Opial inequality in q-calculus

In this article we give q-analogs of the Opial inequality for q-decreasing functions. Using a closed form of the restricted q-integral (see Gauchman in Comput. Math. Appl. 47:281–300, 2004), we establish a new integral inequality of the q-Opial type.

The constant h 4 is the best possible.
Integral inequalities of the form (1) have an interest in itself, and also have important applications in the theory of ordinary differential equations and boundary value problems (see [1,2,4]). In the years thereafter, numerous generalizations, extensions and variations of the Opial inequality have appeared (see [12,14]). The one containing fractional derivatives is investigated as well (see [3,5]).
In the continuous case, the Opial inequity, in its modified form, states that if f (x) is an absolutely continuous function with f (a) = 0, and f ∈ L 2 (a, b) where a and b are finite, then In a recent paper [14], Yang proved the following generalization of the Opial inequality.

Preliminaries
Here we present necessary definitions and facts from the q-calculus. We follow the terminology and notations used in the books [8,9,11,13]. In what follows, q is a real number satisfying 0 < q < 1, and q-natural number is defined by [n] q = 1q n 1q = q n-1 + · · · + q + 1, . . . , n ∈ N.
Definition 2.1 Let f be a function defined on an interval (a, b) ⊂ R, so that qx ∈ (a, b) for all x ∈ (a, b). For 0 < q < 1, we define the q-derivative as In the paper [7], Jackson defined q-integral, which in the q-calculus bears his name.
On this basis, in the same paper, Jackson defined an integral on [a For a positive integer n and a = bq n , using the left-hand side integral of (3), in the paper It is easy to see that if the function f is increasing (decreasing), then it is q-increasing (q-decreasing) too.

Results and discussions
Our main results are contained in three theorems.
Theorem 3.1 Let f ∈ C 1 [0, 1] be q-decreasing function with f (bq 0 ) = 0. Then, for any Proof Using Definition 2.1 and (4) In view of f (bq n ) = n-1 j=0 f (bq j+1 )f (bq j ) and Hölder's inequality, we obtain By elementary calculations, we easily transform the right-hand side of the last inequality into However, because of 0 < q < 1, we have Since n ≥ [n] q = 1-q n 1-q , we have -n p (1q) p ≤ -(1q n ) p , and we arrive at the inequality After interchanging the boundaries in the right-hand side integral, and replacing bq n with a, we find which proves the theorem.
Remark 3.2 In particular, by taking p = 1, the inequality (9) in Theorem 1 reduces to the following Opial inequality in q-calculus: The following theorems are concerned with q-monotonic functions.

Theorem 3.3 If f (x) and g(x) are absolutely continuous q-decreasing functions on (a, b) and f (bq 0 ) = 0 and g(bq
Proof whence, using the Gauchman q-restricted integral, we have Denoting f (bq j ) = f (bq j+1 )f (bq j ) and g(bq j ) = g(bq j+1 )g(bq j ), we can rewrite the last sum in the form of n-1 j=0 [f (bq j ) g(bq j ) + g(bq j+1 ) f (bq j )], and we find b a f (x)D q g(x) + g(qx)D q f (x) d q x = -f bq n g bq n .
Proof For k ∈ N 0 , we have the following identities: From (8) and (9) we observe that From (10) and using the elementary inequality where z, w ≥ 0 and s, t > 0 are real numbers, we find Using Hölder's inequality on the right side of (11) with indices s + t, s+t s+t-1 , we have Summing the inequality (12) from 0 to n -1, we obtain After multiplying the left-hand side of (13) by b(1q)q k , we transform it into the form of 1 b 1 1q n-1 k=0 b(1q)q k |f (bq k )| s |g(bq k )| t q k , and after multiplying the right-hand side of (13) by (b(1q)q k ) s+t-1 , we transform it into the form of ( n 2 ) s+t s + t b(1q) s+t-1 s n-1 k=0 q k s+t-1 |f (bq k )f (bq k+1 )| s+t b s+t-1 (1q) s+t-1 (q k ) s+t-1 + t n-1 k=0 q k s+t-1 |g(bq k )g(bq k+1 )| s+t b s+t-1 (1q) s+t-1 (q k ) s+t-1 .