Volume 8 (2007), Issue 4, Article 106, 13 pp.
OSTROWSKI TYPE INEQUALITIES OVER BALLS AND SHELLS VIA A
TAYLOR-WIDDER FORMULA
GEORGE A. ANASTASSIOU
D EPARTMENT OF M ATHEMATICAL S CIENCES
T HE U NIVERSITY OF M EMPHIS
M EMPHIS , TN 38152, U.S.A.
ganastss@memphis.edu
Received 06 August, 2007; accepted 27 November, 2007
Communicated by S.S. Dragomir
A BSTRACT. The classical Ostrowski inequality for functions on intervals estimates the value
of the function minus its average in terms of the maximum of its first derivative. This result
is extended to higher order over shells and balls of RN , N ≥ 1, with respect to an extended
complete Tschebyshev system and the generalized radial derivatives of Widder type. We treat
radial and non-radial functions.
Key words and phrases: Ostrowski inequality, multivariate inequality, ball and shell, Taylor-Widder formula, extended complete Tschebyshev system.
2000 Mathematics Subject Classification. 26D10, 26D15.
1. I NTRODUCTION
The classical Ostrowski inequality (of 1938, see [12]) is
2 !
Z b
x − a+b
1
1
2
f (y)dy − f (x) ≤
+
(b − a)kf 0 k∞ ,
b−a a
4
(b − a)2
for f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. This was extended to RN , N ≥ 1, over
balls and shells in [5], [6], [4]. Earlier this extension was done over boxes and rectangles, see
[2, p. 507-520], and [3], see also [1]. The produced Ostrowski type inequalities, in the above
mentioned references, were mostly sharp and they involved the first and higher order derivatives
of the engaged function f .
Here we derive a set of very general higher order Ostrowski type inequalities over shells
and balls with respect to an extended complete Tschebyshev system (see [10]) and generalized
derivatives of Widder type (see [15]). The proofs are based on the polar method and the general
Taylor-Widder formula (see [15], 1928). Our results generalize the higher order Ostrowski type
inequalities established in the above mentioned references.
255-07
2
G EORGE A. A NASTASSIOU
2. BACKGROUND
The following are taken from [15]. Let f, u0 , u1 , . . . , un ∈ C n+1 ([a, b]),n ≥ 0, and the
Wronskians
Wi (x) := W [u0 (x), u1 (x), . . . , ui (x)]
:=
u0 (x)
u00 (x)
..
.
(i)
u1 (x)
u01 (x)
(i)
...
...
u0 (x) u1 (x) . . .
ui (x)
u0i (x)
,
i = 0, 1, . . . , n.
(i)
ui (x)
Assume Wi (x) > 0 over [a, b]. Clearly then
φ0 (x) := W0 (x) = u0 (x), φ1 (x) :=
Wi (x)Wi−2 (x)
W1 (x)
,
.
.
.
,
φ
(x)
:=
,
i
(W0 (x))2
(Wi−1 (x))2
are positive on [a, b].
For i ≥ 0, the linear differentiable operator of order i:
W [u0 (x), u1 (x), . . . , ui−1 (x), f (x)]
Li f (x) :=
,
Wi−1 (x)
i = 2, 3, . . . , n
i = 1, . . . , n + 1;
L0 f (x) := f (x), ∀ x ∈ [a, b].
Then for i = 1, . . . , n + 1 we have
d
1
d
1
d
d 1 d f (x)
Li f (x) = φ0 (x)φ1 (x) · · · φi−1 (x)
···
.
dx φi−1 (x) dx φi−2 (x) dx
dx φ1 (x) dx φ0 (x)
Consider also
u0 (t)
u1 (t)
···
ui (t)
u00 (t)
u01 (t)
···
u0i (t)
1
···
···
···
···
,
gi (x, t) :=
Wi (t)
(i−1)
(i−1)
(i−1)
u0 (t) u1 (t) · · · ui (t)
u0 (x)
u1 (x) · · ·
ui (x)
i = 1, 2, . . . , n; g0 (x, t) := uu00(x)
, ∀x, t ∈ [a, b].
(t)
Note that gi (x, t) as a function of x is a linear combination of u0 (x), u1 (x), . . . , ui (x) and it
holds
Z x
Z x1
Z xi−2
Z xi−1
φ0 (x)
gi (x, t) =
φ1 (x)
···
φi−1 (xi−1 )
φi (xi )dxi dxi−1 . . . dx1
φ0 (t) · · · φi (t) t
t
t
t
Z x
1
=
φ0 (s) · · · φi (s)gi−1 (x, s)ds,
i = 1, 2, . . . , n.
φ0 (t) · · · φi (t) t
Example 2.1 ([15]). The sets
{1, x, x2 , . . . , xn },
{1, sin x, − cos x, − sin 2x, cos 2x, . . . , (−1)n−1 sin nx, (−1)n cos nx}
fulfill the above theory.
We mention
Theorem 2.1 (Karlin and Studden (1966), see [10, p. 376]). Let u0 , u1 , . . . , un ∈ C n ([a, b]),
n ≥ 0. Then {ui }ni=0 is an extended complete Tschebyshev system on [a, b] iff Wi (x) > 0 on
[a, b], i = 0, 1, . . . , n.
We also mention
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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O STROWSKI T YPE I NEQUALITIES VIA A TAYLOR -W IDDER F ORMULA
3
Theorem 2.2 (D. Widder, [15, p. 138]). Let the functions
f (x), u0 (x), u1 (x), . . . , un (x) ∈ C n+1 ([a, b]),
and the Wronskians W0 (x), W1 (x), . . . , Wn (x) > 0 on [a, b], x ∈ [a, b]. Then for t ∈ [a, b] we
have
u0 (x)
f (x) = f (t)
+ L1 f (t)g1 (x, t) + · · · + Ln f (t)gn (x, t) + Rn (x),
u0 (t)
where
Z x
Rn (x) :=
gn (x, s)Ln+1 f (s)ds.
t
For example, one could take u0 (x) = c > 0. If ui (x) = xi , i = 0, 1, . . . , n, defined on [a, b],
then
(x − t)i
Li f (t) = f (i) (t) and gi (x, t) =
, t ∈ [a, b].
i!
So under the assumptions of Theorem 2.2, we have
Z x
n
u0 (x) X
(2.1) f (x) = f (y)
+
Li f (y)gi (x, y) +
gn (x, t)Ln+1 f (t)dt, ∀ x, y ∈ [a, b].
u0 (y) i=1
y
If u0 (x) = c > 0, then
(2.2)
f (x) = f (y) +
n
X
Z
x
Li f (y)gi (x, y) +
gn (x, t)Ln+1 f (t)dt,
∀ x, y ∈ [a, b].
y
i=1
We call Li the generalized Widder-type derivative.
We need
Notation 2.1. Let A be a spherical shell ⊆ RN , N ≥ 1, i.e. A := B(0, R2 ) − B(0, R1 ),
0 < R1 < R2 .
Here the ball B(0, R) := {x ∈ RN : |x| < R}, R > 0, where | · | is the Euclidean norm,
2π N/2
also S N −1 := {x ∈ RN : |x| = 1} is the unit sphere in RN with surface area ωN := Γ(N/2)
. For
N
N −1
x ∈ R − {0} one can write uniquely x = rω, where r > 0, ω ∈ S
.
Let f ∈ C n+1 (A), n ≥ 0. If f is radial, i.e., f (x) = g(r), where r = |x|, R1 ≤ r ≤ R2 , then
g ∈ C n+1 ([R1 , R2 ]).
For radial f define
(2.3)
θi f (x) := Li g(r),
Here
g (i) (r) =
∂ i f (x)
,
∂ri
For F ∈ C(A) we have
Z
Z
(2.4)
F (x)dx =
A
i = 1, . . . , n + 1,
all
∀ x ∈ A.
i = 1, . . . , n + 1.
Z
R2
F (rω)r
S N −1
N −1
dr dω.
R1
We notice that
(2.5)
N
N
R2 − R1N
and
V ol(A) =
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
Z
R2
sN −1 ds = 1,
R1
ωN (R2N − R1N )
.
N
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4
G EORGE A. A NASTASSIOU
3. R ESULTS ON
THE
S HELL
Remark 3.1. Here, let u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]), and W0 , W1 , . . . , Wn > 0 on [R1 , R2 ],
0 < R1 < R2 , n ≥ 0 integer, with u0 (r) = c > 0.
Let also f ∈ C n+1 (A). We assume first that f is radial, i.e. there exists g such that f (x) =
g(r), r = |x|, R1 ≤ r ≤ R2 . Clearly g ∈ C n+1 ([R1 , R2 ]).
Let x ∈ A. Then by using the polar method (2.4) we obtain
R
N
(3.1)
= f (x) −
(3.2)
= g(r) −
(3.3)
(3.4)
(3.5)
f (y)dy
V ol(A)
R
R
R2
A
E(x) := f (x) −
S N −1
R1
f (sω)sN −1 ds dω
ωN (R2N − R1N )
R
R
R2
N −1
N S N −1 R1 g(s)s
ds dω
ωN (R2N − R1N )
Z R2
N
= g(r) −
g(s)sN −1 ds
R2N − R1N
R1
Z R2
Z R2
N
N −1
N −1
=
g(r)s
ds −
g(s)s
ds
R2N − R1N
R1
R1
Z R2
N
=
(g(r) − g(s))sN −1 ds =: (∗).
N
N
R2 − R1
R1
Let s, r ∈ [R1 , R2 ], then by generalized Taylor’s formula (2.2) we get
(3.6)
g(s) − g(r) =
n
X
Li g(r)gi (s, r) + Rn (r, s),
i=1
where
Z
(3.7)
Rn (r, s) :=
s
gn (s, t)Ln+1 g(t)dt.
r
But it holds
s
Z
(3.8)
|Rn (r, s)| ≤
|gn (s, t)|dt kLn+1 gk∞,[R1 ,R2 ] .
r
By calling
Z
(3.9)
s
|gn (s, t)|dt ,
Nn (r, s) :=
∀ s, r ∈ [R1 , R2 ],
r
we get
(3.10)
|Rn (r, s)| ≤ Nn (r, s)kLn+1 gk∞,[R1 ,R2 ] ,
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
∀ s, r ∈ [R1 , R2 ].
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O STROWSKI T YPE I NEQUALITIES VIA A TAYLOR -W IDDER F ORMULA
5
Therefore by (3.5), (3.6), we have
#
Z R2 " X
n
N
(3.11) (∗) =
Li g(r)gi (s, r) + Rn (r, s) sN −1 ds
N
N
R2 − R1
R1
i=1
#
"
X
Z R2
Z R2
n
N
≤
Li g(r)gi (s, r)sN −1 ds +
|Rn (r, s)|sN −1 ds
R2N − R1N
R
R
1
1
i=1
" n
Z
R2
by (3.10)
X
N
|L
g(r)|
gi (s, r)sN −1 ds
≤
i
R2N − R1N
R1
i=1
Z R2
N −1
(3.12)
+(kLn+1 gk∞,[R1 ,R2 ] )
Nn (r, s)s
ds .
R1
We have established the following result.
Theorem 3.2. Let u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]), W0 , W1 , . . . , Wn > 0 on [R1 , R2 ], 0 <
R1 < R2 , n ≥ 0 integer, with u0 (r) = c > 0. Let f ∈ C n+1 (A) be radial, i.e. there exists g
such that f (x) = g(r), r = |x|, R1 ≤ r ≤ R2 , x ∈ A.
Then
R
f (y)dy
E(x) := f (x) − A
V ol(A)
Z R2
N
g(s)sN −1 ds
= g(r) −
R2N − R1N
R1
"X
Z R2
n
N
≤
|Li g(r)|
gi (s, r)sN −1 ds
R2N − R1N
R
1
i=1
Z R2
N −1
+(kLn+1 gk∞,[R1 ,R2 ] )
(3.13)
Nn (r, s)s
ds
R1
"X
Z R2
n
N
=
|θi f (x)|
gi (s, |x|)sN −1 ds
N
N
R2 − R1
R1
i=1
Z R2
N −1
+(kθn+1 f k∞,A )
Nn (|x|, s)s
ds .
R1
Corollary 3.3. Let the conditions of Theorem 3.2 hold. Assume further that Li g(r0 ) = 0,
i = 1, . . . , n, for a fixed r0 ∈ [R1 , R2 ]. For all x0 = r0 ω ∈ A, ω ∈ S N −1 , we have
R
f (y)dy
E(x0 ) = f (x0 ) − A
V ol(A)
Z R2
N
= g(r0 ) −
g(s)sN −1 ds
N
N
R2 − R1
R1
Z R2
N
N −1
≤
(kLn+1 gk∞,[R1 ,R2 ] )
Nn (r0 , s)s
ds
R2N − R1N
R1
Z R2
N
N −1
=
(kθn+1 f k∞,A )
(3.14)
Nn (|x0 |, s)s
ds .‘
R2N − R1N
R1
Interesting cases also arise when r0 = R1 or R2 .
We continue Remark 3.1 with
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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6
G EORGE A. A NASTASSIOU
Remark 3.4. Now let f ∈ C n+1 (A), n ≥ 0, x ∈ A, x = rω, r > 0. Clearly for fixed ω ∈ S N −1 ,
since the function f (rω), r ∈ [R1 , R2 ] is radial, it also belongs to C n+1 ([R1 , R2 ]).
By applying the internal inequality (3.13) we get
Z R2
N
(3.15) f (rω) −
f (sω)sN −1 ds
R2N − R1N
R1
"X
Z R2
n
N
≤
|(Li f (·ω))(r)|
gi (s, r)sN −1 ds
R2N − R1N
R1
i=1
Z R2
N −1
+(kLn+1 f (·ω)k∞,[R1 ,R2 ] )
Nn (r, s)s
ds .
R1
For non-radial f we define again
(3.16)
θi f (x) = θi f (rω) := (Li f (·ω))(r),
i = 1, . . . , n + 1, ∀ x ∈ A.
all
Here the involved
∂ i f (rω)
∂ i f (x)
=
, i = 1, . . . , n + 1,
∂ri
∂ri
are the radial derivatives. In a sense θi is a generalized radial derivative of Widder-type.
Hence
"X
n
N
|θi f (rω)|
(3.17) R.H.S.(3.15) ≤
R2N − R1N
i=1
Z R2
Z R2
N −1
N −1
gi (s, r)s
ds + kθn+1 f k∞,A
Nn (|x|, s)s
ds .
R1
R1
Therefore, by (3.15) and (3.17) we have
Z
Z
Γ N2
1
(3.18)
f (rω)dω −
f (y)dy
2π N/2 S N −1
V ol(A) A
( n Z
)
"
Z R2
N
Γ(N/2) X
|θi f (rω)|dω
gi (s, r)sN −1 ds
≤
2π N/2
R2N − R1N
N
−1
S
R1
i=1
Z R2
N −1
+kθn+1 f k∞,A
Nn (|x|, s)s
ds .
R1
We have established the following.
Theorem 3.5. Let u0 , u1 , . . . , un ∈ C n+1 ([R1 , R2 ]); W0 , W1 , . . . , Wn > 0 on [R1 , R2 ], 0 <
R1 < R2 , n ≥ 0 an integer, with u0 (r) = c > 0. Let f ∈ C n+1 (A), x ∈ A; x = rω,
r ∈ [R1 , R2 ], ω ∈ S N −1 .
Then
R
R
N
Γ
f (rω)dω
f
(y)dy
2
sN −1
(3.19) E(x) = f (x) − A
≤ f (x) −
N/2
V ol(A)
2π
"
(
)
Z R2
n Z
X
Γ N2
N
+
|θi f (rω)|dω
gi (s, r)sN −1 ds
2π N/2 i=1
R2N − R1N
N
−1
s
R1
Z R2
N −1
+kθn+1 f k∞,A
Nn (|x|, s)s
ds , ∀ x ∈ A.
R1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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O STROWSKI T YPE I NEQUALITIES VIA A TAYLOR -W IDDER F ORMULA
7
Corollary 3.6. Let the conditions of Theorem 3.5 hold. Assume that
θi f (r0 ω) = 0, i = 1, . . . , n, for a fixed r0 ∈ [R1 , R2 ],
∀ ω ∈ S N −1 ;
also consider all x0 = r0 ω ∈ A for any ω ∈ S N −1 . Then
R
f (y)dy
E(x0 ) = f (x0 ) − A
V ol(A)
R
Γ N2 ) S N −1 f (r0 ω)dω
≤ f (x0 ) −
2π N/2
Z R2
N
N −1
(3.20)
Nn (|x0 |, s)s
ds .
+
kθn+1 f k∞,A
R2N − R1N
R1
When r0 = R1 or R2 is of special interest.
4. R ESULTS ON
THE
S PHERE
Notation 4.1. Let N ≥ 1, B(0, R) := {x ∈ RN : |x| < R} be the ball in RN centered at the
N
origin and of radius R > 0. Note that V ol(B(0, R)) = ωNNR .
Let f from B(0, R) into R and consider f to be radial, i.e. f (x) = g(r), where r = |x|,
0 ≤ r ≤ R. We assume g ∈ C n+1 ([0, R]), n ≥ 0. Clearly then f ∈ C(B(0, R)).
For F ∈ C(B(0, R)) we have
Z R
Z
Z
N −1
(4.1)
F (x)dx =
F (rω)r
dr dω.
S N −1
B(0,R)
0
We notice that
(4.2)
N
RN
Z
R
sN −1 ds = 1.
0
The operator θi in the radial case, is as defined in (2.3), now ∀x ∈ B(0, R). In the non-radial
case, for f ∈ C n+1 (B(0, R)), θi is defined as in (3.16), ∀ x ∈ B(0, R) − {0}.
We make the following remark.
Remark 4.1. Here, let u0 , u1 , . . . , un ∈ C n+1 ([0, R]) and W0 , W1 , . . . , Wn > 0 on [0, R],
R > 0, n ≥ 0 an integer, with u0 (r) = c > 0.
We again first assume that f is radial on B(0, R), i.e. there exists g such that f (x) = g(r),
r = |x|, 0 ≤ r ≤ R. Assume further that g ∈ C n+1 ([0, R]). Let x ∈ B(0, R). Then by using
the polar method (4.1) we obtain
R
R
R
R
n−1
N S N −1 0 g(s)s ds dω
f (y)dy
B(0,R)
(4.3) E(x) := f (x) −
= g(r) −
V ol(B(0, R))
ωN RN
Z R
Z R
by (4.2) N
N
n−1
(4.4)
g(s)s ds
=
(g(r) − g(s))sN −1 ds =: (∗).
= g(r) − N
N
R
R
0
0
Let s, r ∈ [0, R], then by the generalized Taylor’s formula (2.2) we get
(4.5)
g(s) − g(r) =
n
X
Li g(r)gi (s, r) + Rn (r, s),
i=1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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8
G EORGE A. A NASTASSIOU
where
Z
(4.6)
Rn (r, s) :=
s
gn (s, t)Ln+1 g(t)dt.
r
But it holds that
s
Z
(4.7)
|Rn (r, s)| ≤
|gn (s, t)|dt kLn+1 gk∞,[0,R] .
r
By calling
Z
(4.8)
s
|gn (s, t)|dt ,
Nn (r, s) :=
∀ s, r ∈ [0, R],
r
we get
(4.9)
|Rn (r, s)| ≤ Nn (r, s)kLn+1 gk∞,[0,R] ,
∀ s, r ∈ [0, R].
Therefore by (4.4) and (4.5), we have
#
Z R "X
n
N
(∗) = N
Li g(r)gi (s, r) + Rn (r, s) sN −1 ds
R
0
i=1
" n Z
#
Z R
R
N X
(4.10)
≤ N
Li g(r)gi (s, r)sN −1 ds +
|Rn (r, s)|sN −1 ds
R
0
0
i=1
" n
Z
R
by (4.9) N X
≤
|L
g(r)|
gi (s, r)sN −1 ds
i
RN i=1
0
Z R
N −1
(4.11)
+(kLn+1 gk∞,[0,R] )
Nn (r, s)s
ds .
0
We have established the next result.
Theorem 4.2. Let u0 , u1 , . . . , un ∈ C n+1 ([0, R]), W0 , W1 , . . . , Wn > 0 on [0, R], R > 0, n ≥ 0
an integer, with u0 (r) = c > 0. Let f from B(0, R) into R be radial, i.e. there exists g such that
f (x) = g(r), r = |x|, 0 ≤ r ≤ R, ∀ x ∈ B(0, R); further assume that g ∈ C n+1 ([0, R]).
Then
R
Z R
f (y)dy
N
B(0,R)
E(x) := f (x) −
= g(r) − N
g(s)sn−1 ds
V ol(B(0, R))
R
0
" n
#
Z
Z R
R
N X
≤ N
|Li g(r)|
gi (s, r)sN −1 ds + (kLn+1 gk∞,[0,R] )
Nn (r, s)sN −1 ds
R
0
0
i=1
" n
Z
R
N X
= N
|θi f (x)|
gi (s, |x|)sN −1 ds
R
0
i=1
Z R
N −1
(4.12)
Nn (|x|, s)s
ds .
+(kθn+1 f k∞,B(0,R) )
0
The following corollary holds.
Corollary 4.3. Let the conditions of Theorem 4.2 hold. Assume further that Li g(r0 ) = 0,
i = 1, . . . , n, for a fixed r0 ∈ [0, R]; consider all x0 = r0 ω ∈ B(0, R) and any ω ∈ S N −1 .
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O STROWSKI T YPE I NEQUALITIES VIA A TAYLOR -W IDDER F ORMULA
9
Then
R
B(0,R)
E(x0 ) := f (x0 ) −
(4.13)
f (y)dy
V ol(B(0, R))
Z R
N
= g(r0 ) − N
g(s)sn−1 ds
R
0
Z R
N
N −1
≤ N (kLn+1 gk∞,[0,R] )
Nn (r0 , s)s
ds
R
0
Z R
N
N −1
= N (kθn+1 f k∞,B(0,R) )
Nn (|x0 |, s)s
ds .
R
0
Interesting cases especially arise when r0 = 0 or R.
We continue Remark 4.1 with
Remark 4.4. Let f be non-radial. Here assume that f ∈ C n+1 (B(0, R)). Consider x ∈
B(0, R) − {0}, which is written uniquely as x = rω, r ∈ (0, R], ω ∈ S N −1 . Then by using
again the polar method (4.1) we obtain
R
R
f (y)dy
f
(rω)dω
N
−1
B(0,R)
S
(4.14)
−
ωN
V ol(B(0, R))
R
R
R
R
N −1
N
f
(sω)s
ds
dω
S N −1
0
N −1 f (rω)dω
= S
−
(4.15)
ωN
ωN RN
Z
Z R
by (4.2)
N
N −1
=
f (rω)s
ds dω
ωN RN
S N −1
0
Z R
Z
N
N −1
f (sω)s
ds dω
−
ωN RN
S N −1
0
Z R
Z
N
N −1
(4.16)
(f (rω) − f (sω))s
ds dω =: (∗).
=
ωN RN S N −1
0
Clearly here f (·ω) ∈ C n+1 ((0, R]).
Let ρ ∈ (0, R], then by (2.2) we get
(4.17)
f (ρω) − f (rω) =
n
X
((Li (f (·ω)))(r))gi (ρ, r) + Rn (r, ρ),
i=1
where
Z
(4.18)
ρ
gn (ρ, t)(Ln+1 (f (·ω)))(t)dt.
Rn (r, ρ) :=
r
That is
(4.19)
f (ρω) − f (rω) =
n
X
(θi f (rω))gi (ρ, r) + Rn (r, ρ),
i=1
with
Z
(4.20)
Rn (r, ρ) =
ρ
gn (ρ, t)θn+1 f (tω)dt.
r
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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10
G EORGE A. A NASTASSIOU
We further assume that
θ := kθn+1 f k∞,B(0,R)−{0} < +∞.
(4.21)
Therefore we obtain
Z
|Rn (r, ρ)| ≤
(4.22)
ρ
|gn (ρ, t)|dt θ.
r
By calling
ρ
Z
(4.23)
|gn (ρ, t)|dt ,
Nn (r, ρ) :=
∀ ρ, r ∈ [0, R],
r
we get
|Rn (r, ρ)| ≤ Nn (r, ρ)θ.
(4.24)
Hence by (4.19) we obtain
|f (ρω) − f (rω)| ≤
n
X
|θi f (rω)||gi (ρ, r)| + |Rn (r, ρ)|
i=1
(4.25)
≤
n
X
|θi f (rω)||gi (ρ, r)| + Nn (r, ρ)θ.
i=1
By the continuity of f and gi , i = 1, . . . , n,and by taking the limit as ρ → 0 in the external
inequality (4.25), we obtain
(4.26)
|f (0) − f (rω)| ≤
n
X
|θi f (rω)||gi (0, r)| + Nn (r, 0)θ.
i=1
Notice here that gn (ρ, t) is jointly continuous in (ρ, t) ∈ [0, R]2 , hence Nn (r, ρ) is continuous
in ρ ∈ [0, R].
That is, ∀ s ∈ [0, R] we get
(4.27)
|f (sω) − f (rω)| ≤
n
X
|θi f (rω)||gi (s, r)| + Nn (r, s)θ.
i=1
Consequently by (4.16) and (4.27) we find
Z R
Z
N
N −1
(4.28)
(∗) ≤
|f (sω) − f (rω)|s
ds dω
ωN RN S N −1
0
" n Z
Z R
X
N
N −1
(4.29)
≤
|θi f (rω)||gi (s, r)|s
ds dω
ωN RN i=1
S N −1
0
Z R
Z
N −1
+θ
Nn (r, s)s
ds dω
S N −1
0
R
Pn R
R
N −1
|θ
f
(rω)||g
(s,
r)|s
ds
dω
i
i
i=1
S N −1
0
=
V ol(B(0, R))
R
R
N −1
θN 0 Nn (r, s)s
ds
(4.30)
+
RN
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O STROWSKI T YPE I NEQUALITIES VIA A TAYLOR -W IDDER F ORMULA
=
N
X
i=1
(4.31)
11

N −1
|g
(s,
r)|s
ds
i
0


V ol(B(0, R))
R
R
θN 0 Nn (r, s)sN −1 ds
+
.
RN
 R
|θ f (rω)|dω
S N −1 i
R R
That is
R
f (y)dy
f
(rω)dω
N −1
B(0,R)
∆ := S
−
ωN
V ol(B(0, R))

 R
R R
N −1
n
|gi (s, r)|s
ds
X
N −1 |θi f (rω)|dω
0

 S
≤
V
ol(V
(0,
R))
i=1
R
R
θN 0 Nn (r, s)sN −1 ds
+
.
RN
R
(4.32)
The following theorem holds.
Theorem 4.5. Let u0 , u1 , . . . , un ∈ C n+1 ([0, R]), W0 , W1 , . . . , Wn > 0 on [0, R], R > 0, n ≥ 0
an integer, with u0 (r) = c > 0. Let f ∈ C n+1 (B(0, R)). Assume that
θ := kθn+1 f k∞,B(0,R)−{0} < +∞.
(4.33)
Let x ∈ B(0, R) − {0}, which is written uniquely as x = rω, r ∈ (0, R], ω ∈ S N −1 .
Then
R
E(x) := f (x) −
(4.34)
B(0,R)
f (y)dy
V ol(B(0, R))
Z
Γ(N/2)
≤ f (x) −
f (rω 0 )dω 0
2π N/2 S N −1

 R
R R
0
0
N −1
n
|θ
f
(rω
)dω
|g
(s,
r)|s
ds
X
i
i
N −1
0
 S

+
V
ol(B(0,
R))
i=1
R
R
N −1
θN 0 Nn (r, s)s
ds
+
.
RN
We finally give the following corollary.
Corollary 4.6. Let the conditions of Theorem 4.5 hold. Assume further that θi f (r0 ω 0 ) = 0,
∀ ω 0N −1 , for some r0 ∈ (0, R], for all i = 1, . . . , n. Consider all x0 = r0 ω ∈ B(0, R) − {0},
and any ω ∈ S N −1 .
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 106, 13 pp.
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12
G EORGE A. A NASTASSIOU
Then
R
E(x0 ) := f (x0 ) −
B(0,r)
f (y)dy
V ol(B(0, R))
Z
Γ(N/2)
≤ f (x0 ) −
f (|x0 |ω 0 )dω 0
2π N/2 S N −1
R
R
N −1
θN 0 Nn (|x0 |, s)s
ds
(4.35)
+
.
RN
An interesting case is when r0 = R.
5. A DDENDUM
We give
Proposition 5.1. Let f ∈ C 1 (B(0, R)) such that f is radial, i.e. f (x) = g(r), r = |x|,
0 ≤ r ≤ R, ∀ x ∈ B(0, R). Then
(i) ∃ g 0 ∈ C((0, R]),
(ii) ∃ g 0 (0) = 0.
Proof. (i) is obvious.
(ii) Let u be a unit vector, h > 0, then
g(h) − g(0)
f (hu) − f (0)
=
h
h
∇f (0) · hu + o(h)
(by first order multivariate Taylor’s formula) =
h
∇f (0) · hu o(h)
o(h)
=
+
= ∇f (0) · u +
h
h
h
o(h)
h→0
−→ ∇f (0) · u, by
→ 0.
h
The last is true for any unit vector u. Thus ∇f (0) = 0, proving the claim.
By Proposition 5.1, we see that, it may well be that g 0 is discontinuous at zero, if only f ∈
C 1 (B(0, R)).
Therefore the assumption that g ∈ C n+1 ([0, R]) in Theorem 4.2 seems to be the best.
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