Landau--Kolmogorov inequality revisited

The Landau-Kolmogorov problem consists of finding the upper bound $M_k$ for the norm of intermediate derivative $|f^{(k)}|$, when the bounds $|f| \le M_0$ and $|f^{(n)}| \le M_n$, for the norms of the function and of its higher derivative, are given. Here, we consider the case of a finite interval, and when all the norms are the max-norms. Our interest to that particular case is motivated by the fact that there are good chances to add this case to a short list of Landau--Kolmogorov inequalities where a complete solution exists, i.e., a solution that covers all values of $n,k\in\N$ (and, for a finite interval, all values of $\sigma = M_n/M_0$). The main guideline here is Karlin's conjecture that says that, for all $n,k\in\N$ and all $\sigma>0$, the maximum of $|f^{(k)}|$ is attained by a certain Chebyshev or Zolotarev spline. So far, it has been proved only for small $n \ge 4$ with all $\sigma$, and for all $n$ with particular $\sigma = \sigma_n$. Here, we prove Karlin's conjecture in several further subcases: 1) all $n,k\in\N$ and all $0<\sigma \le \sigma_n$ 2) all $n \in \N$, all $\sigma>0$, with $k=1,2$ 3) all $\sigma>0$, with $n<10$ and $0


Introduction
The Landau-Kolmogorov problem consists of finding the upper bound M k for the norm of intermediate derivative f (k) , when the bounds f ≤ M 0 and f (n) ≤ M n , for the norms of the function and of its higher derivative, are given.
Here, we consider the case of a finite interval when f ∈ W n ∞ [−1, 1] and all the norms are the max-norms, · = · L∞[−1 ,1] . Precisely, given n, k ∈ N and σ ≥ 0, we define the functional class Our interest to that particular case is motivated by the fact that there are good chances to add this case to a short list of Landau-Kolmogorov inequalities where a complete solution exists, i.e., a solution that covers all values of n, k ∈ N (and, for a finite interval, all values of σ > 0). The main guideline in finding out how good these chances are is the following conjecture. If (1.1) is true for particular set {n, k, σ}, then the function f ∈ W n ∞ (σ) that provides extremum M k (σ) to the value f (k) over W n ∞ (σ) is the same as the solution to the pointwise problem at the end-point of the interval. The latter solution is however known to be a certain Chebyshev or Zolotarev spline Z n (·, σ) (which is just a polynomial for small σ), and thus we have a characterization of the extremal function. So far, Karlin's conjecture has been proved for small n with all σ, and for all n with particular σ, namely in the following cases: all σ Chui-Smith [1] (σ ≤ σ n ), Landau [5] (σ > σ n ); n = 3, all σ, Sato [8], Zvyagintsev-Lepin [12]; n = 4, all σ, Zvyagintsev [11] (σ ≤ σ n ), Naidenov [7] (σ > σ n ); n ∈ N, σ = σ n , Eriksson [3] .
Here σ n := T (n) n = 2 n−1 n! , where T n is the Chebyshev polynomial of degree n on the interval [−1, 1]. The value σ = σ n serves as a borderline between two types of the extremal Zolotarev functions Z n (·, σ): if σ ≤ σ n , then Z n is a polynomial of degree n, while for σ > σ n it is a perfect spline of degree n with r knots. There are further borderlines σ n,r (with σ n,1 := σ n ) which indicate that the spline Z n (·, σ) has exactly r knots if σ n,r < σ ≤ σ n,r+1 , but that distinction is hardly of any use, since for n > 3 there are no reasonable estimates for perfect splines even with one knot. In this respect, we may apply more or less developed polynomial tools to tackle the problem for σ ≤ σ n , and then may try to use polynomial estimates in the spline case, when σ > σ n .
In this paper, we prove Karlin's conjecture in several further subcases. 1) The first result closes the "polynomial" case and proves that, for σ ≤ σ n , the extremum value of the k-th derivative of f ∈ W n ∞ (σ) is provided by the corresponding Zolotarev polynomial. 2) For the "spline" case, we managed to advance only up to the second derivative. The further advance depends mostly on improving the lower bound for the exact constant C n,k in Landau-Kolmogorov inequality on the half-line:

R+
The existing lower bounds for C n,k , which are due to Stechkin, are not very satisfactory for general n and k > 2.
3) However, for small n, these bounds can be improved, thus leading to one more extension.
In all the cases, the proof is based on comparing the upper bound for the local extrema of the function m k (·, σ) with the lower bound for the value m k (1, σ). The technique we use is not working for the value k = n − 1, what explains restriction in (1.4). In (1.3), i.e., for 0 ≤ σ < σ n , we managed to cover the case k = n − 1 by different means.
The upper bounds are given in terms of Zolotarev polynomials and these estimates may be viewed as a generalization to higher derivatives of Markov-type results of Schur [9] and Erdős-Szegő [2]. These bounds demonstrate once again, if we borrow the words of Shoenberg said about cubic splines, "the brave behaviour of Zolotarev polynomials under difficult circumstances".

Main ingredients of the proof
Karlin's conjecture states that the function m k (·, σ) (which is a positive even function) reaches its maximal value at the end-points of the interval [−1, 1]. To establish this fact it is sufficient to check that, at any point x 0 inside the interval (−1, 1) where m k (·, σ) takes its local maximum, we have If f is the function from W n ∞ (σ) that attains a locally maximal value m k (x 0 , σ), then clearly and it makes sense to introduce the following quantity: The next statement follows immediately.
then Karlin's conjecture is true.
In order to verify inequality (2.1), we split it into two parts and then check whether A ≤ B. So, we need two different estimates: a) a good lower bound for the end-point value m k (1, σ) = sup{|f (k) (1)| : f ∈ W n ∞ (σ)}, b) a good upper bound for |f (k) (x 0 )|, where f is from W n ∞ (σ) and satisfies f (k+1) (x 0 ) = 0. Actually, if x = x 0 stays sufficiently far away from the end-points x = ±1, then a reasonable upper bound for |f (k) (x 0 )| can be established irrespectively of whether f (k+1) (x 0 ) vanishes or not. Therefore, for the upper bounds for |f (k) (x)|, we will consider two cases with an appropriately chosen value ω k .
We will distinguish between the cases σ ≤ σ n and σ > σ n .
1a) Lower estimates for m k (1, σ). Clearly, m k (1, σ) is monotoniously increasing with σ, therefore, we have the trivial estimate However, this estimate is too rough when k = O(n), so we will use a finer one.
On the other hand, by (3) and ( where p is any polynomial of degree n that satisfies conditions (1)- (3) in (2.4).
Let {Z n (·, θ)} be the family of the Zolotarev polynomials parametrized with respect to the value of its highest derivative θ := Z (n) n (·, θ) (see Sect. 3 for details). Given x 0 , our choice for p in (2.5) is the dilated Zolotarev polynomial Z n (·, θ x0 ) such that Z (k+1) n (x 0 , θ x0 ) = 0. An advantage of choosing such a p is that, for x 0 ∈ [ω k , 1], the value of p (k) (x 0 ) can be further bounded in terms of the single Zolotarev polynomial Z n (·, θ k ) such that Namely, as we show in Sects. 3-4, In Sects. 5-6, we provide the estimates for the values appeared here on the right-hand side and, thus, arrive at the following statement.
1b 2 ) Upper estimate for m k (x, σ). We use a technique based on the Lagrange interpolation. Let ℓ ∆ ∈ P n−1 be the polynomial of degree n − 1 that interpolates In Sect. 7, we prove that calculation of the suprema on the right-hand side is reduced to computing the largest local maxima of two specific polynomials and that leads to the following estimate.
The latter estimate is not particularly good for k = 1 and k = 2, so for such k we also use another one 1c) Final step. The constants in estimates (2.3), (2.6) and (2.7) are easy to compare (they are simple functions of t = σ/σ n ) and, in Sect. 8, we prove that if n ∈ N, 1 ≤ k ≤ n−2 and 0 ≤ σ ≤ σ n , then max A n,k (σ), A * n,k (σ) ≤ B n,k (σ) , and that implies 2) The case σ > σ n .
For that case, it is more convenient to reformulate the original problem. Namely, instead of considering functions from the class i.e., functions on a fixed interval I 1 = [−1, 1] with increasing norms f (n) [−1,1] ≤ σ, we will consider functions from the class 2a) Lower estimate for m k (1, I s ). Denote by B + n,k the best constant in the Landau-Kolmogorov inequality on the half-line for the normalized functions: Proof. Clearly, with n and σ n fixed, the spaces defined in (2.9) are embedded into each other, namely W n ∞ (I s ) ⊃ W n ∞ (I t ) for s < t, therefore for the suprema m k (1, I s ) := sup |f (k) (1)| over those spaces, we have the inequalities Letting t = −∞, we obtain (2.11).
2b). Upper estimates for m k (x, I s ) and m * k (x 0 , I s ). Similar arguments show that the upper bounds for m k (x, I s ) and m * k (x, I s ) are majorized by those of m k (x, I 1 ) and m * k (x, I 1 ), respectively. Namely, moving the interval I = [a, b] of length |I| = 2 inside any I s , we see that W n where s 0 is the middle of the interval [−s, 1]. The right-hand sides are equivalent to the values m ( * ) k (x, σ n ) and for those we have the upper estimates (2.6)-(2.7).
2c) Final step. In Sect. 11 we prove that the constants in (2.11)-(2.13) satisfy the inequality and that proves that

Zolotarev polynomials
Here, we remind some facts about Zolotarev polynomials taking some extracts from our survey [10, p.240-242]. Note that we use a slightly different parametrization for Z n .
There are many Zolotarev polynomials, for example the Chebyshev polynomials T n and T n−1 of degree n and n − 1, with n + 1 and n equioscillation points, respectively. One needs one parameter more to get uniqueness. We will use parametrization through the value of the n-th derivative of Z n : As θ traverses the interval [−σ n , σ n ], Zolotarev polynomials go through the following transformations: Zolotarev polynomials subdivide into 3 groups depending on the stucture of the set A := (τ i ) of their alternation points.
1) A contains n + 1 points: then Z n is the Chebyshev polynomial T n .
2) A contains n points but only one of the endpoints: then Z n is a stretched Chebyshev polynomial T n (ax + b), |a| < 1.
3) A contains n points including both endpoints: then Z n is called a proper Zolotarev polynomial and it is either of degree n, or the Chebyshev polynomial T n−1 of degree n − 1.
V. Markov proved that zeros of Z ′ n (·, θ) are monotonically increasing functions of θ ∈ [−σ n , σ n ], with β going through the infinity as θ passes the zero. It follows that, for any θ 1 , θ 2 , zeros of Z ′ n (·, θ 1 ) and Z ′ n (·, θ 2 ) interlace with each other, hence by the Markov interlacing property the same is true for their derivatives of any order. In particular, the following lemma is true.
n−1 in increasing order, and, for any given θ, let Another consequence of the interlacing property is the following observation.
, and let Z n (·, θ k ) be the Zolotarev polynomials whose (k + 1)st derivative vanishes at x = 1, i.e., Further, for a given x 0 ∈ (ω k , 1), let Z n (·, θ x0 ) be the Zolotarev polynomial such that Then Proof. According to our parametrization, we have −T n (x) = Z n (x, −σ n ), and as θ increases from −σ n to −0, the rightmost zero of Z (k+1) n (·, θ) increases from ω k to +∞, passing through the value 1 for some θ := θ k . Therefore 2) Here we give some upper estimates for the values T Proof. The first inequality was proved by Eriksson [3] who actually derived a stronger estimate: The second inequality is due to Erdös-Szegö [2, p.464]. To derive the third one, we note that the function F k (·) has the single minimum at x * = 2k+1 2k+5 = 5 9 , therefore, if x * < ω 2 < 1, then But ω 2 is the largest zero of the third derivative of T n , therefore it is greater than the third largest zero of T ′ n , i.e., ω 2 > cos 3π n , so (3.2) is valid if cos 3π n ≥ 5 9 , and the latter holds for n ≥ 10.
Proof. The values of local maxima of |T (k) n (ξ i )| increase with |ξ i |, and since ω k = max i |ξ i |, we have max n (x)| decreases monotonically from the rightmost maximum T

A generalization of Erdös-Szegó result
By Q n we denote the unit ball in the space P n , i.e., the set of polynomials p ∈ P n such that p ≤ 1. According to the well-known Markov inequality and equality is attained at x = 1 for p = T n .
In 1942, Erdős and Szegő [2] refined Shur's result by showing that the limit λ ∞ = lim n→∞ λ n exists and it is equal to where E, K are the complete elliptic integrals associated with the modulus κ. (They did not improve the uniform bound (4.1) though.) They also showed that, for any x 0 ∈ [−1, 1], the supremum of |p ′ (x 0 | is attained when p is a Zolotarev polynomial Z n (·, θ), and that the maximum over x 0 is attained at x 0 = 1 for n ≥ 4, and at x 0 = 0 for n = 3.
In this section, we generalize these results to the derivatives of order k ≥ 2.
Denote by the best constant in the pointwise Markov inequality, and by the best constant in the pointwise Schur-type inequality. It is clear that and that equality occurs only if µ ′ k (x 0 ) = 0, i.e. if x 0 is a point of local extremum (maximum or minimum) of the function µ k (·) inside (−1, 1).
The next two lemmas are straightfroward extensions of the arguments given in [2, pp.461-462], from k = 1 to k ≥ 2.
Then, for small δ > 0, there is a point n (x 0 ), where Z (k+1) n (x 0 ) = 0 and let p ∈ Q n be the polynomial such that and, for small ǫ, its k-th derivative has a local maximum in the neighbourhood of x 0 (because Z (k) n has). Let x 1 be the point of that maximum, i.e., q (k+1) (x 1 ) = 0. Then q (k) (x 1 ) > q (k) (x 0 ), and respectively , the latter inequality by definition of µ * k (·).  Then max Proof. Let η i be the points of local maxima of of µ * k (·) inside the interval (−1, 1). Then max The corollary shows that, inside (−1, 1), the local maxima of µ * k (·) coincide with the extrema (maxima or minima) of µ k (·). On the other hand, V. Markov proved that the local maxima of µ k (·) coincide with those of |T Further, it is known that the local maxima of |T (k) n | are increasing as |ξ i | increases, i.e, where ω k is the rightmost zero of T  Then max Proof. According to Corollary 2.4, We take p as a dilated Zolotarev polynomial Z n (·, θ x0 ) such that Z (k+1) n (x 0 , θ x0 ) = 0. The latter satisfies conditions (1)-(2), and its highest derivative has the value θ x0 . So, if θ x0 ≥ σ, then condtion (3) is fulfilled with p = Z n (·, θ x0 ), but if θ x0 < σ, then we have to scale Z n to ensure (3). So we set Finally, where the first inequality us due to Theorem 4.4, and the second one is due to Lemma 3.3.

Upper estimates for Z
so we will give some upper estimates for the constant λ k such that n (1) We will get those estimates using the following lemma.
Lemma 5.1 Let p ∈ P n be any polynomial that satisfies the following conditions: Proof. The proof is parallel to the proof of Lemma 2.3, since Z n satisfies Z n ≤ 1. Assuming the contrary to (5.2), we derive that the n-th derivative of h := p − γZ n should change its sign which is impossible as h is a polynomial of degree n 2a) We will construct several p that satisfy (5.1) using alternation properties of T n and T n−1 . We start with the simplest one.

Lemma 5.2 We have
Proof. Take so that p has an n-alternance on [− cos π n , 1] for any c, and where the last equality defines particular c := (1) . Then and since q (m) (1) = mT (m) n (1), it follows that 2b) The next lemma improves the previous estimate for k = O(n).

Lemma 5.3 We have
Proof. Take Then and using we obtain, after some simplifications, and that proves the first inequality (5.4). Using we obtain λ n,k = λ n,k γ ≤ 1 k + 1 n n − k γ = 1 k + 1 n − 1 n − 1 + k and that proves (5.5).
2c) In the next lemma, we get further improvements for k = 1 and k = 2.

Lemma 5.4 We have
Proof. Set ξ := cos π n , and let The polynomial r has an n-alternance on [−1, ξ], so that, after finding r (k) (ξ) we will transform it to the polynomial p(x) := r −1 + (x + 1) 1+ξ 2 , which has an n-alternance on [−1, 1] and satisfies Let us find r (k) (ξ). We have so that setting a k := T (k) n (ξ), we obtain Further, we have a 0 = T n (ξ) = −1, a 1 = T ′ n (ξ) = 0, and, for k ≥ 2, the values a k can be computed from the recurrence relation In particular, we find For k = 1, this gives For k = 2, we obtain One can show that c(n, ξ) = c(n, cos π n ) is increasing with n to its limit value given in (5.6).

Remark 5.5
We checked two other possibilities to construct p.
1) The option which is slightly worse than (5.3).
2) The option is very poor for small k, and for large k = O(n) it is slightly worse than (5.5).
Proof. We will use the following estimate. Let f ∈ W n ∞ (σ n ), i.e., f ≤ 1 and Let (ξ i ) be the zeros of T (k+1) n , and let δ k = max i |ξ i − ξ i+1 |. Set Then Proof. Since (cos πi n ) are zeros of T ′ n , the zeros ξ i of T (k+1) n are located in the intervals cos π(i+k) n < ξ i < cos πi n , and for the distance between two consecutive ξ i we have So, we need to prove that The function f is concave, therefore it is bounded from above by its tangent ℓ at t = 2α, i.e.
So, we are done, once we prove that Both functions are straight lines, so we need to check this inequality only at the end-points. 1) At t = α, we have So, we need the inequality amd the latter is valid for k ≤ n − 2.
Proof. For f ∈ W n ∞ (σ), let l ∈ P n−1 be the Lagrange polynomial of degree n − 1 that interpolates f at the points of local extrema of Z n (·, σ) on the interval [−1, 1], i.e. Then 1) It is known that the extremum value D n−1 (x, σ) (which is a constant, since p (n−1) ≡ const) is attained by the polynomial p ∈ P n−1 such that It is easy to see that, with ω(x, σ) : Indeed, (9.2) is clearly fulfilled, and p is of degree n − 1 because the leading coefficients of both polynomials on the right-hand side are equal to σ/n!. Therefore 2) For Ω n−1 (x, σ) we show below that Thus, from (9.1)-(9.4), we obtain and theorem is proved.
Proof. For Ω n−1 (x, σ) we have the convex majorant so we need to prove that For large n, this inequality is self-evident because the alternation points τ i (σ) are spread sufficiently uniform in the interval [−1, 1], therefore c 1 (σ) < 1 while c 2 → 1. But we need it for all n ≥ 2.