Some Extensions of the Kolmogorov–Stein Inequality

DOI 10.1007/s10013-014-0090-2Some Extensions of the Kolmogorov–Stein Inequality Ha Huy Bang · Vu Nhat Huy Received: 2 September 2013 / Accepted: 14 March 2014 © Vietnam Academy of Scienc

DOI 10.1007/s10013-014-0090-2

Some Extensions of the Kolmogorov–Stein Inequality

Ha Huy Bang · Vu Nhat Huy

Received: 2 September 2013 / Accepted: 14 March 2014

© Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2014

Abstract In this paper, we prove some extensions of the Kolmogorov–Stein inequality for

derivatives in L p ( R) norm to differential operators generated by a polynomial.

Keywords L pspaces· Orlicz spaces · Kolmogorov inequality

Mathematics Subject Classification (2010) 26A24 · 41A17

1 Introduction

Let 1≤ p ≤ ∞, n ≥ 2, and 0 < k < n In the space of all functions, f ∈ C n ( R) such that

f, f, , f (n) belong to L p ( R), consider the following inequality:

f (k)p≤ Af p + Bf (n)p.

It is well known that this inequality is equivalent to the inequality (see [7])

f (k)p≤ A1−k/n B k/n ( f p + f (n)p)

and also equivalent to

f (k)p≤ Cn,kf 1−k/n

p f (n)k/n

p ,

where

C n,k=



A

1− (k/n)

1−(k/n)

B k/n

k/n

.

H H Bang (  )

Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet

Street, Cau Giay, Hanoi, Vietnam

e-mail: hhbang@math.ac.vn

V N Huy

Department of Mathematics, College of Science, Vietnam National University, 334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam

e-mail: nhat huy85@yahoo.com

For p= ∞, Kolmogorov proved in [15] that Cn,k = Kn−k /K (1−k/n)

n , where

K j = 4

π

∞



k=0

( −1) k /(2k + 1) j+1

for even j , while

K j= 4

π

∞



k=0

1/(2k + 1) j+1

for odd j Moreover the constants {Cn,k : n, k ∈ N, k < n} are sharp in the sense that these

cannot be replaced by smaller ones

This result of Kolmogorov has been extended by Stein to L p ( R) norm [20] and in [1]

to any Orlicz norm The Kolmogorov–Stein inequality and its modifications are a prob-lem of interest for many mathematicians and have various applications (see, for example, [1,3,5 12,14, 20–22]) In this paper, we prove some extensions of the Kolmogorov–

Stein inequality for derivatives in L p ( R) norm to differential operators generated by a

polynomial

2 Inequalities forL pSpaces

Denote by C k ( R) the set of all functions f that are continuous in R together with all

deriva-tives f (n) , n ≤ k, and Lp ( R) (1 ≤ p ≤ ∞) the collection of all functions f specified on R

for which the norm

f p=

 [R|f (x)| p dx]1/p , 1≤ p < ∞,

ess sup|f (x)|, p= ∞

is finite

The convolution f ∗ g of two functions f, g ∈ L1(R) is defined as

(f ∗ g)(x) =

Rf (x − y)g(y)dy (x ∈ R). (1)

If f ∈ Lp ( R), 1 ≤ p ≤ ∞, and g ∈ L1( R), then the integral (1) converges for almost all

x∈ R, and we have the following result

Young inequality (see, e.g., [2]) Let 1 ≤ p ≤ ∞, f ∈ Lp ( R), g ∈ L1(R) Then f ∗ g ∈

L p ( R) and

f ∗ gp ≤ f pg1.

Furthermore,S(R) stands for the Schwartz space on R and S( R) for the dual space of

tempered distributions onR Recall that Lp ( R) ⊂ S( R) for any 1 ≤ p ≤ ∞ Let f ∈

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

ˆ

f (ξ )= √1

2π

 +∞

−ixξ f (x)dx.

The Fourier transform of a tempered generalized function f is defined via the formula

Ff, ϕ = f, Fϕ, ϕ ∈ S(R).

Let n ∈ N and P (x) be a polynomial of degree n: P (x) := nk=0a k x k We put

P1(x) := nk=0 |ak|x k , D k f = (−i) k f (k) and then the differential operator P (D) is

obtained from P (x) by substituting x → −i∂/∂x We define the function ρ ∈ C∞( R) as

follows

ρ(x):=

Ce

1

|x|2−1 if|x| < 1,

0 if|x| ≥ 1,

where the constant C satisfies

Rρ(x)dx= 1, and put

ρ1/4(x) = 4ρ(4x), (x) = F(1 [−3/2,3/2] ∗ ρ 1/4)(x).

Clearly, ∈S(R) and ˆ(x) = 1 ∀x ∈ [−1, 1], ˆ(x) = 0 ∀x /∈ (−2, 2) Put λ := 1

We have the following theorem

Theorem 1 Let m, n ∈ N and f ∈ C m +n ( R), 1 ≤ p ≤ ∞, P (x) be a polynomial of

degree n Assume f, D m (P (D)f ) ∈ Lp ( R) Then P (D)f ∈ Lp ( R) and

P (D)f p ≤ λ(Km2m + 1)P1(2 ) f p + Km −m D m (P (D)f )p ∀ > 0,

where K m is the Favard constant.

To obtain Theorem 1, we need the following results

Bohr-Favard inequality (see, e.g., [2,4,12]) Let 1≤ p ≤ ∞, σ > 0, f ∈ C n ( R) and

supp ˆf ∩ (−σ, σ ) = ∅ Assume that f (n) ∈ Lp ( R) Then f ∈ Lp ( R) and

f p ≤ σ −n K nf (n)p, where the Favard constants K nare best possible and have the following properties

1= K0≤ K2< · · · < π4 < · · · < K3≤ K1= π2.

Bernstein inequality (see, e.g., [13,18]) Assume 1≤ p ≤ ∞, σ > 0, f ∈ Lp ( R) and

supp ˆf ⊂ [−σ, σ ] Then

f (n)p≤ σ n f p , n = 1, 2,

The constants 1 are best possible

Proof of Theorem 1 for > 0 we define

 (x) := ( x).

Then, 1= 1= λ and ˆ  (x) = ˆ( x

) = 1 ∀x ∈ [− , ], ˆ  (x) = ˆ( x

) = 0 ∀x /∈

( −2 , 2 ) Therefore, supp ˆ ⊂ [−2 , 2 ] Hence, by applying Bernstein inequality, we

have

D k  1≤ (2 ) k  1= λ(2 ) k , k = 0, 1, , n. (2)

We define the functions g, h as follows

g = f ∗  , h = f − g.

Clearly, suppˆg ⊂ supp ˆ  ⊂ [−2 , 2 ], supp ˆh ⊂ (−∞, − ] ∪ [ , +∞), f = g + h and

D k g = f ∗ (D k  ) It follows from Young inequality that

gp ≤ f p 1= λf p

and then

hp ≤ f p + gp ≤ (λ + 1)f p

Using Young inequality and D k g = f ∗ (D k  ), we obtain D k g ∈ Lp ( R) and

D k gp≤ f pD k  1, k = 0, 1, , n. (3) From (2)–(3), we have the following estimate

D k gp≤ λ(2 ) k f p , k = 0, 1, , n. (4) From (4), we get P (D)g∈ Lp ( R) and

P (D)gp ≤ λP1(2 ) f p (5) Similarly, we have

D m (P (D)g)p≤ (2 ) m P (D)gp ≤ λ(2 ) m P1 (2 ) f p

Therefore,

D m (P (D)h)p ≤ D m (P (D)g)p+ D m (P (D)f )p

≤ λ(2 ) m P1(2 ) f p + D m (P (D)f )p. (6)

By Bohr–Favard inequality and P (D)h ⊂ (−∞, − ] ∪ [ , +∞), we obtain

P (D)hp ≤ Km −m D m (P (D)h)p.

Then it follows from (6) that

P (D)hp ≤ Km −m [λ(2 ) m P1 (2 ) f p + D m (P (D)f )p] (7) From (5), (7) and f = g + h, we have P (D)f ∈ Lp ( R) and

P (D)f p ≤ P (D)gp + P (D)hp (8)

By (5), (7) and (8), we obtain

P (D)f p ≤ λ(Km2m + 1)P1(2 ) f p + Km −m D m (P (D)fp ∀ > 0.

The proof is complete

By Theorem 1 we obtain the following results:

Corollary 1 Let m, n ∈ N and f ∈ C m+n ( R), 1 ≤ p ≤ ∞, P (x) be a polynomial of

degree n Assume f, D m (P (D)f ) ∈ Lp ( R) Then P (D)f ∈ Lp ( R) and

P (D)f p ≤ λ(π2 m−1 + 1)P1(2 )f p+π

2

−m D m (P (D)f )p ∀ > 0.

Corollary 2 Let m, n ∈ N Then, there exists a constant C < ∞ independent of f such that

f + D n fp≤ C f p+ m +n

f  m

p D m f + D m +n fn

p

Proof Put

H ( ) = λ(π2 m−1 + 1)(1 + (2 ) n ) f p+ π

2 m D m f + D m+n fp.

By Theorem 1,

f + D n fp≤ min{H( ) : > 0}.

Therefore, since

min{H( ) : > 0} = H m +n



A D m f + D m+n fp

f p



,

where A = (πm)/(λn(π2 m+n+ 2n+1 )), we obtain that

f + D n fp ≤ H m +n



A D m f + D m+n fp

f p



= λ(π2 m−1+ 1)f p + C1 m +n

f  m

p D m f + D m+n fn

p , where C1 is independent of f The proof is complete.

Let Q(x) := mk=0 b k x k be a polynomial of degree m ≥ 1 with real coefficients,

sat-isfying the condition |Q(x)| = 0 ∀x ∈ R We define Q1(x) := mk=0 |bk|x k , φ(x) =

(1 − ˆ /4(x))/Q(x) Since|Q(x)| = 0 ∀x ∈ R, there exists a constant a > 0 such that

|Q(x)| ≥ a(1 + x2) and then φ(x) ∈ L1( R) Hence ˆφ is well defined on R and we can see

that ˆφ ∈ L1( R) Put λ1:= √ 1

2π  ˆφ1 Further, we have

Theorem 2 Let m, n ∈ N and f ∈ C m+n ( R), 1 ≤ p ≤ ∞, P (x) be a polynomial of degree

n, Q(x) be a polynomial of degree m ≥ 1 with real coefficients Assume |Q(x)| = 0 ∀x ∈ R

and f, (QP )(D)f ∈ Lp ( R) Then P (D)f ∈ Lp ( R) and

P (D)f p ≤ λ1(QP )(D)f p + λ(λ1Q1(2 ) + 1)P1(2 )f p ∀ > 0.

Proof We define the functions  , g, h as in the proof of Theorem 1 Arguing as in the

proof of Theorem 1, we have

(QP )(D)gp ≤ Q1(2 ) P (D)gp ≤ λQ1(2 )P1(2 ) f p (9)

Therefore, since h = f − g, we get

(QP )(D)hp ≤ (QP )(D)gp + (QP )(D)f p

≤ λQ1(2 )P1(2 ) f p + (QP )(D)f p (10) Further, put

t (x) = 1 − ˆ /4(x).

Then t (x) = 1 ∀x ∈ (−∞, − /2] ∪ [ /2, +∞) Hence, it follows from F((QP )(D)h) = Q(x)F(P (D)h) and supp F((QP )(D)h) ⊂ (−∞, − ] ∪ [ , +∞) that

t (x) F((QP )(D)h) = Q(x)F(P (D)h).

Therefore

F(P (D)h) = t (x)

Q(x) F((QP )(D)h).

So

P (D)h= √1

2π ((QP )(D)h)∗F−1

t (x) Q(x)



Therefore, using Young inequality and (QP )(D)h ∈ Lp ( R), we obtain P (D)h ∈ Lp ( R)

and

P (D)hp ≤ √1

2π (QP )(D)hp

F−1

t (x) Q(x)





1

= √1

2π (QP )(D)hp

FQ(x) t (x)  ,

P (D)hp ≤ λ1(QP )(D)hp

From this and (10), we obtain

P (D)hp ≤ λ1(λQ1(2 )P1(2 )f p + (QP )(D)f p ). (11)

Since f = g + h, we have

P (D)f p ≤ P (D)gp + P (D)hp (12) From (5), (11) and (12), we get P (D)f ∈ Lp ( R) and

P (D)f p ≤ λ1(QP )(D)f p + λ(λ1Q1(2 )+ 1)P1(2 )f p ∀ > 0.

The proof is complete

Note that Theorems 1, 2 still hold for generalized derivatives which will be seen in the next section

3 Inequalities for Orlicz Spaces

Let : [0, +∞) → [0, +∞] be an arbitrary Young function, i.e., (0) = 0, (t) ≡ 0 and

is convex Denote by

¯(t) = sup s≥0 {ts − (s)}

the Young function conjugate to and L  ( R) the space of all measurable functions u such

that

|u, v| =

Ru(x)v(x)dx

 < ∞

for all v with ρ(v, ¯ ) <∞, where

ρ(v, ¯ )=



R ¯(|v(x)|)dx.

Then L  ( R) is a Banach space with respect to the Orlicz norm

u= sup

ρ(v, ¯ )≤1



Ru(x)v(x)dx

,

which is equivalent to the Luxemburg norm

f ()= inf



λ >0 : 

R( |f (x)|/λ)dx ≤ 1



< ∞.

Recall that · ()=  · pwhere (t) = t pwith 1≤ p < ∞ and  · ()=  · ∞when

(t) = 0 for 0 ≤ t ≤ 1 and (t) = ∞ for t > 1 So, Lebesgue spaces Lp ( R), 1 ≤ p ≤ ∞

are special cases of Orlicz spaces (see [16,17,19]) Note that L ( R) ⊂ S( R), where

S( R) is the Schwartz space of tempered generalized functions.

Using Theorems 1, 2, and the method investigated in [1 3], we obtain the following theorems

Theorem 3 Let be an arbitrary Young function, m, n ∈ N and P (x) be a polynomial of

degree n Assume f ∈ L ( R) and D m (P (D)f ) ∈ L ( R) in the S-sense Then P (D)f ∈

L  ( R) and

P (D)f () ≤ λ(Km2m + 1)P1(2 ) f () + Km −m D m (P (D)f() ∀ > 0.

Theorem 4 Let be an arbitrary Young function, m, n ∈ N, P (x) be a polynomial of

degree n, Q(x) be a polynomial of degree m with real coefficients, and |Q(x)| = 0 ∀x ∈ R.

Assume f ∈ L ( R) and (QP )(D)f ∈ L ( R) in the S-sense Then P (D)f ∈ L ( R) and

P (D)f () ≤ λ1(QP )(D)f () + λ(λ1Q1(2 )+ 1)P1(2 )f (Φ) ∀ > 0.

Remark 1 Theorems 3 and 4 still hold if we replace Luxemburg’s norm by Orlicz’s norm.

Acknowledgments This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.01-2011.32.

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Proof We define the functions  , g, h as in the proof of Theorem Arguing as in the< /i>

proof of Theorem 1, we have

(QP )(D)gp... is the Schwartz space of tempered generalized functions.

Using Theorems 1, 2, and the method investigated in [1 3], we obtain the following theorems

Theorem Let be an arbitrary... generalization of an inequality of Bohr Math Scand 2, 33–45 (1954)

14 Kofanov, V.A.: On sharp Kolmogorov-type inequalities taking into account the number of sign changes