A SHARP REMEZ INEQUALITY FOR TRIGONOMETRIC ...
A SHARP REMEZ INEQUALITY FOR TRIGONOMETRIC POLYNOMIALS E. NURSULTANOV AND S. TIKHONOV
Abstract. We obtain a sharp Remez inequality for the trigonometric polynomial Tn of degree n on [0, 2π): ³ ° ° ° nβ ´° ¡ ¢, °Tn ° °Tn ° 6 1 + 2 tan2 L∞ ([0,2π)) L∞ [0,2π)\B 4m where 2π is the minimal period of Tn and |B| = β < 2πm is a measurable subset of T. In particular, this m n gives the asymptotics of the sharp constant in the Remez inequality: C(n, β) = 1 + where (0.1)
(nβ)2 + O(β)4 , 8
° ° °Tn ° L ([0,2π)) C(n, β) := sup sup ° ° ∞ ¡ ¢. ° Tn ° |B|=β Tn L∞ [0,2π)\B
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 most n ∈ N, i.e., n o X (1.1) Nn = Tn : Tn (x) = ck eikx , ck ∈ C, x ∈ T . |k|6n
An important question in polynomial approximation is the following one. Let |B| = µ(B) be the linear Lebesgue measure of B. How large can kTn kL∞ ([0,2π)) be if ¯n o¯ ¯ ¯ ¯ x ∈ [0, 2π) : |Tn (x)| > 1 ¯ 6 ε holds for some 0 < ε 6 1? The well-known Remez inequality answers this question. For any Lebesgue measurable set B ⊂ T such that |B| ≡ β < π/2 we have (1.2)
kTn kL∞ ([0,2π)) 6 C(n, |B|)kTn kL∞ ([0,2π)\B) ,
Tn ∈ Nn .
We will write C(n, |B|) in place of C(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 with C(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 Also, another improvement was obtained in [Er2, Th. 3.4]: was sharpened as ¡ C(n, |B|) = exp(2n|B|). ¢ C(n, |B|) = exp n(|B| + 1.75|B|2 ) . In case of B ≡ [a, b] ⊂ [−π, π] a sharp Remez inequality is given by (see [Er1]) ³ 2π − β ´ ³ 2π − β ´´ 1³ (1.3) kTn kL∞ ([0,2π)) 6 tan2n + cot2n kTn kL∞ ([0,2π)\B) , |B| = β, Tn ∈ Nn 2 8 8 and equality in (1.3) holds if and only if ³ cos(x − (a + b)/2) − cos2 ((b − a)/4) ´ , C ∈ R, Tn (x) = C Tn sin2 ((b − a)/4) Date: July 6, 2011. 2000 Mathematics 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
2
E. NURSULTANOV AND S. TIKHONOV
where Tn (x) is the Chebyshev polynomial of degree n, i.e., p p ¢n ¡ ¢n ´ 1 ³¡ Tn (x) = x + x2 − 1 + x − x2 − 1 2 for every x ∈ R \ (−1, 1). Therefore, the best known bounds of the sharp constant in the Remez inequality are given by ³ 2π − β ´ ³ 2π − β ´´ ¡ ¢ 1³ (1.4) tan2n + cot2n 6 C(n, β) 6 exp nβ min{2, (1 + 1.75β)} , 2 8 8 for 0 < β 6 π/2. Moreover, Erd´elyi [Er2] and Ganzburg [Ga] conjectured that ³ 2π − β ´ ³ 2π − β ´´ 1³ (1.5) C(n, β) = tan2n + cot2n 2 8 8 for any β > 0. In case when the measure |B| = β is big, that is, when π/2 < β < 2π, it is known that ³ 17 ´2n C(n, β) 6 C(n, β) = 2π − β (see [Er1, Er2, Ga, Na]). Another upper bound of C(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 small β (see, e.g., [MT]). Precisely, we prove ³ nβ ´ 2πm (1.6) kTn kL∞ ([0,2π)) 6 1 + 2 tan2 kTn kL∞ ([0,2π)\B) , 0 0, ¯ e∈M |e| |e|>t
e
which is called the average function of f on the net M ([NT]). If supe∈M |e| = α and t > α, then f (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 function f defined on T belongs to the Lorentz space Lp,q = Lp,q (T) (see e.g. [BS]) if µZ 2π ³ ´q dt ¶1/q < ∞, 0 < p, q < ∞, kf kLp,q = t1/p f ∗ (t) t 0 and kf kLp,∞ =
sup t∈(0,2π)
t1/p f ∗ (t) < ∞,
0 < p 6 ∞.
A SHARP REMEZ INEQUALITY
3
The scale of net spaces is more general than the scale of Lorentz spaces. To define them, let 0 < p, q 6 ∞, M be a net in T, and a 2π-periodic function f be locally integrable. We say that f belongs to the Net space Np,q (M ) = Np,q (M, T) ([NA, NT]) if µZ kf kNp,q (M ) =
2π
(t1/p f (t, M ))q
0
dt t
¶1/q < ∞,
0 < p, q < ∞,
and kf kNp,∞ (M ) =
t1/p f (t, M ) < ∞,
sup
0 < p 6 ∞.
t∈(0,2π)
The net spaces Np,q (M ) are a natural generalization of the Lorentz spaces Lp,q , since if M is a collection of all compact sets in T, then Np,q (M ) = Lp,q (T) for 1 < p < ∞, 0 < q 6 ∞. Finally, multidimensional variants of Remez and Nikol’skii’s inequalities are obtained in Section 5. In particular, for trigonometric polynomials X X Tn (x) = ··· ck ei(k,x) , ck ∈ C, x ∈ Td , d > 1, |k1 |6n1
|kd |6nd
we obtain the following Remez inequality kTn kL∞ (Td ) 6
1 q Qd cos d |B| j=1
jnj 2
kTn kL∞ (Td \B) ,
0 6 |B| < Qd
πd
j=1
jnj
,
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 6 d 6 2π. Let Tn (t) be the decreasing rearrangement of Tn on its minimal period, i.e., ¯n o¯ ¯ ¯ λmin (σ, Tn ) = ¯ x ∈ [0, d) : |Tn (x)| > σ ¯
and
n o Tn? (t) = inf σ : λmin (σ, Tn ) 6 t .
Remark that this concept describes the order of polynomials more accurately than Tn∗ . For example, ¡
¢∗ ¡ ¢∗ β cos nx (β) = cos x (β) = cos , 4
0 6 β < 2π
but (2.1)
¡
¢? nβ cos nx (β) = cos , 4
06β
0. If d is the minimal period of f , then (a) (b) (c) (d) (e)
we have f ? (β) 6 f ∗ (β) for 0 < β 6 d and f ? (β) < f ∗ (β) for d < 2π; there exists m ∈ N such that d = 2π/m; we have n/m ∈ N and g(x) = f (x/m) is a trigonometric polynomial of degree n/m; f ∗ (β) = g ∗ (β), β ∈ [0, 2π); f ? (β) = f ∗ (mβ) for 0 6 β 6 d.
Proof. Item (a) follows from λmin (σ, f ) 6 λ(σ, f ) and λmin (σ, f ) < λ(σ, f ) for d < 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; then f (x) = f (x + rd), x ∈ R. Using 2π-periodicity, we get f (x) = f (x + 2π) = f (x + α + rd), x ∈ R. Then f (x) = f (x + α), which contradicts α < d.
4
E. NURSULTANOV AND S. TIKHONOV
(c) Let d = 2π/m be the minimal period of f . Putting g(x) = f (x/m), x ∈ [0, 2π), we have Z 2π m−1 m−1 X Z 2π(k+1)/m X Z 2π/m −irx −irx cr (f ) = f (x)e dx = f (x)e dx = f (x)e−ir(x+2πk/m) dx 0
2πk/m
k=0
Z (2.2)
2π/m
=
f (x)e−irx dx
µ m−1 X
0 m−1 P
k=0
0
¶
e−2πikr/m) .
k=0
½
m, for r = ms, s ∈ N, , we get cr (f ) = 0 for r 6= ms. Since 0, otherwise k=0 |cn (f )| + |c−n (f )| > 0, then n/m ∈ N. Therefore, using (2.2) we get Z 2π/m Z 2π cms (f ) = m f (x)e−imsx dx = f (x/m)e−isx dx = cs (g),
Taking into account
e−2πikr/m =
0
0
and g is the trigonometric polynomial of degree n/m. (d) Making use of 2π/m-periodicity, we get ¯n ¯n o¯ o¯ ¯ ¯ ¯ ¯ λ(α, f ) = ¯ x ∈ [0, 2π] : |f (x)| > α ¯ = m ¯ x ∈ [0, 2π/m] : |f (x)| > α ¯ ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ = ¯ x ∈ [0, 2π] : |f (x/m)| > α ¯ = ¯ x ∈ [0, 2π] : |g(x)| > α ¯ = λ(α, g). Then g ∗ (β) = f ∗ (β) for any β ∈ [0, 2π]. (e) We proved that d = 2π m and λmin (σ, f ) =
1 m λ(σ, f ).
Hence, for 0 < β < d =
2π m,
f ? (β) = inf{σ : λmin (σ, f ) 6 β} = inf{σ : λ(σ, f ) 6 mβ} = f ∗ (mβ).
¤
We denote by Mn the set of real trigonometric polynomials, i.e., n o (2.3) Mn = Tn ∈ Nn : Tn (x) ∈ R, x ∈ T . 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. For Tn ∈ Mn such that |Tn (x)| 6 1 and Tn (0) = 1, we have π π (2.4) Tn (x) > cos nx, − 6 x 6 . n n Lemma 2.3. Let Tn ∈ Mn be such that max Tn (x) − min Tn (x) = 2. Then for 0 6 β
(cos nx + α) (β),
(2.5) where α = max Tn (x) − 1 . Proof. Denote
Ξ := max Tn (x) and ξ := min Tn (x). Let d be the minimal period of Tn . Then there exist x1 , x2 ∈ [0, d) such that Tn (x1 ) = Ξ, Tn (x2 ) = ξ and the polynomial Ten (x) = Tn (x1 − x) − Ξ + 1 satisfies all conditions of Lemma 2.2. Hence, for 0 6 |x| 6 nπ , it follows that Ten (x) > cos nx.
(2.6) Moreover, |x1 − x2 | >
π n
since if |x1 − x2 |
cos nx + α,
A SHARP REMEZ INEQUALITY
Similarly, Lemma 2.2 gives (2.8)
Tn (x2 − x) 6 cos n
´ ¡π − x + α = − cos nx + α, n
(here it is enough to consider −Tn ). Assume first that α ∈ [−1, 1] and 0 6 |x| 6 (2.9) Also using (2.8) for
5
arccos(−α) . n
0 6 |x| 6
π n
Then using (2.7) we get
Tn (x1 − x) > cos nx + α > 0. arccos(−α) n
6x6
2π n
³
Tn and therefore
−
arccos(−α) , n
we have ³ ´ π¢ x2 − x − 6 cos nx + α 6 0 n
¯ ³ ³ π ´´¯¯ ¯ ¯Tn x2 − x − ¯ > | cos nx + α|. n
(2.10)
Further, let ³ ³ ³ π arccos(−α) ´ ³ π arccos(−α) ´´ arccos(−α) arccos(−α) ´ A1 := x1 − , x1 + , A2 := x2 − − , x2 + − . n n n n n n Since |x1 − x2 | > nπ , then A1 ∩ A2 = ∅. Moreover, |A1 | + |A2 | = 2π/n 6 d and for γ > 0 ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ ¯ x ∈ [0, d) : |Tn (x)| > γ ¯ > ¯ x ∈ A1 ∪ A2 : |Tn (x)| > γ ¯ ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ = ¯ x ∈ A1 : |Tn (x)| > γ ¯ + ¯ x ∈ A2 : |Tn (x)| > γ ¯ . Then, making use of (2.9) and (2.10) and translation invariance of the distribution function, we have for γ>0 ¯n o¯¯ ³ arccos(−α) arccos(−α) ´ ¯ ¯ , : | cos nx + α| > γ ¯¯ |{x ∈ [0, d) : |Tn (x)| > γ}| > ¯ x ∈ − n n ¯n o¯¯ ³ arccos(−α) 2π arccos(−α) ´ ¯ ¯ +¯ x∈ , − : | cos nx + α| > γ ¯¯ n n n ¯n o¯¯ ³ 2π ´ ¯ ¯ : | cos nx + α| > γ ¯¯ . = ¯ x ∈ 0, n Therefore, for α ∈ [−1, 1] we get the required inequality (2.5) for 0 6 β < 2π n . Let now α > 1 and therefore ξ > 0. Applying (2.5) to Tn − ξ, it follows that ?
?
(Tn − ξ) (β) > (cos nx + 1) (β), Taking into account that
06β
1. For α < −1 it is enough to consider −Tn . Lemma 2.4. Let n ∈ N and 0 6 β < (a)
2π n .
Then
1 + sin2 1+α 6 ? 1 − sin2 06α61 (cos nx + α) (β) sup
nβ 4 nβ 4
;
(b) inf (cos nx + α)? (β) > cos2
06α 1 − α, i.e., α > | cos nx + α|).
(1−cos 2
nβ 2 )
(see Figure 1 showing the graph of the function f (x) =
Figure 1. Then (cos nx + α)? (β) = cos
(2.11) Note that the function
ϕ(α) =
nβ + α. 2
1+α cos nβ 2 +α
is monotone with respect to α and therefore 1+α 1+α (2.12) sup = sup . ? (cos nx + α)? (β) nβ 06α61 (cos nx + α) (β) 06α6(1−cos )/2 2
Next let 0 6 α 6
1−cos 2
nβ 2
, then (cos nx + α)? (β) =
nv cos nu 2 + cos 2 , 2
where 0 6 u 6 v 6 β satisfy the following equations u+v =β (2.13) nv cos nu 2 − α = cos 2 + α (see Figure 2).
Figure 2. Hence, n(β−u) 2 + cos nu 1+α 2 − cos 2 = . n(β−u) (cos nx + α)? (β) cos nu + cos 2 2
A SHARP REMEZ INEQUALITY
7
Let us find the maximum of the right-hand side for u ∈ [0, β/2]. Take f (z) =
2 + cos z − cos(a − z) , cos z + cos(a − z)
z ∈ [0, a/2),
where 0 6 a < π. Then f 0 (z) =
−2 sin a − 2 sin(a − z) + 2 sin z 2
(cos z + cos(a − z))
< 0,
z ∈ [0, a/2).
Therefore, f (z) is non-increasing on [0, a2 ) and has a maximum at z = 0. Then, by (2.12), n(β−u) 2 + cos nu 1+α 2 − cos 2 = max ? n(β−u) nu 06u6 β 06α61 (cos nx + α) (β) cos 2 + cos 2 2
sup
=
3 − cos nβ 2 1 + cos nβ 2
=
nβ 4 sin2 nβ 4
1 + sin2 1−
,
i.e., (a) is verified. To prove (b), let α > 1 and then (cos nx + α)? = (cos nx + 1)? + α − 1 > (cos nx + 1)? . Therefore, in (b), infimum is attained for 0 6 α 6 1. Further we remark that for α >
1−cos 2
nβ 2
we have
nβ + α, 2 which is the non-decreasing function with respect to α. Then (cos nx + α)? (β) = cos
inf
06α
inf 06α6(1−cos
nβ 2 )/2
(cos nx + α)? (β).
Next, for 0 6 α 6 (1 − cos nβ 2 )/2, we have nv cos nu 2 + cos 2 , 0 6 u 6 v 6 β, 2 where u, v, and α are related by (2.13). Taking into account that
(cos nx + α)? (β) =
min
06u6 β 2
cos n(β−u) + cos nu nβ 2 2 = cos2 , 2 4
we get (b). To prove (c), we just use (2.11): (2.14)
(cos nx + α)? (β) = cos
(1 − cos nβ nβ nβ 2 ) + = cos2 . 2 2 4
¤
3. Sharp Remez inequalities First, we note that the classical Remez inequality can be written as an inequality in terms of rearrangements (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) (3.2) (3.3)
f ∗ (0) 6 C(β)f ∗ (β), kf kL∞ (Ω) 6 C(|B|)kf kL∞ (Ω\B)
β > 0;
for any measurable set
¯ ¯ ¯ ¯ if ¯{x ∈ Ω : |f (x)| > 1}¯ < β,
B ⊂ Ω;
then kf kL∞ (Ω) 6 C(β).
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E. NURSULTANOV AND S. TIKHONOV
Proof. Let us verify that (3.1) ⇒ (3.2) ⇒ (3.3) ⇒ (3.1). First assume that (3.1) holds. Let B be a subset of Ω such that |B| = β. Note that for any ε > 0 ¯n o¯ ¯ ¯ A = ¯ x ∈ Ω : |f (x)| > f ∗ (β + ε) ¯ > (β + ε). Then
¯n ¯ o ¯ ¯ ¯ x ∈ Ω : |f (x)| > f ∗ (β + ε) \ B ¯ > 0
and kf kL∞ (Ω\B) > kf kL∞ (A\B) > f ∗ (β + ε). Using (3.1), we get kf kL∞ (Ω) = f ∗ (0) 6 C(β + ε)f ∗ (β + ε) 6 C(β + ε)kf kL∞ (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)| 6 1 for any x ∈ Ω \ B. Then (3.2) gives ¡ ¢ ¡ ¢ ¡ ¢ kf kL∞ (Ω) 6 C |B| kf kL∞ (Ω\B) 6 C |B| 6 C β . Assuming (3.3), consider β > 0 and ψ(x) = ff∗(x) (β) , we get ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯{x ∈ Ω : |ψ(x)| > 1}¯ = ¯{x ∈ Ω : |f (x)| > f ∗ (β)}¯ 6 β < β + ε. By (3.3) we have kψkL∞ (Ω) 6 C(β + ε) for any ε > 0. Since C(·) is right continuous, kψkL∞ (Ω) 6 C(β), i.e., (3.1) follows. ¤ Thus, to investigate the Remez inequality, one can use its rearrangement form (3.4)
Tn∗ (0) 6 C(n, β) Tn∗ (β),
β > 0.
Note that dealing with polynomials with the minimal periods less than 2π allows us to decrease the constant C(n, β). Indeed, if d = 2π/m is a period of Tn , then by Lemma 2.1, Tn (x/m) is a polynomial of degree n/m and ∗ ∗ n (x) := T (x/m), n (β) = T (β), β > 0. Tm Tm n n n Then inequality (3.4) is true with the constant C( m , β). To overcome this, we are going to study Remez’s inequality with the ?-rearrangement, i.e.,
(3.5)
Tn? (0) 6 C(n, β) Tn? (β),
β > 0.
Let us now recall that Nn and Mn are the sets of complex and real polynomials of degree at most n, 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 0 6 β < 2π. Then (3.6)
sup Tn ∈Nn
Tn∗ (0) Tn∗ (0) = sup , ∗ Tn (β) Tn ∈Mn Tn∗ (β)
and (3.7)
sup Tn ∈Nn
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈Mn Tn? (β)
Proof. It is clear that sup Tn ∈Nn
Tn∗ (0) Tn∗ (0) > sup . ∗ Tn (β) Tn ∈Mn Tn∗ (β)
Let now Tn ∈ Nn and |Tn (x0 )| = maxx∈T |Tn (x)|. Suppose α =
Tn (x0 ) |Tn (x0 )| ,
then Re (¯ αTn ) ∈ Mn and
max |Re (¯ αTn ) (x)| = |Tn (x0 )| = Tn∗ (0). x∈T
A SHARP REMEZ INEQUALITY
9
Further, since |Re (¯ αTn (x)) | 6 |Tn (x)|, we get ∗
∗
Tn∗ (0) (Re (¯ αTn )) (0) (Re (¯ αTn )) (0) Tn∗ (0) = 6 6 sup ∗ Tn∗ (β) Tn∗ (β) (Re (¯ αTn )) (β) Tn ∈Mn Tn∗ (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 ∈ N and 0 6 β < (3.8)
2π n .
sup Tn ∈Mn
Then nβ Tn? (0) = 1 + 2 tan2 Tn? (β) 4
and (k)
(3.9)
sup Tn ∈Mn
kn−k Tn kL∞ nβ = 1 + tan2 . Tn? (β) 4
Proof. To prove (3.8), we define two subsets of Mn : n o (3.10) M0n = Tn ∈ Mn : max Tn (x) − min Tn (x) = 2 x
and
x
n o M1n = Tn ∈ M0n : 1 6 max Tn (x) 6 2 .
(3.11)
x
Note that for any Tn 6= const there exists γ > 0 such that γTn ∈ M0n . Then it is clear that sup Tn ∈Mn
Let us now show that sup Tn ∈M0n
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M0n Tn? (β) Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M1n Tn? (β)
Suppose that Tn ∈ M0n but Tn ∈ / M1n and −Tn ∈ / M1n . Denoting Ξ := max Tn (x), note that Ξ > 2. Indeed, 1 assume 0 6 Ξ < 1, then it is clear −Tn ∈ Mn , which contradicts the choice of Tn . The case Ξ < 0 can be reduced to the case Ξ > 0 by taking −Tn . Therefore, Ξ > 2 and ξ := min Tn (x) > 0. Then for Pn (x) := Tn (x) − ξ > 0 we have Pn? (β) = Tn? (β) − ξ, On the other hand, Pn ∈
β > 0.
M1n , ?
Tn? (0) Pn? (0) + ξ (Tn? (0) − ξ) + ξ Pn? (0) = = 6 , ? Tn? (β) Pn? (β) + ξ Pn? (β) (Tn (β) − ξ) + ξ because of Pn? (0) > Pn? (β) and ξ > 0. Hence, we get sup Tn ∈Mn
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M1n Tn? (β)
Therefore, it suffices to verify (3.8) on the class M1n . Let further Tn ∈ M1n , then by Lemma 2.3, ?
Tn? (β) > (cos nx + α) (β), where 0 6 α = Ξ − 1 6 1. Then Ξ 1+α T ? (0) 6 = ? ? T (β) (cos nx + α) (β) (cos nx + α)? (β) and therefore, sup Tn ∈M1n
T ? (0) 1+α 6 sup . T ? (β) 06α61 (cos nx + α)? (β)
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E. NURSULTANOV AND S. TIKHONOV
Making use of Lemma 2.4(a), we get sup Tn ∈Mn
where 0 6 β
n n ¯ β β 2πi |e| ¯ e nβ i=0 − 2m + m
sup e∈Mh.int. |e|=β
and kTn kL∞
sup Tn ∈Mn
Tn (β, M )
kTn kL∞
6 sup Tn ∈M0n
6
Tn (β, Mh.int. )
6 sup Tn ∈M0n
kTn kL∞ ¯R ¯ 1 ¯ supe∈Mh.int. |e| T (y)dy ¯ e n |e|=β
Ξ
sup
nβ 2m
16Ξβ
|e|>β
i.e., inequality (3.18) is sharp. The proof of inequality (3.19) is similar to the proof above using (3.20)
sup Tn ∈M0n
kTn kL∞ Tn (β, Mint. )
6 sup Tn ∈M0n
kTn kL∞ nβ ¯R ¯6 . 1 ¯ ¯ supe∈Mint. |e| e Tn (y)dy 2 sin nβ 2
¤
|e|=β
The next theorem plays an important role in approximation theory (see Stechkin [St] and Nikol’skii [Ni] in case β = π/n). In particular, it generalizes Bernstein’s inequality kTn0 k∞ 6 nkTn k∞ . Corollary 3.10. Let n ∈ N. We have sup Tn ∈Mn
kTn0 kL∞ = kTn (x + β) − Tn (x)kL∞
µ
¶ n , 2 sin βn 2
0m t p (cos nx + Ξ − 1) (t) . t 0 = m1/p
q
t p (T ? (t))q
Using Lemma 2.1(e), we have µZ kTn kLp,q > m1/p n−1/p
2π
0
³ ´q dt ¶1/q q t p (cos x + Ξ − 1)∗ (t) . t
Then sup Tn ∈M0 n Ξ>1
m1/p kTn kL∞ Ξ 6 sup 1/p n kTn kLp,q Ξ>1 k cos(·) + Ξ − 1kLp,q
(4.7)
=
1+α . k cos(·) + αkLp,q 06α cos z + Ξ − 1 > 0, 0 6 z 6 for any (x1 , x2 ) ∈ Qz and then Finally,
π 2
³ ´ Tn∗ µ(Qz ) > cos z + Ξ − 1 > 0.
Ξ 1 Tn∗ (0) 6 sup = ∗ (µ(Q )) T cos z + Ξ − 1 cos z z Ξ>1 Tn ∈Mn (2) n sup
18
E. NURSULTANOV AND S. TIKHONOV
and since µ(Qz ) =
2z 2 n1 n2
= β, we arrive at (5.2).
¤
In the general case (d > 2) the proof is similar. Remark 5.2. Let d > 2, n ∈ Nd , and Tn ∈ Mn (d). Then Tn∗ (0) 6
(5.4) where 0 6 β
2 v u d ³ Y u ¡ ¢ ´ 1 jn j d t 6 exp 0.5 d n1 . . . nd |B| 1/d , q 6 exp |B| Qd jnj 2 cos d |B| j=1 j=1
2
d Y 2 . jn j j=1
|B|
2 and Tn ∈ Mn (d). Then ! 1 µ ¶1/p à Z π ³ ´q dt − q 2 kTn kL∞ (Td ) d d! 6 d t p cos t . (5.6) sup ³ ´1/p 2d t Tn ∈Mn Qd 0 kTn kLp,q (Td ) j=1 nj Proof. Making use of Theorem 5.1 and Remark 5.2 for Tn ∈ Mn , we get ! ÃZ (2π)d ³ ´q dt 1 ∗ kTn kLp,q (Td ) = t p Tn (t) t 0 v q1 q q1 u d πd Z Qd π d ³ Z ´ Q u Y jnj q dt d jnj jnj 1 1 j=1 j=1 d > dt Tn∗ (0) t p cos t t > t p Tn∗ (t) t 2 t 0 0 j=1 −1/p à !1 Z π2 ³ d ´q dt q Y d jn j Tn∗ (0), = d t p cos t 2 t 0 j=1
and we arrive at (5.6).
¤
In some cases the integral on the right-hand side of (5.6) can be simply calculated. For example, for d = 2, we have n1 n2 ¢ kTn kL1 , kTn k∞ 6 ¡ π 4 2 −1 √ 2 n1 n2 kTn k∞ 6 √ kTn kL2 , π2 − 4 √ n1 n2 kTn k∞ 6 3/2 kTn kL2,1 . 2 Similar to Remark 4.4 we can prove the following (Lp , Lp1 )-Nikol’skii inequality for Tn ∈ Mn (d), d > 2, µY ¶ p1 − p1 d 1 1−p/p1 kTn kLp (Td ) 6 C nj kTn kLp (Td ) , 0 < p < p1 < ∞, 1
j=1
where C is the constant in the right-hand side of (5.6) with p = q.
A SHARP REMEZ INEQUALITY
19
Finally, we note that Remez type inequalities hold not only for trigonometric polynomials but also for wider classes of functions. Denote by Eν,µ , ν, µ > 0 the collection of functions f ∈ L∞ (R2 ) such that ∂2f ∈ C(R2 ), α1 + α2 = 2, αi > 0 and α α ∂x1 1 ∂x2 2 ° ° ° ∂2f ° ° ° ° α ° 6 ν α1 µα2 °f °L∞ (R2 ) , α1 + α2 = 2, αi > 0. ° ∂x 1 ∂xα2 ° 1 2 L∞ (R2 ) This class in particular contains all functions of exponential type (ν, µ) with regards to x1 and x2 , respectively. Theorem 5.4. Let ν, µ > 0 and 0 6 β < (5.7)
4 νµ .
For any f ∈ Eν,µ we have µ ¶−1 f ∗ (0) (νµβ) sup ∗ 6 1− . 4 f ∈Eν,µ fν,µ (β)
Proof. Without loss of generality we assume that max |f (x, y)| = f (0, 0) = 1.
(x,y)∈R2
Then since (0, 0) is an extreme point, taking into account the definition of Eν,µ and Taylor’s formula, there exists (c1 , c2 ) ∈ (−|x|, |x|) × (−|y|, |y|) such that µ ¶ 1 ∂2f ∂2f ∂2f 1 2 2 f (x, y) = f (0, 0) + (c , c )x + 2 (c , c )xy + (c , c )y > 1 − (ν|x| + µ|y|)2 . 1 2 1 2 1 2 2 ∂x2 ∂x∂y ∂y 2 Suppose z = |x|ν + |y|µ, then using the same argument as in the proof of Theorem 5.1, √ z2 f (x, y) > 1 − > 0, 0 < z < 2 2 n o 4 for any point (x, y) belonging to Qz = (x, y) : |x|ν + |y|µ 6 z . Finally, for 0 6 β < νµ we have f ∗ (β) > 1 −
νµβ >0 4
and therefore (5.7) follows.
¤
References [An] V. Andrievskii, A note on a Remez-type inequality for trigonometric polynomials. J. Approx. Theory 116(2) (2002), 416–424. [BKP1] V. Babenko, V. Kofanov, S. Pichugov, Inequalities of Kolmogorov type and some their applications in approximation theory, Rendiconti del Circolo Matematico di Palermo Serie II, Suppl. 52 (1998), 223–237. [BKP2] V. Babenko, V. Kofanov, S. Pichugov, Comparison of rearrangements and Kolmogorov-Nagy type inequalities for periodic functions, Approximation Theory: A volume dedicated to Bl. Sendov (B. Bojanov, Ed.), DARBA, Sofia (2002), 24–53. [BS] C. Bennett, R. Sharpley, Interpolation of Operators, Academic Press, 1988. [Be] S. N. Bernstein, On one class of interpolation formulas (Russian), Izv. Ak. Nauk, 9 (1931), 1151–1161. [BE] P. Borwein, T. Erdelyi, Polynomials and polynomial inequalities, Springer, New York, 1995. [DP] Z. Ditzian, A. Prymak, Nikol’skii inequalities for Lorentz spaces, Rocky Mountain Journal of Mathematics, 40(1) (2010), 209–223. elyi, Remez-type inequalities on the size of generalized polynomials, J. London Math. Soc. (2) 45 (1992), 255–264. [Er1] T. Erd´ [Er2] T. Erd´ elyi, Remez-type inequalities and their applications, J. Comput. Appl. Math. 47 (1993), 167–210. elyi, George Lorentz and inequalities in approximation, St. Petersbg. Math. J. 21(3) (2010), 365–405. [Er3] T. Erd´ [EMN] T. Erd´ elyi, A. M´ at´ e, P. Nevai, Inequalities for generalized nonnegative polynomials, Constr. Approx. 8(2) (1992), 241-255. [Ga] M. I. Ganzburg, Polynomial inequalities on measurable sets and their applications, Constr. Approx. 17 (2001), 275–306. [Kr] A. Kro´ o, On Remez-type inequalities for polynomials in Rm and C m , Anal. Math. 27(1) (2001), 55–70. [LGM] G.G. Lorentz, M. von Golitschek, Y. Makovoz, Constructive approximation: advanced problems. Springer, 1996. [MT] G. Mastroianni, V. Totik, Weighted polynomial inequalities with doubling and A∞ weights, Constr. Approx. 16(1) (2000), 37–71. [Na] F. L. Nazarov, Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type, Algebra i Analiz 5(4) (1993), 3–66; translation in St. Petersburg Math. J. 5(4) (1994), 663–717. [Ni] S. M. Nikol’skii, A generalization of one inequality by S. N. Bernstein (Russian), Dokl. Akad. Nauk SSSR, 60 (1948), 1507–1510.
20
E. NURSULTANOV AND S. TIKHONOV
[Nu] E.D. Nursultanov, Net spaces and inequalities of Hardy-Littlewood type, Sb. Math. 189, No.3 (1998), 399–419; translation from Mat. Sb. 189(3) (1998), 83–102. [NA] E.D. Nursultanov, T.U. Aubakirov, Interpolation theorem for stochastic processes, Eurasian Math. J., 1(1) (2010), 8–16. [NT] E. Nursultanov, S. Tikhonov, Net spaces and boundedness of integral operators, J. of Geometric Analysis, 21(4) (2011), 950–981. [Re] E. Remes, Sur une propri´ et´ e extremale des polynˆ omes de Tchebychef (French), Commun. Inst. Sci. Math. et Mecan., Univ. Kharkoff et Soc. Math. Kharkoff, IV. Ser. 13 (1936), 93-95. Online: http://www.math.technion.ac.il/hat/fpapers/remezppr.pdf [St] S. B. Stechkin, A generalization of some inequalities by S. N. Bernstein (Russian), Dokl. Akad. Nauk SSSR, 60 (1948), 1511–1514. [SU] S. B. Stechkin, P. L. Ulyanov, Sequences of convergence for series, Trudy Mat. Inst. Steklov, 86 (1965), 3–83. English translation in Proc. Steklov Inst. Math., (1967), 1–85. E. Nursultanov L.N. Gumilyov Eurasian National University and Kazakh Branch of Moscow State University Munatpasova, 7 010010 Astana Kazakhstan E-mail address: [email protected] S. Tikhonov ` tica ICREA and Centre de Recerca Matema Facultat de Cincies, UAB 08193 Bellaterra Barcelona, Spain E-mail address: [email protected]
Abstract. We obtain a sharp Remez inequality for the trigonometric polynomial Tn of degree n on [0, 2π): ³ ° ° ° nβ ´° ¡ ¢, °Tn ° °Tn ° 6 1 + 2 tan2 L∞ ([0,2π)) L∞ [0,2π)\B 4m where 2π is the minimal period of Tn and |B| = β < 2πm is a measurable subset of T. In particular, this m n gives the asymptotics of the sharp constant in the Remez inequality: C(n, β) = 1 + where (0.1)
(nβ)2 + O(β)4 , 8
° ° °Tn ° L ([0,2π)) C(n, β) := sup sup ° ° ∞ ¡ ¢. ° Tn ° |B|=β Tn L∞ [0,2π)\B
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 most n ∈ N, i.e., n o X (1.1) Nn = Tn : Tn (x) = ck eikx , ck ∈ C, x ∈ T . |k|6n
An important question in polynomial approximation is the following one. Let |B| = µ(B) be the linear Lebesgue measure of B. How large can kTn kL∞ ([0,2π)) be if ¯n o¯ ¯ ¯ ¯ x ∈ [0, 2π) : |Tn (x)| > 1 ¯ 6 ε holds for some 0 < ε 6 1? The well-known Remez inequality answers this question. For any Lebesgue measurable set B ⊂ T such that |B| ≡ β < π/2 we have (1.2)
kTn kL∞ ([0,2π)) 6 C(n, |B|)kTn kL∞ ([0,2π)\B) ,
Tn ∈ Nn .
We will write C(n, |B|) in place of C(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 with C(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 Also, another improvement was obtained in [Er2, Th. 3.4]: was sharpened as ¡ C(n, |B|) = exp(2n|B|). ¢ C(n, |B|) = exp n(|B| + 1.75|B|2 ) . In case of B ≡ [a, b] ⊂ [−π, π] a sharp Remez inequality is given by (see [Er1]) ³ 2π − β ´ ³ 2π − β ´´ 1³ (1.3) kTn kL∞ ([0,2π)) 6 tan2n + cot2n kTn kL∞ ([0,2π)\B) , |B| = β, Tn ∈ Nn 2 8 8 and equality in (1.3) holds if and only if ³ cos(x − (a + b)/2) − cos2 ((b − a)/4) ´ , C ∈ R, Tn (x) = C Tn sin2 ((b − a)/4) Date: July 6, 2011. 2000 Mathematics 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
2
E. NURSULTANOV AND S. TIKHONOV
where Tn (x) is the Chebyshev polynomial of degree n, i.e., p p ¢n ¡ ¢n ´ 1 ³¡ Tn (x) = x + x2 − 1 + x − x2 − 1 2 for every x ∈ R \ (−1, 1). Therefore, the best known bounds of the sharp constant in the Remez inequality are given by ³ 2π − β ´ ³ 2π − β ´´ ¡ ¢ 1³ (1.4) tan2n + cot2n 6 C(n, β) 6 exp nβ min{2, (1 + 1.75β)} , 2 8 8 for 0 < β 6 π/2. Moreover, Erd´elyi [Er2] and Ganzburg [Ga] conjectured that ³ 2π − β ´ ³ 2π − β ´´ 1³ (1.5) C(n, β) = tan2n + cot2n 2 8 8 for any β > 0. In case when the measure |B| = β is big, that is, when π/2 < β < 2π, it is known that ³ 17 ´2n C(n, β) 6 C(n, β) = 2π − β (see [Er1, Er2, Ga, Na]). Another upper bound of C(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 small β (see, e.g., [MT]). Precisely, we prove ³ nβ ´ 2πm (1.6) kTn kL∞ ([0,2π)) 6 1 + 2 tan2 kTn kL∞ ([0,2π)\B) , 0 0, ¯ e∈M |e| |e|>t
e
which is called the average function of f on the net M ([NT]). If supe∈M |e| = α and t > α, then f (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 function f defined on T belongs to the Lorentz space Lp,q = Lp,q (T) (see e.g. [BS]) if µZ 2π ³ ´q dt ¶1/q < ∞, 0 < p, q < ∞, kf kLp,q = t1/p f ∗ (t) t 0 and kf kLp,∞ =
sup t∈(0,2π)
t1/p f ∗ (t) < ∞,
0 < p 6 ∞.
A SHARP REMEZ INEQUALITY
3
The scale of net spaces is more general than the scale of Lorentz spaces. To define them, let 0 < p, q 6 ∞, M be a net in T, and a 2π-periodic function f be locally integrable. We say that f belongs to the Net space Np,q (M ) = Np,q (M, T) ([NA, NT]) if µZ kf kNp,q (M ) =
2π
(t1/p f (t, M ))q
0
dt t
¶1/q < ∞,
0 < p, q < ∞,
and kf kNp,∞ (M ) =
t1/p f (t, M ) < ∞,
sup
0 < p 6 ∞.
t∈(0,2π)
The net spaces Np,q (M ) are a natural generalization of the Lorentz spaces Lp,q , since if M is a collection of all compact sets in T, then Np,q (M ) = Lp,q (T) for 1 < p < ∞, 0 < q 6 ∞. Finally, multidimensional variants of Remez and Nikol’skii’s inequalities are obtained in Section 5. In particular, for trigonometric polynomials X X Tn (x) = ··· ck ei(k,x) , ck ∈ C, x ∈ Td , d > 1, |k1 |6n1
|kd |6nd
we obtain the following Remez inequality kTn kL∞ (Td ) 6
1 q Qd cos d |B| j=1
jnj 2
kTn kL∞ (Td \B) ,
0 6 |B| < Qd
πd
j=1
jnj
,
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 6 d 6 2π. Let Tn (t) be the decreasing rearrangement of Tn on its minimal period, i.e., ¯n o¯ ¯ ¯ λmin (σ, Tn ) = ¯ x ∈ [0, d) : |Tn (x)| > σ ¯
and
n o Tn? (t) = inf σ : λmin (σ, Tn ) 6 t .
Remark that this concept describes the order of polynomials more accurately than Tn∗ . For example, ¡
¢∗ ¡ ¢∗ β cos nx (β) = cos x (β) = cos , 4
0 6 β < 2π
but (2.1)
¡
¢? nβ cos nx (β) = cos , 4
06β
0. If d is the minimal period of f , then (a) (b) (c) (d) (e)
we have f ? (β) 6 f ∗ (β) for 0 < β 6 d and f ? (β) < f ∗ (β) for d < 2π; there exists m ∈ N such that d = 2π/m; we have n/m ∈ N and g(x) = f (x/m) is a trigonometric polynomial of degree n/m; f ∗ (β) = g ∗ (β), β ∈ [0, 2π); f ? (β) = f ∗ (mβ) for 0 6 β 6 d.
Proof. Item (a) follows from λmin (σ, f ) 6 λ(σ, f ) and λmin (σ, f ) < λ(σ, f ) for d < 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; then f (x) = f (x + rd), x ∈ R. Using 2π-periodicity, we get f (x) = f (x + 2π) = f (x + α + rd), x ∈ R. Then f (x) = f (x + α), which contradicts α < d.
4
E. NURSULTANOV AND S. TIKHONOV
(c) Let d = 2π/m be the minimal period of f . Putting g(x) = f (x/m), x ∈ [0, 2π), we have Z 2π m−1 m−1 X Z 2π(k+1)/m X Z 2π/m −irx −irx cr (f ) = f (x)e dx = f (x)e dx = f (x)e−ir(x+2πk/m) dx 0
2πk/m
k=0
Z (2.2)
2π/m
=
f (x)e−irx dx
µ m−1 X
0 m−1 P
k=0
0
¶
e−2πikr/m) .
k=0
½
m, for r = ms, s ∈ N, , we get cr (f ) = 0 for r 6= ms. Since 0, otherwise k=0 |cn (f )| + |c−n (f )| > 0, then n/m ∈ N. Therefore, using (2.2) we get Z 2π/m Z 2π cms (f ) = m f (x)e−imsx dx = f (x/m)e−isx dx = cs (g),
Taking into account
e−2πikr/m =
0
0
and g is the trigonometric polynomial of degree n/m. (d) Making use of 2π/m-periodicity, we get ¯n ¯n o¯ o¯ ¯ ¯ ¯ ¯ λ(α, f ) = ¯ x ∈ [0, 2π] : |f (x)| > α ¯ = m ¯ x ∈ [0, 2π/m] : |f (x)| > α ¯ ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ = ¯ x ∈ [0, 2π] : |f (x/m)| > α ¯ = ¯ x ∈ [0, 2π] : |g(x)| > α ¯ = λ(α, g). Then g ∗ (β) = f ∗ (β) for any β ∈ [0, 2π]. (e) We proved that d = 2π m and λmin (σ, f ) =
1 m λ(σ, f ).
Hence, for 0 < β < d =
2π m,
f ? (β) = inf{σ : λmin (σ, f ) 6 β} = inf{σ : λ(σ, f ) 6 mβ} = f ∗ (mβ).
¤
We denote by Mn the set of real trigonometric polynomials, i.e., n o (2.3) Mn = Tn ∈ Nn : Tn (x) ∈ R, x ∈ T . 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. For Tn ∈ Mn such that |Tn (x)| 6 1 and Tn (0) = 1, we have π π (2.4) Tn (x) > cos nx, − 6 x 6 . n n Lemma 2.3. Let Tn ∈ Mn be such that max Tn (x) − min Tn (x) = 2. Then for 0 6 β
(cos nx + α) (β),
(2.5) where α = max Tn (x) − 1 . Proof. Denote
Ξ := max Tn (x) and ξ := min Tn (x). Let d be the minimal period of Tn . Then there exist x1 , x2 ∈ [0, d) such that Tn (x1 ) = Ξ, Tn (x2 ) = ξ and the polynomial Ten (x) = Tn (x1 − x) − Ξ + 1 satisfies all conditions of Lemma 2.2. Hence, for 0 6 |x| 6 nπ , it follows that Ten (x) > cos nx.
(2.6) Moreover, |x1 − x2 | >
π n
since if |x1 − x2 |
cos nx + α,
A SHARP REMEZ INEQUALITY
Similarly, Lemma 2.2 gives (2.8)
Tn (x2 − x) 6 cos n
´ ¡π − x + α = − cos nx + α, n
(here it is enough to consider −Tn ). Assume first that α ∈ [−1, 1] and 0 6 |x| 6 (2.9) Also using (2.8) for
5
arccos(−α) . n
0 6 |x| 6
π n
Then using (2.7) we get
Tn (x1 − x) > cos nx + α > 0. arccos(−α) n
6x6
2π n
³
Tn and therefore
−
arccos(−α) , n
we have ³ ´ π¢ x2 − x − 6 cos nx + α 6 0 n
¯ ³ ³ π ´´¯¯ ¯ ¯Tn x2 − x − ¯ > | cos nx + α|. n
(2.10)
Further, let ³ ³ ³ π arccos(−α) ´ ³ π arccos(−α) ´´ arccos(−α) arccos(−α) ´ A1 := x1 − , x1 + , A2 := x2 − − , x2 + − . n n n n n n Since |x1 − x2 | > nπ , then A1 ∩ A2 = ∅. Moreover, |A1 | + |A2 | = 2π/n 6 d and for γ > 0 ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ ¯ x ∈ [0, d) : |Tn (x)| > γ ¯ > ¯ x ∈ A1 ∪ A2 : |Tn (x)| > γ ¯ ¯n o¯ ¯n o¯ ¯ ¯ ¯ ¯ = ¯ x ∈ A1 : |Tn (x)| > γ ¯ + ¯ x ∈ A2 : |Tn (x)| > γ ¯ . Then, making use of (2.9) and (2.10) and translation invariance of the distribution function, we have for γ>0 ¯n o¯¯ ³ arccos(−α) arccos(−α) ´ ¯ ¯ , : | cos nx + α| > γ ¯¯ |{x ∈ [0, d) : |Tn (x)| > γ}| > ¯ x ∈ − n n ¯n o¯¯ ³ arccos(−α) 2π arccos(−α) ´ ¯ ¯ +¯ x∈ , − : | cos nx + α| > γ ¯¯ n n n ¯n o¯¯ ³ 2π ´ ¯ ¯ : | cos nx + α| > γ ¯¯ . = ¯ x ∈ 0, n Therefore, for α ∈ [−1, 1] we get the required inequality (2.5) for 0 6 β < 2π n . Let now α > 1 and therefore ξ > 0. Applying (2.5) to Tn − ξ, it follows that ?
?
(Tn − ξ) (β) > (cos nx + 1) (β), Taking into account that
06β
1. For α < −1 it is enough to consider −Tn . Lemma 2.4. Let n ∈ N and 0 6 β < (a)
2π n .
Then
1 + sin2 1+α 6 ? 1 − sin2 06α61 (cos nx + α) (β) sup
nβ 4 nβ 4
;
(b) inf (cos nx + α)? (β) > cos2
06α 1 − α, i.e., α > | cos nx + α|).
(1−cos 2
nβ 2 )
(see Figure 1 showing the graph of the function f (x) =
Figure 1. Then (cos nx + α)? (β) = cos
(2.11) Note that the function
ϕ(α) =
nβ + α. 2
1+α cos nβ 2 +α
is monotone with respect to α and therefore 1+α 1+α (2.12) sup = sup . ? (cos nx + α)? (β) nβ 06α61 (cos nx + α) (β) 06α6(1−cos )/2 2
Next let 0 6 α 6
1−cos 2
nβ 2
, then (cos nx + α)? (β) =
nv cos nu 2 + cos 2 , 2
where 0 6 u 6 v 6 β satisfy the following equations u+v =β (2.13) nv cos nu 2 − α = cos 2 + α (see Figure 2).
Figure 2. Hence, n(β−u) 2 + cos nu 1+α 2 − cos 2 = . n(β−u) (cos nx + α)? (β) cos nu + cos 2 2
A SHARP REMEZ INEQUALITY
7
Let us find the maximum of the right-hand side for u ∈ [0, β/2]. Take f (z) =
2 + cos z − cos(a − z) , cos z + cos(a − z)
z ∈ [0, a/2),
where 0 6 a < π. Then f 0 (z) =
−2 sin a − 2 sin(a − z) + 2 sin z 2
(cos z + cos(a − z))
< 0,
z ∈ [0, a/2).
Therefore, f (z) is non-increasing on [0, a2 ) and has a maximum at z = 0. Then, by (2.12), n(β−u) 2 + cos nu 1+α 2 − cos 2 = max ? n(β−u) nu 06u6 β 06α61 (cos nx + α) (β) cos 2 + cos 2 2
sup
=
3 − cos nβ 2 1 + cos nβ 2
=
nβ 4 sin2 nβ 4
1 + sin2 1−
,
i.e., (a) is verified. To prove (b), let α > 1 and then (cos nx + α)? = (cos nx + 1)? + α − 1 > (cos nx + 1)? . Therefore, in (b), infimum is attained for 0 6 α 6 1. Further we remark that for α >
1−cos 2
nβ 2
we have
nβ + α, 2 which is the non-decreasing function with respect to α. Then (cos nx + α)? (β) = cos
inf
06α
inf 06α6(1−cos
nβ 2 )/2
(cos nx + α)? (β).
Next, for 0 6 α 6 (1 − cos nβ 2 )/2, we have nv cos nu 2 + cos 2 , 0 6 u 6 v 6 β, 2 where u, v, and α are related by (2.13). Taking into account that
(cos nx + α)? (β) =
min
06u6 β 2
cos n(β−u) + cos nu nβ 2 2 = cos2 , 2 4
we get (b). To prove (c), we just use (2.11): (2.14)
(cos nx + α)? (β) = cos
(1 − cos nβ nβ nβ 2 ) + = cos2 . 2 2 4
¤
3. Sharp Remez inequalities First, we note that the classical Remez inequality can be written as an inequality in terms of rearrangements (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) (3.2) (3.3)
f ∗ (0) 6 C(β)f ∗ (β), kf kL∞ (Ω) 6 C(|B|)kf kL∞ (Ω\B)
β > 0;
for any measurable set
¯ ¯ ¯ ¯ if ¯{x ∈ Ω : |f (x)| > 1}¯ < β,
B ⊂ Ω;
then kf kL∞ (Ω) 6 C(β).
8
E. NURSULTANOV AND S. TIKHONOV
Proof. Let us verify that (3.1) ⇒ (3.2) ⇒ (3.3) ⇒ (3.1). First assume that (3.1) holds. Let B be a subset of Ω such that |B| = β. Note that for any ε > 0 ¯n o¯ ¯ ¯ A = ¯ x ∈ Ω : |f (x)| > f ∗ (β + ε) ¯ > (β + ε). Then
¯n ¯ o ¯ ¯ ¯ x ∈ Ω : |f (x)| > f ∗ (β + ε) \ B ¯ > 0
and kf kL∞ (Ω\B) > kf kL∞ (A\B) > f ∗ (β + ε). Using (3.1), we get kf kL∞ (Ω) = f ∗ (0) 6 C(β + ε)f ∗ (β + ε) 6 C(β + ε)kf kL∞ (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)| 6 1 for any x ∈ Ω \ B. Then (3.2) gives ¡ ¢ ¡ ¢ ¡ ¢ kf kL∞ (Ω) 6 C |B| kf kL∞ (Ω\B) 6 C |B| 6 C β . Assuming (3.3), consider β > 0 and ψ(x) = ff∗(x) (β) , we get ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯{x ∈ Ω : |ψ(x)| > 1}¯ = ¯{x ∈ Ω : |f (x)| > f ∗ (β)}¯ 6 β < β + ε. By (3.3) we have kψkL∞ (Ω) 6 C(β + ε) for any ε > 0. Since C(·) is right continuous, kψkL∞ (Ω) 6 C(β), i.e., (3.1) follows. ¤ Thus, to investigate the Remez inequality, one can use its rearrangement form (3.4)
Tn∗ (0) 6 C(n, β) Tn∗ (β),
β > 0.
Note that dealing with polynomials with the minimal periods less than 2π allows us to decrease the constant C(n, β). Indeed, if d = 2π/m is a period of Tn , then by Lemma 2.1, Tn (x/m) is a polynomial of degree n/m and ∗ ∗ n (x) := T (x/m), n (β) = T (β), β > 0. Tm Tm n n n Then inequality (3.4) is true with the constant C( m , β). To overcome this, we are going to study Remez’s inequality with the ?-rearrangement, i.e.,
(3.5)
Tn? (0) 6 C(n, β) Tn? (β),
β > 0.
Let us now recall that Nn and Mn are the sets of complex and real polynomials of degree at most n, 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 0 6 β < 2π. Then (3.6)
sup Tn ∈Nn
Tn∗ (0) Tn∗ (0) = sup , ∗ Tn (β) Tn ∈Mn Tn∗ (β)
and (3.7)
sup Tn ∈Nn
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈Mn Tn? (β)
Proof. It is clear that sup Tn ∈Nn
Tn∗ (0) Tn∗ (0) > sup . ∗ Tn (β) Tn ∈Mn Tn∗ (β)
Let now Tn ∈ Nn and |Tn (x0 )| = maxx∈T |Tn (x)|. Suppose α =
Tn (x0 ) |Tn (x0 )| ,
then Re (¯ αTn ) ∈ Mn and
max |Re (¯ αTn ) (x)| = |Tn (x0 )| = Tn∗ (0). x∈T
A SHARP REMEZ INEQUALITY
9
Further, since |Re (¯ αTn (x)) | 6 |Tn (x)|, we get ∗
∗
Tn∗ (0) (Re (¯ αTn )) (0) (Re (¯ αTn )) (0) Tn∗ (0) = 6 6 sup ∗ Tn∗ (β) Tn∗ (β) (Re (¯ αTn )) (β) Tn ∈Mn Tn∗ (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 ∈ N and 0 6 β < (3.8)
2π n .
sup Tn ∈Mn
Then nβ Tn? (0) = 1 + 2 tan2 Tn? (β) 4
and (k)
(3.9)
sup Tn ∈Mn
kn−k Tn kL∞ nβ = 1 + tan2 . Tn? (β) 4
Proof. To prove (3.8), we define two subsets of Mn : n o (3.10) M0n = Tn ∈ Mn : max Tn (x) − min Tn (x) = 2 x
and
x
n o M1n = Tn ∈ M0n : 1 6 max Tn (x) 6 2 .
(3.11)
x
Note that for any Tn 6= const there exists γ > 0 such that γTn ∈ M0n . Then it is clear that sup Tn ∈Mn
Let us now show that sup Tn ∈M0n
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M0n Tn? (β) Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M1n Tn? (β)
Suppose that Tn ∈ M0n but Tn ∈ / M1n and −Tn ∈ / M1n . Denoting Ξ := max Tn (x), note that Ξ > 2. Indeed, 1 assume 0 6 Ξ < 1, then it is clear −Tn ∈ Mn , which contradicts the choice of Tn . The case Ξ < 0 can be reduced to the case Ξ > 0 by taking −Tn . Therefore, Ξ > 2 and ξ := min Tn (x) > 0. Then for Pn (x) := Tn (x) − ξ > 0 we have Pn? (β) = Tn? (β) − ξ, On the other hand, Pn ∈
β > 0.
M1n , ?
Tn? (0) Pn? (0) + ξ (Tn? (0) − ξ) + ξ Pn? (0) = = 6 , ? Tn? (β) Pn? (β) + ξ Pn? (β) (Tn (β) − ξ) + ξ because of Pn? (0) > Pn? (β) and ξ > 0. Hence, we get sup Tn ∈Mn
Tn? (0) Tn? (0) = sup . Tn? (β) Tn ∈M1n Tn? (β)
Therefore, it suffices to verify (3.8) on the class M1n . Let further Tn ∈ M1n , then by Lemma 2.3, ?
Tn? (β) > (cos nx + α) (β), where 0 6 α = Ξ − 1 6 1. Then Ξ 1+α T ? (0) 6 = ? ? T (β) (cos nx + α) (β) (cos nx + α)? (β) and therefore, sup Tn ∈M1n
T ? (0) 1+α 6 sup . T ? (β) 06α61 (cos nx + α)? (β)
10
E. NURSULTANOV AND S. TIKHONOV
Making use of Lemma 2.4(a), we get sup Tn ∈Mn
where 0 6 β
n n ¯ β β 2πi |e| ¯ e nβ i=0 − 2m + m
sup e∈Mh.int. |e|=β
and kTn kL∞
sup Tn ∈Mn
Tn (β, M )
kTn kL∞
6 sup Tn ∈M0n
6
Tn (β, Mh.int. )
6 sup Tn ∈M0n
kTn kL∞ ¯R ¯ 1 ¯ supe∈Mh.int. |e| T (y)dy ¯ e n |e|=β
Ξ
sup
nβ 2m
16Ξβ
|e|>β
i.e., inequality (3.18) is sharp. The proof of inequality (3.19) is similar to the proof above using (3.20)
sup Tn ∈M0n
kTn kL∞ Tn (β, Mint. )
6 sup Tn ∈M0n
kTn kL∞ nβ ¯R ¯6 . 1 ¯ ¯ supe∈Mint. |e| e Tn (y)dy 2 sin nβ 2
¤
|e|=β
The next theorem plays an important role in approximation theory (see Stechkin [St] and Nikol’skii [Ni] in case β = π/n). In particular, it generalizes Bernstein’s inequality kTn0 k∞ 6 nkTn k∞ . Corollary 3.10. Let n ∈ N. We have sup Tn ∈Mn
kTn0 kL∞ = kTn (x + β) − Tn (x)kL∞
µ
¶ n , 2 sin βn 2
0m t p (cos nx + Ξ − 1) (t) . t 0 = m1/p
q
t p (T ? (t))q
Using Lemma 2.1(e), we have µZ kTn kLp,q > m1/p n−1/p
2π
0
³ ´q dt ¶1/q q t p (cos x + Ξ − 1)∗ (t) . t
Then sup Tn ∈M0 n Ξ>1
m1/p kTn kL∞ Ξ 6 sup 1/p n kTn kLp,q Ξ>1 k cos(·) + Ξ − 1kLp,q
(4.7)
=
1+α . k cos(·) + αkLp,q 06α cos z + Ξ − 1 > 0, 0 6 z 6 for any (x1 , x2 ) ∈ Qz and then Finally,
π 2
³ ´ Tn∗ µ(Qz ) > cos z + Ξ − 1 > 0.
Ξ 1 Tn∗ (0) 6 sup = ∗ (µ(Q )) T cos z + Ξ − 1 cos z z Ξ>1 Tn ∈Mn (2) n sup
18
E. NURSULTANOV AND S. TIKHONOV
and since µ(Qz ) =
2z 2 n1 n2
= β, we arrive at (5.2).
¤
In the general case (d > 2) the proof is similar. Remark 5.2. Let d > 2, n ∈ Nd , and Tn ∈ Mn (d). Then Tn∗ (0) 6
(5.4) where 0 6 β
2 v u d ³ Y u ¡ ¢ ´ 1 jn j d t 6 exp 0.5 d n1 . . . nd |B| 1/d , q 6 exp |B| Qd jnj 2 cos d |B| j=1 j=1
2
d Y 2 . jn j j=1
|B|
2 and Tn ∈ Mn (d). Then ! 1 µ ¶1/p à Z π ³ ´q dt − q 2 kTn kL∞ (Td ) d d! 6 d t p cos t . (5.6) sup ³ ´1/p 2d t Tn ∈Mn Qd 0 kTn kLp,q (Td ) j=1 nj Proof. Making use of Theorem 5.1 and Remark 5.2 for Tn ∈ Mn , we get ! ÃZ (2π)d ³ ´q dt 1 ∗ kTn kLp,q (Td ) = t p Tn (t) t 0 v q1 q q1 u d πd Z Qd π d ³ Z ´ Q u Y jnj q dt d jnj jnj 1 1 j=1 j=1 d > dt Tn∗ (0) t p cos t t > t p Tn∗ (t) t 2 t 0 0 j=1 −1/p à !1 Z π2 ³ d ´q dt q Y d jn j Tn∗ (0), = d t p cos t 2 t 0 j=1
and we arrive at (5.6).
¤
In some cases the integral on the right-hand side of (5.6) can be simply calculated. For example, for d = 2, we have n1 n2 ¢ kTn kL1 , kTn k∞ 6 ¡ π 4 2 −1 √ 2 n1 n2 kTn k∞ 6 √ kTn kL2 , π2 − 4 √ n1 n2 kTn k∞ 6 3/2 kTn kL2,1 . 2 Similar to Remark 4.4 we can prove the following (Lp , Lp1 )-Nikol’skii inequality for Tn ∈ Mn (d), d > 2, µY ¶ p1 − p1 d 1 1−p/p1 kTn kLp (Td ) 6 C nj kTn kLp (Td ) , 0 < p < p1 < ∞, 1
j=1
where C is the constant in the right-hand side of (5.6) with p = q.
A SHARP REMEZ INEQUALITY
19
Finally, we note that Remez type inequalities hold not only for trigonometric polynomials but also for wider classes of functions. Denote by Eν,µ , ν, µ > 0 the collection of functions f ∈ L∞ (R2 ) such that ∂2f ∈ C(R2 ), α1 + α2 = 2, αi > 0 and α α ∂x1 1 ∂x2 2 ° ° ° ∂2f ° ° ° ° α ° 6 ν α1 µα2 °f °L∞ (R2 ) , α1 + α2 = 2, αi > 0. ° ∂x 1 ∂xα2 ° 1 2 L∞ (R2 ) This class in particular contains all functions of exponential type (ν, µ) with regards to x1 and x2 , respectively. Theorem 5.4. Let ν, µ > 0 and 0 6 β < (5.7)
4 νµ .
For any f ∈ Eν,µ we have µ ¶−1 f ∗ (0) (νµβ) sup ∗ 6 1− . 4 f ∈Eν,µ fν,µ (β)
Proof. Without loss of generality we assume that max |f (x, y)| = f (0, 0) = 1.
(x,y)∈R2
Then since (0, 0) is an extreme point, taking into account the definition of Eν,µ and Taylor’s formula, there exists (c1 , c2 ) ∈ (−|x|, |x|) × (−|y|, |y|) such that µ ¶ 1 ∂2f ∂2f ∂2f 1 2 2 f (x, y) = f (0, 0) + (c , c )x + 2 (c , c )xy + (c , c )y > 1 − (ν|x| + µ|y|)2 . 1 2 1 2 1 2 2 ∂x2 ∂x∂y ∂y 2 Suppose z = |x|ν + |y|µ, then using the same argument as in the proof of Theorem 5.1, √ z2 f (x, y) > 1 − > 0, 0 < z < 2 2 n o 4 for any point (x, y) belonging to Qz = (x, y) : |x|ν + |y|µ 6 z . Finally, for 0 6 β < νµ we have f ∗ (β) > 1 −
νµβ >0 4
and therefore (5.7) follows.
¤
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[Nu] E.D. Nursultanov, Net spaces and inequalities of Hardy-Littlewood type, Sb. Math. 189, No.3 (1998), 399–419; translation from Mat. Sb. 189(3) (1998), 83–102. [NA] E.D. Nursultanov, T.U. Aubakirov, Interpolation theorem for stochastic processes, Eurasian Math. J., 1(1) (2010), 8–16. [NT] E. Nursultanov, S. Tikhonov, Net spaces and boundedness of integral operators, J. of Geometric Analysis, 21(4) (2011), 950–981. [Re] E. Remes, Sur une propri´ et´ e extremale des polynˆ omes de Tchebychef (French), Commun. Inst. Sci. Math. et Mecan., Univ. Kharkoff et Soc. Math. Kharkoff, IV. Ser. 13 (1936), 93-95. Online: http://www.math.technion.ac.il/hat/fpapers/remezppr.pdf [St] S. B. Stechkin, A generalization of some inequalities by S. N. Bernstein (Russian), Dokl. Akad. Nauk SSSR, 60 (1948), 1511–1514. [SU] S. B. Stechkin, P. L. Ulyanov, Sequences of convergence for series, Trudy Mat. Inst. Steklov, 86 (1965), 3–83. English translation in Proc. Steklov Inst. Math., (1967), 1–85. E. Nursultanov L.N. Gumilyov Eurasian National University and Kazakh Branch of Moscow State University Munatpasova, 7 010010 Astana Kazakhstan E-mail address: [email protected] S. Tikhonov ` tica ICREA and Centre de Recerca Matema Facultat de Cincies, UAB 08193 Bellaterra Barcelona, Spain E-mail address: [email protected]
