On the best approximation of the differentiation operator Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, Vol. 1, No. 1, 2015

ON THE BEST APPROXIMATION OF THE DIFFERENTIATION OPERATOR1

Vitalii V. Arestov

Institute of Mathematics and Computer Science, Ural Federal University; Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences,

Yekaterinburg, Russia, vitalii.arestov@urfu.ru

Abstract: 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.

Key words: Differentiation operator, Stechkin's problem, Kolmogorov inequality

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 [1-6, 8-10] and the bibliography therein).

Let C = C(-to, to) 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 (-to, to) with the norm equal to the total variation V ^ of a measure n, and let Lr, 1 < r < to, be the space of measurable functions with the (finite) norm

\\x\\r = ^y \x(t)\r dt

The Fourier transform x of a function x £ L1 is defined by the formula

9(t) = / x(n)

In this case the inverse Fourier transform has the form

m = j m dn.

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 9 £ S' on the function x £ S by (d,x). The Fourier transform 9 of a functional 9 £ S' is the functional 9 £ S acting according to the rule (9,x) = (9,9). If 9 £ LY, 1 < y < 2, then 9 £ Ly, 1 + I = 1.

r

1The paper was originally published in a hard accessible collection of articles Approximation of Functions

by Polynomials and Splines (UNTs AN SSSR, Sverdlovsk, 1985), p. 3-14 (in Russian).

Denote by Fn, n > 1, the set of functions x E C whose derivatives x(n of order n are continuous functions such that their Fourier transforms are measures, i. e.

x<->(t) = / ^ *,<„), <u

where n = /j.x = x(n) E M. We will denote the total variation V n of a measure n in (1) by \\x(n \\V. We will consider the subclass Qn = {x E Fn : \\x(n)\\V < 1} in Fn. We study the problem of the best approximation of the differentiation operator of order k (0 < k < n) on the class Qn by the set L(N) of linear bounded operators T in the space C with the norm \\T\\ = \\T\\c< N. In other words, we study the quantity

e(N) = ekn(N) = inf {u(T) : T E L(N)}, (2)

where

u(T) = uk,n(T) = sup {\\x(k) - Tx\\c : x E Qn}. (3)

Our main results are the following two statements.

Theorem 1. For each h > 0 we have

k

ek ANk,n(h)) = ~hn-k, (4)

n

where

16 ^ 1

N,2(") = -n g (2JTT)3 ■ (5)

Nk,n(h) = ~~h-k, n > 3, 1 < k < n - 1. (6)

Theorem 2. Functions of the class Fn satisfy the sharp inequality

\\x(k)\\c < Kk,n \\x\\Cf \\x(n)\\V, (7)

and the smallest possible constant in this inequality is

^ (32 ^ 1 \2 1

'2 =V~3 h >1

Kkn = 1, n > 3, 1 < k <n. (8)

The fact that functions from the set Fn 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.

Proof 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

w(^) = sup{\\x(k)\\c : x E Qn, \\x\\c < $}, 0. (9)

It follows from the homogenity of u(S) (see, e.g. [11, p. 116]) that

u(5) = K5a, a = (10)

with K = w(1). This fact implies inequality (7), and the smallest possible constant in (7) is K = Kkn = w(1). Using S. B. Stechkin's method [8], one can show that e(N) > u(5) - N5 = K5a - N5

for each N > 0 and 5 > 0, that is, K < N51 a + e(N)5 a. Minimizing the latter expression with respect to 5 > 0, we obtain the inequality

K" < - (n—k rk (^ I (11)

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 £ Fn) 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 n be an odd 2n-periodic function which is defined on [0,n] by the formula n(t) = t — nt2. We have

8 1

= £ c, sm(^ + 1)t, Q = -2 -—-3. (12)

1=0

It is not difficult to see that the operator T = T1>2 defined by the formula

1 ro

2v (h)

(Ti,2x)(t) = — £ * {x(t + (27 + 1)v (h)) — x(t — (21 + 1)v (h))}, (13)

t=0

where v = v (h) = nr, is a linear bounded operator in C and

1 16 —' 1

|\TWc~c = ^E C = nhX = Ni2<h). <14)

1=0 1=0 v '

Introduce the function ^>(t) = (t — n(t))t 2. To determine its norm in the space C, we notice that t > n(t) > t — 112 for all t > 0, and thus |^(t)| < 1. Furthermore, if t £ [0,n], then ^(t) = 1. Consequently,

M\c = (15)

Now let us prove that the representation

x'(t) — (Ti,2x)(t) = —iv(h) J e2ntTi p(2nrv(h)) d^x(r) (16)

holds for functions x £ F2, where i = ¡x = x" is the measure from representation (1). First assume that a function x and its derivative x" both belong to L2. In this case, the function y = x' — Tx belongs to L2 as well, and it is easy to see that the Fourier transform of the function y has the form y(t) = —iv p(2ntv) x''(t). Taking the inverse Fourier transform, we obtain the expression

y(t) = x'(t) — (Tx)(t) = —iv j p(2nrv) xi'(r) e2nitT dr, (17)

which is representation (16) in this particular case.

Now let x be an arbitrary function from the class F2. Introduce the functions

c (t) = (1 +12)-1, at) = Z (et), z = z£ = x(£.

Obviously, z and z" belong to L2, and z" can be written as z" = z0 + z1 + z2, where z0 = x''(£, zi = 2x'C, z2 = xC. By (17),

z'(0) — (Tz)(0) = —iv(h) J Mr) {z0(T) + Zi(r) + Z2(r)} dr, (18)

with ^>o(r) = ^>(2nvr). We will take the limit of this relation as e ^ 0. Obviously, z'(0) = x'(0), and (Tz)(0) ^ (Tx)(0) as e ^ 0. Consider the integrals Jj (e) = f ^0(r) Zj (r) dr constituting the right-hand side of (18). The function ^>0 belongs to L2, thus, using the Holder inequality and the Parseval equality, we obtain

|Ji(e)| < \\^o\\2 \Z1\2 = \N\\2 \\zi\2 < 2 \\^o\\2 Wx'Wc \\Ce\\2 = 2 \\^o\\2 \\x'\\c \\C'\\2e1.

We see that J1(e) ^ 0 as e ^ 0. In a similar way one can show that J2(e) ^ 0 as e ^ 0. Now let us investigate the behaviour of J0(e). The Fourier transform of the function z0 = x"(£ is the convolution

Zo(r) = / (e(r - t) dy,x(t).

It follows that

Jo(e) = J <fo(r) z0(r) dr = J (e(t) J Po(t + r) d^(r) dt.

The family of the functions (£ is 5-shaped, consequently, Jo(e) tends to f ^o(r) d^(r) as e ^ 0. Thus, the limit of (18) as e ^ 0 is

x'(0) - (Tx)(0) = -iv J <p0(r) dßx(r), x e F2.

This is equivalent to the fact that representation (16) holds for each function x E F2.

Using (16) and (15), one can estimate quantity (3) for operator (13) from above, namely,

h

ui,2(Ti,2) < v(h) M\c = ~.

By (14), this yields

h

ei,2(Ni , 2(h)) < 2. (19)

Moreover, inequalities (11) and (19) give the estimate

32 ^ 1

K2 < I £ (27+17 <2°)

for the best constant in the inequality

\\x \\c

< K (\\x\\ c HAW)2 (21)

which is a particular case of (7).

Now we will derive statements converse to (20) and (19). For, consider the function

1 rn n _ u

X(t) = Xi , 2(t) = - / —-sin utdu. (22)

2 o sin u

Obviously, X is an odd entire function. Furthermore, since

m-i .2 -i r,

Esm2 mu 1 - cos 2mu

sin (2j + 1)u =-=-,

sin u 2 sin u

j=o

we have

, . 1 fn n - u , V-^ , , 1 fn n - u , , ,

X(t) = — I -sin utdu = > (t) + - / -sin ut cos2mudu, (23)

AW 2 Jo sin u j=o 2 Jo sin u ' v '

where

Vj (t) = (n - u) sin (2j + 1)u sin utdu. o

Each of the functions Vj is entire and it is easy to check that

n2

Vj (2j + 1) = —,

1 + cos tn ( 1 1 1 ,

Vj= 2 \ (2j + 1 -1)2 - (2j + 1 +yj • t = 2j + L (24)

For a fixed t, the value of the last integral in (23) tends to zero as m ^ o, therefore

X(t) - gVj(t) = i+T^ g{ - jW}. (25)

It follows from (24) that Vj(t) > 0 for t > 0 and Vj(2m + 1) = 0 for j = m. Hence, the function x is non-negative on the half-line (0, o), and

n2

X(2j + 1) = y, j = 0,1,... . (26)

Using the well-known identity

1 1 ^ ^

1 1 A 1

we obtain

sin2 nt n2 (t - k)2'

n2 1 + cos tn A ( 1 1

^ I (2 j + 1 - t)2 + (2 j + 1 + t)2 ( ' (27)

4 2 (2j + 1 -1)2 (2j + 1+1)2

It follows from relations (25)-(27) that

n2

0 < x(t) < T, t > 0,

n2

\\x\\c(-«,,«,) = X(2j + 1) = y, j > 0. (28)

Further on, using (25) we find that

. 1 u(n - uK , ^ 1 / x

x'(0) = - -- du = 4> ---rrr. (29)

2 Jo sin u ¿o (2! + 1)3 v ;

Now let us calculate the integral

J =1 tn u2(n- u) du

2 o sin u Taking n - u as a new variable, we obtain

1 fn u2(n - u) + (n - u)2u , n C u(n - u) , ^ 1

J = 7/ —-^-— du = -/ -J- du = 2n> ——.

4 Jo sinu 4 Jo sinu ^ (2! + 1)3

Denote by y the odd function which vanishes for u > n and is y(u) = nnU for u E (0, n). The inverse Fourier transform z = y of this function

1 fn n — u

z(t) = y(t) = - —-sin 2ntudu

2 o sin u

is equal to ix(2nt). Therefore,

n2

NIc = T,

z'(0)= i2nx/(0) = 8n^ -—-3 ,

e=o v ;

rn X 1

\z"Wi = (2n)2yo u2y(u) du = (2n)3 Y, {2£ + 1)3 •

Thus, the function z belongs to F2 and provides the following estimate from below for the best constant K in (21):

K2 > |z'(0)|2 32 ^ 1 , .

K = n3 g (30)

Inequalities (30), (20), (19), (11) imply the relations

K22 = I g (2!^- * 2h)) = h-

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

Qn(S) = {x e S : ||XW||i < 1}C Qn

was, in fact, given, and the value of the best constant Kk,n(S) in inequality (7) on the set of functions x e 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 n be a 2n-periodic odd function which is defined on [0, n] by the formulae

1 /2\n-1 n n(t) = t - - - tn, t e n \ n !

0-2

n(t) = n(n -1), t e

TT

L2

Using the function n, we define a function p on the real line by p(t) = (t — n(t)) t n. The functions n and p satisfy the following properties (see [3, proof of Theorem 4.1]):

<x

n(t) = £ ce sin (2! + 1)t, (—1)% > 0, ! > 0,

1=0

nn — 1

C

= n

2n

1=0

n N 1 ( 2 xn-1

c = Plol 1

V2/ n \n

Now take h > 0, put v = v(h) = ^ and define an operator T = T1 n in C by the formula

1 °° 2v (h)

(T1 ,nx)(t) = —— Y ce {x(t + (2^ + 1)v) - x(t - (21 + 1)v)}• (31)

t=o

It is clear that Ti>n is a linear bounded operator in C and

1 ^ 1 \\Ti,n\\c= M = ^ ■

(32)

£=o

As in the proof for n = 2 above, one can show that the representation

c'(t) - (Ti,nx)(t) = (-iv)^ e2ntTi V(2nvr) dx(n)(r)

holds for all functions x E Fn. It follows from this representation that

ui,n(Ti,n) < v

ni

h

ni

c=

n

(33)

(34)

In the case n = 3, k = 2, denote by n the even 2n-periodic function, defined on [0,n] by the formulae

n(t) = t2 - ^t3, t E n(t) = n(n - t), t E

01

n

L2 ,n

We have

It follows that

n(t) = £ c£ (1 - cos 2!t), ct =-n3

i=i

n3

2 E ce = 2 E C2j+i

j=o

|c = n

t=i

.2) 12

Moreover, it is easy to see that the function V(t) = (t2 - n(t))t-3 satisfies the property

mc = v (n) =

Now we define a bounded linear operator T2,3 in the space C by the formula

(T2,3x)(t) = -^ c£{x(t + 2!v) - 2x(t) + x(t - 2tv)}. For this operator we have

e=i

2 ^ h-2

\T2,3\c^c < N = -3-

t=i

For each x E F3 we have the representation

x<« - -vj^ V(2nrv) ^);

it follows from this representation that

2h

u2,3(T2,3) < V \\v\\c = Y. Now we define an operator Tk,n for arbitrary n > 3, 1 < k < n by the formula

Tk,n — Tm,n-k+m Tk-m,m , 0 < m < k < n.

(35)

(36)

(37)

(38)

(39)

For example, we can take m = k — 1, then (39) takes the form

Tk,n = Tk-1,n-1 T1,n- (40)

and if the operator Tk,n is defined for all 3 < n < n, 1 < k < n — 1 (and n = n, k = 1), then using (40) we define Tk,n for n = n, k = 2,... ,n — 1. Let us check that

IlTk,nII< — h-k, (41) n

k

ukn(Tk,n) < n hn-k (42)

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

IITk nH < ITm,n-k+m H |Tk-n,m|. (43)

For x e Fn, write x(k) — Tk nx in the form

dm x(k-m)

X ^ Tk,nx — „nm Tm,n-k+mX ^ + Tm,n-k+m(X ^ Tk-m,nx) •

This representation gives the estimate uk,n (Tk

Inequalities (41) and (42) follow from estimates (43) and (44) by induction.

Statements (41), (42) imply the following estimate from above for quantity (2) for n > 3, 1 < k < n:

n— k k k n k

ek n -h—k) <- hn—k, h > 0. (45)

nn

Inequality (11) gives the estimate from above

Kk n < 1 (46)

for the best constant in inequality (7).

On the other hand, the function ^(t) = sin t belongs to Fn for each n, and

IWIc = I№(k)IIc = II^(n) IIv = 1.

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

IITk n^c^c = Nk, n(h)-

k

ek,n(Nk,n(h)) = Uk,n(Tk,n(h)) = - hn—k

n

for all n > 2, 1 < k < n, i. e., the operators Tk,n are extremal operators in problem (2). Moreover, the sine function for n > 3, 1 < k < n and the function x 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 Wrn the set of functions x e Lr n L2 such that their derivatives x(n 1) are locally absolutely continuous, and x(n) e L2.

Consider the subclass Qrn = {x E Wn : \\x(n)\\2 < 1} in the set Wn. For a linear bounded operator T in Lr consider the quantity

U(T) = sup {\\x(k) - Tx\\2 : x E Qrn}.

We are interested in the quantity

Ek,n(N)r = inf {U(T) : T ELr (N)} (47)

of the best approximation of the differentiation operator of order k in the space L2 on the class Qrn by the set Lr(N) of linear bounded operators in Lr with the norm \\T\\ = \\T\\Lr~^Lr < N; for r = o we consider the space C of continuous functions in the place of Lr.

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 Tf, n which provides the lower bound in (47). The author's paper [3] gives a solution of problem (47) for 1 < r < o and n > 3 (1 < k < n). Namely, it is shown that

n- k k k n k

Ekn\ -h-^ = - hn-k, h > 0, (48)

' V n )r n

and an extremal operator is the one defined by formulae (31), (35), (40); this operator differs from the operator Tj!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 Ei , 2(N)r, in general, depends on r, namely, Ei , 2(N> Ei , 2(N)2. We will see that ei , 2(N) = Ei , 2(N)^ and extremal operators in these problems coincide, so that problem (2) and problem (47) for r = o 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 E Lr(N) which are shift-invariant. The following statement holds.

Theorem 3. If n = 2, k = 1, r = o, then for each h > 0 we have

h 16 1

EI,2(NI,2(-))^ = -, NI,2(-) = ^ £ (^+1)3 , (49)

and the operator Ti,2 defined in (13) is extremal.

Proof. Representation (17) holds for functions x E W|°. Therefore,

\\x' - Ti,2x\2 = V \\Vo xi2 < V \\v\\c \W'\\2 = V \\v\\c \\x"h, and, consequently, U(Ti>2) < |. Moreover, \\Ti)2\c^c = Ni;2(h). Hence,

h

Ei,2(NI,2(-))^ < 2. (50)

It follows from Theorem 3.1 in [3] that (cf. (11))

2(NEi,2(N)M)2 > K(S), (51)

where K(S) is the best constant in inequality (21) on the set S. Let us prove that K(S) = Ki>2. Consider the family of the functions Xe(t) = e-£2t2 x(t), where the function x is defined by (22). It is easy to see that xe E S, xe(0) = x'(0), and \\xe\c ^ \\x\c, wx'! \\x''\i as e 0. From these facts we conclude that K(S) > Ki>2, and, consequently, K(S) = Ki>2. This yields an inequality converse to (50) and thus proves Theorem 3. □

Remark. The operator Ti>2 is also extremal in problem (47) for r = 2, but

\\Ti,2\\L2^L2 = — < \\Ti,2\\c^c. (52)

One can conjecture that the operator Ti>2 is extremal for all r (1 < r < o).

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