NEARLY COMONOTONE APPROXIMATION. Abstract. We discuss the degree of approximation by polynomials of a function f that is

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1 NEARLY COMONOTONE APPROIMATION D. Leviatan I. A. Shevchuk Abstract. We discuss the degree of approximation by polynomials of a function f that is piecewise monotone in [?; ]. We would like to approximate f by polynomials which are comonotone with it. We show that by relaxing the requirement for comonotonicity in small neighborhoods of the points where changes in monotonicity occur near the endpoints, we can achieve a higher degree of approximation. We show here that in that case the polynomials can achieve the rate of!3. On the other h, we show in another paper, that no relaxing of the monotonicity requirements on sets of measures approaching 0, allows!4 estimates. x. Introduction Let I := [?; ], for s let Y := fy i g s i=0 ;? = y s < < y < y 0 =. Finally let () (Y ) be the set of continuous functions f on I, such that f is nondecreasing on [y i ; y i? ], when i is odd it is nonincreasing on [y i ; y i? ], when i is even, set A polynomial P n (x) := s? Y i= (x? y i ): is said to be comonotone with f 2 () (Y ) on the set E I, if P 0 n(x)(x) 0; x 2 E. Note that if f 2 C (?; ), then f 0 (x)(x) 0; x 2 (?; ), if only if f 2 () (Y ). A. S. Shvedov [0] proved that for each Y there exists a constant c(y ), such that for every f 2 () (Y ) all n an algebraic polynomial P n, of degree n, which is comonotone with f on I, exists satisfying (.) kf? P n k C(I) c(y )! 2 (f; =n); 99 Mathematics Subject Classication. 4A0, 4A25, 4A29. Key words phrases. Approximation by polynomials, Shape preserving approximation. Part of this work was done while the authors were guests of Prof. R. A. DeVore at the University of South Carolina. They acknowledge partial support by ONR grant N by DoD grant N

2 where! k (f; ) denotes the modulus of smoothness of order k, of f. (Earlier DeVore [2] proved (.) for the case s =, which is the case where f is monotone of course the dependence of c(y ) on Y is meaningless, i.e., c is an absolute constant.) More recently, R. A. DeVore. M. Yu [4] G. A. Dzyubenko [6] have shown that one can get also pointwise estimates, namely, (.') jf(x)? P n (x)j c(y )! 2 (f; n (x)); where n (x) := p? x 2 =n + =n 2. On the other h it is known (see [0]) that in (.) (.'), one cannot replace! 2 by! k with any k 3. It is quite natural to ask whether one can strengthen (.) in the sense of being able to replace! 2 by moduli of smoothness of higher order, if one is willing to allow P n not to be comonotone with f on a rather "small" subset of I. This indeed turns out to be possible for! 3, as we show in Theorem. However, even this improvement comes to a halt, it cannot be extended to! 4, thus not to! k for any k > 3. We devote a separate paper [3] to proving this assertion when f is monotone. Here we will only state the result in the general case (see Theorem 4 below) the proof is a modication of [3]. We begin with some notation needed for the statement of Theorem. Let x? :=, x n+ :=? for eac = 0; : : : ; n, set x j := x j;n := cos (j=n), I j := I j;n := [x j ; x j? ], := ;n := ji j j := x j?? x j. For later reference we need the following well known relations (see e.g. [7]) (.2) n (x) < < 5 n (x); x 2 I j ; (.3) < 3 ; j = ; : : : ; n; (.4) 2 n(y) < 4 n (x)(jx? yj + n (x)); x; y 2 I; which in turn implies (.5) 2(jx? yj + n (x)) > jx? yj + n (y); x; y 2 I; 2

3 (.6) c jx? x j j + 2 C n (x) jx? x j j + n (x) ; =2 jx? x j j + : Given Y, let O i := O i;n (Y ) := (x j+ ; x j?2 ); if y i 2 [x j ; x j? ); O(n; Y ) := s? [ i= O i ; n ; O(0; Y ) := [?; ]; O (n; Y ) := O(n; Y ) [ I [ I n ; n ; O (0; Y ) := [?; ]: We rst prove Theorem. There are constants c = c(s) C(s) for which, if f 2 () (Y ), then for every n >, a polynomial P n of degree not exceeding n, which is comonotone with f on I n O ([n=c]; Y ) exists, such that (.7) jf(x)? P n (x)j C(s)! 3 (f; n (x)): We are able to obtain estimates involving moduli of higher orders for classes of dierentiable functions, namely, Theorem 2. Let k 2 be xed. Then there are constants c = c(s; k) C(s; k) for which, if f 2 () (Y ) \ C [?; ], then for each n k?, a polynomial P n of degree not exceeding n, which is comonotone with f on I n O([n=c]; Y ), exists such that (.8) jf(x)? P n (x)j C(s; k) n (x)! k? (f 0 ; n (x)); x 2 I: It is interesting to note that the dierentiablity of f, without giving up some small neighborhoods of the points Y, in general does not allow statements like (.8). Indeed among others, it is shown in [8] that there is an f 2 () (Y ) \ C [?; ], with s > number of chnages of monotonicity, which is thrice dierentiable in (?; ) such that 3

4 (? x 2 ) 3=2 f (3) (x) is bounded there, yet the least distance (in the sup-norm) between f polynomials which are truly comonotone with it, can be made as large as one wishes. (There are other interesting phenomena for truly comonotone approximation, the interested reader is referred to [8].) Observe that O(; Y ) = O (; Y ) = [?; ], thus it is clear that for n < 2c, we place no monotonicity constraint on the approximating polynomials. Therefore Theorems 2 follow from the well-known unconstrained Timan{Dzjadyk{Freud{Brudnyi estimates we only have to prove them for larger n. We shall make no attempt to estimate how small the constants c in the above theorems can be. Obviously, the smaller they are the smaller the neighborhoods O ([n=c]; Y ) O([n=c]; Y ) are, thus the stronger the results are. However, we feel it is important to point out that c cannot be too small or the above theorems become false for s > 2. To this end we prove the following result in Section 5. Theorem 3. For each A n 60A, there exists a collection Y n := fy i g 3 i=0, a function f = f n 2 () (Y n ), such that any polynomial P n of degree not exceeding n which satises Pn(x)f 0 0 (x) 0; x =2 O (8n; Y n ); necessarily satises also (.9) f? P n > A:! (f; n ()) Note that the collection Y n depends on A that if we stated Theorems 2 with constants that depend on Y, then obviously we would not have the analogue of Theorem 3. Also, evidently when A increases, n is taken bigger bigger. Indeed, for small (xed) n, it is possible to take c in Theorems 2 as small as we wish if we are willing to pay by enlarging C. Furthermore, if s = or 2, then it is possible to take c arbitrarily small (at the expense of increasing C). Finally one should note that for any s, the neighborhoods of the endpoints in Theorem, can be taken to be of length of arbitrarily small () proportion of =n 2 while allowing C := C(s; ). 4

5 To conclude this section, we state without proof (the proof will be given elsewhere) the following result which asserts that Theorem cannot be valid for higher moduli of smoothness (see [3]). To this end, given > 0 a function f 2 () (Y ), we denote E () n (f; ) := inf kf? P n k C(I) ; P n where the inmum is taken over all polynomials P n of degree not exceeding n satisfying meas(fx : P 0 n (x)(x) 0g \ I) 2? : Theorem 4. For each sequence = f n g n=, of nonnegative numbers tending to 0, there exists a function f := f 2 () (Y ), such that (.0) lim sup n! Throughout this paper we take k 2. E () n (f; n )! 4 (f; =n) = : In the sequel we will have constants which depend on s k. If they are independent of any other term, then we will not explicitly write this dependence. However, we will use the notation c C to denote such constants which are of no signicance to us may dier on dierent occurrences, even in the same line; we will have constants with indices c ; c 2 ; : : : C ; C 2 ; : : : when we have a reason to keep trace of them in the computations that we have to carry in the proofs. x2. Auxiliary lemmas Since we deal with functions f, which are piecewise monotone, then f 0 exists a.e. in I. We will use the max-norm of f as well as the norm of f 0 in L (when applicable). Thus, for any interval J I, let us denote kfk J := kfk L (J); which is obviously compatible with the max-norm whenever f is continuous. Throughout this section, n is going to be xed so that we would not have to carry n as an index for the intervals etc. First we prove 5

6 Lemma. Let H 0 := 0 < H < H 2 < H 3, j := H j? H j?, j = ; 2; 3, be such that 3 < 2= j < 3, j = ; 3; let f 2 C[0; H 3 ] be nondecreasing in [0; H 3 ]. Then there is a quadratic polynomial P 2, interpolating f at H H 2, such that kf? P 2 k [H ;H 2 ] c! 3 (f; 2 ; [0; H 3 ]); P 0 2(x) 0; x 2 [H ; H 2 ]: Proof. Let L 0 (x) := L(x; f; 0; H ; H 2 ) L 2 (x) := L(x; f; H ; H 2 ; H 3 ) be the Lagrange polynomials of degree 2, interpolating f at the points 0; H ; H 2 H ; H 2 ; H 3, respectively let L (x) := L(x; f; H ; H 2 ) be the linear function which interpolates f at H H 2. If L 0 0(x) 0, or L 0 2(x) 0, for x 2 [H ; H 2 ], then the assertion follows from Whitney's inequality. Otherwise, we have whence L 00 0(x) 0 L 00 2(x) 0; x 2 [H ; H 2 ]; L 2 (x) L (x) L 0 (x); x 2 [H ; H 2 ]: Applying Whithey's inequality we get, for x 2 [H ; H 2 ]: L (x)? f(x) L 0 (x)? f(x) kl 0? fk [H ;H 2 ] kl 0? fk [0;H3 ] c! 3 (f; 2 ; [0; H 3 ]); f(x)? L (x) f(x)? L 2 (x) kf? L 2 k [H ;H 2 ] kf? L 2 k [0;H3 ] c! 3 (f; 2 ; [0; H 3 ]): Hence Next we have kl? fk [H ;H 2 ] c! 3 (f; 2 ): 6

7 Lemma 2. If f 2 C [0; h] f 0 (x) 0 for x 2 [0; h], then there is a polynomial P k? of degree k? such that (2.) kf? P k? k [0;h] ch! k? (f 0 ; h; [0; h]); (2.2) f(0) = P k? (0); f(h) = P k? (h) (2.3) P 0 k? (x) 0; x 2 [0; h]: Proof. If f is a polynomial of degree < k, then obviously there is nothing to prove. Otherwise let P be the polynomial of the best uniform approximation of f 0 k?2 in [0; h], denote E k?2 := kf 0? Pk?2k [0;h] > 0: Set P k? (x) := f(0) + hr 0 f(h)? f(0) (P k?2 (u) + E k?2) du Z x 0 (P k?2 (u) + E k?2) du; where it is readily seen that the denominator is not zero that (2.3) follows by P k?2 (u) + E k?2 f 0 (u) 0; u 2 [0; h]: Also, it is evident that (2.2) holds thus we only have to prove (2.). x 2 [0; h] we have To this end, for f(x)? P k? (x) = Z x 0? (f 0 (u)? P k?2(u)? E k?2 ) du Z h 0 (f 0 (u)? P k?2(u)? E k?2 ) du 7 xr 0 hr 0 (P (u) + E k?2 k?2) du : (P k?2 (u) + E k?2) du

8 Hence kf? P k? k [0;h] 2hE k?2 : Now (2.) follows by Whitney's inequality. Denote by k the collection of continuous piecewise polynomials of degree k with the knots at the x j 's. Thus, S 2 k is dierentiable in I except perhaps at the x j 's. We denote this derivative by S 0. Let ' 2 k, i.e., '(0+) = 0 '(t) is nondecreasing while t?k '(t) is nonincreasing on (0; ). We will use the ordinary notation f 2 H ' k f 2 W H ' k, respectively, for f with! k (f; ) ' for dierentiable f with! k (f 0 ; ) '. Then, Lemmas 2 readily imply the following Lemmas 3 4, respectively. Lemma 3. Let ' 2 3. If f 2 H ' 3 f 2 () (Y ); then there is an S 2 2 such that kf? Sk Ij c'( ); j = ; : : : ; n; S 0 (x)(x) 0 in I n O T Lemma 4. Let! 2 k? '(t) := t!(t). If f 2 W H! k? (Y ), then there is an S 2 k? such that kf? Sk Ij c'( ); j = ; : : : ; n; S 0 (x)(x) 0 in I n O: The following lemma is proved very much like [7] Lemma 5.4 (see also a simpler variant [8] Theorem ). Lemma 5. Let ' 2 k suppose that f is locally absolutely continuous, that kf 0 k Ij '( ) j = ; : : : ; n that f 0 (x)(x) 0; a.e. x =2 O or x =2 O : 8

9 Then there is a polynomial V n such that kf? V n k Ij c'( ); j = ; : : : ; n Vn(x)(x) 0 0; x =2 O or x =2 O ; respectively. Now let I i;j be the smallest interval containing I i I j denote h i;j := ji i;j j. For S 2 k?, put (2.4) a i;j = a i;j (S; ') := jjp i? p j jj Ii '( ) hj h i;j k ; i; j = ; : : : ; n; where p i is the polynomial dened by p i j Ii := Sj Ii. Finally for any E I let a k (S; '; E) := max a i;j (S; '); where the maximum is taken over all i; j such that I i \ E 6= ; I j \ E 6= ;, where J denotes the interior of J; a k := a k (S; '; I): We have Lemma 6. There is a constant c, depending only on k, such that for any f 2 H ' k S 2 k?, if (2.5) kf? Sk Ij '( ); j = ; : : : ; n; then (2.6) a k c: Proof. We divide I j into k subintervals of equal lengths by setting x j;0 := x j < x j; < < x j;k? < x j? we let L k be the Lagrange polynomial of degree k? interpolating f at x j;l, l = 0; : : : ; k?. Then by Whitney's theorem (2.7) kf? L k k Ij c! k (f; ) c'( ): 9

10 Hence by (2.5) kp j? L k k Ij c'( ); which implies (2.8) kp j? L k k Ii c hi;j k '( ): At the same time, (2.7) implies (see [9], p. 5 (4.5)) (2.9) kf? L k k Ii c hi;j k '( ): Combining (2.5) with (2.8) (2.9) we obtain kp i? p j k Ii c hi;j k '( ) + '(h i ) c hi;j k '( ); where if > h i we used the inequality '(h i ) '( ), if h i, then due to ' 2 k we have, hi;j k hi;j k '(h i ) '(h i ) '( ): h i x3. The main lemmas We begin with a well-known partition of unity by polynomials which goes back to G. Freud Yu. A. Brudnyi (see e.g. Dzjadyk [5] p. 273{277.) For each xed integer r, a collection f j;n g n j=, of polynomials of degree n, exists such that (3.) for q = 0; ; : : : we have (3.2) j (q) j;n (x)j C q+ n (x) n j= j;n (x) ; r+ n (x) ; x 2 I; jx? x j j + n (x) where C depends on q r. (Inequality (3.2) for q = 0 follows from [5] p. 277 (3), by (.4) (.6); for higher q by induction. Actually, we only need q = 0;.) First we prove 0

11 Lemma 7. Let r 3k, ' 2 k, S 2 k?. For n n, with n divisible by n, the polynomial (3.3) D n (x) := n i= p i (x) :I ;n I i ;n (x); satises (3.4) js(x)? D n (x)j C 0 a k '( n (x)); x 2 I; (3.5) js 0 (x)? D 0 '( n (x)) n (x)j C 0 a k ; x 2 I: n (x) Moreover, for each 0 < <, (3.6) js 0 (x)? D 0 n (x)j C a k (S; '; (x? ; x + ) \ I) '( n(x)) n (x) + C 2 a k n (x) n (x) + r+?3k '( n (x)) ; x 2 I; n (x) where C l = C l (k; r). Proof. Recall that throughout the paper we assume k 2. (Since for k =, S is a constant, Lemma 7 is valid also for k =.) We will only prove (3.6), the proof of (3.4) being similar, evidently, (3.5) being an immediate consequence of (3.6). We x j n, x 2 I j to save in writing we set := n (x), := n (x). Since p j? p i is a polynomial of degree not exceeding k?, then Hence by (.6) (2.4), kp j? p i k Ij c hi;j h i k? kp j? p i k Ii : (3.7) kp j? p i k Ij c hi;j h i k? hi;j k '( )a i;j ca i;j '( ) hi;j 3k?2 =: c i;j ;

12 which in turn implies (3.8) kp 0 j? p 0 ik Ij c i;j : (Note that for u 2 I i, (.2) (.3) imply that h i;j ju? xj +, that is, there are constants 0 < c < C independent of i, j n, for which ch i;j < ju? xj + < Ch i;j.) Now, if we write jx? x i j := minfjx? x i j; jx? x i? jg, then it follows by (3.7) (3.8) that (3.9) jp j (x)? p i (x)j c jx? x ij i;j ; Indeed if i = j, there is nothing to prove; if i 6= j, then (3.9) is an immediate consequence of (3.7) the inequality (see (.3)) jx? x i j > =3; if i = j + (a similar proof applies to i = j? ), then jp j (x)? p i (x)j = j Z x since jx? x j+ j > jx? x j j. Thus, if we denote x j (p 0 j (u)? p0 i (u)) duj jx? x jj c i;j = c jx? x ij i;j ; i (x) := then by (3.2) with q = 0, we have for i 6= j, (3.0) :I ;n I i ;n (x); h j i (x)j c ;nr ( + jx? x ;n j) r+ :I ;n I i r c ( + jx? x i j) r+ = ch i r ( + jx? x i j) r+ : In the same way (with q = ) we get for i 6= j, (3.) j 0 i (x)j ch i r? ( + jx? x i j) : r+ 2 :I ;n I i h ;n

13 Now by (3.), S 0 (x)? D 0 n (x) = i6=j (pj (x)? p i (x)) 0 i(x) + (p 0 j(x)? p 0 i(x)) i (x) =: i6=j i (x); by virtue of (3.8) through (3.), Hence js 0 (x)? D 0 n (x)j j i (x)j ch i r? i;j ( + jx? x i j) r r+2?3k h i ca i;j : + jx? x i j = n i=0;i6=j i:jx?x ij< j i (x)j j i (x)j + i:jx?x ij j i (x)j ca k (S; '; (x? ; x + ) \ I) h i ( + jx? x i= i j) 2 r+2?3k + ca k h i + jx? x i j ca k () i6=j:jx?x ij Z? + ca k r+?3k 2 ca k () du ( + jx? uj) 2 Z + ca k n du ( + u) r+2?3k r+?3k ; + where x being xed, we used the shorter notation a k () := a k (S; '; (x? ; x + )). This concludes the proof of (3.6). The following lemma is crucial to our proof. Lemma 8. Let the interval E consist of l 2s of the intervals I j, let J be a subcollection of l=4 of those intervals we write J := [J. Then for each ' 2 k, there exists a polynomial Q n (x) = Q n (x; E; J ; '), of degree not exceeding 30ksn, satisfying (3.2) Q 0 n(x)(x) 0; x =2 E n (O [ J); 3

14 (3.3) Q 0 n(x) sgn(x)? ; x 2 E n (O [ J); (3.4) Q 0 l n(x) sgn(x) c 3 ; x 2 J n O; where we may assume that c 3 ; (3.5) Q 0 n (x) sgn(x) c l 3 36sk ; + dist (x; E) x 2 I n (E [ O); 3 (3.6) jq n (x)j Cl k+ 3 kl jej ; x 2 I; jej + dist (x; E) where jej denotes the length of E. Proof. We begin by quoting some results from [7] (see Lemma 5.3 there the denitions above it). We put b := max(6s; 8k? s + ), for eac such that I j \ O = ;, (which we denote by j 2 H), we let the polynomials T j (x) = T j;n (x; b; Y ) Tj (x) := Tj;n (x; b; Y ) of degree (b + )(4n? 2) + s + 2 be those dened there. We recall that T j () = Tj () = ; that by [7] (5.5) (5.6) we have, (3.7) T 0 j (x)(x) sgn(x j) 0; x 2 I; (3.8) T 0 j (x)(x) sgn(x j ) 0; x 2 I n I j : Since [7] (5.9) implies j x? y i x j? y i j + j x? x j x j? y i j < 4; x 2 I j ; j T 0 j (x)j c ; x 2 I j; 4

15 whence, (3.9) '( )j T 0 j (x)j c ; x 2 I j: If j (x) := [xj ;](x), then by [7] (5.23) (5.24), we have j j (x)? T j (x)j c j j (x)? T j (x)j c Hence, by virtue of (.5), it follows that '( )j j (x)? T j (x)j c'( ) 8k+ jx? x j j + 8k+ jx? x j j + ; x 2 I; : x 2 I: jx? x j j + 4 ; x 2 I; (3.20) '( )j j (x)? Tj (x)j c'( ) jx? x j j + 4 ; x 2 I: Similarly, by (.6), we have (see also [7] (5.29) (5.30)). (3.2) '( )j j (x)? T j (x)j 4 ; x 2 I; jx? x j j + (3.22) '( )j j (x)? Tj (x)j 4 ; x 2 I: jx? x j j + In view of (.2), it follows from [7] (5.28) that (3.23) '( )jt 0 j(x)j c ; x 2 I j; observing that j x? y i x j? y i j jx? y i j jx? x j j + jx? y i j =3 jx? x j j + ; x 2 I n O; 5

16 then nally (.6) [7] (5.22) (see also (5.27)) yield, (3.24) '( )jtj(x)j 0 c 4b+s+k?2 ; x 2 I n O: jx? x j j + We are ready to proceed with the proof. We write j 2 H(E) if j 2 H (dened at the beginning of the proof) I j E. And we denote by j 0 j 0, the biggest the smallest indices, respectively in H(E). Note that (3.25) jx? x j 0j + c(dist(x; E) + ); x x j 0; jx? x j0 j + c(dist(x; E) + ); x x j0 : Similarly we write H(J ) := H(E) \ J H(J 0 ) := H(J ) [ fj 0 ; j 0 g. Denote by l the number of all indices in H(E) by the number of indices in H(J 0 ). Then (3.26) l? 3s l l; + 2: We set j := sgn(x j ) consider two cases. First assume that j has the same sign, for all j 2 H(E) let P (3.27) a := l j2h(j 0 '(h ) j) Pj2H(E)nH(J '(h ) j) : We estimate the numerator by (3.28) j2h(j 0 ) '( ) ( + 2) max '( ): j2h(e) By (3.26), the number of elements in H(E) n H(J ) is at least l? l? 3s? l=2: Since ' 2 k, we can prove in a similar way to that of [9] Lemma 7., that j2h(e)nh(j ) '( ) cl max '( ): j2h(e) 6

17 Therefore we have established that a dened above is bounded, say by c, independently of E J. Hence by (3.9) (3.29) a'( )j T 0 j(x)j c ; x 2 I j: Put Q n (x) := c l j2h(j 0 ) '( )T j (x)? a j2h(e)nh(j ) '( ) Tj (x) A : We will show that Q n has the required properties. First, (3.2) follows immediately from (3.7) (3.8). Also, by virtue of (3.7) (3.8) we have, (3.30) Q 0 n(x) sgn(x)? a c '( )j T 0 j(x)j; x 2 I j E n (O [ J); (3.3) (3.32) (3.33) Q 0 n(x) sgn(x) c l jt 0 j(x)j; x 2 I j J n O; Q 0 n(x) sgn(x) c l '( 0 )jt 0 j 0 (x)j; x x j0?; Q 0 n(x) sgn(x) c l '( 0)jT 0 j 0(x)j; x x j 0: Now we obtain (3.3) from (3.29) (3.30); (3.4) follows by virtue of (3.23) (3.3); (3.32), (3.33), (3.23), (3.24) (3.25) are combined to yield (3.5). Thus, to complete the proof we have to prove (3.6). To this end we rewrite Q n as c Q n (x) = l 0 l j2h(j 0 ) j2h(j 0 ) =: A(x) + B(x); say: '( )(T j (x)? j (x))? a '( ) j (x)? a j2h(e)nh(j ) j2h(e)nh(j ) '( ) j (x) '( )( T j (x)? j (x)) If x 2 I n E, then all j (x) wit 2 H(E), have the same value so that (3.27) implies that B(x) = 0. On the other h, if x 2 E, then by virtue of (3.27) (3.28), A A (3.34) jb(x)j 2l( + 2=) max '( ) cl max '( ) j2h(e) j2h(e) k=2 jej cl cl k+ 3 kl : 7

18 In order to estimate A(x) we rst assume that jej. Then we apply (3.2) (3.22) get (3.35) Z h ja(x)j cl 3 j (jx? x j j + ) du 4 cl3 E (jx? uj + ) 4 j2h(e) 3 cl 3 (dist(x; E) + ) cl jej : 3 jej + dist(x; E) If on the other h, jej <, then by (3.20) we have (3.36) ja(x)j cl j2h(e) cl '(jej) h 5 j '( ) (jx? x j j + ) 4 j2h(e) cljej 3 '(jej) jej cljej 3 cljej 3 Z jej 4 (jx? x j j + jej) 4 j2h(e) du (jx? uj + jej) 4 (jx? x j j + jej) 4 E (dist(x; E) + jej) cl 3 3 jej : jej + dist(x; E) This concludes the proof of (3.6). In the second case, we assume that there are j 2 H(E) j 2 2 H(E), such that j j2 < 0. (In this case we do not make use of Tj, hence we can even strengthen (3.3) by replacing its right-h P side by zero.) Depending on the sign of j2h(j 0 '(h ) j) j we take b 0 so that for Q n (x) := l j2h(j 0 ) '( )T j (x) j + b'(i )T ji (x) ji ; i = or i = 2; we have Q n () = 0. Note that this implies that (3.37) b l P j2h(j 0 ) '(i ) 8

19 Now we proceed as in the rst case readily obtain Q 0 n(x) sgn(x) 0 x 2 I; Q 0 n(x) sgn(x) l jtj(x)j; 0 x 2 I j J n O; Q 0 n(x) sgn(x) l '( 0 )jt 0 j 0 (x)j; x x j0?; Q 0 n(x) sgn(x) l '( 0)jT 0 j 0(x)j; x x j 0: Finally, we rewrite Q n (x) = l 0 l j2h(j 0 ) '( )(T j (x)? j (x)) j + b'(i )(T ji (x)? ji (x)) ji A j2h(j 0 ) =: A(x) + B(x); say: '( ) j (x) j + b'(i ) ji (x) ji A As before, if x 2 I n E, then B(x) = 0, if x 2 E, then by (3.37) (see (3.34)) jb(x)j 2l + 2 max '( ) cl k+ 3 kl : j2h(e) The estimate of ja(x)j is done in the same way as (3.35) (3.36), where again we employ (3.37) x4 Proof of Theorems 2 The crux of the proof is a lemma the idea of which goes back to the seminal paper by DeVore []. Lemma 9. Let ' 2 k S 2 k?. Assume that (4.) a k (S; ') ; that S 0 (x)(x) 0; for x 2 I n O or for x 2 I n O : 9

20 Then there is a polynomial P n of degree cn such that (4.2) jp n (x)? S(x)j C'( n (x)); (4.3) P 0 n(x)(x) 0; for x 2 I n O or for x 2 I n O ; respectively: Proof. Set r = 3k? + 36sk, let c := C (k; r) of (3.6). We x an integer c 4 so that (4.4) c 4 max(8k=c 3 ; 2s); where c 3 is the constant of (3.4). Without loss of generality we are going to assume that n is divisible by c 4, i.e., n =: Nc 4, where this denes N. We divide I into N intervals, E q := [x qc4 ; x (q?)c4 ] = I qc4 [ [ I (q?)c4 +; q = ; : : : ; N: We will write j 2 UC (for \Under Control") if there is an x 2 I j such that (4.5) js 0 (x)j 5c ; we will say that q 2 G (for \Good"), if E q contains at least 2k? 3 intervals I j wit 2 UC. Note that if q 2 G, then (4.) implies that any of the polynomials p j, (q? )c 4 + j qc 4 satises (4.6) kp 0 j k E q c: Indeed, let i 2 UC let x 2 I i be such that (4.5) holds for x. Then by virtue of (4.), jp 0 j(x )j jp 0 j(x )? p 0 i(x )j + jp 0 i(x )j c kp j? p i k Ii + c '( n(x )) h i n (x ) c h i hij k c ; where we used the fact that when (q? )c 4 + i; j qc 4, then h i h ij. Since there are at least k? intervals with i 2 UC, which are not adjacent to each other, 20

21 p 0 j '( ) is of degree k? 2, we conclude that p0 j is bounded in E q by the same bound. Again for any x 2 E q. This proves (4.6). In particular, (4.7) js 0 (x)j c ; x 2 E q: We remark that once (4.7) holds in E q, then it holds (with perhaps a bigger constant) in E q+ [ E q [ E q?. Given any set A I denote (4.8) A e := [ Ij \A6=;I j ; A 2e := (A e ) e A 3e := ((A e ) e ) e : Now set (4.9) E := [ q =2G E q ; decompose S into a \small" part a \big" one by setting S 0 (x); if x =2 E e s (x) := 0; if x 2 E e ; s 2 := S 0? s, nally putting S (x) := We will show that Z x s (u) du + S(?); S 2 (x) := Z x?? s 2 (u) du: (4.0) a k (S ; ') c; which by virtue of (4.) implies, (4.) a k (S 2 ; ') c + < [c + 2] =: c 5 : To this end, put then we will prove that, (4.2) js 0 (x)? p 0 j(x)j c p j := S j Ij ; jx? xj j + k? ; x 2 I: 2

22 We rst observe that either p 0 0 or j p0 = j p0 j, in the latter case (4.7) implies Hence we always have, (4.3) jp 0 j (x)j c jp 0 j (x)j c ; x 2 I j : jx? xj j + k?2 ; x 2 I: Next we note that (4.7) is valid for S every x 2 I, i.e., (4.4) js 0 (x)j c ; x 2 I: Now, if, then '( ) (.2) through (.5) yield c (jx? x h 2 j j + ); j if, then?k h?k j '( ), (.2) (.4) imply Hence (4.4) yields k? h k j c jx? xj j + (k?)=2 : (4.5) js 0 (x)j c jx? xj j + k? ; x 2 I: Combining (4.3) (4.5) we have (4.2), by it for x 2 I i, which is (4.0). js (x)? p j (x)j = j Z x (S 0 (u)? p0 j (u)) duj x j c hij k? jx? x j j c'( ) hij k ; The set E is a union of disjoint intervals F p = [a p ; b p ], between any two of which there is an interval E q with q 2 G. We may assume that n > c 4 c 5 we will write p 2 AG 22

23 (for "Almost Good") if F p consists of no more than c 5 intervals E q, i.e., if it consists of no more than c 4 c 5 intervals I j. Set let F := [ p=2ag F p ; S 0 (x); if x 2 F e s 4 (x) := 0; otherwise; s 3 := S 0? s 4. (For the denition of F e see (4.8).) Now put S 3 (x) := Z x? s 3 (u) du + S(?); S 4 (x) := Z x? s 4 (u) du: Evidently S 3 S 4 are comonotone with S in I, proceeding as we did above we get, (4.6) js 0 3(x)j c ; x 2 I; (4.7) a k (S 4 ; ') < c 6 : Now (4.6) together with Lemma 5 implies the existence of a polynomial V n comonotone with S on I n O or on I n O, as the case may be, such that which is (4.8) js 3 (x)? V n (x)j c; x 2 I: Since s 4 (x) = S 0 (x); x 2 F e ; then by (4.) we have for p =2 AG, (4.9) a k (S 4 ; '; F e p ) a k (S; '; F e p ) a k (S; ') : Also for such p, Hence from (4.), S 0 4(x) = S 0 2(x); x 2 F 3e p : (4.20) a k (S 4 ; '; Fp 3e ) = a k (S 2 ; '; Fp 3e ) a k (S 2 ; ') < c 5 : 23

24 We still have to approximate S 4. To this end we construct three polynomials Q n M n of degree < 30ksn D n (S 4 ; ) of degree n. We begin with Q n. For each q for which E q F, let J q be the collection of intervals I j E q wit 2 UC. Recall that q =2 G, therefore by (4.4), the number of such intervals is at most 2k? 4 < c 4 =4, the total number of intervals in E q is c 4. Thus Lemma 8 is applicable for each E q if we set Q n := q: E q F Q n (; E q ; J q ; '); where on the right-h side are the polynomials guaranteed by Lemma 0, denote J := [ q: Eq F J q ; J = [J ; then we conclude that Q n satises (4.2) Q 0 n(x)(x) 0; x 2 (I n F ) [ O [ J; (4.22) Q 0 n(x) sgn(x)? ; x 2 F n (O [ J); by (4.4), (4.23) Q 0 n(x) sgn(x) 4 ; x 2 J n O: Note that (4.2), (4.22) (4.23) follow since for any given x all relevant Q 0 n (x; E q; J q ; '), except perhaps one, have the same sign. Finally it follows from (3.6) that (4.24) jq n (x)j c; x 2 I: Next we dene the polynomial M n. For each F p with p =2 AG, let J p? denote the collection of three intervals in the left side of F 3e p n F p, let J p+ be the collection of three interval in the right side of F 3e p n F p. Similarly, let F p? F p+ be closed intervals each consisting of l := c 4 c 5 intervals I j such that J p? := [J p? F p? F 3e p J p+ := [J p+ F p+ Fp 3e. Now we set M n := (Q n (; F p+ ; J p+ ; ') + Q n (; F p? ; J p? ; ')) : p=2ag 24

25 Since l = c 4 c 5 = 3, it follows by (4.4) that c 3 l= 4c 5. Again we have by Lemma 8, (4.25) M 0 n(x) sgn(x)?2 ; x 2 F n O; M 0 n(x)sgn(x) 0; x 2 O; (4.26) M 0 n (x) sgn(x) 4c 5 ; x 2 F 3e n (F [ O); (4.27) M 0 n(x) sgn(x) c 7 Finally, it readily follows by virtue of (3.6) that 36sk ; x =2 F 2e [ O: dist(x; F e ) (4.28) jm n (x)j c; x 2 I: The third auxiliary polynomial the properties of which we need to recall is D n := D n (S 4 ; ). By the choice of r by (4.7), Lemma 7 yields (4.29) js 4 (x)? D n (x)j C 0 c 6 ; x 2 I; for any > 0, (4.30) js 0 (x)? 4 D0 n (x)j c a k (S 4 ; '; (x? ; x + )) 36ks n (x) + c 8 ; x 2 I; where c 8 := C 2 c 6. Noting that n (x) we are going to prescribe n = cn so big that n n ; (4.3) c 8 n n 36ks c min(; 3c 5 ; c 7 ): 25

26 Now we write R n := D n + c Q n + c M n ; by virtue of (4.24), (4.28) (4.29), we have js 4 (x)? R n (x)j c; x 2 I: In view of (4.8), this proves (4.2) for P n := R n + V n. Thus in order to conclude the proof of Lemma 9, we should prove that (4.3) holds for our P n. To this end, we recall that V n is comonotone with S where it is required so that we only have to deal with R n. Since (4.30) holds with an arbitrary, we will prescribe dierent ones as needed. As long as x 2 F 2e, it suces to take :=, while we recall (.2) the fact that both x + n (x) x? n (x) are increasing in I n (I [ I n ). First assume that x 2 F, so that (x? ; x + ) F e. If x 2 J n O, then S4(x) 0 sgn(x) 0, we obtain by (4.23), (4.25), (4.9), (4.30) (4.3), that R 0 n(x) sgn(x) c Q 0 n(x) sgn(x) + S4(x) 0 sgn(x) (4.32) + c Mn(x) 0 sgn(x)? js4(x) 0? D 0 n (x)j 4c? 2c? c n (x)? c 8 (c? c 8 (n=n ) 36ks ) 0: 36ks If, on the other h, x 2 F n(j [O), then (4.5) is violated by virtue of (4.22), (4.25), (4.9), (4.30) (4.3), we get R 0 n (x) sgn(x) S0 (x) sgn(x) + 4 c Q 0 n (x) sgn(x) (4.33) + c M 0 n(x) sgn(x)? js 0 4(x)? D 0 n (x)j 5c? c 8? c? 2c? c n 36ks (x) (c? c 8 (n=n ) 36ks ) 0: 26

27 Now assume that x 2 F 2e n (F [ O) so that (x? ; x + ) F 3e. S 0 4 (x) sgn(x) 0, by (4.2), (4.26), (4.20), (4.30) (4.3), we obtain R 0 n (x) sgn(x) c Q 0 n (x) sgn(x) + S0 4 (x) sgn(x) Again we have (4.34) + c Mn(x) 0 sgn(x)? js4(x) 0? D 0 n (x)j 4c c 5? c n (x) c 5? c 8 (3c c 5? c 8 (n=n ) 36ks ) 0: 36ks Finally, if x =2 F 2e [ O, then we set := dist(x; F e ), which implies that S 0 4 vanishes on (x? ; x + ). Hence a k (S 4 ; '; (x? ; x + )) = 0, so by (4.2), (4.27), (4.30) (4.3), we conclude that R 0 n(x) sgn(x) c Q 0 n(x) sgn(x) + S4(x) 0 sgn(x) (4.35) + c M 0 n (x) sgn(x)? js0 (x)? 4 D0 n (x)j 36ks n (x) c c 7? c8 36ks 36ks n 36ks! (x)) c c 7? c 8 36ks (c c 7? c 8 (n=n ) 36ks ) 0: Combining (4.32) through (4.35) we have constructed a polynomial satisfying (4.2) (4.3). The proofs of Theorems 2 now follow from Lemmas 3, 4 9, except that in Lemma 9 the polynomial is of degree cn. This is easily rectied. First we may assume that c k? then we replace n by [n=c], observe that [n=c] (x) 4c 2 n (x); x 2 I; thus '( [n=c] (x)) '(4c 2 n (x)) 4 k c 2k '( n (x)); where we applied the fact that ' 2 k. Hence Theorems 2 hold for n c, while for smaller n, see the remark after the statement of Theorem 2. 27

28 x5 A counterexample In this section we prove Theorem 3 by providing an example (see a similar example in [7] Example.). Let y := =(20n) y 2 :=?=(20n) dene f(x) := 8 >< >: ; x y 2 ;?20nx; jxj < 20n ;?; x y : Then of course f is nondecreasing in [?; y ] [y 2 ; ]; nonincreasing in [y ; y 2 ]. Note that O (8n; Y n ) [y? =4n; y + =4n] [ [y 2? =4n; y 2 + =4n] [ [?;? + =n 2 ] [ [? =n 2 ; ] =: ~ O(n; Y n ). For t < =(0n), we readily see that!(f; t) = 20nt. Let P n be comonotone with f outside ~ O(n; Y n ) contrary to (.9) assume that jf(x)? P n (x)j A!(f; n (x)); x 2 [?; ]: Since n ( =n 2 ) < 3=n 2 =0n, this implies jf( =n 2 )? P n ( =n 2 )j < 60A=n : By virtue of the denition of f, we thus obtain P n ( =n 2 ) < 0. Now, P n is nondecreasing in [? + =n 2 ;?=4n] [ [=4n;? =n 2 ]. Therefore we conclude that the norm of P n in the interval [? + =n 2 ;? =n 2 ], is attained in [?=4n; =4n]. Note that P n is positive at?=4n negative at =4n, hence it vanishes somewhere inside, say at. If jp n ()j = kp n k := kp n k [?+=n 2 ;?=n 2 ], then? < =2n, whence there exists such that jpn()j 0 > P n() =2n = 2nkP nk: On the other h, for n 2, 2=(2? 2=n 2 ) = 4=3 p p? 2? =6n 2 > 7=8, thus by Bernstein's inequality for the interval [? + =n 2 ;? =n 2 ] we obtain jp 0 32n n ()j 2 kp nk; which is a contradiction. This completes the proof of Theorem 3. 28

29 References. R. A. DeVore, Monotone approximation by polynomials, SIAM J. Math. Anal. 8 (977), R. A. DeVore, Degree of approximation, in Approximation theory II (G. G. Lorentz, C. K. Chui, L. L. Schumaker, eds.), Acad. Press, New York, 976, pp R. A. DeVore, D. Leviatan I. A. Shevchuk, Approximation of monotone functions: a counter example, in Curves Surfaces with Applications in CAGD (A. Le Mehaute, C. Rabut L. L. Schumaker, eds.), Proceedings of the Chamonix Conference 996, Verbilt University Press, Nashville TN, 997, pp R. A. DeVore. M. Yu, Pointwise estimates for monotone polynomial approximation, Constr. Approx. (985), V. K. Dzjadyk, Introduction to the theory of uniform approximation of functions by polynomials, (in Russian), \Nauka" Moscow, G. A. Dzyubenko, Pointwise estimate of comonotone approximation, Ukrain. Mat. Zh. 46 (994), G. A. Dzyubenko, J. Gilewicz I. A. Shevchuk, Piecewise monotone pointwise approximation, Constr. Approx. (to appear). 8. D. Leviatan I. A. Shevchuk, Some positive results counter examples in comonotone approximation I, J. Approx. Theory (to appear). 9. I. A. Shevchuk, Polynomial approximation traces of functions continuous on a segment, (in Russian), "Naukova Dumka", Kiev, A. S. Shvedov, Coapproximation of piecewise{monotone functions by polynomials, Mat. Zametki, 30 (98), School of Mathematical Sciences, Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv 69978, Israel Institute of Mathematics, National Academy of Sciences of Ukraine, Kyiv 25260, Ukraine 29