A note on Whitney's theorem

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the uniform metric by polynomials from πm and ωk(f, [a, b]) is the k-th modulus of smoothness of f in C[a, b]. Whitney's theorem states that for any f ∈ C[a, b] and.
A note on Whitney’s theorem Petar P. Petrov

Department of Mathematics, University of Sofia, Blvd. James Boucher 5, 1164 Sofia, BULGARIA E-mail: [email protected] Abstract Let πm be the space of all algebraic polynomials of degree less than or equal to m. For a given function f ∈ C[a, b], Em (f ) denotes the best approximation to f in the uniform metric by polynomials from πm and ωk (f, [a, b]) is the k-th modulus of smoothness of f in C[a, b]. Whitney’s theorem states that for any f ∈ C[a, b] and k ≥ 1, Ek−1 (f ) ≤ Cωk (f, [a, b]), where C is a constant independent on f . In this note we prove that the exact constant in the above inequality for k-monotone functions is equal to 1/2, namely, sup k [a,b]\π f ∈M+ k−1

1 Ek−1 (f ) = , ωk (f, [a, b]) 2

k where M+ [a, b] ⊂ C[a, b] is the cone of all k-monotone functions on [a, b].

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Introduction

Let πm be the space of all algebraic polynomials of degree less than or equal to m. For any function f ∈ C[a, b] we denote by ∆kh f (x) :=

k X

!

(−1)k−j

j=0

k f (x + jh), {x, x + kh} ⊂ [a, b], j

its k-th finite difference with step h > 0. The k-th modulus of smoothness of f in C[a, b] is defined by ωk (f, [a, b]) := sup |∆kh f (x)|. x,x+kh∈[a,b]

Let Em (f ) := min max |f (x) − P (x)| P ∈πm x∈[a,b]

be the best uniform approximation to f by polynomials from πm . Whitney [17] was the first who proved in 1957 that (1)

Wk :=

sup f ∈C[a,b]\πk−1

1

Ek−1 (f ) 0 such that {x, x + kh} ⊂ [a, b]. For example, f is non-negative if k = 0, monotone non-decreasing if k = 1 and convex if k = 2. An equivalent definition is: A function f ∈ C[a, b] is k-monotone if all k-th order divided differences f [x0 , x1 , . . . , xk ] :=

k X

f (xj ) k Y

j=0

,

a ≤ x0 < x1 < . . . < xk ≤ b,

(xj − xi )

i=0 i6=j

are non-negative. Denote by M+k [a, b] ⊂ C[a, b] the cone of all k-monotone functions on [a, b].

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Main result

The main result of this note is the following Theorem 1. If f ∈ M+k [a, b], then (3)

1 Ek−1 (f ) ≤ ωk (f, [a, b]). 2

k Moreover, the constant 1/2 in (3) is exact, i.e., there exist functions {fm }∞ m=1 ⊂ M+ [a, b]\πk−1 such that Ek−1 (fm ) 1 lim = . m→∞ ωk (fm , [a, b]) 2

Proof. Let Lk−1 (f, x0 , x1 , . . . , xk−1 ; x) ∈ πk−1 be the Lagrange polynomial, interpolating f at the points a ≤ x0 < x1 < . . . < xk−1 ≤ b. Let us denote by tj := a + jh,

j = 0, 1, . . . , k, 2

h := (b − a)/k,

the equidistant points in [a, b] with step h and set P1 (x) := Lk−1 (f, t0 , t1 , . . . , tk−1 ; x), P2 (x) := Lk−1 (f, t1 , t2 , . . . , tk ; x). Let f be an arbitrary function from M+k [a, b]. By the error representation formula f (x) − Lk−1 (f, x0 , x1 , . . . , xk−1 ; x) = f [x, x0 , . . . , xk−1 ](x − x0 ) . . . (x − xk−1 ) and the fact that f is k-monotone we get (−1)j (f (x) − P1 (x)) ≥ 0,

x ∈ [tk−1−j , tk−j ]

and (−1)j+1 (f (x) − P2 (x)) ≥ 0,

x ∈ [tk−1−j , tk−j ]

for j = 0, 1, . . . , k − 1. Therefore P1 (x) ≤ f (x) ≤ P2 (x),

x ∈ [tj−1 , tj ],

j = k, k − 2, . . . ,

P2 (x) ≤ f (x) ≤ P1 (x),

x ∈ [tj−1 , tj ],

j = k − 1, k − 3, . . . ,

and f (x) − P1 (x) + P2 (x) ≤ 1 |P2 (x) − P1 (x)|. 2 2

(4)

Polynomials P1 and P2 can be represented by the Newton interpolation formula as follows: P1 (x) = f [t1 ] + f [t1 , t2 ](x − t1 ) + . . . +f [t1 , t2 , . . . , tk−1 ](x − t1 ) . . . (x − tk−2 ) +f [t0 , t1 , . . . , tk−1 ](x − t1 ) . . . (x − tk−1 ) and P2 (x) = f [t1 ] + f [t1 , t2 ](x − t1 ) + . . . +f [t1 , t2 , . . . , tk−1 ](x − t1 ) . . . (x − tk−2 ) +f [t1 , t2 , . . . , tk ](x − t1 ) . . . (x − tk−1 ). From here,

(5)

|P2 (x) − P1 (x)| = khf [t0 , t1 , . . . , tk ]|(x − t1 ) . . . (x − tk−1 )| ∆k f (t0 ) = kh h k |(x − t1 ) . . . (x − tk−1 )| k!h ≤ ∆kh f (t0 ) ≤ ωk (f, [a, b]),

since max |(x − t1 ) . . . (x − tk−1 )| = (tk − t1 ) . . . (tk − tk−1 ) = (k − 1)!hk−1 .

x∈[a,b]

Now, inequality (3) is a direct consequence by (4) and (5). Further, consider the functions 

fm (x) := −Tk−1

2x b − 1/m + a − b − 1/m − a b − 1/m − a 3



+ C(k, m)(x − b + 1/m)k+ ,

m > 1/h,

where Tk−1 is the (k − 1)-st Chebyshev polynomial of the first kind and the constant k

C(k, m) = m





1 + Tk−1

b + 1/m − a b − 1/m − a



is chosen so that fm (b) = 1. Since (k) fm (x) = k!C(k, m)(x − b + 1/m)0+

is non-negative and monotone non-decreasing, then fm ∈ M+k [a, b] ∩ M+k+1 [a, b]. By a well known result for k-monotone functions (see, e.g., [1, p. 61]) it follows that (6)

ωk (fm , [a, b]) =

∆kh fm (a)



= 1 + Tk−1

b + 1/m − a b − 1/m − a



On the  other hand, since fm takes the values ±1 at k + 1 points (the extremal points of 2x Tk−1 b−1/m−a ee-Pousin’s theorem − b−1/m+a b−1/m−a and the point b) with alternating signs, Vall´ ([16]) implies that Ek−1 (fm ) ≥ 1.

(7) By (3), (6) and (7) we obtain

Ek−1 (fm ) 1 1  . ≥ ≥ 2 ωk (fm , [a, b]) 1 + Tk−1 b+1/m−a b−1/m−a Therefore, lim

m→∞

1 Ek−1 (fm ) = . ωk (fm , [a, b]) 2

The proof is complete. 2

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