(k, s)-Riemann-Liouville fractional integral and applications Mehmet Zeki SARIKAYA∗† , Zoubir DAHMANI‡, Mehmet Ey¨ up KIRIS§ and ¶ Farooq AHMAD
Abstract In this paper, we introduce a new approach on fractional integration, which generalizes the Riemann-Liouville fractional integral. We prove some properties for this new approach. We also establish some new integral inequalities using this new fractional integration.
Keywords: Riemann-Liouville fractional integrals, synchronous function, Chebyshev inequality, H¨older inequality. 2000 AMS Classification: 26A33;26A51;26D15.
1. Introduction Fractional calculus and its widely application have recently been paid more and more attentions. For more recent development on fractional calculus, we refer the reader to [7, 12, 15, 16, 19]. There are several known forms of the fractional integrals of which two have been studied extensively for their applications [5, 10, 11, 14, 21]. The first is the Riemann-Liouville fractional integral of α ≥ 0 for a continuous function f on [a, b] which is defined by Z x 1 α−1 α Ja f (x) = (x − t) f (t)dt, α ≥ 0, a < x ≤ b. Γ(α) a This integral is motivated by the well known Cauchy formula: Z x Z x Z t1 Z tn−1 1 n−1 dt1 (x − t) f (t)dt, n ∈ N∗ . dt2 ... f (tn )dtn = Γ(n) a a a a The second is the Hadamard fractional integral introduced by Hadamard [9]. It is given by: Z x³ 1 x ´α−1 dt Jaα f (x) = log f (t) , α > 0, x > a. Γ(α) a t t ∗Department of Mathematics, Faculty of Science and Arts, D¨ uzce University, D¨ uzce, Turkey, Email:
[email protected] † Corresponding Author. ‡ Laboratory of Pure and Applied Mathematics, UMAB, University of Mostaganem, Algeria, Email:
[email protected] § Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University, Afyon-TURKEY, Email:
[email protected],
[email protected] ¶ Centre for Advanced Studies in Pure and Applied Mathematics, Bahauddin Zakariya University, Multan, 60800, Pakistan, Email:
[email protected]
2
The Hadamard integral is based on the generalization of the integral Z tn−1 Z x Z Z x³ dt1 t1 dt2 f (tn ) 1 1 x ´n−1 dt ... dtn = log f (t) t1 a t2 tn Γ(α) Γ(α) a t t a a for n ∈ N∗ . In [10], Katugampola gave a new fractional integration which generalizes both the Riemann-Liouville and Hadamard fractional integrals into a single form. This generalization is based on the observation that, for n ∈ N∗ , Z x Z t1 Z tn−1 Z ¢n−1 s (s + 1)1−n x ¡ s+1 s s s t1 dt1 t f (t)dt, t2 dt2 ... tn f (tn )dtn = x − ts+1 Γ(α) a a a a which gives the following fractional version Z ¢α−1 s (s + 1)1−n x ¡ s+1 s α Ja f (x) = x − ts+1 t f (t)dt, Γ(α) a where α and s 6= −1 are real numbers. Recently, in [6], Diaz and Pariguan have defined new functions called k-gamma and k-beta functions and the Pochhammer k-symbol that is respectively generalization of the classical gamma and beta functions and the classical Pochhammer symbol: x
n!k n (nk) k −1 , k > 0, n→∞ (x)n,k
Γk (x) = lim
where (x)n,k is the Pochhammer k-symbol for factorial function. It has been tk
shown that the Mellin transform of the exponential function e− k is the k-gamma function, explicitly given by Z ∞ tk Γk (x) = tx−1 e− k dt, x > 0. 0 x
Clearly, Γ(x) = lim Γk (x), Γk (x) = k k −1 Γ( xk ) and Γk (x + k) = xΓk (x). Furtherk→1
more, k-beta function is defined as follows Z y 1 1 x −1 Bk (x, y) = t k (1 − t) k −1 dt, k 0 k (y) so that Bk (x, y) = k1 B( xk , ky ) and Bk (x, y) = ΓΓk (x)Γ . k (x+y) Later, under the above definitions, in [13], Mubeen and Habibullah have introduced the k-fractional integral of the Riemann-Liouville type as follows: Z x α 1 −1 α (x − t) k f (t)dt, α > 0, x > 0. k J f (x) = kΓk (α) 0
Note that when k → 1, then it reduces to the classical Riemann-liouville fractional integral.
2. (k, s)-Riemann-Liouville fractional integral In this section, we present the (k, s) fractional integration which generalizes all of the above Riemann-Liouville fractional integrals as follows:
3
2.1. Definition. Let f be a continuous function on on a the finite real interval [a, b]. Then (k, s)-Riemann-Liouville fractional integral of f of order α > 0 is defined by: α Z ¢ α −1 (s + 1)1− k x ¡ s+1 s α (2.1) x − ts+1 k ts f (t)dt, x ∈ [a, b], k Ja f (x) := kΓk (α) a where k > 0, s ∈ R\{−1}. In the following theorem, we prove that the (k, s) fractional integral is well defined: 2.2. Theorem. Let f ∈ L1 [a, b], s ∈ R\{−1} and k > 0. Then sk Jaα f (x) exists for any x ∈ [a, b], α > 0. h¡ ¢ α −1 i Proof. Let ∆ := [a, b] × [a, b] and P : ∆ → R ; P (x, t) = xs+1 − ts+1 k ts . It clear to see that P = P+ + P− , where ½ ¡ ¢ α −1 xs+1 − ts+1 k ts , a ≤ t ≤ x ≤ b P+ (x, t) := 0 ,a ≤ x ≤ t ≤ b and
½ ¡ P− (x, t) :=
ts+1 − xs+1 0
¢ αk −1
xs
,a ≤ t ≤ x ≤ b , a ≤ x ≤ t ≤ b.
Since P is measurable on ∆, then we can write Z b Z x Z x ¡ s+1 ¢ α −1 ¢α k ¡ s+1 P (x, t)dt = P (x, t)dt = x − ts+1 k ts dt = x − as+1 k . α a a a By using the repeated integral, we obtain ! ÃZ ! Z b ÃZ b Z b b P (x, t) |f (x)| dt dx = |f (x)| P (x, t)dt dx a
a
a
k α
=
b
!
Z
P (x, t) |f (x)| dt dx = a
a
Z
b
¡ s+1 ¢α x − as+1 k |f (x)| dx
a
¢α k ¡ s+1 b − as+1 k α
≤ That is Z b ÃZ
a
ÃZ
b
|f (x)| a
b
Z
b
|f (x)| dx. a
! P (x, t)dt dx
a
¢α k ¡ s+1 b − as+1 k kf (x)kL1 [a,b] < ∞. α Therefore, the function Q : ∆ → R; Q(x, t) := P (x, t)f (x) is integrable over ∆ by Rb Tonelli’s theorem. Hence, by Fubini’s theorem a P (x, t)f (x)dx is an integrable function on [a, b], as a function of t ∈ [a, b]. That is, sk Jaα f (x) exists. ¤ ≤
Now, we prove the commutativity and the semigroup properties of the (k, s)Riemann-Liouville fractional integral. We have:
4
2.3. Theorem. Let f be continuous on [a, b], k > 0 and s ∈ R\{−1}. Then, £ ¤ s α s β s α+β f (x) = sk Jaβ [ sk Jaα f (x)] , k Ja k Ja f (x) = k Ja for all α > 0, β > 0, x ∈ [a, b]. Proof. Thanks to Definition 1 and by Dirichlet’s formula, we have Z ¢ α −1 £ ¤ (s + 1)1− αk x ¡ s+1 s α s β x − ts+1 k ts ks Jaβ f (t)dt k Ja k Ja f (x) = kΓk (α) a · ¸ β α 1− ¢α ¢β R t ¡ s+1 (s+1)1− k R x ¡ s+1 s+1 k −1 s (s+1) k s+1 k −1 s = kΓk (α) a x t τ f (τ )dτ dt. −t t −τ kΓk (β) a α
(s+1)1− k kΓk (α)
=
Rx¡ a
s+1
x
−t
s+1
¢ αk −1
· t
β
(s+1)1− k kΓk (β)
s
Rt¡ a
s+1
t
−τ
s+1
¢ βk −1
¸ s
τ f (τ )dτ dt.
That is (2.2)
¸ ·Z x α+β Z x ¡ ¢β ¡ s+1 ¢α (s + 1)2− k s+1 k −1 s s+1 k −1 s s+1 = 2 t t −τ dt dτ. τ f (τ ) x −t k Γk (α)Γk (β) a τ ¡ ¢ ¡ ¢ Using the change of variable y = ts+1 − τ s+1 / xs+1 − τ s+1 , we can write s α k Ja
(2.3)
Z
x
£
¤
s β k Ja f (x)
¢ β −1 ¡ s+1 ¢ α −1 ¡ s+1 (xs+1 −τ s+1 ) t − τ s+1 k ts dt = x − ts+1 k s+1
τ
(xs+1 −τ s+1 )
=
α+β −1 k
s+1
R1 0
(1 − y)
α k −1
y
β k −1
dy
=
α+β −1 k
R1 0
(xs+1 −τ s+1 )
α
α+β −1 k
s+1
kBk (α, β).
According to the k-beta function and by (2.2) and (2.3), we obtain s α k Ja
£
¤
s β k Ja f (x)
α+β
= =
(s + 1)1− k kΓk (α + β)
Z
x
¡
xs+1 − τ s+1
¢ α+β k −1
τ s f (τ )dτ
a
s α+β f (x). k Ja
This completes the proof of the Theorem 2.3. 2.4. Theorem. Let α, β > 0, k > 0 and s ∈ R\{−1}. Then, we have (2.4)
s α k Ja
h¡ ¢ β −1 i = xs+1 − as+1 k
¡ s+1 ¢ α+β −1 Γk (β) , x − as+1 k α (s + 1) k Γk (α + β)
where Γk denotes the k-gamma function.
β
(1 − y) k −1 y k −1 dy
¤
5
¡ ¢ ¡ ¢ Proof. By Definition 1 and using the change of variable y = xs+1 − ts+1 / xs+1 − as+1 ; x ∈ ]a, b], we get α Z h¡ ¢β i ¢ α −1 ¡ ¢ α+β −1 (s + 1)1− k x ¡ s+1 s α s+1 s+1 k −1 J x − a = x − ts+1 k ts ts+1 − as+1 k dt k a kΓk (α) a =
¢ α+β −1 Z 1 α ¡ (s + 1)− k xs+1 − as+1 k β α (1 − y) k −1 y k −1 dy kΓk (α) 0 ¡
=
¢ α+β −1 xs+1 − as+1 k Bk (α, β). α (s + 1) k Γk (α)
The case a = x is trivial. The proof of the Theorem 2.4 is complete.
¤
2.5. Remark. (i :) Taking s = 0, k > 0 in (2.4), we obtain h i β α+β Γk (β) −1 α k −1 (2.5) = (x − a) k . k Ja (x − a) Γk (α + β) (ii :) The formula (2.4) for s = 0, k = 1 becomes h i Γ(β) β−1 α+β−1 Jaα (x − a) = (x − a) . Γ(α + β) 2.6. Corollary. Let k > 0 and s ∈ R\{−1}. Then the formula ¡ s+1 ¢ α −2 1 s α x − as+1 k (2.6) α k Ja (1) = (s + 1) k Γk (α + k) is valid for any α > 0. 2.7. Remark. (a :) For s = 0, k > 0 in (2.6), we get (2.7)
α k Ja (1)
=
α 1 −2 (x − a) k . Γk (α + k)
(b :) For s = 0, k = 1 we have Jaα (1) =
1 α−2 (x − a) . Γ(α + 1)
3. Some new (k, s)-Riemann-Liouville fractional integral inequalities Chebyshev inequalities can be represented in (k, s)-fractional integral forms as follows: 3.1. Theorem. Let f, g be two synchronous on [0, ∞), then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold: (3.1) (3.2)
s α k Ja f g(t)
≥
1 s α J f (t) sk Jaα g(t) α Ja (1) k a
s α s β s β s α k Ja f g(t) k Ja (1)+ k Ja f g(t) k Ja (1)
≥
s α s β s α s β k Ja f (t) k Ja g(t)+ k Ja g(t) k Ja f (t).
6
Proof. Since the functions f and g are synchronous on [0, ∞), then for all x, y ≥ 0, we have (f (x) − f (y)) (g (x) − g(y)) ≥ 0. Therefore (3.3)
f (x) g (x) + f (y) g (y) ≥ f (x) g (y) + f (y) g (x)
1− α ¡ ¢ α −1 k Multiplying both sides of (3.3) by (s+1) ts+1 − xs+1 k xs , then integrating kΓk (α) the resulting inequality with respest to x over (a, t), we obtain α
(s + 1)1− k kΓk (α)
Z
t
¡
ts+1 − xs+1
¢ αk −1
xs f (x) g (x) dx
a
α Z ¢ α −1 (s + 1)1− k t ¡ s+1 t − xs+1 k xs f (y) g (y) dx kΓk (α) a Z t 1− α ¡ s+1 ¢ α −1 (s + 1) k ≥ t − xs+1 k xs f (x) g (y) dx kΓk (α) a α Z t ¡ s+1 ¢ α −1 (s + 1)1− k + t − xs+1 k xs f (y) g (x) dx, kΓk (α) a
+
i.e. (3.4)
s α k Ja f g(t)
+ f (y) g (y) sk Jaα (1) ≥ g (y) sk Jaα f (t) + f (y) sk Jaα g(t).
1− α ¡ ¢ α −1 k Multiplying both sides of (3.3) by (s+1) ts+1 − y s+1 k y s , then integrating kΓk (α) the resulting inequality with respest to y over (a, t), we obtain α
+ 1)1− k kΓk (α)
(s s α k Ja f g(t) + ≥
1− α k
(s s α k Ja f (t)
+ 1) kΓk (α)
+
(s s α k Ja g(t)
¡
t
¡ s+1 ¢ α −1 t − y s+1 k y s f (y) g (y) dy
ts+1 − y s+1
t
≥
a
Z
t
¡
ts+1 − y s+1
a
1 s α J f (t) sk Jaα g(t) Jaα (1) k a
and the first inequality is proved.
y s dy
¡ s+1 ¢ α −1 t − y s+1 k y s g(y)dy
that is s α k Ja f g(t)
¢ αk −1
a
Z
1− α k
+ 1) kΓk (α)
t
a
Z
α
+ 1)1− k kΓk (α)
(s s α k Ja (1)
Z
¢ αk −1
y s f (y) dy,
7 β 1− ¢ β −1 k ¡ s+1 Multiplying both sides of (3.3) by (s+1) t − y s+1 k y s , then integratkΓk (α) ing the resulting inequality with respest to y over (a, t), we obtain β
+ 1)1− k kΓk (α)
(s s α k Ja f g(t) + sk Jaα (1)
(s + 1) kΓk (α)
(s s α k Ja f (t)
+ 1) kΓk (α)
≥
+ sk Jaα g(t)
¡
t
¡ s+1 ¢ β −1 t − y s+1 k y s f (y) g (y) dy
ts+1 − y s+1
¢ βk −1
y s dy
a
Z
1− β k
(s + 1) kΓk (α)
t
a
Z
1− β k
1− β k
Z
t
¡ s+1 ¢ β −1 t − y s+1 k y s g(y)dy
a
Z
t
¡
ts+1 − y s+1
¢ βk −1
y s f (y) dy,
a
that is s α s β s β s α k Ja f g(t) k Ja (1)+ k Ja f g(t) k Ja (1)
≥ sk Jaα f (t) sk Jaβ g(t)+ sk Jaα g(t) sk Jaβ f (t)
and the second inequality is proved. The proof is completed.
¤
3.2. Theorem. Let f, g be two synchronous on [0, ∞), h ≥ 0, then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold: 1 β k
(s + 1) Γk (β + k) + ≥
¡ s+1 ¢ β −2 t − as+1 k
s α k Ja f gh(t)
¡ s+1 ¢ α −2 1 t − as+1 k (s + 1) Γk (α + k) α k
s α s β k Ja f h(t) k Ja g(t)
+
s α s β k Ja gh(t) k Ja f (t)
+ sk Jaα f (t) sk Jaβ gh(t) +
s β k Ja f gh(t)
−
s α s β k Ja h(t) k Ja f g(t)
−
s α s β k Ja f g(t) k Ja h(t)
s α s β k Ja g(t) k Ja f h(t).
Proof. Since the functions f and g are synchronous on [0, ∞) and h ≥ 0, then for all x, y ≥ 0, we have (f (x) − f (y)) (g (x) − g(y)) (h (x) + h (y)) ≥ 0. By opening the above, we get (3.5)
f (x) g (x) h (x) + f (y) g (y) h (y) ≥
f (x) g (y) h (x) + f (y) g (x) h (x) − f (y) g (y) h (x) −f (x) g (x) h (y) + f (x) g (y) h (y) + f (y) g (x) h (y) .
8 1− α ¡ ¢ α −1 k Multiplying both sides of (3.5) by (s+1) ts+1 − xs+1 k xs , then integrating kΓk (α) the resulting inequality with respest to x over (a, t), we obtain
α
(s + 1)1− k kΓk (α)
Z
t
¢ α −1 ¡ s+1 t − xs+1 k xs f (x) g (x) h (x) dx
a
α Z ¢ α −1 (s + 1)1− k t ¡ s+1 +f (y) g (y) h (y) t − xs+1 k xs dx kΓk (α) a α Z t ¡ s+1 ¢ α −1 (s + 1)1− k ≥ g (y) t − xs+1 k xs f (x) h (x) dx kΓk (α) a Z 1− α ¢ α −1 (s + 1) k t ¡ s+1 +f (y) t − xs+1 k xs g (x) h (x) dx kΓk (α) a α Z ¢ α −1 (s + 1)1− k t ¡ s+1 t − xs+1 k xs h (x) dx −f (y) g (y) kΓk (α) a α Z t ¡ s+1 ¢ α −1 (s + 1)1− k −h (y) t − xs+1 k xs f (x) g (x) dx kΓk (α) a α Z ¢ α −1 (s + 1)1− k t ¡ s+1 +g (y) h (y) t − xs+1 k xs f (x) dx kΓk (α) a α Z t ¡ s+1 ¢ α −1 (s + 1)1− k +f (y) h (y) t − xs+1 k xs g (x) dx. kΓk (α) a
i.e,
(3.6) ≥
s α k Ja f gh(t)
+ f (y) g (y) h(y) sk Jaα (1)
g (y) sk Jaα f h(t) + f (y) sk Jaα gh(t) − f (y) g (y) sk Jaα h(t) − h(y) sk Jaα f g(t) +g (y) h (y) sk Jaα f (t) + f (y) h (y) sk Jaα g(t).
9 β 1− ¢ β −1 k ¡ s+1 Multiplying both sides of (3.6) by (s+1) t − y s+1 k y s , then integrating kΓk (β) the resulting inequality with respest to y over (a, t), we obtain β Z ¢ β −1 (s + 1)1− k t ¡ s+1 s α J f gh(t) t − y s+1 k y s dy k a kΓk (β) a
Z
β
+sk Jaα (1) ≥
(s + 1)1− k kΓk (β)
¡
Z
t
¡
β
Z
t
(s + 1)1− k kΓk (β) β
− sk Jaα h(t)
(s + 1)1− k kΓk (β)
y s f (y) g (y) h(y)dy
β
(s + 1)1− k kΓk (β)
+ sk Jaα g(t)
(s + 1)1− k kΓk (β)
β
¢ βk −1
y s g (y) dy
¡ s+1 ¢ β −1 t − y s+1 k y s f (y) dy
t
¡ s+1 ¢ β −1 t − y s+1 k y s f (y) g (y) dy
a
(s + 1)1− k kΓk (β)
+ sk Jaα f (t)
ts+1 − y s+1
a
Z
β
− sk Jaα f g(t)
¢ βk −1
a β
+ sk Jaα gh(t)
ts+1 − y s+1
a
+ 1)1− k kΓk (β)
(s s α k Ja f h(t)
t
Z
t
¡ s+1 ¢ β −1 t − y s+1 k y s h(y)dy
a
Z
t
¡ s+1 ¢ β −1 t − y s+1 k y s g (y) h (y) dy
t
¡ s+1 ¢ β −1 t − y s+1 k y s f (y) h (y) dy,
a
Z a
that is s α s β k Ja f gh(t)k Ja (1)
+sk Jaα (1)sk Jaβ f gh(t) ≥
s α s β k Ja f h(t) k Ja g(t)
+
s α s β k Ja gh(t) k Ja f (t)
− sk Jaα h(t) sk Jaβ f g(t) −
s α s β k Ja f g(t) k Ja h(t)
+ sk Jaα f (t) sk Jaβ gh(t) +
s α s β k Ja g(t) k Ja f h(t)
which this completes the proof.
¤
3.3. Corollary. Let f, g be two synchronous on [0, ∞), h ≥ 0, then for all t > a ≥ 0, α > 0, the following inequalities for (k, s)-fractional integrals hold: ¢ α −2 s α ¡ s+1 1 t − as+1 k α k Ja f gh(t) k (s + 1) Γk (α + k) ≥
s α s α k Ja f h(t) k Ja g(t)
+
s α s α k Ja gh(t) k Ja f (t)
−
s α s α k Ja h(t) k Ja f g(t).
3.4. Theorem. Let f, g and h be three monotonic functions defined on [0, ∞) satisfying the following (f (x) − f (y)) (g (x) − g(y)) (h (x) − h (y)) ≥ 0
10
for all x, y ∈ [a, t] , then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold: ¡ s+1 ¢ β −2 s α 1 t − as+1 k β k Ja f gh(t) (s + 1) k Γk (β + k) − ≥
¡ s+1 ¢ α −2 1 t − as+1 k (s + 1) Γk (α + k) α k
s α s β k Ja f h(t) k Ja g(t)
s α s β k Ja gh(t) k Ja f (t)
+
−sk Jaα f (t) sk Jaβ gh(t) −
s β k Ja f gh(t)
−
s α s β k Ja h(t) k Ja f g(t)
+
s α s β k Ja f g(t) k Ja h(t)
s α s β k Ja g(t) k Ja f h(t).
Proof. The proof is similar to that given in Theorem 3.2.
¤
3.5. Theorem. Let f and g be two functions on [0, ∞), then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold: ¡ s+1 ¢ β −2 s α 2 1 t − as+1 k (3.7) β k Ja f (t) (s + 1) k Γk (β + k) +
¡ s+1 ¢ α −2 1 t − as+1 k (s + 1) Γk (α + k) α k
s β 2 k Ja g (t)
≥ 2 sk Jaα f (t) sk Jaβ g(t) (3.8)
s α 2 s β 2 k Ja f (t) k Ja g (t)
s β 2 s α 2 k Ja f (t) k Ja g (t)
+
≥ 2 sk Jaα f g(t) sk Jaβ f g(t).
Proof. Since, 2
(f (x) − g(y)) ≥ 0 then we have (3.9)
f 2 (x) + g 2 (y) ≥ 2f (x)g(y).
β 1− α ¡ 1− ¢ α −1 ¢ β −1 k k ¡ s+1 Multiplying both sides of (3.9) by (s+1) ts+1 − xs+1 k xs and (s+1) t − y s+1 k y s , kΓk (α) kΓk (β) then integrating the resulting inequality with respest to x and y over (a, t) respectively, we obtain (3.7). On the other hand, since
2
(f (x)g(y) − f (y)g(x)) ≥ 0 then under procedures similar to the above we obtain (3.8).
¤
3.6. Corollary. Let f and g be two functions on [0, ∞), then for all t > a ≥ 0, α > 0, the following inequalities for (k, s)-fractional integrals hold: ¡ s+1 ¢ α −2 £s α 2 ¤ 1 s β 2 t − as+1 k α k Ja f (t) + k Ja g (t) (s + 1) k Γk (α + k) ≥
2 sk Jaα f (t) sk Jaα g(t)
s α 2 s α 2 k Ja f (t) k Ja g (t)
2
≥ [sk Jaα f g(t)] .
11
3.7. Theorem. Let f : R → R and defined by Z x f¯(x) = ts f (t)dt, x > a ≥ 0, s ∈ R\{−1} a
then for α ≥ k > 0 s α k Ja f (x)
=
1 s α−k ¯ J f (x) k k a
Proof. By definition of the (k, s)-fractional integral and by using Dirichlet’s formula, we have Z t α Z ¢α (s + 1)1− k x ¡ s+1 s α¯ s+1 k −1 s us f (u)dudt J (x) = t f x − t k a kΓk (α) a a Z x Z x 1− α ¡ s+1 ¢ α −1 (s + 1) k = us f (u) x − ts+1 k ts dtdu kΓk (α) a u α Z x ¢α ¡ s+1 (s + 1)− k = x − us+1 k us f (u)du Γk (α + k) a =
k sk Jaα+k f (x).
This completes the proof of Theorem 3.7.
¤
We give the generalized Cauchy-Buniakovsky-Schwarz inequality as follows: 3.8. Lemma. Let f, g, h : [a, b] → (0, ∞) be three functions 0 ≤ a < b. Then (3.10)
ÃZ
! ÃZ
b
m
b
x
g (t)h (t)f (t)dt
! n
ÃZ
y
g (t)h (t)f (t)dt
a
≥
a
!2
b
g
m+n 2
(t)h
x+y 2
(t)f (t)dt
.
a
where m, n, x, y arbitrary real numbers. Proof. Z
b
s Z p g m (t)hx (t)f (t)
a
Z
b
Z m
x
n
a
y
n
y
g (t)h (t)f (t)dt + g (t)h (t)f (t) a
(t)h
x+y 2
2 g m (t)hx (t)f (t)dt dt ≥ 0
b
g m (t)hx (t)f (t)dt
a
s m+n 2
b
a
Z
b
g (t)h (t)f (t)
Z
b
(t)f (t) a
≥ 0
g n (t)hy (t)f (t)dt −
s Z g n (t)hy (t)f (t)
p
a
"
−2 g
b
s Z g m (t)hx (t)f (t)dt a
b
g n (t)hy (t)f (t)dt dt
12
ÃZ
! ÃZ
b
2
m
g (t)h (t)f (t)dt
≥ 2
n
y
g (t)h (t)f (t)dt
a
ÃZ
!
b
x
a
! sZ
b
g
m+n 2
(t)h
x+y 2
b
Z
g m (t)hx (t)f (t)dt
(t)f (t)dt
a
s
a
b
g n (t)hy (t)f (t)dt a
which this give the requair inequality.
¤
3.9. Theorem. Let f ∈ L1 [a, b]. Then ³ ´2 ´³ ´ ³ m+n α α n( α −1)+1 p ( −1)+1 r+p s m( k −1)+1 r (3.11) f (x) sk Ja k f (x) ≥ sk Ja 2 k f 2 (x) k Ja for k, m, n, r, p > 0 and α > 1. 1− α ¡ ¢ α −1 s k Proof. By taking g(t) = xs+1 − ts+1 k , f (t) = t (s+1) and h(t) = f (t) in kΓk (α) (3.10), we obtain µ ¶µ ¶ α Z α Z ¢m( αk −1) s r ¢n( α −1) s p (s + 1)1− k x ¡ s+1 (s + 1)1− k x ¡ s+1 x − ts+1 t f (t)dt x − ts+1 k t f (t)dt kΓk (α) kΓk (α) a a µ ≥
α
(s + 1)1− k kΓk (α)
Z
x
¶2 ¡ s+1 ¢ m+n ( α −1) s r+p t f 2 (t)dt x − ts+1 2 k
a
which can be written as (3.11).
¤
3.10. Remark. For k = 1 in (3.11), we get the following inequalities ³ ´³ ´ ³ m+n ´2 (α−1)+1 r+s Jam(α−1)+1 f r (x) k Jan(α−1)+1 f s (x) ≥ k Ja 2 f 2 (x) .
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