t (cid:12)(cid:12)(cid:12) g ( bq j ) − g ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) ≤ (cid:16) n − X j =0 (cid:0) ( q j ) s + t − s + t (cid:1) s + ts + t − (cid:17) s + t − s + t × (cid:16) n − X j =0 q j (cid:12)(cid:12)(cid:12) g ( bq j ) − g ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) s + t (cid:17) s + t . We multiply both inequalities by b (1 − q ), then raise them to the power s + t .Thus, we obtain (cid:16) b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) s + t ≤ b s + t − (1 − q n ) s + t − × b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) f ( bq j ) − f ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) s + t = b s + t − (1 − q n ) s + t − Z ba | D q f ( x ) | s + t d q x, and similarly (cid:16) b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) g ( bq j ) − g ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12)(cid:17) s + t ≤ b s + t − (1 − q n ) s + t − × b (1 − q ) n − X j =0 q j (cid:12)(cid:12)(cid:12) g ( bq j ) − g ( bq j +1 ) bq j − bq j +1 (cid:12)(cid:12)(cid:12) s + t = b s + t − (1 − q n ) s + t − Z ba | D q g ( x ) | s + t d q x, so that, because b (1 − q n ) = b − a , from (10) there follows (9), whereby wecomplete the proof. Remark 2.5.
In the special case when s = t = r and f ( x ) = g ( x ) = h ( x ) , theinequality established in (9) reduces to the q -Wirtinger-type inequality Z ba | h ( x ) | r d q x ≤ ( b − a ) r Z ba ( D q h ( x )) r d q x. References [1] Opial, Sur une inegalite, Ann. Polon. Math., New York, 8, 29 - 32, (1960)[2] Agarwal, Lakshmikantham, Uniqueness and nonuniqueness criteria for or-dinary differential equations, 1993.[3] Agarwal, Pang, Opial inequalities with applications in differential and dif-ference equations, Dordrecht:Kluwer Acad. Publ. (1995)[4] Anastassiou, J. Appl. Func.l Anal, Vol. 9 Issue 1/2 (2014) 230–238.75] Bainov, Simeonov, Integral inequalities and applications, Dordrecht:Kluwer Acad. Publ. (1992)[6] Caputo, M. Fabrizio, A new definition of fractional derivative without sin-gular kernel, Progr. Fract. Differ. Appl 1, 73-85, 2015.[7] Gauchman, Integral inequalities in q -calculus, Computers and Mathematicswith Applications, 47, 281–300 (2004)[8] Jackson, On a q -Definite Itegrals. Quarterly Journal of Pure and AplliedMathematics 41 (1910) 193 - 203.[9] T.Z. Mirkovic, S.B. Triˇckovi´c, M.S. Stankovi´c, Opial inequality in qq