OF THE SPACE OF CONTINUOUS FUNCTIONS

VOL. 75 1998 NO. 1

ON APPROXIMATION BY LAGRANGE INTERPOLATING POLYNOMIALS FOR A SUBSET

OF THE SPACE OF CONTINUOUS FUNCTIONS

BY

S. P. Z H O U (HANGZHOU)

We construct a C

piecewise differentiable function that is not C

piece- wise analytic and satisfies a Jackson type estimate for approximation by Lagrange interpolating polynomials associated with the extremal points of the Chebyshev polynomials.

1. Introduction. Let C

be the class of functions which have k continuous derivatives on [−1, 1], in particular, C

= C

be the class of continuous functions on [−1, 1]. For a function f ∈ C

, let L

(f, x) be the nth Lagrange interpolating polynomial of f associated with the extremal points {x

} = {cos(j − 1)π/n}

of the Chebyshev polynomials in [−1, 1].

An explicit formula for L

(f, x) is L

(f, x) = ω

(x)

n

 −f (1) 2(x − 1) +

n

X

l=2

(−1)

f (x

) x − x

+ (−1)

f (−1) 2(x + 1)

 ,

where ω

(x) = √

1 − x

sin(n arccos x).

It is well known that L

(f, x) does not converge for all f ∈ C

. However, Mastroianni and Szabados [MS] considered a subset of C

and proved

Theorem 1. If f ∈ C

and if there is a partition of [−1, 1], −1 = a

< a

< . . . < a

= 1, such that each f |

(j = 0, 1, . . . , s) is a polynomial , then, for |x| ≤ 1, as n → ∞,

|f (x) − L

(f, x)| = O

 √ 1 − x

n

 min



1, 1

n min

|x − a

|

 . Recently, [Li] improved and generalized the above result to

1991 Mathematics Subject Classification: Primary 41A05.

[1]

Theorem 2. If f is C

piecewise analytic on [−1, 1] with singular points a

, . . . , a

, then, as n → ∞,

(1) |f (x) − L

(f, x)| = O  |ω

(x)|

n

 min



1, 1

n min

|x − a

|



holds uniformly for x ∈ [−1, 1].

[Li] uses the following definition: A function f defined on [−1, 1] is called C

piecewise analytic on [−1, 1] if f ∈ C

and if there is a partition of [−1, 1], −1 = a

< a

< . . . < a

= 1, such that each f |

(j = 0, 1, . . . , s) has an analytic continuation to [−1, 1]. The points a

, . . . , a

are called the singular points of f .

At the end of the paper, Li wrote: “We . . . note that our term ‘piecewise analytic’ . . . has a different meaning from the usual one: We require that each analytic piece has an analytic continuation to [−1, 1]. The technical requirement is needed because of our method. We do not know if there exists a function that is not C

piecewise analytic but satisfies the estimation (1).”

In the present paper we construct a C

piecewise differentiable function that is not C

piecewise analytic and satisfies (1). Finally, based on some observations, we raise an open question.

2. Result

Definition. Let k ≥ 0. A function f defined on [−1, 1] is called C

piecewise differentiable on [−1, 1] with singular points a

, . . . , a

if f ∈ C

and if there is a partition of [−1, 1], −1 = a

< a

< . . . < a

= 1, such that each f |

(j = 0, 1, · · · , s) has k + 1 continuous derivatives on [a

, a

], but f does not have the (k + 1)th derivative at the singular points a

, . . . , a

.

Theorem 3. Let k ≥ 0. There is a C

piecewise differentiable function f (x) on [−1, 1] with singular point zero which does not have the (k + 2)th derivative at the endpoints ±1 and for which

(2) |f (x) − L

(f, x)| = O  |ω

(x)|

n

 min

 1, 1

nx



holds uniformly for x ∈ [−1, 1].

P r o o f. Let T

(x) = cos(n arccos x) be the Chebyshev polynomial of degree n, and

S(x) =

 x

, x ≥ 0,

0, x < 0.

Set n

= 2, and n

= n

for j = 1, 2, . . . Define f (x) = T (x) + S(x) :=

∞

X

j=1

n

T

(x) + S(x).

We show this is the desired function.

Obviously f ∈ C

. The argument that f |

and f |

have k + 1 continuous derivatives on [−1, 0] and [0, 1] respectively is also trivial. Also, we note that T (x) evidently has 2k + 3 continuous derivatives at zero and S(x) does not have the (k + 1)th derivative there; consequently, f (x) does not have the (k + 1)th derivative at zero. All those facts mean that f is C

piecewise differentiable on [−1, 1] with singular point zero.

Let n

≤ n < n

, j = 1, 2, . . . For any x ∈ [−1, 1] we have T (x) − L

(T, x) = J (x) − L

(J, x),

where J (x) = P

l=j+1

n

T

(x). Now for l ≥ j + 1, T

(x) − L

(T

, x)

= ω

(x) n

 T

(1) − T

(x) 2(x − 1) +

n

X

m=2

(−1)

(T

(x) − T

(x

)) x − x

+ (−1)

(T

(x) − T

(−1)) 2(x + 1)

 .

Hence there are ξ

between x and x

and an absolute constant C > 0 such that

|T

(x) − L

(T

, x)|

≤ |ω

(x)|

n    

−T

(ξ

)

2 +

n

X

m=2

(−1)

T

(ξ

) + (−1)

T

(ξ

) 2

≤ Cn

|ω

(x)| (by the Markov inequality).

Then

|J (x) − L

(J, x)| ≤ C|ω

(x)|

∞

X

l=j+1

n

≤ Cn

|ω

(x)| ≤ Cn

|ω

(x)|.

Together with the known result (see [Li]) that

|S(x) − L

(S, x)| = O  |ω

(x)|

n

 min

 1, 1

nx



holds uniformly for x ∈ [−1, 1], we have thus finished the proof of (2).

Denote by t

the largest zero of T

(x). We know that (3) n

≤ 1 − t

≤ 1 − cos (k + 2)π

n

≤ (k + 2)

π

2n

since the zeros of T

(x) interlace those of T

(x) for l = 0, 1, . . . We also notice that (T

(1) − T

(x))/(1 − x) = T

(ξ) > 0, ξ ∈ [x, 1], for x ∈ [t

, 1], so that by (3),

T

(1) − T

(x)

1 − x > T

(1) − T

(t

) 1 − t

(4)

≥ 2n

(k + 2)

π

T

(1) ≥ C

n

for x ∈ [t

, 1], where C

is a positive constant only depending upon k.

Write

T

(1) − T

(t

) 1 − t

= 1

1 − t

j

X

l=1

n

(T

(1) − T

(t

))

+ 1

1 − t

∞

X

l=j+1

n

(T

(1) − T

(t

)) =: I

+ I

. We check that, by (4),

I

≥ C

j,

while by inequality (3), the Markov inequality and the definition of {n

},

|I

| = O(1)n

∞

X

l=j+1

n

kT

l

k = O(1)n

n

= O(1)n

. Therefore

T

(1) − T

(t

) 1 − t

≥ C

j.

As lim

t

= 1, we conclude that T (x) does not have the (k + 2)th

derivative at the endpoint 1. The same conclusion holds for the other end-

point −1. By noting that S(x) has derivatives of any order at ±1, we have

thus proved that f (x) does not have the (k + 2)th derivative at ±1. The

proof of Theorem 3 is complete.

3. Remark. Based on observations and calculations, we have reasons to believe that C

piecewise differentiable functions might achieve the required Jackson type estimate (1). Precisely, we raise the following question:

Problem. Let k ≥ 0. If f is a C

piecewise differentiable function on [−1, 1] with singular points a

, . . . , a

, does it hold that , for |x| ≤ 1, as n → ∞,

|f (x) − L

(f, x)| = O  |ω

(x)|

n

 min



1, 1

n min

|x − a

|



?

REFERENCES

[MS] G. M a s t r o i a n n i and J. S z a b a d o s, Jackson order of approximation by Lagrange interpolation, in: Proc. Second Internat. Conf. in Functional Analysis and Ap- proximation Theory (Aquafredda di Maratea, 1992), Rend. Circ. Mat. Palermo (2) Suppl. 1993, no. 33, 375–386.

[Li] X. L i, On the Lagrange interpolation for a subset of C

functions, Constr. Approx.

11 (1995), 287–297.

Department of Mathematics Hangzhou University Hangzhou, Zhejiang 310028 China

Received 11 December 1996