The classical inequalities such as the inequalities of Bernstein, Bohr, Nikol’skii and others play an important role in Analysis, the Approximation Theory and Applications (see 8,14 16). Recall now the BohrFavard inequality: Let σ > 0, 1 ≤ p ≤ ∞, f ∈ C m(R), supp ˆf ∩ (−σ, σ) = ∅, where ˆf is the Fourier transform of f, and Dmf ∈ Lp(R). Then f ∈ Lp(R) and (see 2, 5, 6, 7)
Trang 1HA HUY BANG & VU NHAT HUY
Abstract In this paper we give a new inequality of the Bohr-Nikol’skii type.
1 Introduction The classical inequalities such as the inequalities of Bernstein, Bohr, Nikol’skii and others play an important role in Analysis, the Approximation Theory and Ap-plications (see [8],[14] - [16]) Recall now the Bohr-Favard inequality: Let σ > 0,
1 ≤ p ≤ ∞, f ∈ Cm(R), supp ˆf ∩ (−σ, σ) = ∅, where ˆf is the Fourier transform of
f , and Dmf ∈ Lp(R) Then f ∈ Lp(R) and (see [2, 5, 6, 7]):
kf kp ≤ σ−mKmkDmf kp, where the Favard constant Km are best possible and have the following properties
1 = K0 ≤ K2 < < 4
π < < K3 ≤ K1 = π
2. The Bohr-Favard inequality has applications to the Approximation Theory: For example, let g ∈ Cm(R), Djg ∈ Lp(R), j = 0, 1, , m and we approximate g by
a function h in some subclass of Lp(R) Putting kg − hkp = and applying the Bohr-Favard inequality, we get the following estimates for the errors if inf{|x| : x ∈ suppF (g − h)} = σ > 0:
kDjg − Djhkp ≥ σjKj−1, j = 1, 2, , m
In this paper we give a new inequality, which combines the inequality of Bohr-Favard and the Nikol’skii idea of inequality for functions in different metrics
2 main result Let f ∈ L1(R) and ˆf = F f be its Fourier transform
ˆ
f (ξ) = √1
2π
Z +∞
−∞
e−ixξf (x)dx
The Fourier transform of a tempered generalized function f is defined via the formula
ϕ ∈ S(R)
Let K be an arbitrary compact set in R and > 0 Denote by Dmf = (−i)mf(m) and K := {x ∈ R : ∃ξ ∈ K : |x − ξ| ≤ }
Now, we state our main theorem
Key words and phrases L p - spaces, Bohr-Favard inequality, Nikol’skii inequality.
2010 AMS Subject Classification 26D10.
1
Trang 2Theorem 1 Let 1 ≤ q < p ≤ ∞, m ≥ 3, f ∈ Lp(R), σ > 0 and supp ˆf ∩(−σ, σ) = ∅ Then there exists a constant C > 0 not depending on f, m, σ such that
kDmf kq ≥ Cmλσm−λkf kp, (1)
where
λ = 1
q − 1
p > 0.
Before giving the proof of Theorem 1, we need the following result (see [17]): Young inequality Let f ∈ Lp(R) and g ∈ Lq(R), where 1 ≤ p, q ≤ ∞ and
1
p +1q ≥ 1 Then f ∗ g ∈ Lr(R) and
kf ∗ gkr≤ kf kpkgkq, where
1
r =
1
p+
1
q − 1
Proof of Theorem 1 Let us first prove (1) for the case σ = 1 Indeed, put K := (−∞, −1] ∪ [1, +∞) and
η(ξ) =
(
C1e
1 ξ2−1 if |ξ| < 1,
0 if |ξ| ≥ 1, where C1 is chosen such that R
Rη(ξ)dξ = 1 We define the sequence of functions (φm(ξ))m∈N via the formula
φm(ξ) = (1K3/(4m)∗ η1/(4m))(ξ), where
η1/(4m)(ξ) = 4mη(4mξ)
Then η1/(4m)(ξ) = 0 for all ξ 6∈ [−1/(4m), 1/(4m)], R
Rη1/(4m)(ξ)dξ = 1 Hence, for all m ∈ N we have φm(ξ) ∈ C∞(R), and φm(ξ) = 1 ∀ξ ∈ K1/(2m), φm(ξ) = 0 ∀ξ /∈
K1/m So, it follows from supp [Dmf ⊂ K that φm(−ξ) [Dmf = [Dmf Therefore, since
[
Dmf = ξmf ,ˆ
we get
φm(−ξ) [Dmf = ξmf ,ˆ and then
[
Dmf φm(−ξ)/ξm = ˆf Hence, for m ≥ 3
f = (2π)−1/2(Dmf ) ∗ F−1(φm(−ξ)(ξ)m) = (2π)−1/2(Dmf ) ∗ F (φm(ξ)/(−ξ)m) Therefore, since Young inequality, we have the following estimate for m ≥ 3
kf kp = (2π)−1/2k(Dmf ) ∗ F (φm(ξ)/ξm)kp ≤ (2π)−1/2kDmf kqkF (φm(ξ)/ξm)kr, (2)
where
1
p =
1
q + 1
r − 1
Trang 3Since 1 ≤ q < p ≤ ∞, we have 1 < r ≤ ∞ We define for m ≥ 3
km := 1 + 1
m, gm(ξ) = φm(kmξ), Φm(ξ) = φm(ξ) − gm(ξ).
Hence,
(F (gm(ξ)/ξm))(x) = (km)m(F (φm(kmξ)/(kmξ)m)(x) = (km)m−1(F (φm(ξ)/ξm))(x/km) So,
F (gm(ξ)/ξm)
r
= (km)m−1+1r F (φm(ξ)/ξm)
r
Then it follows from (km)m−1+1 ≥ (km)m−1 = (1 + m1)m−1 ≥ 3
2 that
F (gm(ξ)/ξm)
r≥ 3
2 F (φm(ξ)/ξ
m)
r Therefore, since Φm(ξ) = φm(ξ) − gm(ξ) we get
F (Φm(ξ)/ξm)
r ≥ F (gm(ξ)/ξm)
r− F (φm(ξ)/ξm)
r
(3)
≥ 1
2 F (φm(ξ)/ξ
m
)
r From (2)-(3) we obtain
kf kp ≤ 2(2π)−1/2kDmf kqkF (Φm(ξ)/ξm)kr (4)
Now, we will prove that there exists a constant C such that for all m ∈ N, m ≥ 3
kF (Φm(ξ)/ξm)kr ≤ m−1+1rC(2π)1/2/2
Indeed, put C2 = max{kη(j)k1, j ≤ 3} Since η1/(4m)(x) = 4mη(4mx), we obtain
η(j)1/(4m)(x) = (4m)j+1η(j)(4mx) and then
kη1/(4m)(j) k1 = (4m)jkη(j)k1 ≤ C2(4m)j, ∀j ≤ 3
Therefore,
φ(j)m
∞ =
(1K3/(4m) ∗ η(j)1/(4m))
∞ ≤ η1/(4m)(j)
1
≤ (4m)jC2, ∀j ≤ 3
(5)
Note that φm(ξ) = 1 ∀ξ ∈ (−∞, −1 + (1/2m)] ∪ [1 − (1/2m), +∞) and φm(ξ) =
0 ∀ξ ∈ [−1 + (1/m), 1 − (1/m)]
So, if |ξ| < 1−(3/m) then |ξ| < |kmξ| < 1−(1/m) and then φm(ξ) = φm(kmξ) = 0, which implies Φm(ξ) = 0 Further, if |ξ| > 1 then |kmξ| > |ξ| > 1 and then
φm(ξ) = φm(kmξ) = 1, which implies Φm(ξ) = 0 From these we have
suppΦm ⊂ [1 − (3/m), 1] ∪ [−1, (3/m) − 1]
(6)
Now, for ξ ∈ [1 − (3/m), 1] ∪ [−1, (3/m) − 1] we get
ξ − kmξ
=
ξ m
≤ 1
m. (7)
Trang 4From (5) and (7) we have the following estimate for ξ ∈ [1−(3/m), 1]∪[−1, (3/m)−1]
Φm(ξ)
=
φm(ξ) − gm(ξ)
=
φm(ξ) − φ(kmξ)
(8)
≤
ξ − kmξ
φ0m
∞ ≤ 1
m4mC2 = 4C2, and
Φ0m(ξ)
=
φ0m(ξ) − gm0 (ξ)
=
φ0m(ξ) − φ0m(kmξ)
(9)
=
φ0m(ξ) − kmφ0m(kmξ)
≤
φ0m(ξ) − φ0m(kmξ)
+
(1 − km)φ0m(kmξ)
≤
ξ − kmξ
φ00m
∞+
1 − km
φ0m
∞
≤ 1
m(4m)
2C2+
1 − km
4mC2
≤ 20mC2 Put Hm(x) = (F (Φm(ξ)/ξm))(x) Then
Hm(x) = √1
2π Z
R
e−ixξΦm(ξ)/ξmdξ
Therefore, since (6), we have
sup
x∈R
Hm(x)
≤ √1 2π Z
R
Φm(ξ)/ξm
dξ = √1
2π Z
1−m3≤|ξ|≤1
Φm(ξ)/ξm
dξ and it follows from (5) that
sup
x∈R
Hm(x)
≤ 6
m√ 2π supξ∈R
Φm(ξ)
(1 − 3
m)
−m ≤ 24e
4C2
m√ 2π. (10)
We also obtain that
sup
x∈R
xHm(x)
= √1 2πsupx∈R
Z
R
e−ixξ(mΦm(ξ)
ξm+1 −Φ
0
m(ξ)
ξm )dξ
≤ √1 2π Z
R
mΦm(ξ)
ξm+1 − Φ
0
m(ξ)
ξm
dξ
Therefore, since (5)-(6), we have
sup
x∈R
xHm(x)
≤ √1 2π Z
1−m3≤|ξ|≤1
mΦm(ξ)
ξm+1 −Φ
0
m(ξ)
ξm
dξ (11)
m√ 2π1−3sup
m ≤|ξ|≤1
mΦm(ξ)
ξm+1 − Φ
0
m(ξ)
ξm
m√ 2π
h sup
ξ∈R
Φm(ξ)
m(1 − 3
m)
−m−1
+ sup
ξ∈R
Φ0m(ξ)
(1 − 3
m)
−mi
Trang 5≤ 6
m√ 2π
h 4C2me4+ 20C2me4i = 144e
4C2
√ 2π .
We see for 1 < r < ∞
Hm
r
r= Z
|x|≤m
Hm(x)
r
dx + Z
|x|≥m
Hm(x)
r
dx (12)
≤ sup
x∈R
Hm(x)
r Z
|x|≤m
1dx + sup
x∈R
xHm(x)
r Z
|x|≥m
1
xr
dx
= 2m sup
x∈R
Hm(x)
r
+2m
−r+1
r − 1 supx∈R
xHm(x)
r
From (10)-(12), we obtain that
Hm
r
r≤ 2m24e
4C2
m√ 2π
r
+2m
−r+1
r − 1
144e4C2
√ 2π
r
= 2m−r+1e
4C2
√ 2π
r
(144
r
r − 1 + 96
r), and then for 1 < r < ∞
Hm
r
≤ e
4C2
√ 2π(
144r2
r − 1 + 96
r2)1m−1+1 = m−1+1/C
(13)
where C =√
2π/e4C2(144r2
r−1 + 96r2)1r Further, if r = ∞ then Hm ∞= sup
x∈R
Hm(x) and so,
Hm ∞≤ 24e4C2/(m√
2π)
(14)
Using (13) and (14), we obtain for 1 < r ≤ ∞ a constant C > 0 not depending on
f, m, σ such that
kF (Φm(ξ)/ξm)kr = kHmkr ≤ m−1+1(2π)1/2/(2C)
(15)
From (4) and (15) we have
kf kp ≤ m1p − 1
kDmf kq/C and then
kDmf kq ≥ Cm1−1pkf kp
So, (1) have been proved for the case σ = 1
Next, we prove (1) for any σ > 0 Put
g(x) = f (x
σ).
Then it follows from supp ˆf ∩ (−σ, σ) = ∅ that suppˆg ∩ (−1, 1) = ∅ Therefore,
kDmgkq ≥ Cm1q − 1
pkgkp (16)
Since g(x) = f (σx), we have
kgkp = σp1kf kp, kDmgkq = σ−m+1kDmf kq Hence, it follows from (16) that
σ−m+1qkDmf kq ≥ Cm1q − 1
pσ1pkf kp
Trang 6kDmf kq≥ Cm1q −1p
σm+1p −1qkf kp
Corollary 2 Let 1 ≤ q < p ≤ ∞, f ∈ Lp(R), σ > 0 and supp ˆf ∩ (−σ, σ) = ∅ Then for all a < 1q −1
p we have the following limit lim
m→∞kDmf kq/(maσm) = +∞
In comparison, using Bohr-Favard inequality, we get that the sequence
{kDmf kq/σm}∞
m=1 is separated with the origin while by Corollary 2 we have the stronger result: limm→∞kDmf kq/(maσm) = +∞ for all a < 1q − 1
p
In the following theorem, we give a result for the sequence of Lp(R)−norm of primitives of a function (see the notion of the primitive of functions in Lp(R) in [3], [17]):
Theorem 3 Let 1 ≤ q < p ≤ ∞, σ > 0, f ∈ Lq(R), supp ˆf ∩ (−σ, σ) = ∅ and {Imf }∞m=0 ⊂ Lp(R), where I0f = f , Imf is a primitive of Im−1f for m = 1, 2, Then for a < 1q −1
p we have the following limits
lim
m→∞maσmkImf kp = 0,
In particular,
lim
m→∞σmkImf kp = 0
Proof It was shown in [3] that
supp dImf = supp ˆf ∀m ∈ N
Therefore, supp dImf ∩ (−σ, σ) = ∅ and then it follows from Theorem 1 that
kf kq ≥ Cm1q − 1
pσm+1p − 1
qkImf kp Hence,
lim
m→∞maσmkImf kp = 0
From Theorem 1 and the Bohr-Favard inequality we have the following result: Corollary 4 Let 1 ≤ q < p ≤ ∞, σ > 0 Denote
Nσ,p:= {f ∈ Lp(R) : supp ˆf ⊂ (−∞, −σ] ∪ [σ, +∞)}
and
γm := inf
f ∈N σ,p
kDmf kq
σmkf kp. Then γm ≤ π
2γm+1 and
lim
m→∞γm = ∞
Trang 7Note that if p = q then γm = Km ∀m ∈ N and the conclusion limm→∞γm = ∞ does not hold
Consecutively applying Theorem 1 to each variable, we get the following result for the n-dimensional case:
Theorem 5 Let 1 ≤ q < p ≤ ∞, σj > 0, α = (α1, , αn) ∈ Zn
+, αj ≥ 3 (j =
1, , n), f ∈ Lp(Rn) and supp ˆf ⊂Qn
j=1((−∞, −σj] ∪ [σj, +∞)) Then there exists
a constant C > 0 not depending on f, α, σ such that
kDαf kq ≥ C
n
Y
j=1
αλjσαj −λ
j kf kp, where
λ = 1
q − 1
p > 0, σ = (σ1, σ2, , σn).
Acknowledgement
This research is funded by Vietnam National Foundation for Science and Tech-nology Development (NAFOSTED) under grant number 101.01-2011.32 A part of this work was done when the authors were working at the Vietnam Institute for Advanced Study in Mathematics (VIASM) The authors would like to thank the VIASM for providing a fruitful research environment and working condition
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HA HUY BANG
Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam
E-mail address: hhbang@math.ac.vn
VU NHAT HUY
Department of Mathematics, College of Science, Vietnam National University, 334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam
E-mail address: nhat huy85@yahoo.com
...m. (7)
Trang 4From (5) and (7) we have the following estimate for ξ ∈ [1−(3/m), 1]∪[−1,...
∞ =
(1K3/(4m) ∗ η(j)1/(4m))
∞ ≤ η1/(4m)(j)
1
≤...
1
≤ (4m)jC2, ∀j ≤
(5)
Note that φm(ξ) = ∀ξ ∈ (−∞, −1 + (1/2m)] ∪ [1 − (1/2m), +∞) and φm(ξ) =
0 ∀ξ