DSpace at VNU: A Study of the Sequence of Norm of Derivatives (or Primitives) of Functions Depending on Their Beurling Spectrum

DOI 10.1007/s10013-015-0146-yA Study of the Sequence of Norm of Derivatives or Primitives of Functions Depending on Their Beurling Spectrum Ha Huy Bang 1 · Vu Nhat Huy 2 Received: 22 Oct

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DOI 10.1007/s10013-015-0146-y

A Study of the Sequence of Norm of Derivatives

(or Primitives) of Functions Depending on Their

Beurling Spectrum

Ha Huy Bang 1 · Vu Nhat Huy 2

Received: 22 October 2014 / Accepted: 8 December 2014

Abstract In this paper, we characterize the behavior of the sequence of norm of derivatives

(or primitives) of functions by their Beurling spectrum in Banach spaces The Bernstein inequality for Banach spaces is also obtained

Keywords Banach Spaces· Beurling spectrum

Mathematics Subject Classification (2010) 26D10· 46E30

1 Introduction

The following result was proved in [9]: Let 1 ≤ p ≤ ∞ and f (m) ∈ L p ( R), m =

0, 1, 2, Then there always exists the following limit

lim

m→∞f (m)1/m

p

m→∞f (m)1/m

p = σ f = sup{|ξ| : ξ ∈ supp ˆ f },

where ˆ f is the Fourier transform of f

The result shows that the behavior of the sequencef (m)1/m

p is wholly characterized by

the spectrum of f , and it reminds us of the following well-known spectral radius formula in

Ha Huy Bang

hhbang@math.ac.vn

Vu Nhat Huy

nhat huy85@yahoo.com

1 Institute of Mathematics, Vietnamese Academy of Science and Technology,

18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam

2 Department of Mathematics, College of Science, Vietnam National University,

334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam

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the Theory of Banach algebras: Let A be a unital Banach algebra and x ∈ A Then the limit

limm→∞x m1/malways exists and satisfies

lim

m→∞x m1/m = r A (x),

where r A (x) = sup{|λ| : λ ∈ σ A (x) } is the spectral radius of x, σ A (x) is the spectrum of x which is the set of all λ ∈ C such that x − λ1 is not invertible in A and 1 is the unit in A (see

[18,22]) That is why σ f is called the local spectral radius of the differential operator D,

and this result was studied and developed by many authors (see [1 12,15–20,26–29]) It is natural to ask what will happen when we replace derivatives by integrals To this question,

Tuan answered for p= 2 in [29] and we answered in [13] for 1≤ p ≤ ∞, and in [14] for

a more general case (but with the assumption that the spectrum must be compact) In this paper, we extend the results in [9,13,14,29] to Banach spaces, we also extend the results

in [17] to a general case and the Bernstein inequality is also obtained

2 Main Results

Let f ∈ L1( R) and ˆ f = Ff be its Fourier transform

ˆ

f (ξ )= √1

2π

+∞

−∞ e

−ixξ f (x)dx,

and ˇf = F−1fbe its inverse Fourier transform

ˇ

f (ξ )= √1

2π

+∞

−∞e

ixξ f (x)dx.

Let ( X, · X) denote a complex Banach space and let BC( R → X) denote the set of all X valued bounded continuous functions f : R → X For a given function f ∈ BC(R → X),

we definef ∞ = sup{f (t)X: t ∈ R} Then (BC(R → X), · ∞)itself is a Banach

space We define the derivative Df of f ∈ BC(R → X), as usual,

Df (s)= lim

δ→0

f (s + δ) − f (s)

Note that for every λ ∈ C \ iR the operator λ − D is invertible Hence, the spectrum of the differential operator is iR

Clearly, the inverse of λ − D is a bounded operator on (BC(R → X), · ∞) Moreover,

+∞

−∞ ϕ(t)(λ − D) −n f (t)dt =

+∞

−∞

(λ − D) −n ϕ(t)

f (t)dt

for any f ∈ BC(R → X) The convolution ϕ ∗ f of f with a Schwartz function is defined

by letting

ϕ ∗ f (s) =

+∞

−∞ ϕ(s − t)f (t)dt.

Proposition 1 (Young inequality for Banach spaces) Let f ∈ BC(R → X), ϕ ∈ S(R) Then ϕ ∗ f ∈ BC(R → X) and

ϕ ∗ f ∞≤ f ∞ϕ1.

The Beurling spectrum Spec(f ) of a function f ∈ BC(R → X) is defined by

Spec(f ) = {ξ ∈ R : ∀ > 0, ∃ϕ ∈ S(R) : supp ϕ ⊂ (ξ − , ξ + )), ϕ ∗ f ≡ 0}.

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The Beurling spectral radius ρ(f ) of f is defined by

ρ(f ) = sup{|ξ| : ξ ∈ Spec(f )}.

Note that Spec(f ) is always a closed subset ofR Moreover,

Proposition 2 We have the following properties of Beurling spectrum:

– ifϕ = 0 on Spec(f ) then ϕ ∗ f = 0;

– ifϕ = 1/√2π on Spec(f ) then ϕ ∗ f = f ;

– Spec((λ − D)−1f ) = Spec(f ) for every f ∈ BC(R → X) and λ ∈ C \ iR.

See [21] for more details Now, we defineD = −iD and the differential operator P (D)

is obtained from P (x) by substituting x → −iD (for P (x) =n

k=0a k x k then P ( D)f =

n

k=0a k D k f)

Theorem 1 Let P (x) be a polynomial and f ∈ BC(R → X), f ≡ 0 Assume that Spec(f )

is a compact set Then

lim

m→∞P m (D)f 1/m

∞ = sup

ξ ∈Spec(f ) |P (ξ)|. (1)

Proof Let us first show that

lim

m→∞P m ( D)f 1/m

∞ ≥ sup

ξ ∈Spec(f ) |P (ξ)|. (2)

Indeed, let σ ∈ Spec(f ) satisfy P (σ ) = 0 For a small enough positive number , we have

P (x) = 0 for all x ∈ B(σ, ) From the definition of Beurling spectrum, there exists a function ϕ ∈ C∞( R), supp ϕ ⊂ B(σ, ) such that ϕq∗f ≡ 0 We put

ϕ m = F−1(ϕ(x)/P m (x)).

Then ϕ mis well defined and it follows from

D k ϕ m = F−1(x kϕ m (x)) = F−1(x k ϕ(x)/P m (x)) ∀k ∈ Z+

that

P m ( D)ϕ m = F−1(P m (x)ϕ(x)/P m (x))=ϕ.q (3)

Since ϕ m ∗ (D k f ) = (D k ϕ m ) ∗ f for all k ∈ Z+, we have

ϕ m ∗ (P m ( D)f ) = (P m ( D)ϕ m ) ∗ f.

Thus, by (3), we get

0 <

q

ϕ ∗ f∞=ϕ m ∗ P m ( D)f∞.

Applying Young inequality for Banach spaces, we obtain

0 <

q

ϕ ∗ f

∞=ϕ m ∗ P m ( D)f

∞≤ P m ( D)f ∞ϕ m1

and then

lim

m→∞P m (D)f 1/m

∞ ≥ lim

m→∞ϕ m1/m

1

−1

According to [14] and ϕ(x) = 0 ∀x /∈ B(σ, ), we conclude that

lim

m→∞ϕ m1/m

1 ≤ sup

ξ ∈B(σ,) |1/P (ξ)|.

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Therefore, by (4), we obtain

lim

m→∞P m ( D)f 1/m

∞ ≥ inf

ξ ∈B(σ,) |P (ξ)|.

Letting → 0, we get

lim

m→∞P m ( D)f 1/m

Because (5) holds for any σ ∈ Spec(f ), P (σ ) = 0, we arrive at (2)

Finally, we show

lim

m→∞P m ( D)f 1/m

∞ ≤ sup

where K = Spec(f ) Indeed, for > 0, we can choose a function h ∈ C∞( R) satisfying the conditions h(ξ ) = 1/√2π for ξ ∈ K /2and h(ξ ) = 0 for ξ ∈ K Using Proposition 2,

we have f = ˇh ∗ f Clearly, D k f = D k ( ˇ h ∗ f ) = (D k ˇh) ∗ f for all k ∈ Z+ That gives

P m ( D)f = (P m ( D) ˇh) ∗ f.

Then, since P m ( D) ˇh = F−1(h(x)P m (x)), we get

P m ( D)f = F−1(h(x)P m (x)) ∗ f.

Applying Young inequality, we obtain

P m ( D)f ∞≤ f ∞F−1(h(ξ )P m (ξ ))1.

Hence,

lim

m→∞P m ( D)f 1/m

∞ ≤ limm→∞F (h(ξ)P m (ξ ))1/m

We have known in [11] that

lim

m→∞F (h(ξ)P m (ξ ))1/m

1 = limm→∞g m1/m

1 ≤ sup

ξ ∈K 

|P (ξ)|. (8) Combining (7) and (8), we get

lim

m→∞P m (D)f 1/m

∞ ≤ sup

ξ ∈K 

|P (ξ)|

and then (6) by letting → 0 Combing (2) and (6), we arrive at (1)

The proof is complete

We define I λ = (λ − D)−1, where λ ∈ C \ iR The integral operator P (I λ )is obtained

from P (x) by substituting x → I λ We have the following theorem:

Theorem 2 Let P (x) be a polynomial, f ∈ BC(R → X), f ≡ 0 and λ ∈ C \ iR Assume

that Spec(f ) is a compact set Then there always exists the following limit

lim

m→∞P m (I λ )f1/m

∞

and

lim

m→∞P m (I λ )f1/m

∞ = sup

ξ ∈Spec(f ) |P (1/(λ − iξ))|.

Proof Now we show that

lim

m→∞P m (I λ )f1/m

∞ ≥ sup

ξ ∈Spec(f ) |P (1/(λ − iξ))|. (9)

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Indeed, let σ be an arbitrary element in Spec(f ) satisfying P (1/(λ − iσ )) = 0 For a small enough positive number , we have P m ( 1/(λ − iξ)) = 0 for all ξ ∈ B(σ, ) From the definition of Beurling spectrum, there exists ϕ ∈ C∞

0 ( R), supp ϕ ⊂ B(σ, ) such that

q

ϕ ∗ f ≡ 0 Put

ϕ m = F−1 ϕ(ξ )/P m ( 1/(λ − iξ)).

Then, ϕ m is well defined, ϕ m∈S(R) and P m (I λ )ϕ m =ϕq Clearly, ϕ m ∗(I k

λ f ) = (I k

λ ϕ m ) ∗f for all k∈ Z+ That gives ϕ m ∗ (P m (I λ )f ) = (P m (I λ )ϕ m ) ∗ f and then ϕ m ∗ P m (I λ )f = q

ϕ ∗ f So, applying Young inequality, we obtain

0 <

q

ϕ ∗ f

∞=ϕ m ∗ P m (I λ )f

∞≤P m (I λ )f

∞ϕ m1. (10) Clearly,

sup

x∈R|(1 + x2)ϕ m (x)|

≤√1

2π

ξ ∈B(σ,)

D2 ϕ(ξ )/P m ( 1/(λ − iξ)) + ϕ(ξ )/P m ( 1/(λ − iξ)) dξ.

So

sup

x∈R

(1 + x2)ϕ m (x) ≤ C

1m2 sup

ξ ∈B(σ,)

1/P m+2( 1/(λ − iξ)) , (11) where

C1 := √2

2π

ξ ∈K 

|(D2ϕ)(ξ )P2( 1/(λ − iξ))| + 2|(Dϕ)(ξ)D(P (1/(λ − iξ)))

×P (1/(λ − iξ))| + |ϕ(ξ)D2(P ( 1/(λ − iξ)))P (1/(λ − iξ))

+|ϕ(ξ)(D(P (1/(λ − iξ))))2| + |ϕ(ξ)P2( 1/(λ − iξ))|dξ.

Further, we have

R|ϕ m (x) |dx ≤ π sup

x∈R|(1 + x2)ϕ m (x) |. (12) Combining (11)–(12), we obtain

lim

m→∞ϕ m1/m

1 ≤ sup

ξ ∈B(σ,) |1/P (1/(λ − iξ))|. (13) Relations (10) and (13) imply

lim

m→∞P m (I λ )f1/m

∞ ≥ inf

ξ ∈B(σ,) |P (1/(λ − iξ))|.

Letting → 0, we get

lim

m→∞P m (I λ )f1/m

∞ ≥ |P (1/(λ − iσ ))|. (14) Because (14) holds for any σ ∈ Spec(f ), P (1/(λ − iσ )) = 0, we confirm (9)

Put K = Spec(f ) Now we claim that

lim

m→∞P m (I λ )f1/m

∞ ≤ sup

ξ ∈K |P (1/(λ − iξ))|. (15)

Indeed, for any > 0, there exists h ∈ C∞( R) such that h(x) = 1/√2π if x ∈ K /2 and

h(x) = 0 if x /∈ K Note that λ − iξ = 0 for all ξ ∈ K , so the following function is well defined:

H m = F−1(h(ξ )P m ( 1/(λ − iξ))).

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Using Proposition 2, we have f = ˇh∗f and observe from I k

λ f = (I k

λ ˇh) ∗ f ∀k ∈ Z+

that

P m (I λ )f = (P m (I λ ) ˇ h) ∗ f. (16)

From I λ k ˇh = F−1(h(ξ )/(λ − iξ) k ) , we conclude P m (I λ ) ˇ h = F−1(h(ξ )P m ( 1/(λ − iξ))) =

H m Therefore, by using (16), we have P m (I λ )f = H m ∗ f Hence, it follows from

Proposition 1 that

P m (I λ )f∞≤ f ∞H m1.

So

lim

m→∞P m (I λ )f1/m

∞ ≤ limm→∞H m1/m

Clearly,

√

2π sup

x∈R|(1 + x2)H m (x)|

≤

ξ ∈K 

|D2(h(ξ )P m ( 1/(λ − iξ)))| + |h(ξ)P m ( 1/(λ − iξ))| dξ

≤

ξ ∈K 

+m|h(ξ)D2(P ( 1/(λ − iξ)))P m−1( 1/(λ − iξ))| + |h(ξ)P m ( 1/(λ − iξ))|dξ.

So

sup

x∈R|(1 + x2)H m (x) | ≤ C2m2 sup

ξ ∈K 

|P m−2( 1/(λ − iξ))|, (18)

where C2does not depend on m Further, we have

R|H m (x) |dx ≤ π sup

x∈R|(1 + x2)H m (x) |. (19) Combining (18)–(19), we obtain

lim

m→∞H m1/m

1 ≤ sup

ξ ∈K 

|P (1/(λ − iξ))|. (20)

We combine inequality (17) with inequality (20) to conclude that

lim

m→∞P m (I λ )f1/m

∞ ≤ sup

ξ ∈K 

|P (1/(λ − iξ))|.

Letting → 0, we confirm (15) Combining (15) and (9), we have

lim

m→∞P m (I λ )f1/m

∞ = sup

ξ ∈Spec(f ) |P (1/(λ − iξ))|.

Based on the results about Bernstein inequality in L p ( R)-spaces (see [23–25]), we obtain Bernstein inequality for Banach spaces in the following theorem:

Theorem 3 Let σ > 0, f ∈ BC(R → X) and Spec(f ) ⊂ [−σ, σ ] Then there exists a

constant C not depending on f, m, σ such that

D m f∞≤ Cσ m f ∞, m = 1, 2, (21)

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Proof Let us first prove (21) for the case σ = 1 Indeed, put K := [−1, 1] and the function

is defined as follows

=

C1e 1/(ξ2−1) if|ξ| < 1,

0 if|ξ| ≥ 1, where C1is chosen such that

R = 1 We define the sequence of functions φ m (ξ ) m≥1

via the formula

φ m (ξ ) = (1 K 3/(4m) 1/(4m) )(ξ ),

where

1/(4m) (ξ )

1/(4m) (ξ ) = 0 for all ξ ∈ [−1/(4m), 1/(4m)] andR 1/(4m) (ξ )dξ = 1 Hence, for

any m ≥ 1, we have φ m (ξ ) ∈ C∞

0 ( R), and φ m (ξ ) = 1 ∀ξ ∈ K 1/(2m) , φ m (ξ ) = 0 ∀ξ /∈

K 1/m So, it follows from Spec(f ) ⊂ K that f = (2π) −1/2 F (φ

m ) ∗ f Hence,

D m f∞= (2π) −1/2 F (φ m (ξ )ξ m ) ∗ f ∞.

Therefore, applying Young inequality (Proposition 1), we obtain the estimate

D m f∞= (2π) −1/2 F (φ m (ξ )ξ m ) ∗ f ∞≤ (2π) −1/2 f ∞F (φ m (ξ )ξ m )1. (22) Define

k m:= 1 + 1

m , g m (ξ ) = φ m

ξ

k m

, m (ξ ) = φ m (ξ ) − g m (ξ ).

Then

(F (g m (ξ )ξ m ))(x) = (k m ) m

F

φ m

ξ

k m

ξ

k m

m

(x)

= (k m ) m+1(F (φ

m (ξ )ξ m ))(k m x).

So

F (g m (ξ )ξ m )

1= (k m ) mF (φ m (ξ )ξ m )

1.

Then, it follows from (k m ) m = (1 + 1

m ) m ≥ 2 that

F (g m (ξ )ξ m )

1≥ 2F (φ m (ξ )ξ m )

1.

Therefore, since m (ξ ) = φ m (ξ ) − g m (ξ )we get

F ( m (ξ )ξ m )

1≥F (g m (ξ )ξ m )

1−F (φ m (ξ )ξ m )

1≥F (φ m (ξ )ξ m )

1. (23) From (22)–(23), we obtain

D m f∞≤ (2π) −1/2 f ∞F ( m (ξ )ξ m )1. (24)

Now, we show the existence of a constant C such that

F ( m (ξ )ξ m )1≤ (2π) 1/2 C

for all m ≥ 1 Indeed, we put C2 (j )1, j 1/(4m) (x)

(j )

1/(4m) (x) = (4m) j+1 (j ) ( 4mx) and then

(j )

1/(4m)1= (4m) j (j )1≤ C2( 4m) j ∀j ≤ 3.

Therefore,

φ (j )

m

∞=(1

K 3/(4m) 1/(4m) (j ) )

∞≤ (j )

1/(4m)

1≤ (4m) j C2 ∀j ≤ 3. (25)

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Note that φ m (ξ ) = 1 ∀ξ ∈ (−1 − (1/2m), 1 + (1/2m)), and φ m (ξ ) = 0 ∀ξ ∈ (−∞, −1 −

( 1/m)) ∪(1+(1/m), +∞) So, if |ξ| < 1 then |ξ/k m | < |ξ| < 1 and φ m (ξ ) = φ m (ξ /k m )=

1, i.e., m (ξ )= 0

Further, if|ξ| > 1+(3/m) then |ξ| > |ξ/k m | > 1+(1/m) and then φ m (ξ ) = φ m (ξ /k m )=

0, i.e., m (ξ )= 0

So we have

supp m ⊂ [1, 1 + (3/m)] ∪ [−1 − (3/m), −1]. (26)

Now, if ξ ∈ [1, 1 + (3/m)] ∪ [−1 − (3/m), −1] then

ξ − k ξ m = (k m − 1)ξ

k m

= mk ξ m ≤ m4. (27) From (25) and (27), we get the following estimate for ξ ∈ [1, 1+(3/m)]∪[−1−(3/m), −1]

| m (ξ ) | = |φ m (ξ ) − g m (ξ )| =

φ m (ξ ) − φ

ξ

k m

≤

ξ − ξ

k m

·φ

m

∞≤ 4

m (ξ ) = φ

m (ξ ) − g m(ξ ) = φ

m (ξ )−

φ m

ξ

k m

=

φ

m (ξ )− 1

k m

φ

m

ξ

k m

≤ φ

m (ξ ) − φ m

ξ

k m

+ 1− 1

k m

φ

m

ξ

k m

≤

ξ − ξ

k m

·φ

m

∞+

1 − 1

k m

·φ

m

∞

≤ 4

m ( 4m)

2C2+

1 −k1m 4mC2≤ 68mC2, (29) and

m (ξ ) = φ

m (ξ ) − gm (ξ ) = φ

m (ξ )−

φ m

ξ

k m

=

φ

m (ξ )− 1

k2

m

φ

m

ξ

k m

≤ φ

m (ξ ) − φ m

ξ

k2

m

+ 1− 1

k2

m

φ

m

ξ

k m

≤

ξ − ξ

k2

m

· φ

m∞+

1 − 1

k2

m

·φ

m

∞

≤ 8

m ( 4m)

3C2+

1 −k12

m

(4m)2C2≤ 528m2C2. (30)

Put H m (x) = (F ( m (ξ )ξ m ))(x) Then

H m (x)= √1

2π

Re

−ixξ

m (ξ )ξ m dξ.

Therefore, from (26), we obtain

sup

x∈R|H m (x)| ≤√1

2π

R

m (ξ )ξ m dξ = 1

√

2π

1≤|ξ|≤1+ 3

m

m (ξ )ξ m dξ and it follows from (25) that

sup

x∈R|H m (x)| ≤ 6

m√

2π ξsup∈R| m (ξ )|

1+ 3

m

≤ 96e3C2

m√

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We also have

x2H m (x) = 1

√

2π

Re −ixξ (

m (ξ )m(m − 1)ξ m−2+

m (ξ ) 2mξ m−1+

m (ξ )ξ m )dξ

≤ √1

2π

R

e −ixξ ( m (ξ )m(m − 1)ξ m−2+ m (ξ ) 2mξ m−1+ m ξ m)dξ dξ and then

sup

x∈R

x2H m (x)

≤ √1

2π

1≤|ξ|≤1+ 3

m

| m (ξ )m(m − 1)ξ m−2+ (ξ ) 2mξ m−1+ (ξ )ξ m |dξ

m√

2π

sup

ξ∈R|(ξ)|m(m − 1)

1+ 3

m

m−2

+ sup

ξ∈R|(ξ ) |2m

1+ 3

m

m−1

+ sup

ξ∈R|(ξ )|

1+ 3

m

.

So, from (28)–(30), we get

sup

x∈R

x2H m (x) ≤ 8

m√

2π 16C2m(m − 1)e3+ 68mC22me3+ 528m2C2e3

≤ 5280e3C2m

√

We have

H m1 =

|x|≤m |H m (x) |dx +

|x|≥m |H m (x) |dx

≤ sup

x∈R|H m (x)|

|x|≤m 1dx+ sup

x∈R|x2H m (x)|

|x|≥m

x12

dx

= 2m sup

x∈R|H m (x)| + 2

m xsup∈R|x2H m (x) |. (33) From (31)–(33), we obtain

H m1≤ 2m

96e3C2

m√

2π

+ 2

m

5280e3C2m

√

2π .

Therefore, from (24), we can choose a constant C such that

D m f∞≤ Cf ∞.

So, (21) has been proved with σ= 1

Next, we prove (21) for any σ > 0 Indeed, put

g(x) = fx

σ

.

Since Spec(f ) ⊂ [−σ, σ ], Spec(g) ⊂ [−1, 1] Therefore

Since g(x) = f ( x

σ ), we have

g∞= f ∞, D m g∞= σ −m D m f∞.

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Hence, it follows from (34) that

σ −m D m f∞≤ Cf ∞.

We conclude that

D m f∞≤ Cσ m f ∞.

The minimum of C satisfying the above inequalities is called the Bernstein constant.

Theorem 4 Let f ∈ BC(R → X), σ ∈ R+ Assume that Spec(f ) ⊂ [−σ, σ ] Then

lim

m→∞σ −m D m f∞≤ C lim

λ→1 −f (·) − f (λ·)∞, (35)

where C is the Bernstein constant.

Proof Fix λ ∈ (0, 1) Put f λ (x) = f (λx) Then Spec(f λ ) = λSpec(f ) Then it follows from Spec(f ) ⊂ [−σ, σ ] that

Applying the Bernstein inequality for Beurling spectrum, we get

D m f λ∞≤ C(λσ ) m f λ∞ (37)

for all m ∈ N Using (36) and Spec(f ) ⊂ [−σ, σ ], we obtain Spec(f − f λ ) ⊂ [−σ, σ ].

Then it follows from the Bernstein inequality for Beurling spectrum that

D m (f − f λ )∞≤ Cσ m f − f λ∞. (38) Therefore, using (37) and (38), we have

σ −m D m f∞ ≤ σ −m D m (f − f λ )∞+ σ −m D m f λ∞

≤ Cf − f λ∞+ Cλ m f λ∞

= Cf − f λ∞+ Cλ m f ∞

for all m∈ N Hence,

lim

m→∞σ −m D m f∞≤ Cf − f λ∞.

Letting λ→ 1−, we get lim

m→∞σ −m D m f∞≤ C lim λ→1 −f (·)−f (λ·)∞ The proof

is complete

Corollary 1 Let f ∈ BC(R → X), σ ∈ R+ Assume that Spec(f ) ⊂ [−σ, σ ] Then the

sequence {σ −m D m f∞} is bounded Moreover, if lim λ→1 −f (·) − f (λ·)∞= 0 then

lim

m→∞σ

−m D m f∞= 0.

Acknowledgments This research is funded by Vietnam National Foundation for Science and Technol-ogy Development (NAFOSTED) under grant number 101.01-2011.32 The authors would like to thank the referees for the exact corrections.

References

1 Abreu, L.D.: Real Paley–Wiener theorems for the Koornwinder–Swarttouw q-Hankel transform J Math.

Anal Appl 334, 223–231 (2007)

Trang 3

The Beurling spectral radius ρ(f ) of f is defined by

ρ(f... class="page_container" data-page="4">

Therefore, by (4), we obtain

lim

m→∞P m ( D)f 1/m

∞... chosen such that

R = We define the sequence of functions φ m (ξ ) m≥1

via the formula

φ m (ξ )

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