ON THE BEST APPROXIMATION OF THE DIFFERENTIATION OPERATOR 1

In this paper we give a solution of the problem of the best approximation in the uniform norm of the differentiation operator of order k by bounded linear operators in the class of functions with the property that the Fourier transforms of their derivatives of order n (0 < k < n) are finite measures. We also determine the exact value of the best constant in the corresponding inequality for derivatives.

This paper is devoted to studying the best approximation in the uniform norm on the real line of the differentiation operator of order k by bounded linear operators in the class of functions with the property that the Fourier transforms of their derivatives of order n (0 < k < n) are finite measures.S. B. Stechkin [8] was the first who studied the problem of the best approximation of the differentiation operator (or, more generally, of an unbounded operator) by bounded ones.In particular, he noticed that this problem is connected to the best constant in an inequality between the norms of the derivatives.Later these questions were studied by Yu.N. Subbotin, L. V. Taikov, V. N. Gabushin, A. P. Buslaev, the author, and others (see [1][2][3][4][5][6][8][9][10] and the bibliography therein).
Let C = C(−∞, ∞) be the space of continuous bounded (complex-valued) functions on the real line with the uniform norm, let M be the space of finite (complex) Borel measures on (−∞, ∞) with the norm equal to the total variation µ of a measure µ, and let L r , 1 ≤ r < ∞, be the space of measurable functions with the (finite) norm .
The Fourier transform x of a function x ∈ L 1 is defined by the formula In this case the inverse Fourier transform has the form x(t) = e 2πtηi x(η) dη.
Further on, let S be the space of infinitely differentiable, rapidly decreasing functions on the real line, and let S be the corresponding dual space of generalized functions.We will denote the value of a functional θ ∈ S on the function x ∈ S by θ, x .The Fourier transform θ of a functional θ ∈ S is the functional θ ∈ S acting according to the rule θ, Denote by F n , n ≥ 1, the set of functions x ∈ C whose derivatives x (n) of order n are continuous functions such that their Fourier transforms are measures, i. e.
where µ = µ x = x (n) ∈ M .We will denote the total variation µ of a measure µ in (1) by x (n) V .We will consider the subclass We study the problem of the best approximation of the differentiation operator of order k (0 < k < n) on the class Q n by the set L(N ) of linear bounded operators T in the space C with the norm T = T C→C ≤ N .In other words, we study the quantity where Our main results are the following two statements.

Theorem 1.
For each h > 0 we have where Theorem 2. Functions of the class F n satisfy the sharp inequality and the smallest possible constant in this inequality is The fact that functions from the set F n satisfy inequality (7) with some finite constant follows from a result of A. N. Kolmogorov [7], for x (n) C ≤ x (n) V .However, one cannot obtain the smallest possible constant in (7) using this approach.P r o o f of the both theorems will be done simultaneously following the scheme which was developed by S. B. Stechkin [8] and later used by other authors (see, e. g., [1,4,5,9,10]).Consider It follows from the homogenity of ω(δ) (see, e.g.[11, p. 116]) that with K = ω(1).This fact implies inequality (7), and the smallest possible constant in (7) is for each N > 0 and δ > 0, that is, K ≤ N δ 1−α + e(N )δ −α .Minimizing the latter expression with respect to δ > 0, we obtain the inequality Consequently, an upper estimate for e(N ) (a concrete operator) gives an upper estimate for K, and a lower estimate for K (a concrete function x ∈ F n ) gives a lower estimate for e(N ).We start the concrete realization of this scheme by considering the case n = 2, k = 1.First we obtain an upper bound for e(N ) using a concrete operator.Let η be an odd 2π-periodic function which is defined on [0, π] by the formula η(t) = t − 1 π t 2 .We have It is not difficult to see that the operator T = T 1,2 defined by the formula where ν = ν(h) = πh 2 , is a linear bounded operator in C and Introduce the function ϕ(t) = (t − η(t))t −2 .To determine its norm in the space C, we notice that t ≥ η(t) ≥ t − 1 π t 2 for all t ≥ 0, and thus Now let us prove that the representation holds for functions x ∈ F 2 , where µ = µ x = x is the measure from representation (1).First assume that a function x and its derivative x both belong to L 2 .In this case, the function y = x − T x belongs to L 2 as well, and it is easy to see that the Fourier transform of the function y has the form y(t) = −iν ϕ(2πtν) x (t).Taking the inverse Fourier transform, we obtain the expression which is representation (16) in this particular case.Now let x be an arbitrary function from the class F 2 .Introduce the functions Obviously, z and z belong to L 2 , and z can be written as with ϕ 0 (τ ) = ϕ(2πντ ).We will take the limit of this relation as ε → 0. Obviously, z (0) = x (0), and (T z)(0) → (T x)(0) as ε → 0. Consider the integrals J j (ε) = ϕ 0 (τ ) z j (τ ) dτ constituting the right-hand side of (18).The function ϕ 0 belongs to L 2 , thus, using the Hölder inequality and the Parseval equality, we obtain We see that J 1 (ε) → 0 as ε → 0. In a similar way one can show that J 2 (ε) → 0 as ε → 0. Now let us investigate the behaviour of J 0 (ε).The Fourier transform of the function It follows that The family of the functions ζ ε is δ-shaped, consequently, J 0 (ε) tends to ϕ 0 (τ ) dµ(τ ) as ε → 0. Thus, the limit of (18) as ε → 0 is This is equivalent to the fact that representation (16) holds for each function x ∈ F 2 .Using ( 16) and ( 15), one can estimate quantity (3) for operator (13) from above, namely, By (14), this yields Moreover, inequalities (11) and (19) give the estimate for the best constant in the inequality which is a particular case of ( 7).Now we will derive statements converse to (20) and (19).For, consider the function Obviously, χ is an odd entire function.Furthermore, since we have where ϕ j (t) = π 0 (π − u) sin (2j + 1)u sin ut du.
Each of the functions ϕ j is entire and it is easy to check that For a fixed t, the value of the last integral in (23) tends to zero as m → ∞, therefore It follows from (24) that ϕ j (t) ≥ 0 for t ≥ 0 and ϕ j (2m + 1) = 0 for j = m.Hence, the function χ is non-negative on the half-line (0, ∞), and Using the well-known identity we obtain It follows from relations (25)-( 27) that Further on, using (25) we find that Now let us calculate the integral Taking π − u as a new variable, we obtain Denote by y the odd function which vanishes for u > π and is y(u) = π−u 4 sin u for u ∈ (0, π).The inverse Fourier transform z = ŷ of this function Thus, the function z belongs to F 2 and provides the following estimate from below for the best constant K in (21): Inequalities ( 30), ( 20), ( 19), ( 11) imply the relations This proves Theorems 1 and 2 for n = 2, k = 1.
In the author's paper [3], the solution of problem (2) for the class was, in fact, given, and the value of the best constant K k,n (S) in inequality (7) on the set of functions x ∈ S was determined for n ≥ 3, 1 ≤ k ≤ n − 1.One could use these results to prove Theorems 1 and 2 for n ≥ 3.However, we give here a different proof, or, more exactly, a sketch of the proof.Now assume that n ≥ 3, k = 1.Let η be a 2π-periodic odd function which is defined on [0, π] by the formulae Using the function η, we define a function ϕ on the real line by ϕ(t) = (t − η(t)) t −n .The functions η and ϕ satisfy the following properties (see [3, proof of Theorem 4.1]): Now take h > 0, put ν = ν(h) = πh 2 and define an operator T = T 1,n in C by the formula It is clear that T 1,n is a linear bounded operator in C and As in the proof for n = 2 above, one can show that the representation holds for all functions x ∈ F n .It follows from this representation that In the case n = 3, k = 2, denote by η the even 2π-periodic function, defined on [0, π] by the formulae We have Moreover, it is easy to see that the function ϕ(t) = (t 2 − η(t))t −3 satisfies the property Now we define a bounded linear operator T 2,3 in the space C by the formula For this operator we have For each x ∈ F 3 we have the representation it follows from this representation that Now we define an operator T k,n for arbitrary n ≥ 3, 1 ≤ k < n by the formula For example, we can take m = k − 1, then (39) takes the form and if the operator T k,n is defined for all 3 ≤ n ≤ n, 1 ≤ k ≤ n − 1 (and n = n, k = 1), then using (40) we define T k,n for n = n, k = 2, . . ., n − 1.Let us check that for all n ≥ 3, 1 ≤ k < n.For k = 1, n ≥ 3 and for k = 2, n = 3 these relations coincide with (38), (36), (34), (32).For the other values of the parameters k, n we have This representation gives the estimate Inequalities ( 41) and (42) follow from estimates (43) and (44) by induction.Statements (41), (42) imply the following estimate from above for quantity (2) Inequality (11) gives the estimate from above for the best constant in inequality (7).
On the other hand, the function ψ(t) = sin t belongs to F n for each n, and This function provides estimates that are converse to (45), (46).Thus, equalities (4), (8) are valid for all n ≥ 3, 1 ≤ k < n.This completes the proofs of Theorems 1, 2.
Remark.We also have proved that for all n ≥ 2, 1 ≤ k < n, i. e., the operators T k,n are extremal operators in problem (2).Moreover, the sine function for n ≥ 3, 1 ≤ k < n and the function χ defined by ( 22) for n = 2, k = 1 are extremal in inequality (7), i. e., inequality (7) turns into an equality for them.Problem ( 2) is connected to one further similar problem.Denote by W r n the set of functions x ∈ L r ∩ L 2 such that their derivatives x (n−1) are locally absolutely continuous, and

Arestov
Consider the subclass Q r n = {x ∈ W r n : x (n) 2 ≤ 1} in the set W r n .For a linear bounded operator T in L r consider the quantity of the best approximation of the differentiation operator of order k in the space L 2 on the class Q r n by the set L r (N ) of linear bounded operators in L r with the norm T = T Lr→Lr ≤ N ; for r = ∞ we consider the space C of continuous functions in the place of L r .
For r = 2 and all n ≥ 2, 1 ≤ k ≤ n − 1, problem (47) was solved by Yu.N. Subbotin and L. V. Taikov [9]; in particular, they gave an extremal operator T 0 k,n which provides the lower bound in (47).The author's paper [3] gives a solution of problem (47 and an extremal operator is the one defined by formulae (31), ( 35), (40); this operator differs from the operator T 0 k,n from [9] and does not depend on r.According to a result from [9] for r = 2, formula (48) is also valid for n = 2, k = 1.In what follows we will show that, in contrast to the case when n ≥ 3, the quantity E 1,2 (N ) r , in general, depends on r, namely, E 1,2 (N ) ∞ > E 1,2 (N ) 2 .We will see that e 1,2 (N ) = E 1,2 (N ) ∞ and extremal operators in these problems coincide, so that problem (2) and problem (47) for r = ∞ coincide for all n ≥ 2, 1 ≤ k ≤ n − 1.The reason for this behaviour has been explained in the author's papers [2,3]; it is, in particular, connected to the fact that, in (47), it is enough to consider only operators T ∈ L r (N ) which are shift-invariant.The following statement holds.and the operator T 1,2 defined in (13) is extremal.
Remark.The operator T 1,2 is also extremal in problem (47) for r = 2, but One can conjecture that the operator T 1,2 is extremal for all r (1 ≤ r ≤ ∞).