sup |B|=βsup Tn °°Tn

E. NURSULTANOV AND S. TIKHONOV

Abstract. We obtain a sharp Remez inequality for the trigonometric polynomialTnof degreenon [0,2π):

°°Tn

°°

L_∞([0,2π))6³

1 + 2 tan² nβ 4m

´°°Tn

°°

L_∞¡

[0,2π)\B¢,

where ^2π_m is the minimal period ofTnand|B|=β < ^2πm_n is a measurable subset ofT. In particular, this gives the asymptotics of the sharp constant in the Remez inequality:

C(n, β) = 1 +(nβ)²

8 +O(β)⁴, where

C(n, β) := sup

|B|=βsup

Tn

°°Tn

°°

L_∞([0,2π))

°°Tn

°°

L_∞¡

[0,2π)\B¢. (0.1)

We also obtain sharp Nikol’skii’s inequalities for the Lorentz spaces and Net spaces. Multidimensional variants of Remez and Nikol’skii’s inequalities are investigated.

1. Introduction

Consider the space of (complex) trigonometric polynomials of degree at mostn∈N, i.e., Nn=

n

Tn : Tn(x) = X

|k|6n

cke^ikx, ck ∈C, x∈T o

. (1.1)

An important question in polynomial approximation is the following one. Let |B| = µ(B) be the linear Lebesgue measure ofB. How large cankTnk_{L∞([0,2π))} be if

¯¯

¯ n

x∈[0,2π) :|Tn(x)|>1o¯¯

¯6ε

holds for some 0< ε61? The well-known Remez inequality answers this question.

For any Lebesgue measurable set B⊂Tsuch that |B| ≡β < π/2we have (1.2) kTnkL∞([0,2π))6C(n,|B|)kTnkL∞([0,2π)\B), Tn∈Nn.

We will writeC(n,|B|) in place ofC(n,|B|) while dealing with the sharp constant in this estimate, see (0.1).

Investigation of Remez’s inequalities is a well developed topic. For algebraic polynomials sharp inequality was proved by Remez [Re]. In [BE] and [Er1] (1.2) was proved withC(n,|B|) = exp(4n|B|); the history of the question can be found in e.g. [Er3, Sec. 2], [Ga, Sec. 3], and [LGM]. In [Ga, Th. 3.1], the constant was sharpened as C(n,|B|) = exp(2n|B|). Also, another improvement was obtained in [Er2, Th. 3.4]:

C(n,|B|) = exp¡

n(|B|+ 1.75|B|²)¢ .

In case ofB ≡[a, b]⊂[−π, π] a sharp Remez inequality is given by (see [Er1]) (1.3) kTnkL∞([0,2π)) 61

2

³ tan²ⁿ

³2π−β 8

´

+ cot²ⁿ

³2π−β 8

´´

kTnkL∞([0,2π)\B), |B|=β, Tn∈Nn

and equality in (1.3) holds if and only if Tn(x) =CTn

³cos(x−(a+b)/2)−cos²((b−a)/4) sin²((b−a)/4)

´

, C∈R,

Date: July 6, 2011.

2000Mathematics Subject Classification. Primary 41A17, 41A44; Secondary 46E30.

Key words and phrases. Remez and Nikol’skii’s inequalities, Trigonometric polynomials, Lorentz and Net spaces.

This research was partially supported by MTM 2011-27637, 2009 SGR 1303, RFFI 09-01-00175, and NSH-3252.2010.1.

1

whereTn(x) is the Chebyshev polynomial of degreen, i.e., Tn(x) = 1

2

³¡x+p

x²−1¢_n +¡

x−p

x²−1¢_n´ for everyx∈R\(−1,1).

Therefore, the best known bounds of the sharp constant in the Remez inequality are given by

(1.4) 1

2

³ tan²ⁿ

³2π−β 8

´

+ cot²ⁿ

³2π−β 8

´´

6C(n, β)6exp¡

nβmin{2,(1 + 1.75β)}¢ , for 0< β 6π/2. Moreover, Erd´elyi [Er2] and Ganzburg [Ga] conjectured that

(1.5) C(n, β) = 1

2

³

tan²ⁿ³2π−β 8

´

+ cot²ⁿ³2π−β 8

´´

for anyβ >0.

In case when the measure|B|=β is big, that is, whenπ/2< β <2π, it is known that C(n, β)6C(n, β) =

³ 17 2π−β

´_2n

(see [Er1, Er2, Ga, Na]). Another upper bound ofC(n, β),β >0, was given by Andrievskii [An].

In this paper we investigate Remez’s inequalities for the case of 0< β < ^2π_n. Note that in applications in fact we are usually interested in the case of smallβ (see, e.g., [MT]). Precisely, we prove

(1.6) kTnkL∞([0,2π)) 6

³

1 + 2 tan² nβ 4m

´

kTnkL∞([0,2π)\B), 0< β < 2πm n , where ^2π_m is the minimal period ofT_n∈N_n. Equality in (1.6) holds forT_n(x) = cosnx+¹₂³

1−cos^β₂´ . This improves the right-hand side bound of the sharp constant in the Remes inequality:

(1.7) C(n, β)61 + 2 tan²nβ

4 , 0< β < π n. Combining the left-hand side of (1.4) and (1.7) we show (see Corollary 3.5) that

(1.8) C(n, β) = 1 +(nβ)²

8 +O(β⁴).

This answers Stechkin-Ulyanov’s question ([SU]) on the rate of decrease ofC(n, β). Moreover, C(1, β) = 1 + 2 tan²β

4, which proves conjecture (1.5) forn= 1.

The paper is organized as follows. In section 2, we prove several basic auxiliary results for rearrangements.

We recall that the distribution of a measurable function f on T is defined by λ(σ, f) = µ{x ∈ [0,2π] :

|f(x)|> σ}.Then f^∗(t) = inf{σ:λ(σ, f)6t} is the decreasing rearrangement off. We will also consider the decreasing rearrangement of functions on their minimal periods denoting it byf^?.

In section 3, we prove sharp Remez and Remez-Bernstein’s inequalities in terms of rearrangements and averages which are defined as follows. Any fixed familyM of measurable subsets ofTis called anet inT.

For a 2π-periodic locally integrable functionf we define the function f(t, M) = sup

e∈M

|e|>t

1

|e|

¯¯

¯¯ Z

e

f(x)dx

¯¯

¯¯, t>0,

which is calledthe average function off on the netM ([NT]). If sup_e∈M|e|=αandt > α, thenf(t, M) = 0.

In section 4, we obtain sharp Bernstein and Bernstein-Nikol’skii’s inequalities in the Lorentz spaces, in particular Lebesgue spaces, and in the Net spaces. As usual, a measurable functionf defined onTbelongs to the Lorentz spaceLp,q=Lp,q(T) (see e.g. [BS]) if

kfkLp,q= µZ _2π

0

³

t^1/pf^∗(t)

´_qdt t

¶1/q

<∞, 0< p, q <∞, and

kfkLp,∞= sup

t∈(0,2π)

t^1/pf^∗(t)<∞, 0< p6∞.

The scale of net spaces is more general than the scale of Lorentz spaces. To define them, let 0< p, q6∞, M be a net inT, and a 2π-periodic functionf be locally integrable. We say thatf belongs to the Net space Np,q(M) =Np,q(M,T) ([NA, NT]) if

kfk_Np,q(M)= µZ _2π

0

(t^1/pf(t, M))^qdt t

¶1/q

<∞, 0< p, q <∞, and

kfkNp,∞(M)= sup

t∈(0,2π)

t^1/pf(t, M)<∞, 0< p6∞.

The net spacesNp,q(M) are a natural generalization of the Lorentz spacesLp,q, since ifM is a collection of all compact sets inT, thenN_p,q(M) =L_p,q(T) for 1< p <∞,0< q6∞.

Finally, multidimensional variants of Remez and Nikol’skii’s inequalities are obtained in Section 5. In particular, for trigonometric polynomials

Tn(x) = X

|k1|6n1

· · · X

|kd|6nd

cke^i(k,x), ck∈C, x∈T^d, d>1, we obtain the following Remez inequality

kTnk_L∞(T^d₎6 1 cos ^d

q

|B|Q_d

j=1 jnj

2

kTnk_L∞(T^d_\B), 06|B|< π^d Q_d

j=1jnj

,

cf. [DP] and [Kr].

2. Basic lemmas on rearrangements

Let Tn ∈Nn and d =d(Tn) be the minimal period of Tn. Note that ^2π_n 6d 62π. Let T_n^?(t) be the decreasing rearrangement ofTn on its minimal period, i.e.,

λmin(σ, Tn) =

¯¯

¯ n

x∈[0, d) :|Tn(x)|> σo¯¯

¯ and

T_n^?(t) = inf n

σ:λmin(σ, Tn)6t o

.

Remark that this concept describes the order of polynomials more accurately thanT_n^∗. For example,

¡cosnx¢_∗ (β) =¡

cosx¢_∗

(β) = cosβ

4, 06β <2π but

¡cosnx¢_?

(β) = cosnβ

4 , 06β < 2π n . (2.1)

Moreover, the following result holds.

Lemma 2.1. Let n∈N andf ∈Nn be such that |cn(f)|+|c−n(f)|>0. If d is the minimal period of f, then

(a) we have f^?(β)6f^∗(β)for0< β6dandf^?(β)< f^∗(β) ford <2π;

(b) there existsm∈Nsuch thatd= 2π/m;

(c) we have n/m∈Nandg(x) =f(x/m) is a trigonometric polynomial of degreen/m;

(d) f^∗(β) =g^∗(β), β∈[0,2π);

(e) f^?(β) =f^∗(mβ)for06β 6d.

Proof. Item (a) follows fromλmin(σ, f)6λ(σ, f) andλmin(σ, f)< λ(σ, f) ford <2π.

(b). If d 6= 2π/m, m ∈ N, then we take an integer r such that rd 6 2π < (r+ 1)d. Suppose that α= 2π−rd; thenf(x) =f(x+rd),x∈R. Using 2π-periodicity, we getf(x) =f(x+ 2π) =f(x+α+rd), x∈R. Thenf(x) =f(x+α), which contradictsα < d.

(c) Letd= 2π/mbe the minimal period off. Puttingg(x) =f(x/m),x∈[0,2π),we have cr(f) =

Z _2π

0

f(x)e^−irxdx=

m−1X

k=0

Z _2π(k+1)/m

2πk/m

f(x)e^−irxdx=

m−1X

k=0

Z _2π/m

0

f(x)e−ir(x+2πk/m)dx

= Z _2π/m

0

f(x)e^−irxdx µ_m−1X

k=0

e^−2πikr/m)

¶ . (2.2)

Taking into account ^m−1P

k=0

e^−2πikr/m =

½ m, for r=ms,s∈N,

0, otherwise , we get cr(f) = 0 for r 6= ms. Since

|cn(f)|+|c−n(f)|>0, thenn/m∈N.Therefore, using (2.2) we get cms(f) =m

Z _2π/m

0

f(x)e^−imsxdx= Z _2π

0

f(x/m)e^−isxdx=cs(g), andg is the trigonometric polynomial of degreen/m.

(d) Making use of 2π/m-periodicity, we get λ(α, f) =

¯¯

¯ n

x∈[0,2π] : |f(x)|> α o¯¯

¯=m

¯¯

¯ n

x∈[0,2π/m] : |f(x)|> α o¯¯

¯

=

¯¯

¯ n

x∈[0,2π] : |f(x/m)|> α o¯¯¯=

¯¯

¯ n

x∈[0,2π] : |g(x)|> α

o¯¯¯=λ(α, g).

Theng^∗(β) =f^∗(β) for anyβ∈[0,2π].

(e) We proved thatd=^2π_m andλmin(σ, f) =_m¹λ(σ, f). Hence, for 0< β < d= ^2π_m,

f^?(β) = inf{σ:λmin(σ, f)6β}= inf{σ:λ(σ, f)6mβ}=f^∗(mβ). ¤ We denote byMn the set of real trigonometric polynomials, i.e.,

Mn= n

Tn ∈Nn : Tn(x)∈R, x∈T o

. (2.3)

Let us prove two auxiliary results on rearrangements. First, we recall the following known lemma. The proof can be found in the papers [Be] by S. Bernstein and [St] by S. Stechkin.

Lemma 2.2. ForTn∈Mn such that |Tn(x)|61 andTn(0) = 1, we have

(2.4) Tn(x)>cosnx, −π

n 6x6 π n. Lemma 2.3. Let Tn∈Mn be such that

maxTn(x)−minTn(x) = 2.

Then for 06β < ^2π_n,

T_n^?(β)>(cosnx+α)^?(β), (2.5)

whereα= maxTn(x)−1 . Proof. Denote

Ξ := maxTn(x) and ξ:= minTn(x).

Letdbe the minimal period of Tn. Then there exist x1, x2 ∈[0, d) such thatTn(x1) = Ξ, Tn(x2) =ξand the polynomial

Ten(x) =Tn(x1−x)−Ξ + 1

satisfies all conditions of Lemma 2.2. Hence, for 06|x|6 ^π_n, it follows that

(2.6) Ten(x)>cosnx.

Moreover,|x1−x2|> ^π_n since if|x1−x2|< ^π_n, then

Ten(x1−x2) =Tn(x2)−Ξ + 1 =−1<cosnx, which contradicts (2.6). Inequality (2.6) can be written as follows

(2.7) Tn(x1−x)>cosnx+α,

whereα= Ξ−1 =ξ+ 1.

Similarly, Lemma 2.2 gives

(2.8) Tn(x2−x)6cosn¡π n−x´

+α=−cosnx+α, 06|x|6 π n (here it is enough to consider−Tn).

Assume first thatα∈[−1,1] and 06|x|6 ^{arccos(−α)}_n . Then using (2.7) we get (2.9) Tn(x1−x)>cosnx+α>0.

Also using (2.8) for ^{arccos(−α)}_n 6x6 ^2π_n −^{arccos(−α)}_n , we have Tn

³ x2−

³ x−π

n

¢´6cosnx+α60 and therefore

(2.10)

¯¯

¯Tn

³ x2−

³ x−π

n

´´¯¯

¯>|cosnx+α|.

Further, let A1:=

³

x1−arccos(−α)

n , x1+arccos(−α) n

´ , A2:=

³ x2−

³π

n−arccos(−α) n

´ , x2+

³π

n−arccos(−α) n

´´

. Since|x1−x2|> ^π_n, thenA1∩A2=∅. Moreover,|A1|+|A2|= 2π/n6dand forγ >0

¯¯

¯ n

x∈[0, d) :|Tn(x)|> γ o¯¯

¯>

¯¯

¯ n

x∈A1∪A2:|Tn(x)|> γ o¯¯

¯

=

¯¯

¯ n

x∈A1:|Tn(x)|> γ o¯¯¯+

¯¯

¯ n

x∈A2:|Tn(x)|> γ o¯¯¯.

Then, making use of (2.9) and (2.10) and translation invariance of the distribution function, we have for γ >0

|{x∈[0, d) :|Tn(x)|> γ}|>

¯¯

¯¯ n

x∈³

−arccos(−α)

n ,arccos(−α) n

´

:|cosnx+α|> γo¯¯

¯¯

+

¯¯

¯¯ n

x∈³arccos(−α) n ,2π

n −arccos(−α) n

´

:|cosnx+α|> γo¯¯¯

¯

=

¯¯

¯¯ n

x∈

³ 0,2π

n

´

:|cosnx+α|> γo¯¯

¯¯.

Therefore, forα∈[−1,1] we get the required inequality (2.5) for 06β < ^2π_n. Let nowα >1 and therefore ξ >0. Applying (2.5) toTn−ξ, it follows that

(T_n−ξ)^?(β)>(cosnx+ 1)^?(β), 06β < 2π n . Taking into account that

T_n^?(β) = (Tn−ξ)^?(β) +ξ and

(cosnx+α)^?(β) = (cosnx+ξ+ 1)^?(β) = (cosnx+ 1)^?(β) +ξ,

we get (2.5) forα >1. Forα <−1 it is enough to consider −Tn. ¤ Lemma 2.4. Let n∈Nand06β < ^2π_n. Then

(a)

sup

06α61

1 +α

(cosnx+α)^?(β) 6 1 + sin^2nβ₄ 1−sin^2nβ₄ ; (b)

06α<∞inf (cosnx+α)^?(β)>cos²nβ 4 ;

(c) µ

cosnx+1−cos^nβ₂ 2

¶_?

(β) = cos²nβ 4 .

Proof. Let cos^nβ₂ +α>1−α, i.e.,α> ^(1−cos₂ ^{nβ2 )} (see Figure 1 showing the graph of the functionf(x) =

|cosnx+α|).

Figure 1.

Then

(2.11) (cosnx+α)^?(β) = cosnβ

2 +α.

Note that the function

ϕ(α) = 1 +α cos^nβ₂ +α is monotone with respect toαand therefore

(2.12) sup

06α61

1 +α

(cosnx+α)^?(β)= sup

06α6(1−cos^nβ₂ )/2

1 +α (cosnx+α)^?(β). Next let 06α6 ^1−cos₂ ^nβ2 , then

(cosnx+α)^?(β) = cos^nu₂ + cos^nv₂

2 ,

where 06u6v6β satisfy the following equations (2.13)





u+v=β

cos^nu₂ −α= cos^nv₂ +α (see Figure 2).

Figure 2.

Hence,

1 +α

(cosnx+α)^?(β) = 2 + cos^nu₂ −cos^n(β−u)₂ cos^nu₂ + cos^n(β−u)₂ .

Let us find the maximum of the right-hand side foru∈[0, β/2]. Take f(z) = 2 + cosz−cos(a−z)

cosz+ cos(a−z) , z∈[0, a/2), where 06a < π. Then

f⁰(z) =−2 sina−2 sin(a−z) + 2 sinz

(cosz+ cos(a−z))² <0, z∈[0, a/2).

Therefore,f(z) is non-increasing on [0,^a₂) and has a maximum atz= 0. Then, by (2.12), sup

06α61

1 +α

(cosnx+α)^?(β) = max

06u6^β₂

2 + cos^nu₂ −cos^n(β−u)₂ cos^nu₂ + cos^n(β−u)₂

=3−cos^nβ₂

1 + cos^nβ₂ =1 + sin^2nβ₄ 1−sin^2nβ₄ , i.e., (a) is verified.

To prove (b), letα >1 and then

(cosnx+α)^?= (cosnx+ 1)^?+α−1>(cosnx+ 1)^?.

Therefore, in (b), infimum is attained for 06α61. Further we remark that forα > ^1−cos₂ ^nβ2 we have (cosnx+α)^?(β) = cosnβ

2 +α, which is the non-decreasing function with respect toα. Then

06α<∞inf (cosnx+α)^?(β)> inf

06α6(1−cos^nβ₂ )/2(cosnx+α)^?(β).

Next, for 06α6(1−cos^nβ₂ )/2, we have

(cosnx+α)^?(β) = cos^nu₂ + cos^nv₂

2 , 06u6v6β, whereu, v, andαare related by (2.13). Taking into account that

min

06u6^β₂

cos^n(β−u)₂ + cos^nu₂

2 = cos²nβ

4 , we get (b).

To prove (c), we just use (2.11):

(2.14) (cosnx+α)^?(β) = cosnβ

2 +(1−cos^nβ₂ )

2 = cos²nβ

4 . ¤

3. Sharp Remez inequalities

First, we note that the classical Remez inequality can be written as an inequality in terms of rearrange- ments (see also [DP]).

Remark 3.1. Let Ω be a measurable subset of R, f ∈ L∞(Ω) and a right continuous function C(·) be nondecreasing. Then the following statements are equivalent:

(3.1) f^∗(0)6C(β)f^∗(β), β >0;

(3.2) kfkL∞(Ω)6C(|B|)kfkL∞(Ω\B) for any measurable set B ⊂Ω;

(3.3) if

¯¯

¯{x∈Ω :|f(x)|>1}

¯¯

¯< β, then kfkL∞(Ω)6C(β).

Proof. Let us verify that (3.1)⇒(3.2)⇒ (3.3)⇒(3.1). First assume that (3.1) holds. LetB be a subset of Ω such that|B|=β. Note that for any ε >0

A=

¯¯

¯ n

x∈Ω : |f(x)|>f^∗(β+ε) o¯¯

¯>(β+ε).

Then ¯

¯¯ n

x∈Ω : |f(x)|>f^∗(β+ε)o

\B

¯¯

¯>0 and

kfk_L∞(Ω\B)>kfk_L∞(A\B)>f^∗(β+ε).

Using (3.1), we get

kfkL∞(Ω)=f^∗(0)6C(β+ε)f^∗(β+ε)6C(β+ε)kfkL∞(Ω\B), and therefore (3.2) holds.

Let now (3.2) hold. Denoting B :={x∈Ω :|f(x)| >1} and assuming|B|< β, we have |f(x)|61 for anyx∈Ω\B. Then (3.2) gives

kfkL∞(Ω)6C¡

|B|¢

kfkL∞(Ω\B)6C¡

|B|¢ 6C¡

β¢ . Assuming (3.3), considerβ >0 andψ(x) = _f^f(x)∗(β), we get

¯¯

¯{x∈Ω :|ψ(x)|>1}

¯¯

¯=

¯¯

¯{x∈Ω :|f(x)|> f^∗(β)}

¯¯

¯6β < β+ε.

By (3.3) we havekψk_L∞(Ω)6C(β+ε) for anyε >0. SinceC(·) is right continuous,kψk_L∞(Ω)6C(β), i.e.,

(3.1) follows. ¤

Thus, to investigate the Remez inequality, one can use its rearrangement form T_n^∗(0)6C(n, β)T_n^∗(β), β >0.

(3.4)

Note that dealing with polynomials with the minimal periods less than 2πallows us to decrease the constant C(n, β). Indeed, ifd= 2π/mis a period ofT_n, then by Lemma 2.1,T_n(x/m) is a polynomial of degreen/m and

T_mⁿ(x) :=Tn(x/m), T^∗n

m(β) =T_n^∗(β), β >0.

Then inequality (3.4) is true with the constantC(_mⁿ, β).

To overcome this, we are going to study Remez’s inequality with the?-rearrangement, i.e., (3.5) T_n^?(0)6 C(n, β)T_n^?(β), β >0.

Let us now recall that Nn and Mn are the sets of complex and real polynomials of degree at mostn, respectively. The next remark shows that to study the Remez inequality in the form (3.4) or in the form (3.5), it suffices to deal with real polynomials.

Remark 3.2. Let 06β <2π.Then

(3.6) sup

Tn∈Nn

T_n^∗(0)

T_n^∗(β) = sup

Tn∈Mn

T_n^∗(0) T_n^∗(β), and

(3.7) sup

Tn∈Nn

T_n^?(0)

T_n^?(β) = sup

Tn∈Mn

T_n^?(0) T_n^?(β). Proof. It is clear that

sup

Tn∈Nn

T_n^∗(0)

T_n^∗(β) > sup

Tn∈Mn

T_n^∗(0) T_n^∗(β). Let nowTn∈Nn and|Tn(x0)|= maxx∈T|Tn(x)|.Supposeα= _|T^Tn(x0)

n(x0)|,thenRe(¯αTn)∈Mn and maxx∈T |Re(¯αTn) (x)|=|Tn(x0)|=T_n^∗(0).

Further, since|Re(¯αTn(x))|6|Tn(x)|, we get T_n^∗(0)

T_n^∗(β) = (Re(¯αTn))^∗(0)

T_n^∗(β) 6 (Re(¯αTn))^∗(0)

(Re(¯αTn))^∗(β) 6 sup

Tn∈Mn

T_n^∗(0) T_n^∗(β)

and (3.6) follows. Similarly, we can show (3.7). ¤

3.1. Sharp Remez inequalities in terms of rearrangements. The main result of this section is the following sharp Remez type inequalities.

Theorem 3.3. Let n, k∈Nand06β < ^2π_n. Then

(3.8) sup

Tn∈Mn

T_n^?(0)

T_n^?(β)= 1 + 2 tan²nβ 4 and

(3.9) sup

Tn∈Mn

kn^−kTn^(k)kL_∞

T_n^?(β) = 1 + tan²nβ 4 . Proof. To prove (3.8), we define two subsets ofMn:

(3.10) M⁰_n=

n

Tn∈Mn: max

x Tn(x)−min

x Tn(x) = 2 o

and

(3.11) M¹_n =n

Tn∈M⁰_n : 16max

x Tn(x)62o .

Note that for anyTn 6=constthere existsγ >0 such thatγTn ∈M⁰_n. Then it is clear that sup

Tn∈Mn

T_n^?(0)

T_n^?(β) = sup

Tn∈M⁰_n

T_n^?(0) T_n^?(β). Let us now show that

sup

Tn∈M⁰_n

T_n^?(0)

T_n^?(β) = sup

Tn∈M¹_n

T_n^?(0) T_n^?(β).

Suppose that Tn ∈M⁰_n but Tn ∈/ M¹_n and−Tn ∈/ M¹_n. Denoting Ξ := maxTn(x), note that Ξ>2. Indeed, assume 06Ξ<1, then it is clear −Tn ∈M¹_n, which contradicts the choice of Tn. The case Ξ<0 can be reduced to the case Ξ>0 by taking−Tn.

Therefore, Ξ>2 andξ:= minTn(x)>0. Then forPn(x) :=Tn(x)−ξ>0 we have P_n^?(β) =T_n^?(β)−ξ, β>0.

On the other hand,Pn∈M¹_n, T_n^?(0)

T_n^?(β) = (T_n^?(0)−ξ)^?+ξ

(Tn(β)−ξ)^?+ξ = P_n^?(0) +ξ

P_n^?(β) +ξ 6 P_n^?(0) P_n^?(β), because ofP_n^?(0)>P_n^?(β) andξ >0.Hence, we get

sup

Tn∈Mn

T_n^?(0)

T_n^?(β) = sup

Tn∈M¹_n

T_n^?(0) T_n^?(β). Therefore, it suffices to verify (3.8) on the classM¹_n.

Let furtherTn∈M¹_n,then by Lemma 2.3,

T_n^?(β)>(cosnx+α)^?(β), where 06α= Ξ−161.Then

T^?(0)

T^?(β) 6 Ξ

(cosnx+α)^?(β) = 1 +α (cosnx+α)^?(β) and therefore,

sup

Tn∈M¹_n

T^?(0)

T^?(β)6 sup

06α61

1 +α (cosnx+α)^?(β).

Making use of Lemma 2.4(a), we get sup

Tn∈Mn

T_n^?(0)

T_n^?(β)= sup

Tn∈M¹_n

T^?(0)

T^?(β) 6 1 + sin^2βn₄

1−sin^2βn₄ = 1 + 2 tan²βn 4 , where 06β < ^2π_n.The constant in (3.8) is sharp, take

(3.12) Tn(x) = cosnx+1−cos^nβ₂

2 and use Lemma 2.4(c).

Next, let us show the accuracy of (3.9). Remark that sup

Tn∈Mn

kn^−kTn^(k)kL∞

T_n^?(β) = sup

Tn∈M¹_n

kn^−kTn^(k)kL∞

T_n^?(β) .

LetTn∈M¹_n, then 16Ξ62, maxx|Tn(x)−(Ξ−1)|= 1. Using Bernstein’s inequality, we have sup

Tn∈M¹_n

kn^−kTn^(k)kL∞

T_n^?(β) = sup

Tn∈M¹_n

kn^−k¡

Tn−(Ξ−1)¢_(k) kL∞

T_n^?(β) 6 sup

Tn∈M¹_n

1 T_n^?(β). By Lemma 2.3,

sup

Tn∈M¹_n

kn^−kTn^(k)kL∞

T_n^?(β) 6 sup

Tn∈M¹_n

³ 1

cosnx+ Ξ−1´?

(β)

= 1

inf06c<∞

³

cosnx+c´?

(β) .

Also, taking into account Lemma 2.4 (b), we get sup

Tn∈Mn

kn^−kTn^(k)kL∞

T_n^?(β) = sup

Tn∈M¹_n

kn^−kTn^(k)kL∞

T_n^?(β) 6 1 cos^2nβ₄ .

Sharpness in (3.9) can be verified by considering the function (3.12); see Lemma 2.4(c). ¤ Note that first applying Bernstein’s inequality and then Remez’s inequality (3.8) does not give the sharp constant in the Remez-Bernstein inequality (3.9).

Corollary 3.4. Let n, k∈N,Tn ∈Mn,2π/mbe the minimal period ofTn, and06β < ^2πm_n .Then

(3.13) T_n^∗(0)6

³

1 + 2 tan² nβ 4m

´ T_n^∗(β) and

(3.14) kn^−kT_n^(k)kL∞ 6

³

1 + tan² nβ 4m

´ T_n^∗(β).

Inequalities (3.13) and (3.14) are sharp; they become equalities forTn(x) = cosnx+¹₂³

1−cos^β₂´ .

Proof. Let 06β < ^2πm_n andk∈N.Then Theorem 3.3, where (3.8) and (3.9) are applied at the pointβ/m, and Lemma 2.1(e) give

T_n^∗(0) =T_n^?(0)6 ³

1 + 2 tan² nβ 4m

´ T_n^?¡

β/m¢

= ³

1 + 2 tan² nβ 4m

´ T_n^∗(β) and

kn^−kT_n^(k)kL∞ 6 ³

1 + tan² nβ 4m

´

T_n^?(β/m) = ³

1 + tan² nβ 4m

´

T_n^∗(β). ¤

The next result provides the asymptotics of the sharp constantC(n, β) in the Remez inequality.

Corollary 3.5. We have

(3.15) C(n, β) = 1 +(nβ)²

8 +O(β⁴) and

(3.16) C(1, β) = 1 + 2 tan²β

4.

Proof. It follows from (1.4) and (1.7) that An(β) := 1

2

³ tan²ⁿ

³2π−β 8

´

+ cot²ⁿ

³2π−β 8

´´

6C(n, β)61 + 2 tan²nβ

4 , 0< β <2π n . Letx=β/8. Then

An(x)−1 = 1 2

h ¡cot²ⁿ(π/4−x)−1¢ +¡

tan²ⁿ(π/4−x)−1¢ i

= 1 2

"Ã

cos(π/2−2x) sin²(π/4−x)

n−1X

k=0

cot^2k(π/4−x)

!

−

Ãcos(π/2−2x) cos²(π/4−x)

n−1X

k=0

tan^2k(π/4−x)

!#

= sin 2x 2

"

1 sin²(π/4−x)

n−1X

k=0

¡cot^2k(π/4−x)−1¢

− 1

cos²(π/4−x)

n−1X

k=0

¡tan^2k(π/4−x)−1¢

+ n

sin²(π/4−x)− n cos²(π/4−x)

¸

= sin 2x 2

"

1 sin²(π/4−x)

n−1X

k=0

cos(π/2−2x) sin²(π/4−x)

k−1X

m=0

cot^2m(π/4−x)

+ 1

cos²(π/4−x)

n−1X

k=0

cos(π/2−2x) cos²(π/4−x)

k−1X

m=0

tan^2m(π/4−x) + ncos(π/2−2x) sin²(π/4−x) cos²(π/4−x)

#

= sin²2x 2

"

1 sin²(π/4−x)

n−1X

k=0

1 sin²(π/4−x)

k−1X

m=0

cot^2m(π/4−x)

+ 1

cos²(π/4−x)

n−1X

k=0

1 cos²(π/4−x)

k−1X

m=0

tan^2m(π/4−x) + n

sin²(π/4−x) cos²(π/4−x)

# .

SinceAn(x) is even, we get An(x)−1 = sin²2x

2(^√₂²)⁴ Ã

2

n−1X

k=1

k + n+O(x²)

!

= 2 sin²β 4

³

n²+O(β²)

´

= (nβ)²

8 +O(β⁴) or

An(β) = 1 + (nβ)²

8 +O(β⁴).

On the other hand,

1 + 2 tan²nβ

4 = 1 + (nβ)²

8 +O(β⁴), which implies (3.15).

Further, (3.16) follows from (n= 1 andx=β/8) An(β)−1 = 1

2

h ¡cot²(π/4−x)−1¢ +¡

tan²(π/4−x)−1¢ i

=sin²2x 2

· 1

sin²(π/4−x) cos²(π/4−x)

¸

= 2 tan²2x= 2 tan²β

4. ¤

Remark 3.6. Corollary 3.4 yields the following Remez inequality with the constant written in the expo- nential form. Ifd= 2π/mis the minimal period of Tn ∈Mn, then

kTnkL∞[0,2π)6exp

³(nβ)² 2m²

´

kTnkL∞([0,2π)\B), |B|=β < mπ n ; compare with the known bound exp

³

nβmin{2,(1 + 1.75β)}

´

, cf. (1.4). This follows from (3.13) and the inequality 1 + 2 tan²x/46exp¡_x2

2

¢for 0< x < π.

Next we provide the following result on the Remez inequality inLp (see also [Er2] and [EMN]).

Corollary 3.7. Let p∈(0,∞)and 2π/mbe a minimal period ofTn∈Mn. Then Z _2π

0

¯¯

¯Tn(t)

¯¯

¯^pdt6

· 1 +

µ

1 + 2 tan²n|B|

2m

¶_p¸ Z

T\B

¯¯

¯Tn(t)

¯¯

¯^pdt,

where|B|<^πm_n .

Proof. Using Corollary 3.4, by (3.13), we get Z

B

¯¯

¯Tn(x)

¯¯

¯^pdx6 Z _|B|

0

(T_n^∗(t))^pdt 6|B|

µ

1 + 2 tan²n|B|

2m

¶_p

(T_n^∗(2|B|))^p 6

µ

1 + 2 tan²n|B|

2m

¶_pZ _2|B|

|B|

(T_n^∗(t))^pdt 6

µ

1 + 2 tan²n|B|

2m

¶_pZ _2π

|B|

(T_n^∗(t))^pdt 6

µ

1 + 2 tan²n|B|

2m

¶_pZ

T\B

|Tn(x)|^pdx. ¤

3.2. Sharp Remez inequalities in terms of averages. Remez’s inequality allows us to control how fast polynomials of degree n change on the period. Another inequality that characterizes such properties of polynomials is the following one:

(3.17) kTnkL∞ 6C(n, β)Tn(β, M).

Let [a, b] be an interval ofTof the measure6d. We say thatQis a harmonic intervalif Q=

[r

i=0

³

[a, b] +id´

, r∈N

(see e.g. [Nu]). By Mh.int. we denote the collection of all harmonic intervals in T. Let also Mint. be the collection of all intervals inT.

Theorem 3.8. Let M be a net that contains Mh.int.. Let Tn ∈Mn and ^2π_m be the minimal period of Tn. Then

(3.18) kTnkL∞ 6 nβ

2msin_2m^nβ Tn(β, M), 0< β < πm n , and

(3.19) kTnkL∞ 6 nβ

2 sin^nβ₂ Tn(β, Mint.), 0< β <π n. Inequalities (3.18) and (3.19) are sharp; they become equalities forTn(x) = cosnx.

Remark 3.9. An analogue of (3.19) for a netM that containsMint. also holds, i.e., kTnkL∞ 6 nβ

2 sin^nβ₂ Tn(β, M), 0< β < π n. However, unlike (3.18) this inequality is not sharp.

Proof of Theorem 3.8. Since _T^kTnkL∞

n(β,M) is invariant under multiplication byγ∈R\ {0}, we get sup

Tn∈Mn

kTnkL∞

Tn(β, M) = sup

Tn∈M⁰_n

kTnkL∞

Tn(β, M).

Let then Tn ∈M⁰_n. Without loss of generality we assume that Ξ = maxx∈[−π,π)Tn(x) =Tn(0)>1. Then Lemma 2.2 implies

Tn(x)>cosnx+ (Ξ−1), x∈£

− π 2n, π

2n

¤.