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VOL. LXVIII 1995 FASC. 1

ON M ¨ UNTZ RATIONAL APPROXIMATION IN MULTIVARIABLES

BY

S. P. Z H O U (EDMONTON, ALBERTA)

The present paper shows that for any s sequences of real numbers, each with infinitely many distinct elements, {λ j n }, j = 1, . . . , s, the rational com- binations of x λ

1 m1

1 x λ

2 m2

2 . . . x λ

s

s

ms

are always dense in C I

s

.

1. Introduction. Let C [0,1] be the class of all real continuous functions in [0, 1]. For f ∈ C [0,1] ,

ω(f, t) = max

0<h<t,x∈[0,1−h]

|f (x + h) − f (x)|, kf k = max

x∈[0,1] |f (x)|.

Given a subspace S of C [0,1] , let

R(S) = {P (x)/Q(x) : P (x) ∈ S, Q(x) ∈ S, Q(x) > 0, x ∈ (0, 1]}, where we assume that lim x→0+ P (x)/Q(x) = P (0)/Q(0) is finite in the case Q(0) = 0. For a sequence of real numbers Λ = {λ n } n=0 , write

R(Λ) = R(span{x λ

n

}).

From M¨ untz’s theorem (cf. [2]), it is well-known that the combinations of x λ

n

for

(1) 0 = λ 0 < λ 1 < λ 2 < . . . are dense in C [0,1] if and only if

X

n=1

1 λ n

= ∞.

As to the rational case, in 1976, Somorjai [6] showed a beautiful result that under (1), R(Λ) is always dense in C [0,1] . In 1978, Bak and Newman [1] proved that if λ n is a sequence of distinct positive numbers, then R(Λ) is dense in C [0,1] as well. Recently, our work [7] showed that the above result

1991 Mathematics Subject Classification: 41A20, 41A30, 41A63.

[39]

(2)

also holds for any sequence of real numbers with infinitely many distinct elements.

On the other hand, S. Ogawa and K. Kitahara [5] gave a generalization of M¨ untz’s theorem to multivariable cases. They proved ( 1 ) that for two given positive monotone sequences {α i }, {β j }, the set {1} ∪ {x α

i

} ∪ {y β

j

} is complete in C I

2

if and only if P ∞

i=1 1/α i and P ∞

j=1 1/β j diverge, where I s = {X = (x 1 , . . . , x s ) : 0 ≤ x j ≤ 1, 1 ≤ j ≤ s},

and C I

s

is the class of all continuous functions on I s .

For many reasons, it is quite reasonable to conjecture that the conclusion corresponding to that of [7] will hold for M¨ untz rational approximation in the multivariable case, that is, for any s sequences of real numbers {λ j n }, j = 1, . . . , s, each with infinitely many distinct elements, the rational com- binations of {x λ

1 m1

1 x λ

2 m2

2 . . . x λ

s

s

ms

} are always dense in C I

s

. Since rational combinations are not linear, it is not a trivial work.

The present paper will prove that this is true.

2. Result and proof

Theorem. Let Λ j = {λ j n }, j = 1, . . . , s, be s sequences of real numbers, each with infinitely many distinct elements. Then R(Λ 1 × . . . × Λ s ) is dense in C I

s

.

We need the following lemmas from the univariable case, the first two of which are due to Somorjai [6] and the author [7]. We will, however, give the sketch of proofs here for the sake of completeness.

Lemma 1 (Somorjai [6]). Let {λ n } be a sequence of real numbers such that λ n → +∞ as n → ∞. Given N ≥ 1, for any f ∈ C [0,1] , there are an integer n N and an operator

N

X

k=0

f  k N

 Z k (x)

Z(x) =: r 1 N (f, x) ∈ R({λ j } n j=0

N

) with

0 ≤ Z k (x) ∈ span{x λ

j

} n j=0

N

, Z(x) =

N

X

k=0

Z k (x) such that

kf − r 1 N (f )k = O(ω(f, N −1 )).

P r o o f. We select a sequence {λ n

j

} n j=1

N

from Λ by induction. Let λ n

0

be any element from Λ, and Z 0 (x) = x λ

n0

. Choose λ n

j+1

with the following properties:

( 1 ) For convenience, we only state their result for two variables.

(3)

Z j+1 (x) =

 N

j + 1 x

 λ

nj+1

≤ N −1 Z j (x) for x < j N , Z j+1 (x) > N Z j (x) for x > j + 2

N . Define

r N 1 (f, x) =

N

X

k=0

f  k N

 Z k (x) P N

v=0 Z v (x) for f ∈ C [0,1] . Then by calculation

f (x) − r 1 N (f, x) = O(ω(f, N −1 )).

Lemma 2 (Zhou [7]). Let {λ n } be a sequence of real numbers such that λ n → −∞ as n → ∞. Given N ≥ 1, for any f ∈ C [0,1] , there are an integer n N and an operator

N

X

k=0

f  k N

 C k (x)

C(x) =: r N 2 (f, x) ∈ R({λ j } n j=0

N

) with

0 ≤ C k (x) ∈ span{x λ

j

} n j=0

N

, C(x) =

N

X

k=0

C k (x) such that

kf − r 2 N (f )k = O(ω(f, N −1 )).

P r o o f. Similar to Lemma 1, let λ n

1

be any element from Λ, and C 1 (x) = x λ

n1

. Choose λ n

j+1

satisfying

C j+1 (x) =

 N

N − j x

 λ

nj+1

≥ N C j (x) for x < N − j − 1

N ,

C j+1 (x) < N −1 C j (x) for x > N − j + 2

N .

For f ∈ C [0,1] , define r N 2 (f, x) =

N

X

k=1

f  N − k + 1 N

 C k (x) P N

v=1 C v (x) . Then the required result follows.

Lemma 3. Let {λ n } be a sequence of real numbers with infinitely many distinct elements such that λ n → l as n → ∞ with −∞ < l < ∞. Given N ≥ 1 and ε > 0, there are an integer n N and an operator

N

X

k=0

f (e 1−N/k ) D k (x)

D(x) =: r 3 N (f, x) ∈ R({λ j } n j=0

N

)

(4)

with D k (x), D(x) ∈ span{x λ

j

} n j=0

N

such that

kf − r N 3 (f )k ≤ 2ω(g, N −1/2 ) + kf kε, where g(u) = f (e 1−1/u ). Precisely, we have

(2) D k (x)

D(x) = G k (x) + H k (x),

(3) G k (x) = N

k

 −1

ln(x/e)

 k 

1 + 1

ln(x/e)

 N −k

, and

(4) |H k (x)| ≤ ε

N + 1 .

P r o o f. There are two possibilities: (i) There is a subsequence {λ n

k

} of {λ n } which strictly increases to λ < +∞ (in symbols λ n

k

% λ < +∞) as k → ∞; (ii) there is a subsequence {λ n

k

} which strictly decreases to λ > −∞ (in symbols λ n

k

& λ > −∞) as k → ∞. We will prove Lemma 3 in these two cases separately.

C a s e (i). For convenience, we still write λ n

k

as λ n . So under the hy- pothesis, λ n % λ < +∞ as n → ∞. Let α 0 < α 1 < . . . , and let P k (x) denote the kth divided difference of (x/e) α at α = α 2N −1 , α 2N −2 , . . . , α 2N −k−1 for k = 0, 1, . . . , N − 1, that is,

P 0 (x) = P 0 (x, α 2N −1 ) = (x/e) α

2N −1

,

P 1 (x) = P 1 (x, α 2N −1 , α 2N −2 ) = (x/e) α

2N −1

− (x/e) α

2N −2

α 2N −1 − α 2N −2 , in general,

P k (x) = P k (x, α 2N −1 , . . . , α 2N −k−1 )

= P k−1 (x, α 2N −1 , . . . , α 2N −k ) − P k−1 (x, α 2N −2 , . . . , α 2N −k−1 )

α 2N −1 − α 2N −k−1 ,

0 ≤ k ≤ N − 1, and

P N (x) = P N (x, α N , . . . , α 0 ).

By the mean value theorem (5) P k (x) = (x/e) η

k

ln k (x/e)

k! ,

α 2N −k−1 ≤ η k ≤ α 2N −1 , k = 0, 1, . . . , N − 1, (6) P N (x) = (x/e) η

N

ln N (x/e)

N ! , α 0 ≤ η N ≤ α N .

(5)

Now let f ∈ C [0,1] . Then g(u) = f (e 1−1/u ) ∈ C [0,1] . Write B N (f, x) =

N

X

k=0

f  k N

N k



x k (1 − x) N −k

=

N

X

k=0

f  k N

N k

 N −k X

j=0

(−1) j N − k j

 x j+k

=

N

X

k=0

f  k N

N k

 N X

j=k

(−1) j−k N − k j − k

 x j . For given N ≥ 1, the well-known Bernstein theorem implies that

kg(u) − B N (g, u)k < 3 2 ω(g, N −1/2 ), that is,

(7) kf (x) − B N (g, −1/ ln(x/e))k < 3 2 ω(g, N −1/2 ).

Choose sufficiently large m such that for k ≥ m, 0 < λ − λ k < ε/(4 N (N + 1)).

Set α k = λ m+k , k = 0, 1, . . . , 2N − 1. Define r 3 N (f, x) =

N

X

k=0

f (e 1−N/k ) N k

 P N

j=k (−1) 2j−k (N − j)! N −k j−k P N −j (x)

N !P N (x) .

Then r N 3 (f, x) is a rational combination of {x λ

j

} m+2N −1 j=m , and by (5), (6), r 3 N (f, x) =

N

X

k=0

f (e 1−N/k ) N k

 N X

j=k

(−1) j−k N − k j − k

 −1

ln(x/e)

 j

(x/e) η

j

with η 0 = 0, 0 < η j ≤ λ m+N − λ m ≤ λ − λ m , j = 1, . . . , N . Now write

N

X

j=k

(−1) j−k N − k j − k

 −1

ln(x/e)

 j

(x/e) η

j

=

N

X

j=k

(−1) j−k N − k j − k

 −1

ln(x/e)

 j

+

N

X

j=k

(−1) j−k N − k j − k

 −1

ln(x/e)

 j

((x/e) η

j

− 1)

=

 −1 ln(x/e)

 k 

1 + 1

ln(x/e)

 N −k

+ Σ 1 .

(6)

Since for η > 0,

1 − (x/e) η ln(x/e)

≤ η, we have

1 − (x/e) η

k

ln k (x/e)

≤ η k ≤ ε 4 N (N + 1) for k ≥ 1. Consequently,

1 | ≤ 4 −N (N + 1) −1 ε

N

X

j=k

N − k j − k



≤ ε

2 N (N + 1) ,

and thus (2)–(4) are proved. Now from (4), (7), together with H k (x) =

N k Σ 1 ,

kf (x) − r 3 N (f, x)k ≤ kf (x) − B N (g, −1/ ln(x/e))k + kf k

N

X

k=0

|H k (x)|

3 2 ω(g, N −1/2 ) + kf kε, that is, Lemma 3 holds true in Case (i).

C a s e (ii). We may assume that λ n & λ > −∞ as n → ∞. Take P k (x) = P k (x, λ m , . . . , λ m+k ), 0 ≤ k ≤ N − 1,

P N (x) = P N (x, λ m+N −1 , . . . , λ m+2N −1 ), and

r 3 N (f, x) =

N

X

k=0

f (e 1−N/k ) N k

 P N

j=k (−1) 2j−k (N − j)! N −k j−k P N −j (x)

N !P N (x) .

Similar to Case (i), for given ε > 0 and N ≥ 1, we can prove (2)–(4) and for sufficiently large m,

kf (x) − r N 3 (f, x)k ≤ 3 2 ω(g, N −1/2 ) + kf kε.

The proof of Lemma 3 is now complete.

P r o o f o f t h e T h e o r e m. Given a sequence with infinitely many distinct elements {λ n }, there are three possibilities: (i) {λ n } has at least one finite cluster point; (ii) one cluster point of {λ n } is +∞; (iii) one cluster point of {λ n } is −∞. Without loss of generality, we may assume

λ j n → +∞ as n → ∞, 1 ≤ j ≤ r, λ j n → −∞ as n → ∞, r + 1 ≤ j ≤ t, and

λ j n → l as n → ∞, t + 1 ≤ j ≤ s, −∞ < l < +∞.

(7)

For given N ≥ 1 and ε > 0, from Lemmas 1–3, we may select a common n N

such that

kf (X) − r N 1 (f, x j )k = O(ω x

j

(f, N −1 )), 1 ≤ j ≤ r, kf (X) − r N 2 (f, x j )k = O(ω x

j

(f, N −1 )), r + 1 ≤ j ≤ t, and

kf (X) − B N (g, −1/ ln(x j /e))k = O(ω x

j

(g, N −1/2 )), t + 1 ≤ j ≤ s, hold at the same time, where

ω x

j

(f, δ)

= max

0≤h≤δ |f (x 1 , . . . , x j + h, x j+1 , . . . , x s ) − f (x 1 , . . . , x j , x j+1 , . . . , x s )|.

Define

r N (X) = X

0≤j

1

≤N

. . . X

0≤j

s

≤N

f  j 1

N , . . . , j t

N , . . . , e 1−N/j

t+1

, . . . , e 1−N/j

s



× Z j

1

(x 1 )

Z(x 1 ) . . . Z j

r

(x r ) Z(x r )

C j

r+1

(x r+1 )

C(x r+1 ) . . . C j

t

(x t ) C(x t )

× D j

t+1

(x t+1 )

D(x t+1 ) . . . D j

s

(x s ) D(x s ) . Evidently,

r N (X) ∈ R(span{x λ

1i

} n i=0

N

× span{x λ

2i

} n i=0

N

× . . . × span{x λ

si

} n i=0

N

).

From (2), f (X) − r N (X)

= X

0≤j

1

≤N

. . . X

0≤j

s

≤N



f (X) − f  j 1

N , . . . , j t

N , e 1−N/j

t+1

, . . . , e 1−N/j

s



× Z j

1

(x 1 )

Z(x 1 ) . . . Z j

r

(x r ) Z(x r )

C j

r+1

(x r+1 )

C(x r+1 ) . . . C j

t

(x t ) C(x t )

× G j

t+1

(x t+1 ) . . . G j

s

(x s ) + Σ 3 := Σ 2 + Σ 3 , where by Lemma 3,

(8) |Σ 3 | ≤ 4 s kf kε.

Because f (X) ∈ C I

s

, there is a δ > 0 such that for |X − Y | = q P s

j=1 (x j − y j ) 2 < δ,

|f (X) − f (Y )| < ε,

(8)

while for |X − Y | ≥ δ,

|f (X) − f (Y )| ≤ 2δ −2 kf k

s

X

j=1

(x j − y j ) 2 , therefore in any case

|f (X) − f (Y )| ≤ ε + 2δ −2 kf k

s

X

j=1

(x j − y j ) 2 . Now

2 | ≤ ε + 2δ −2 kf k

 X

0≤j

1

≤N

. . . X

0≤j

s

≤N t

X

i=1

 x i − j i

N

 2

×

s

X

i=t+1

(x i − e 1−N/j

i

) 2 Z j

1

(x 1 )

Z(x 1 ) . . . Z j

r

(x r ) Z(x r )

× C j

r+1

(x r+1 )

C(x r+1 ) . . . C j

t

(x t )

C(x t ) G j

t+1

(x t+1 ) . . . G j

s

(x s )



= ε + 2δ −2 kf k

 r X

i=1 N

X

j

i

=0

 x i − j i

N

 2

Z j

i

(x i ) Z(x i )

×

t

X

i=r+1 N

X

j

i

=0

 x i − j i

N

 2

C j

i

(x i ) C(x i )

s

X

i=t+1 N

X

j

i

=0

(x i − e 1−N/j

i

) 2 G j

i

(x i )

 . Noting that

N

X

k=0

 x − k

N

 2

Z k (x) Z(x) = 2x

 x −

N

X

k=0

k N

Z k (x) Z(x)



 x 2

N

X

k=0

 k N

 2

Z k (x) Z(x)

 , from Lemma 1 we deduce that

N

X

k=0

 x − k

N

 2

Z k (x)

Z(x) = O(N −1/2 ).

The same results also hold for P N

k=0 (x − k/N ) 2 C k (x)/C(x) and for P N

k=0 (x − e 1−N/k ) 2 G k (x) by applying Lemmas 2 and 3. Altogether we have

(9) |Σ 2 | ≤ ε + 2δ −2 kf kO(N −1/2 ), thus combining (8) and (9) we get

|f (X) − r N (X)| = O(ε) + O(N −1/2 ),

which is the required result.

(9)

REFERENCES

[1] J. B a k and D. J. N e w m a n, Rational combinations of x λ

k

, λ k ≥ 0 are always dense in C [0,1] , J. Approx. Theory 23 (1978), 155–157.

[2] E. W. C h e n e y, Introduction to Approximation Theory , McGraw-Hill, 1966.

[3] G. G. L o r e n t z, Bernstein Polynomials, Toronto, 1953.

[4] D. J. N e w m a n, Approximation with Rational Functions, Amer. Math. Soc., Provi- dence, R.I., 1978.

[5] S. O g a w a and K. K i t a h a r a, An extension of M¨ untz’s theorem in multivariables, Bull. Austral. Math. Soc. 36 (1987), 375–387.

[6] G. S o m o r j a i, A M¨ untz-type problem for rational approximation, Acta Math. Acad.

Sci. Hungar. 27 (1976), 197–199.

[7] S. P. Z h o u, On M¨ untz rational approximation, Constr. Approx. 9 (1993), 435–444.

DEPARTMENT OF MATHEMATICS UNIVERSITY OF ALBERTA EDMONTON, ALBERTA CANADA T6G 2G1

Re¸ cu par la R´ edaction le 18.8.1993;

en version modifi´ ee le 30.3.1994

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