The Bojanov-Naidenov problem for trigonometric polynomials and periodic splines

For given $n, r \in \mathbb{N}$; $p, A > 0$ and any fixed interval $[a,b] \subset \mathbb{R}$ we solve the extremal problem $\int\limits_a^b |x(t)|^q dt \rightarrow \sup$, $q \geqslant p$, over sets of trigonometric polynomials $T$ of order $\leqslant n$ and $2\pi$-periodic splines $s$ of order $r$ and minimal defect with knots at the points $k\pi / n$, $k \in \mathbb{Z}$, such that $\| T \| _{p, \delta} \leqslant A \| \sin n (\cdot) \|_{p, \delta} \leqslant A \| \varphi_{n,r} \|_{p, \delta}$, $\delta \in (0, \pi / n]$, where $\| x \|_{p, \delta} := \sup \{ \| x \|_{L_p[a,b]} \colon a, b \in \mathbb{R}, 0 < b - a < \delta\}$ and $\varphi_{n, r}$ is the $(2\pi / n)$-periodic spline of Euler of order $r$. In particular, we solve the same problem for the intermediate derivatives $x^{(k)}$, $k = 1, ..., r-1$, with $q \geqslant 1$.

For given n, r ∈ N; p, A > 0 and any fixed interval [a, b] ⊂ R we solve the extremal problem´b a |x(t)| q dt → sup, q ≥ p , over sets of trigonometric polynomials T of order ≤ n and 2π-periodic splines s of order r and minimal defect with knots at the points kπ/n, k ∈ Z, such that T p,δ ≤ A sin n(·) p,δ , s p,δ ≤ A ϕ n,r p,δ , δ ∈ (0, π/n], where x p, δ := sup{ x Lp[a, b] : a, b ∈ R , 0 < b − a ≤ δ} and ϕ n,r is the (2π/n)-periodic spline of Euler of order r. In particular, we solve the same problem for the intermediate derivatives Key words: Bojanov-Naidenov problem, polynomial, spline, rearragement, comparison theorem.
MSC 2010: Pri 41A17, Sec 41A44 1. Introduction. Let G denote the real line R or a finite interval [a, b] or the unite circle I 2π wich is realized as the interval [0, 2π] with coinsident endpoints. We shall concider the spaces L p (G), 0 < p ≤ ∞, of all measurable functions x : G → R such that x p = x Lp(G) < ∞, where For r ∈ N and p, s ∈ (0, ∞] let L r p,s be the space of all functions x ∈ L p (R) for which x (r−1) is locally absolutely continuous and x (r) ∈ L s (R). We shall write x p instead of x Lp(R) and L r ∞ instead of L r ∞,∞ . It is well known (see for example [1, page 47]) that the problem of finding the best constant C in the Kolmogorov-Nagy type inequality over the class of functions x ∈ L r p,s , where α = r−k+1/q−1/s r+1/p−1/s , q, p, s ≥ 1, r ∈ N, k ∈ N 0 := N {0}, k < r, α ≤ (r − k)/r, is equivalently reduced to the following extremal problem: over the class of functions x ∈ L r p,s satisfying where A 0 , A r are the fix positive numbers. Many mathematicians investigated the problem of finding the best constant in (1.1). There are only few cases in which sharp inequalities of the form (1.1) are known for any r ∈ N and any k < r. The survey of the results in this directions are given in [1] - [3]. For arbitrary interval [a, b] ⊂ R Bojanov and Naidenov have solved the problem , positive on (0, ∞) and such that Φ(t)/t is non-decreasing and Φ(0) = 0. In particularly, Bojanov and Naidenov have solved the problem of Erdös [5] on characterization of the trigonometric polynomial of fixed uniform norm that has maximal arc length over [a, b].
We shall consider the class W of the continuous, nonnegative and convex functions Φ definsd on [0, ∞) and such that Φ(0) = 0. For p > 0 set [6] L The following modification of the Bojanov and Naidenov problem was solved in [7]: over the class of functions x ∈ L r ∞ satisfying As a special case was solved the problem over the same class of functions x ∈ L r ∞ . The generalizations of the results of the article [7] are given in [8], [9]. Denote by ϕ r (t), r ∈ N, the r th 2π-periodic integral with zero mean value on a period of the function ϕ 0 (t) = sgn sin t and for λ > 0 set ϕ λ,r (t) := λ −r ϕ r (λt).
The solution of the problems (1.5) and (1.7) was given in [10] over the class of functions x ∈ L r ∞ satisfying Note that the value x p, δ for p ≥ 1 is the norm but the value L(x) p is not. Let n, r ∈ N and p, A > 0. Denote by T n the set of all trigonometric polynomials of order ≤ n. Let S n,r be the set of 2π-periodic polynomial splines of order r with knots at the points kπ/n, k ∈ Z. Set where the value x p, δ is defined by (1.8).
We solve in this paper the problem (1.5) and (1.7) over the classes T p n (A) and S p n, r (A) (Theorems 1 and 3). As a special case we solve the problem Erdös over the same classes.
2. Preliminaries. For n, r ∈ N and p, A > 0 set Lemma 2. Let n, r ∈ N and p, A > 0. If K = T p n (A) or K = S p n, r (A) then, for any function x ∈ K, the following inequality holds true Proof. Fix a function x ∈ K and arbitrary a ∈ R. Let m be a point of maximum ψ n, r (K, t). Note that ψ n, r (K, ·) p p, δ = m+δ/2 m−δ/2 ψ p n, r (K, t)dt, δ ∈ (0, π/n]. So we conclude from (1.9), (1.10) and (2.1) that wich is equivalent to (2.2). Lemma 2 is proved. We Lemma 3. Let n, r ∈ N and p, A > 0. If K = T p n (A) or K = S p n, r (A) then the function ψ n, r (K, t) is the comparison function for any function x ∈ K.
Proof. Fix a function x ∈ K. By Lemma 2 the inequality (2.2) holds true. If K = T p n (A) then it follows from (2.2) (see, for example, the proof of Theorem 8.1.1 [1]) that the function ψ n, r (K, t) = A sin nt is the comparison function for x.
Let now K = S p n, r (A). Then ψ n, r (K, t) = Aϕ n, r (t). So applying Tikhomirov inequality (see, for example, [1, Lemma 8.2.1]) we conlude from (2.2) that x (r) ∞ ≤ A. Hence, in view of (2.2) the function x satiesfies the conditions of Kolmogorov comparison Theorem [13]. By this Theorem the function ψ n, r (K, t) = Aϕ n, r (t) is the comparison function for x.
Lemma 3 is proved.
The rearrangment of the function |x| is denoted by r(x, t) (see, for example, [14, §1.3]). We also set r(x, t) = 0 for t > b − a.

3)
where m is a point of maximum of the function ψ n, r (K, t) and the number Θ is such that ψ n,r (m − Θ) = ψ n,r (m + Θ), 2Θ = b − a.

4)
where c is a zero of the function ψ n, r (K, t).
Proof. First, we show that the difference δ(t) := r(x, t) − r(ψ, t) has at most one change of sign on [0, ∞) (from -to +). Note that in view of Lemma 2. Set such that z = |x(t i )| = |ψ(y j )|. (2.8) By Lemma 3 ψ n,r is the comparison function for x. So for the points t i and y j , satisfying (2.8), the following inequality holds true Hence, if the points Θ 1 , Θ 2 > 0 are chosen such that then by the theorem about differentiation of rearrengement (see, for example, [14, It follows that the difference δ(t) := r(x, t) − r(ψ, t) has at most one change of sign on [0, ∞) (from -to +). It is exactly the same for difference δ p (t) := r p (x, t) − r p (ψ, t).
Let us consider the integral It is clear that I p (0) = 0. Taking into account the definitions (1.9) and (1.10) of the classes T p n (A) and S p n, r (A) we have Besides, the derivative I p (t) = δ p (t) has at most one change of sign on [0, ∞) (from -to +). Therefore, I p (ξ) ≤ 0 for all ξ ≥ 0 and the inequality (2.6) is proved.
Theorem 1. Let n, r ∈ N; p, A > 0. If K = T p n (A) or K =S p n, r (A) then, for any function Φ ∈ W and arbitrary interval [a, b] ⊂ R, where the number τ is defined by (3.2). In particularly, for q ≥ p, Proof. Fix any function x ∈ K and arbitrary interval [a, b] ⊂ R. Let us write the length of [a, b] in the form (3.1). Set a k := a + kπ/n, k = 0, 1, ..., l. By Lemma 4 where c is a zero, m is a point of maximum of the function ψ n, r and the number Θ is defined by (3.1). Therefore, Moreover, the equality is realized here for the function x(t) = ψ n,r (t+τ ). First statement of Theorem 1 is proved. Setting Φ(t) = t q/p in it we have second statement.
Theorem 1 is proved. Foe n, r ∈ N and A, p > 0 set  9) and (3.3) . Proof. Fix any function x ∈ K ( where K = T n (A, p) or K = S n,r (A, p)) and the number δ ∈ (0, π/n]. Let us prove the inequality x p, δ ≤ ψ n,r p, δ . (3.5) For a function x ∈ T n (A, p) or x ∈ S n,r (A, p) and arbitrary interval [a, b] satisfying b − a ≤ π/n it was proved [9, теоремы 7, 9] where m is a point of maximum of the function ψ n,r and 2Θ = b − a . Setting Φ(t) = t in the inequality (3.6) we have the estimate (3.5). It follows from (3.5) in view of the definitions (1.9), (1.10) and (3.3) both of inclusions. Theorem 2 is proved. Remark 1. It follows from Theorem 2 that the problem (1.5) is proved in Theorem 1 over wider classes than in Theorems 7 and 9 in [9] where this problem has been proved over the classes T n (A, p) and S n, r (A, p)).
Next Theorem contains a solution of the problem (1.7) over this more wide classes T p n (A) andS p n, r (A)). Let n, k, r ∈ N; A, p > 0; [a, b] ⊂ R and the length of the interval [a, b] is presented in the fom (3.1). Choose τ k ∈ R such that where the function ψ n, r (K, t) is defined by (2.1). Besides, let k ≤ r for K =S p n, r (A)).
Theorem 3. Let n, k, r ∈ N; A, p > 0. If K = T p n (A) or K =S p n, r (A) and k ≤ r then, for any function Φ ∈ W and arbitrary interval [a, b] ⊂ R, we have where the number τ k is define by (3.7). In particularly, for any q ≥ 1, Proof. Fix any function x ∈ K and arbitrary interval [a, b] ⊂ R. Let us prove the first statement of Theorem 3. By Lemma 2 x ∞ ≤ ψ n,r ∞ . (3.8) It follows that where i ∈ N if K = T p n (A) and i = 1, 2, ..., r if K =S p n, r (A). Really, ψ n, r (t) = A sin nt if K = T p n (A) and the inequality (3.9) follows from (3.8) and well known Bernstein inequality x (i) ∞ ≤ n i · x ∞ , i ∈ N, for trigonometric polynomials x ∈ T n . If K =S p n, r (A) then ψ n,r (t) = Aϕ n,r (t) and (3.9) follows from (3.8) and Tikhomirov inequality (see, for example, [1, Theorem 8.2.1]) for splines x ∈ S n,r . So for any interval [α, β] satisfying |x (k) (t)| > 0, t ∈ (α, β), we have where the value L(x) p is defined by (1.4). It follows that L(x (k) ) 1 ≤ L(ψ (k) n,r ) 1 . For a function x ∈ T n or x ∈ S n,r satisfying last inequality and for arbitrary innterval [a, b] it was proved [9, теоремы 7, 9] where τ k is defined by (3.7). The equality is odtained here for the function x(t) = ψ n,r (t + τ k ). The first statement of Theorem 3 is proved. Setting in it Φ(t) = t q we have the second statement. Theorem 3 is proved. It is well known that arc length l[a, b] over [a, b] of a function x ∈ L 1 [a, b] is given by formula l[a, b] =´b a 1 + x (t) 2 dt. It is clear that Φ 0 ∈ W for the function Φ 0 (t) = √ 1 + t 2 . Consequently, setting Φ = Φ 0 , k = 1 in the first statement of Theorem 3 we obtain the solution of the problem of Erdös on characterisation of trigonometric polynomials T ∈ T p n (A) that has maximal arc length over [a, b]. Besides, we solve the same problem over the space of splinesS p n, r (A). Corollary 2. Let n, r ∈ N; A, p > 0; [a, b] ⊂ R. Among all trigonometric polynomials T ∈ T p n (A) the maximal arc length over [a, b] has the polynomial T = A sin n(t + τ 1 ), where τ 1 is the same as in Theorem 3.
Among all splines s ∈S p n, r (A) the maximal arc length over [a, b] has the spline Aϕ n,r (t + τ 1 ), where τ 1 is the same as in Theorem 3.