MATEMATIQKI VESNIK
originalni nauqni rad
research paper
66, 4 (2014), 387–396
December 2014
UNIVARIATE FRACTIONAL POLYA TYPE
INTEGRAL INEQUALITIES
George A. Anastassiou
Abstract. Here we establish a series of various fractional Polya type integral inequalities
with the help of generalized right and left fractional derivatives. We give an application to complex
valued functions defined on the unit circle.
1. Introduction
We mention the following famous Polya’s integral inequality, see [12], [13,
p. 62], [14] and [15, p. 83].
Theorem 1. Let f (x) be differentiable and not identically a constant on [a, b]
with f (a) = f (b) = 0. Then there exists at least one point ξ ∈ [a, b] such that
Z b
4
|f 0 (ξ)| >
f (x) dx.
(1)
(b − a)2 a
In [16], Feng Qi presents the following very interesting Polya type integral
inequality (2), which generalizes (1).
Theorem 2. Let f (x) be differentiable and not identically constant on [a, b]
with f (a) = f (b) = 0 and M = supx∈[a,b] |f 0 (x)|. Then
¯Z b
¯
¯
¯ (b − a)2
¯
¯≤
M,
(2)
f
(x)
dx
¯
¯
4
a
where
(b−a)2
4
in (2) is the best constant.
The above motivate the current paper. In this article, we present univariate
fractional Polya type integral inequalities in various cases, similar to (2). For this
purpose we need the following fractional calculus background.
2010 Mathematics Subject Classification: 26A33, 26D10, 26D15
Keywords and phrases: Polya integral inequality; fractional derivative.
387
388
G. A. Anastassiou
Let α > 0, m = [α], β = α − m, 0 < β < R1, f ∈ C([a, b]), [a, b] ⊂ R,
∞
x ∈ [a, b]. The gamma function Γ is given by Γ(α) = 0 e−t tα−1 dt. We define the
left Riemann-Liouville integral
Z x
1
a+
(x − t)α−1 f (t) dt,
(Jα f )(x) =
Γ(α) a
α
a ≤ x ≤ b. We define the subspace Ca+
([a, b]) of C m ([a, b]):
©
ª
a+ (m)
α
Ca+
([a, b]) = f ∈ C m ([a, b]) : J1−β
f
∈ C 1 ([a, b]) .
α
For f ∈ Ca+
([a, b]), we define the left generalized α-fractional derivative of f over
[a, b] as
¡ a+ (m) ¢0
α
f := J1−β
,
Da+
f
see [1, p. 24]. Canavati in [5], first introduced the above over [0, 1]. Notice that
α
Da+
f ∈ C([a, b]).
We need the following left fractional Taylor’s formula, see [1, pp. 8–10], and in
[5] the same formula over [0, 1] that appeared first.
α
([a, b]).
Theorem 3. Let f ∈ Ca+
(i) If α ≥ 1, then
(x − a)2
(x − a)m−1
f (x) = f (a) + f 0 (a)(x − a) + f 00 (a)
+ · · · + f (m−1) (a)
2
(m − 1)!
Z x
1
α
f )(t) dt, for all x ∈ [a, b].
+
(x − t)α−1 (Da+
Γ(α) a
(ii) If 0 < α < 1, we have
Z x
1
α
f (x) =
(x − t)α−1 (Da+
f )(t) dt, for all x ∈ [a, b].
Γ(α) a
Let again α > 0, m = [α], β = α − m, f ∈ C([a, b]), call the right RiemannLiouville fractional integral operator by
Z b
1
α
(Jb−
f )(x) :=
(t − x)α−1 f (t) dt,
Γ(α) x
x ∈ [a, b], see also [2, 8–10, 17]. Define the subspace of functions
©
ª
1−β (m)
α
Cb−
([a, b]) := f ∈ C m ([a, b]) : Jb−
f
∈ C 1 ([a, b]) .
Define the right generalized α-fractional derivative of f over [a, b] as
¡ 1−β (m) ¢0
α
,
Db−
f = (−1)m−1 Jb−
f
0
α
see [2]. We set Db−
f = f . Notice that Db−
f ∈ C([a, b]).
We need the following right Taylor fractional formula from [2].
389
Univariate fractional Polya type integral inequalities
α
Theorem 4. Let f ∈ Cb−
([a, b]), α > 0, m := [α]. Then
(i) If α ≥ 1, we get
f (x) =
m−1
X
k=0
f (k) (b)
α
α
(x − b)k + (Jb−
Db−
f )(x),
k!
all x ∈ [a, b].
(ii) If 0 < α < 1, we get
α
α
f (x) = Jb−
Db−
f (x) =
1
Γ(α)
Z
b
x
α
(t − x)α−1 (Db−
f )(t) dt,
all x ∈ [a, b].
Definition 5. [3] Let f ∈ C([a, b]), x ∈ [a, b], α > 0, m := [a]. Assume that
a+b
α
α
f ∈ Cb−
([ a+b
2 , b]) and f ∈ Ca+ ([a, 2 ]). We define the balanced Canavati type
fractional derivative by
(
α
Db−
f (x), for a+b
2 ≤ x ≤ b,
α
D f (x) :=
α
Da+ f (x), for a ≤ x < a+b
2 .
In [4], we proved the following fractional Polya type integral inequality without
any boundary conditions.
α
([a, a+b
Theorem 6. Let 0 < α < 1, f ∈ C([a, b]). Assume f ∈ Ca+
2 ]) and
a+b
α
f ∈ Cb− ([ 2 , b]). Set
© α
ª
α
M1 (f ) = max kDa+
f k∞,[a, a+b ] , kDb−
f k∞,[ a+b ,b] .
(3)
2
2
Then
¯Z b
¯ Z b
¯
¯
¯
f (x) dx¯¯ ≤
|f (x)| dx ≤
¯
a
a
¢
¡ α
α
f k∞,[ a+b ,b] ³ b − a ´α+1
kDa+ f k∞,[a, a+b ] + kDb−
2
Γ(α + 2)
2
2
≤ M1 (f )
(b − a)α+1
. (4)
Γ(α + 2)2α
Inequalities (4) are sharp, namely they are attained by
½
(x − a)α , x ∈ [a, a+b
2 ]
f∗ (x) =
, 0 < α < 1.
a+b
α
(b − x) , z ∈ [ 2 , b]
(5)
Clearly here non zero constant functions f are excluded.
The last result also motivates this work.
Remark 7. ([4]) When α ≥ 1, thus m = [α] ≥ 1, and by assuming that
f (k) (a) = f (k) (b) = 0, k = 0, 1, . . . , m − 1, we can prove the same statements (4),
(5) as in Theorem 6. If we set there α = 1 we derive exactly Theorem 2. So we
have generalized Theorem 2. Again here f (m) cannot be a constant different than
zero, equivalently, f cannot be a non-trivial polynomial of degree m.
We continue here with other interesting univariate fractional Polya type inequalities.
390
G. A. Anastassiou
2. Main results
We present our first main result.
α
Theorem 8. Let α ≥ 1, m = [α], f ∈ C([a, b]). Assume f ∈ Ca+
([a, a+b
2 ])
a+b
α
(k)
(k)
and f ∈ Cb− ([ 2 , b]), such that f (a) = f (b) = 0, k = 0, 1, . . . , m − 1. Set
© α
ª
α
M2 (f ) = max kDa+
f kL1 ([a, a+b ]) , kDb−
f kL1 ([ a+b ,b]) .
(6)
2
Then
¯Z
¯
¯
¯
b
a
¯
¯
f (x) dx¯¯ ≤
2
³ b − a ´α
1
M2 (f )
kDα f kL1 ([a,b]) ≤
(b − a)α .
Γ(α + 1)
2
Γ(α + 1)2α−1
Here f cannot be a non-trivial polynomial of degree m.
Proof. By the assumption and Theorem 3, we have
Z x
h a + bi
1
α
f (x) =
(x − t)α−1 (Da+
f )(t) dt, for all x ∈ a,
.
Γ(α) a
2
Also it holds, by the assumption and Theorem 4, that
Z b
ha + b i
1
α
f (x) =
(t − x)α−1 (Db−
f )(t) dt, for all x ∈
,b .
Γ(α) x
2
By (7) we get
1
|f (x)| ≤
Γ(α)
Z
x
a
(7)
(8)
α
(x − t)α−1 |(Da+
f )(t)| dt
(x − a)α−1
≤
Γ(α)
Z
a+b
2
a
h a + bi
α
|(Da+
f )(t)| dt, for all x ∈ a,
.
2
By (8) we derive
|f (x)| ≤
1
Γ(α)
Z
b
x
α
(t − x)α−1 |(Db−
f )(t)| dt
(b − x)α−1
≤
Γ(α)
Z
b
a+b
2
α
|(Db−
f )(t)| dt, for all x ∈
ha + b i
,b .
2
Consequently we have
Z a+b
2
|f (x)| dx ≤
µZ a+b
¶
2
1
α
(x − a)α−1 dx kDa+
f kL1 ([a, a+b ])
2
Γ(α) a
³ b − a ´α
1
α
=
kDa+
f kL1 ([a, a+b ]) ,
2
Γ(α + 1)
2
a
and
Z
b
a+b
2
1
|f (x)| dx ≤
Γ(α)
=
µZ
b
a+b
2
1
Γ(α + 1)
α−1
(b − x)
³ b − a ´α
2
(9)
¶
α
dx kDb−
f kL1 ([ a+b ,b])
2
α
kDb−
f kL1 ([ a+b ,b]) .
2
(10)
391
Univariate fractional Polya type integral inequalities
Therefore we obtain by adding (9) and (10) that
Z
³ b − a ´α £
¤
1
α
α
kDa+
f kL1 ([a, a+b ]) + kDb−
f kL1 ([ a+b ,b])
2
2
Γ(α + 1)
2
³ b − a ´α
1
=
kDα f kL1 ([a,b])
Γ(α + 1)
2
© α
ª (b − a)α
α
≤ max kDa+
f kL1 ([a, a+b ]) , kDb−
f kL1 ([ a+b ,b])
,
2
2
Γ(α + 1)2α−1
b
|f (x)| dx ≤
a
proving the claim.
We continue with
Theorem 9. Let p, q > 1 : p1 + 1q = 1, α > 1q , m = [α], f ∈ C([a, b]).
a+b
α
α
(k)
Assume f ∈ Ca+
([a, a+b
(a) = f (k) (b) = 0,
2 ]) and f ∈ Cb− ([ 2 , b]), such that f
1
k = 0, 1, . . . , m − 1. When q < α < 1, the last boundary conditions are void. Set
© α
ª
α
M3 (f ) = max kDa+
f kLq ([a, a+b ]) , kDb−
f kLq ([ a+b ,b]) .
2
(11)
2
Then
Z
b
|f (x)| dx ≤
a
³ b − a ´α+ p1
1
×
2
Γ(α)(p(α − 1) + 1) (α + p1 )
£ α
¤
α
f kLq ([a, a+b ]) + kDb−
× kDa+
f kLq ([ a+b ,b])
1
p
2
≤
2
M3 (f )
1
p
Γ(α)(p(α − 1) + 1) (α +
1
1 α− q1
p )2
(b − a)α+ p .
Again here f cannot be a non-trivial polynomial of degree m.
Proof. By Theorem 3 we have
Z x
1
α
|f (x)| ≤
(x − t)α−1 |(Da+
f )(t)| dt
Γ(α) a
µZ x
¶ p1 µZ x
¶ q1
1
p(α−1)
α
q
(x − t)
dt
|(Da+ f )(t)| dt
≤
Γ(α) a
a
h a + bi
p
1 (x − a)
α
≤
for all x ∈ a,
.
1 kDa+ f kLq ([a, a+b ]) ,
2
Γ(α) (p(α − 1) + 1) p
2
(p(α−1)+1)
That is
h a + bi
1 (x − a)α−1+ p
α
|f (x)| ≤
for all x ∈ a,
1 kDa+ f kLq ([a, a+b ]) ,
2
Γ(α) (p(α − 1) + 1) p
2
1
(12)
392
G. A. Anastassiou
Similarly, from Theorem 4 we get
Z b
1
α
|f (x)| ≤
(t − x)α−1 |(Db−
f )(t)| dt
Γ(α) x
µZ b
¶ p1 µZ b
¶ q1
1
α
≤
(t − x)p(α−1) dt
|(Db−
f )(t)|q dt
Γ(α) x
x
ha + b i
1 (b − x)(α−1)+ p
α
kD
,
for
all
x
∈
f
k
,b .
a+b
b−
Lq ([ 2 ,b])
Γ(α) (p(α − 1) + 1) p1
2
1
≤
That is
ha + b i
(b − x)α−1+ p
1
α
,b .
|f (x)| ≤
for all x ∈
1 kDb− f kLq ([ a+b ,b]) ,
2
Γ(α) (p(α − 1) + 1) p
2
1
(13)
Consequently we obtain by (12) that
Z
a+b
2
|f (x)| dx
a
≤
=
µZ
1
a+b
2
1
Γ(α)(p(α − 1) + 1) p
1
(x − a)
1
α−1+ p
a
³ b − a ´α+ p1
1
p
Γ(α)(p(α − 1) + 1) (α + p1 )
2
¶
α
dx kDa+
f kLq ([a, a+b ])
2
α
kDa+
f kLq ([a, a+b ]) .
2
(14)
Similarly it holds by (13) that
Z b
|f (x)| dx
a+b
2
≤
=
µZ
1
1
Γ(α)(p(α − 1) + 1) p
1
1
p
b
a+b
2
¶
1
α
(b − x)α−1+ p dx kDb−
f kLq ([ a+b ,b])
Γ(α)(p(α − 1) + 1) (α + p1 )
2
³ b − a ´α+ p1
2
α
f kLq ([ a+b ,b]) .
kDb−
2
Adding (14) and (15) we have
Z b
³ b − a ´α+ p1
1
×
|f (x)| dx ≤
1
2
a
Γ(α)(p(α − 1) + 1) p (α + p1 )
£ α
¤
α
× kDa+
f kLq ([a, a+b ]) + kDb−
f kLq ([ a+b ,b])
2
2
© α
ª
α
max kDa+
f kLq ([a, a+b ]) , kDb−
f kLq ([ a+b ,b])
1
2
2
(b − a)α+ p ,
≤
1
1
α−
1
Γ(α)(p(α − 1) + 1) p (α + p )2 q
proving the claim.
(15)
Univariate fractional Polya type integral inequalities
393
Combining Theorem 6, Remark 7, Theorem 8 and Theorem 9, we obtain
Theorem 10. Let any p, q > 1 : p1 + 1q = 1, α ≥ 1, m = [α], f ∈ C([a, b]).
a+b
α
α
(k)
Assume f ∈ Ca+
([a, a+b
(a) = f (k) (b) = 0,
2 ]) and f ∈ Cb− ([ 2 , b]), such that f
k = 0, 1, . . . , m − 1. Then
Z b
|f (x)| dx
a
¢
½ ¡kDα f k
α
+ kDb−
f k∞,[ a+b ,b] ³ b − a ´α+1
a+
∞,[a, a+b
]
2
2
≤ min
,
Γ(α + 2)
2
³ b − a ´α
1
kDα f kL1 ([a,b]) ,
Γ(α + 1)
2
¤
£ α
α
kDa+ f kLq ([a, a+b ]) + kDb−
f kLq ([ a+b ,b]) ³ b − a ´α+ p1 ¾
2
2
1
2
½
(b − a)α+1
M2 (f )
≤ min M1 (f )
,
(b − a)α ,
Γ(α + 2)2α Γ(α + 1)2α−1
¾
1
M3 (f )
α+ p
(b
−
a)
,
1
1
Γ(α)(p(α − 1) + 1) p (α + p1 )2α− q
Γ(α)(p(α − 1) + 1) p (α + p1 )
where M1 (f ) as in (3), M2 (f ) as in (6) and M3 (f ) as in (11). Here f cannot be
a non-trivial polynomial of degree m.
Corollary 11. Under the assumptions of Theorem 10, we have
¯
¯
Z b
Z b
¯ 1
¯
¯
¯≤ 1
f
(x)
dx
|f (x)| dx
¯b − a
¯ b−a
a
a
¢
¡
½ kDα f k
α
+ kDb−
f k∞,[ a+b ,b]
a+
]
∞,[a, a+b
2
2
(b − a)α ,
≤ min
Γ(α + 2)2α+1
1
(b − a)α−1 kDα f kL1 ([a,b]) ,
2α Γ(α + 1)
£ α
¤
¾
α
kDa+ f kLq ([a, a+b ]) + kDb−
f kLq ([ a+b ,b])
1
α+ p
−1
2
2
(b
−
a)
1
1
Γ(α)(p(α − 1) + 1) p (α + p1 )2α+ p
½
(b − a)α
M2 (f )
≤ min M1 (f )
,
(b − a)α−1 ,
Γ(α + 2)2α Γ(α + 1)2α−1
¾
1
M3 (f )
α+ p
−1
(b
−
a)
.
1
1
Γ(α)(p(α − 1) + 1) p (α + p1 )2α− q
In 1938, A. Ostrowski [11] proved the following important inequality.
Theorem 12. Let f : [a, b] → R be continuous on [a, b] and differentiable
on (a, b) whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e., kf 0 k∞ :=
394
G. A. Anastassiou
supt∈(a,b) |f 0 (t)| < +∞. Then
#
¯
¯ "
Z b
2
¯ 1
¯
1 (x − a+b
2 )
¯
¯
f (t) dt − f (x)¯ ≤
+
· (b − a)kf 0 k∞ ,
¯b − a
4
(b − a)2
a
for any x ∈ [a, b]. The constant
1
4
(16)
is the best possible.
In (16), for x = a+b
2 we get
¯
Z b
³ a + b ´¯¯ ³ b − a ´
¯ 1
¯
¯≤
f (t) dt − f
kf 0 k∞ .
¯b − a
¯
2
4
a
We have proved the following
Theorem 13. Let f ∈ C 1 ([a, b]), with f ( a+b
2 ) = 0. Then
¯
¯Z b
¯
¯ (b − a)2 0
¯
¯≤
f
(t)
dt
kf k∞ ,
¯
¯
4
a
where the constant
1
4
is the best possible.
So we have proved once again (2) with only one initial condition.
3. Application
Inequalities for complex valued functions defined on the unit circle were studied
extensively by S. Dragomir, see [6, 7]. We give here our version for these functions
involved in a Polya type inequality, by applying a result of this article.
Let t ∈ [a, b] ⊆ [0, 2π), the unit circle arc A = {z ∈ C : z = eit , t ∈ [a, b]}, and
f : A → C be a continuous function. Clearly here there exist functions u, v : A → R
continuous, the real and the complex part of f , respectively, such that
f (eit ) = u(eit ) + iv(eit ).
So that f is continuous, iff u, v are continuous.
Call g(t) = f (eit ), l1 (t) = u(eit ), l2 (t) = v(eit ), t ∈ [a, b]; so that g : [a, b] → C
and l1 , l2 : [a, b] → R are continuous functions in t.
If g has a derivative with respect to t, then l1 , l2 have also derivatives with
respect to t. In that case
ft (eit ) = ut (eit ) + ivt (eit ),
(i.e. g 0 (t) = l10 (t) + il20 (t)), which means
ft (cos t + i sin t) = ut (cos t + i sin t) + ivt (cos t + i sin t).
Let us call x = cos t, y = sin t. Then
ut (eit ) = ut (cos t + i sin t) = ut (x + iy) = ut (x, y)
=
∂u ∂x ∂u ∂y
∂u(eit )
∂u(eit )
+
=
(− sin t) +
cos t.
∂x ∂t
∂y ∂t
∂x
∂y
Univariate fractional Polya type integral inequalities
395
Similarly we find that
∂v(eit )
∂v(eit )
(− sin t) +
cos t.
∂x
∂y
Rb
Since g is continuous on [a, b], then a f (eit ) dt exists. Furthermore it holds
vt (eit ) =
Z
b
Z
f (eit ) dt =
a
b
Z
u(eit ) dt + i
a
b
v(eit ) dt.
a
We have here that
¯Z b
¯ Z b
Z b
Z b
¯
¯
it
it
it
¯
¯≤
f
(e
)
dt
|f
(e
)|
dt
≤
|u(e
)|
dt
+
|v(eit )| dt
¯
¯
a
a
a
a
Z b
Z b
=
|l1 (t)| dt +
|l2 (t)| dt.
a
a
We give the following application of Theorem 10.
Theorem 14. Let f ∈ C(A, C), [a, b] ⊆ [0, 2π); any p, q > 1 : p1 + 1q = 1,
a+b
α
α
α ≥ 1, m = [α]. Assume l1 , l2 ∈ Ca+
([a, a+b
2 ]) and l1 , l2 ∈ Cb− ([ 2 , b]), such that
(k)
(k)
(k)
(k)
l1 (a) = l2 (a) = l1 (b) = l2 (b) = 0, k = 0, 1, . . . , m − 1. Then
¯Z b
¯ Z b
¯
¯
it
¯
f (e ) dt¯¯ ≤
|f (eit )| dt
¯
a
a
¢
½ ¡kDα l k
α
+ kDb−
l1 k∞,[ a+b ,b] ³ b − a ´α+1
a+ 1 ∞,[a, a+b
]
2
2
≤ min
,
Γ(α + 2)
2
³ b − a ´α
1
kDα l1 kL1 ([a,b]) ,
Γ(α + 1)
2
£ α
¤
α
kDa+ l1 kLq ([a, a+b ]) + kDb−
l1 kLq ([ a+b ,b]) ³ b − a ´α+ p1 ¾
2
2
1
2
Γ(α)(p(α − 1) + 1) p (α + p1 )
¡
¢
½ kDα l k
α
+ kDb−
l2 k∞,[ a+b ,b] ³ b − a ´α+1
a+ 2 ∞,[a, a+b
2 ]
2
,
+ min
Γ(α + 2)
2
³ b − a ´α
1
kDα l2 kL1 ([a,b]) ,
Γ(α + 1)
2
£ α
¤
α
kDa+ l2 kL ([a, a+b ]) + kDb−
l2 kL ([ a+b ,b]) ³ b − a ´α+ p1 ¾
q
q
2
1
2
2
½
α+1
(b − a)
M2 (l1 )
,
(b − a)α ,
≤ min M1 (l1 )
α
Γ(α + 2)2 Γ(α + 1)2α−1
¾
1
M3 (l1 )
α+ p
(b
−
a)
1
1
Γ(α)(p(α − 1) + 1) p (α + p1 )2α− q
Γ(α)(p(α − 1) + 1) p (α + p1 )
396
G. A. Anastassiou
½
(b − a)α+1
M2 (l2 )
,
(b − a)α ,
α
Γ(α + 2)2 Γ(α + 1)2α−1
¾
1
M3 (l2 )
α+ p
,
1
1 (b − a)
Γ(α)(p(α − 1) + 1) p (α + p1 )2α− q
+ min M1 (l2 )
where M1 (li ) as in (3), M2 (li ) as in (6) and M3 (li ) as in (11), i = 1, 2.
Here l1 , l2 cannot be non-trivial polynomials of degree m.
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(received 14.04.2013; in revised form 02.11.2013; available online 15.11.2013)
Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A.
E-mail: ganastss@memphis.edu