Markov Sonin Gaussian rule for singular functions

Journal of Computational and Applied Mathematics 169 (2004) 197 212 www.elsevier.com/locate/cam Markov Sonin Gaussian rule for singular functions G.Mastroianni, D.Occorsio Dipartimento di Matematica, Universita della Basilicata, C. da Macchia omana, Potenza 85100, Italy eceived 21 November 2002; received in revised form 10 December 2003 Abstract The paper deals with the approximation of integrals fw, where w is a Markov Sonin weight and f can be singular at the origin. Gaussian-type quadrature rules, having a better behaviour w.r.t. the ordinary Gaussian rule, are introduced.error estimates in weighted L 1 norm and some numerical tests are given. c 2004 Elsevier B.V. All rights reserved. Keywords: Quadrature; Orthogonal polynomials; Polynomial approximation 1. Introduction and preliminary results We start with some remarks on the Gauss Jacobi rule.denote by v (x) = x ; 1; a Generalized Jacobi weight, the corresponding Gauss quadrature rule is 1 m f(x) x dx = k (v )f(x k )+e m (f; v ); (1) 1 k=1 where k (v ) are the Christoel numbers, x k are the zeros of the mth Generalized Jacobi polynomial and e m (f; v ) is the remainder term.if the function f is such that f (r) v m; 1, with v m; =( x +1=m) and r 1, then [10, Theorem 2, p.117] e m (f; v ) 6 C m r f(r) v m; 1 ; (2) where the positive constant C is independent of m and f and 1 is the usual L 1 -norm.ecalling that, under the same assumptions on f, the following estimate holds [1, p.134]: E m (f) v;1 = inf P P m (f P)v 1 6 C m r f(r) v m; 1 ; (3) Corresponding author. E-mail addresses: mastroianni@unibas.it (G.Mastroianni), occorsio@unibas.it (D.Occorsio). 0377-0427/$ - see front matter c 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2003.12.020

198 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 we can conclude that the Gauss Jacobi rule, for the previous class of functions, behaves like the best L 1 approximation. The situation is quite dierent if we use the Gaussian rule for integrals on unbounded intervals. Consider the following Gaussian rule: m f(x)w (x)dx = k (w )f(x k )+e m (f; w ); (4) k=1 where w (x)=e x2 x ; 1, { k (w )} m k=1 are the Christoel numbers, {x k} m k=1 are the zeros of p m (w ) and e m (f; w ) is the error term.the weight w is known as Markov Sonin or Generalized Hermite weight (see [8,9]).An estimate of e m (f; w ) is given in the following proposition. To this end let be L 1 w the ordinary weighted L 1 space (i.e., f L 1 w fw 1 = f(x) w (x)dx ).We claim as follows. Proposition 1.1. Assuming f w 1, then e m (f; w ) 6 C m 1=6 f w m; 1 ; where the constant C is independent of m and f and ( w m; (x)=e x2 x + 1 ) : m (5) (6) Moreover, for everym there exists a function f = f m s.t. 0 f mw m; 1 and e m ( f m ;w ) C m f mw 1=6 m 1 ; where the constant C is independent of m and f. (7) From the converse inequality (7) follows that estimate (5) is sharp.however, since under the same assumption on f E m (f) w ;1 = inf P P m (f P)w 1 6 C m f w m; 1 ; (8) we conclude that E m (f) w ;1 is better than e m (f; w ), for the considered space of functions. Similar results hold also for Freud weights in [5] and Laguerre weights in + [3]. Note that f w 1 implies (5) and (8), but not necessarily lim m E m (f) w ;1 =lim m e m (f; w )= 0.For instance it happens when 0 and f(x) = log x.obviously, the above two limits hold if sup m f w ;m 1. In this paper we want to construct a quadrature formula simple like the Gaussian rule and with the same rate of convergence of the error of the best approximation in L 1 w.moreover, we will consider special quadrature rules of Gaussian type for the computation of integrals of functions having a singularity at the origin.

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 199 2. Truncated Gaussian rule In order to construct our quadrature rules we need some notations and known results. In the following C denotes positive constants and we shall write C C(a;b;:::;) to denote that the constant C is independent of the parameters a;b;:::, and we use C = C(a;b;:::;) to say that C depends on the parameters a;b;:::. If A and B are two positive quantities depending on some parameters, we write A B if and only if (A=B) ±1 6 C, where C is a positive constant independent of the above parameters. We recall the following inequalities. For any polynomial P m P m and 0 p we have ( P m (x) w (x) dx) p 1=p 6 C ( am a m P m (x) w (x) dx) p 1=p ; C C(m; P m ); (9) where a m = 2m +][ is the Mhaskar ahmanov Sa number w.r.t. the weight w (see [15]), and ]a[, for a +, denotes the smallest integer bigger or equal than a. Moreover for any polynomial P m P m and for any real xed 0, the following bound holds [16, p.335]: P m (x) w (x)dx6ce Am P m (x) w (x)dx; (10) x (1+)a m where the positive constants C;A depend on and are independent of m; P m. Denoting by AC() the set of all absolutely continuous functions on, the Sobolev space Wr 1 is dened in the usual way Wr 1 ()=Wr 1 = {f L 1 w ;f (r 1) AC() : f (r) w 1 }; r 1 with the norm fw W 1 r = fw 1 + f (r) w 1 : For functions in this space the following Favard estimate holds [13]: E m (f) w ;1 6 C ( m) r f(r) w m; 1 ; (11) where w m; is dened in (6) (see [13]). Now we want to introduce a special truncation of a function f L 1 w.to this end, let {p m (w )} m be the sequence of orthonormal polynomials with positive leading coecients and (slightly changing our previous notation of formula (4)) denote by x i ; 0 6 i 6 1 the nonnegative zeros of p m (w ) and set x i = x i ;x 0 = 0.It is well known that x i a m ; 0 6 i 6 (see [8,12]). For a xed (0; 1) and m suciently large (say m m 0 ), dene the integer j = j(m) as x j = min k {x k a m }: (12) 1 [a], for a +, denotes the largest integer smaller than or equal to a.

200 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 We observe that distance of two consecutive zeros x i = x i+1 x i satises the estimate 1 x i m 1 x i = 2m +1=m 2=3 for x i ( a m ;a m ) and x i 1 m for x ( a m ;a m ) (see [12]). Now, by using an arbitrary non decreasing function C () such that { 0; x6 0; (x)= 1; x 1; we introduce the truncated function f j = f j(m) ; m m 0, dened as ( ) x xj f j = f(1 j ); j (x)= : (13) x j+1 x j By denition f j = f in [x j ;x j ] and f j is zero if x x j+1 (cf.[5]).of course f j Wr 1 too. Using a result in [11] (see also [5]), for any xed 0 1 and for all f L 1 w, we have (f f j(m) )w 1 6 C[E M (f) w ;1 +e Am fw 1 ]; (14) where M =[(=(1 + )) 2 m] m, and A = A();C= C() are independent of m and f. Now we can write f(x)w (x)dx = [f(x) f j (x)]w (x)dx + f j (x)w (x)dx: Approximating the second integral by the ordinary Gaussian rule we obtain j f(x)w (x)dx = k (w )f(x k )+r m (f; w ); k= j where now k (w ) corresponds to x k (cf.(4)) r m (f; w )= (f(x) f j (x))w (x)dx + e m (f j ;w ): The rule j k (w )f(x k ) (15) k= j

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 201 will be here called truncated Gaussian rule.as we can see it is the ordinary Gaussian rule in which we have dropped [cm], 0 c 1, terms.now we prove that the truncated Gaussian rule converges to the integral fw faster than the ordinary Gaussian rule.indeed, the following theorem holds: Theorem 2.1. For any f Wr 1 () j [ ] f(x)w (x)dx k (w )f(x k ) 6 C 1 ( m) r f(r) w m; 1 +e Am fw 1 ; (16) k= j where w m; is dened in (6) and the constants A = A();C= C() are independent of m and f. emark. Comparing the error estimate (16) with (5 7) and taking into account (11), the truncated Gaussian rule behaves better than the classical one and generally it converges to the integral with the same rate of convergence as the best approximation of f Wr 1. Now we want to apply a similar procedure for integrals of functions f having a singularity in 0 or continuous in 0 but having discontinuous derivatives.assuming 0 and r 1 and denoting by AC( {0}) the set of all absolutely continuous functions on {0}, dene W 1 r() := W 1 r = {f L 1 w ;f (r 1) AC( {0}): f (r) w 1 } equipped with the norm fw W = fw 1 r 1 + f (r) w 1 ; w (x)=e x2 x : We dene the following two spaces: D 1 (r; )={f W 1 r(); +1 r}; i.e., the function f can be singular in 0 and f (r) w 1. D 2 (r; )={f W 1 r(); not an integer r 1; and f (r ][ 1) AC()}; i.e., functions in D 2 (r; ) have the r ][ 1 absolutely continuous derivative and the next derivative can be discontinuous in 0, but f (r) w 1. In both cases, the error of the best approximation can be estimated by (see [13]) E m (f) w ;1 6 C ( m) r [ f(r) w 1 + fw 1 ]: (17) We complete formula (17) observing that whenever +16r and is an integer, in estimate (17) a log m factor appears in the right-hand side [13]. In order to give some examples of functions in D 1 (r; ) and D 2 (r; ), consider sin (1=x 1=2 ) ; x 0; f(x)=e x2 =2 1 x ; x 0: 1=2

202 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 Then f W 1 s (w s ), for every s =1; 2;:::.Choosing =7, f belongs to D 1 (7; 7).As second example, consider f(x)=e x2 =2 arctan(x) 7=2 D 2 (5; 1 2 ): The function f W 1 5 (w 1=2) and f belongs to D 2 (5; 1 ).Indeed f has the third derivative continuous 2 in 0 while the fourth derivative is discontinuous in zero, but f (5) w 0:5 is in L 1. In order to approximate integrals of kind fw, where f D 1 (r; ) orf D 2 (r; ), we use two dierent quadrature rules. If f D 1 (r; ) we will use f(x)w (x)dx = 16 k 6j If f D 2 (r; ) we will use f(x)w (x)dx = 06 k 6j k (w )f(x k )+r 1 (f): (18) k (w )f(x k )+r 2 (f): (19) In the rst rule the knot zero is excluded, while in the second this is included. About these rules, the following theorem holds. Theorem 2.2. Let f D 1 (r; ). f(x)w (x)dx Assuming f D 2 (r; ) f(x)w (x)dx 16 k 6j 06 k 6j k (w )f(x k ) 6 C ( [ ] fw m) r 1 + f (r) w 1 : (20) k (w )f(x k ) 6 C ( [ ] fw m) r 1 + f (r) w 1 ; (21) where the constants A = A();C= C() are independent of m and f. Before concluding this section, we propose some numerical examples.we compare the absolute value of the errors of the truncated rule (15) (in column T.G.) using 2j knots with those obtained by using the ordinary Gaussian (column O.G.) on m knots.details about the computation of zeros and Christoel numbers w.r.t. the weight w can be found in [2] (see also [14]), where they are computed by means of suitable relations with generalized Laguerre case.finally, we point out that, here and later, the computations have been performed in 16-digits arithmetic.consider { (e x 2 =2 = x 1=2 ); x 0; f 2 (x)= e x2 =2 sin (1= x) ; x 0; I 2 = f 2 (x) x 7 e x2 dx;

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 203 Fig.1.f 2(x)e x2. Table 1 Values for I 2 m O.G. T.G. 2j 128 57.5318902 57.5318902 102 256 57.53189025 57.53189025 204 512 57.531890252 57.531890252 412 1024 Overow 57.531890252 672 Table 2 Values I 3 m O.G. T.G. 2j 64 45.4098 45.4098 58 128 45.40985 45.40985 102 256 45.409854 45.409854 204 512 45.4098546 45.4098546 336 1024 Overow 45.4098546 672 and { (e x 2 =2 = x); x 0; f 3 (x)= I 3 = f 3 (x) x 5 e x2 dx: e x2 =2 (x 2 1); x6 0; (Fig. 1) Since f 2 W 1 7 (w 7) and f 3 W 1 5 (w 5), i.e., f 2 D 1 (7; 7) and f 3 D 1 (5; 5) we use (18).A comparison between the truncated rule with the ordinary one are given in Tables 1 and 2, respectively. For each test we give the signicant digits. As we can see in both cases the numerical error of the truncated Gaussian rule agrees with the estimate given by formula (20).We remark that, dierently from the ordinary Gaussian rule, the

204 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 Table 3 Values for I 4 m O.G. T.G. 2j 64 1.1320 1.1320 38 128 1.13204 1.13204 78 256 1.132048 1.132048 156 512 1.1320485 1.1320485 312 1024 Overow 1.1320485 322 truncated rule can be used also for m odd and only to give a comparison between the rules we have chosen m even. At last we consider f 4 (x)=e x2 =2 arctan(x) 7=2 I 4 = f 4 (x) x 1=2 e x2 dx: Here f 4 W 1 5 (w 1=2) and we use (19) (f 4 D 2 (5; 1 )).(Table 3) 2 3. The proofs Proof of Proposition 1.1. To prove the theorem we use an argument similar in [5].Dene the auxiliary function g as f(x ); x6 x ; g(x)= f(x); x 6 x ; f(x ); x x : We get e m (f; w )=e m (g; w )+e m (f g; w ) { x } = e m (g; w )+ + (f(t) g(t))w (t)dt: (22) x Since g AC(), by using Peano s Theorem [2, p.286] x e m (g; w )= g (t)e m ( t ;w )dt = f (t)e m ( t ;w )dt; x (23) where { 0; x6 t; t (x)= 1; x t: Since [7, p.105] e m ( t ;w ) 6 m (w ;t)

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 205 and m (w ;t)= 1 k=0 p2 k (w ;t) w m;(t) 2m t2 + m ; t 6 2m; (24) 1=3 m 1 where w m; (t) is dened in (6), by (23) it follows e m (g; w ) 6 C f (t) w m 1=6 m; (t)dt: (25) Moreover, ( ) t [f(t) g(t)]w (t)dt x = f (x)dx w (t)dt x x = f (x) w (t)dt dx x x 6 f (x) e x2 =2 x e t2 =2 dt dx x x = f (x) e x2 =2 x d(e t2 =2 ) dt dx x x t 6 f (x) w (x) dx 6 C f (t) w (t)dt: x x m Similarly, we obtain x [f(t) g(t)]w (t)dt 6 C m f (t) w (t)dt: (26) Since w (t) 6 w m; (t), combining two last inequalities and (25) with (22), (5) follows. To prove (7) we consider the function f = f m dened as 0; x6 c m = x 1 ; m f(x)= x c m ; c m 6 x 6 x ; 1 ; x x : m

206 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 We have that 0 f w 1 = x c m w (x)dx.by using (23) it follows e m ( f; w )= f (t)e m ( t ;w )dt = = = x c m x c m x c m =:I 1 I 2 : f (t) t (x)w (x)dx k= [ + ] f (t) w (x)dx (w ) dt t + f x (t) w (x)dxdt (w ) t c m Now we evaluate the order of I 1 I 2.Let us consider I 1. x [ I 1 = f 1 + ] (t)w (t) w (x)dx dt w (t) c m t k (w ) t (x k ) dt f (t)dt and for m large, the function x e x2 =2 is decreasing for c m 6 t 6 x.therefore 1 + x e x2 =2 dx 6 t e t2 =2 e x2 =2 dx 6 2 w (t) t w (t) t t 6 C ; m since c m = x 1= m m 1=m 1=6 1= m m.then (27) I 1 6 C m f w 1 : To estimate I 2 we use (24) in which we replace t with the last zero x and we have Therefore I 2 = (w ) x c m f (t)dt Cm 1=6 w (x ) e m ( f; w ) = I 2 I 1 C m 1=6 f w 1 and (7) follows. Proof of Theorem 2.1. We start from f(x)w (x)dx f(x k ) k (w )= k 6j x c m f (t)dt C m 1=6 f w 1 : [f(x) f j (x)]w (x)dx + e m (f j ;w ): (28) ecalling (14) and taking into account (11) we have [ ] 1 (f f j )w 1 6 C ( m) r f(r) w m; 1 +e Am fw 1 : (29)

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 207 To estimate e m (f j ;w ) we use, as above, [7, p.105] e m (f j ;w ) 6 f j(t) m (w )(t)dt; (30) whence by (24) it follows e m (f j ;w ) 6 C xj+1 f m j(t) w m; (t)dt: x ( j+1) By the denition of f j,weget [ e m (f j ;w ) 6 C xj+1 ] xj+1 f (t)(1 j (t)) w m; (t)dt + f(t) j (t) w m; (t)dt : m x ( j+1) x ( j+1) By denition 0 6 j 6 1 and j (t) 0, only for t A j := [x (j+1) ;x j ] [x j ;x j+1 ] and for t A j j (t) 6 (2 max t (t) )=x j.therefore [ e m (f j ;w ) 6 C xj+1 ] f 1 (t) w m; (t)dt + f(t) w m; (t)dt : m x ( j+1) x j+1 x j A j Since x j+1 x j 1= m [ 1 xj+1 ] e m (f j ;w ) 6 C f (t) w m; (t)dt + f(t) w m; (t)dt : m x ( j+1) A j The second integral is dominated by f(t) w m; (t)dt 6 C[E M (f) w ;1 +e Am fw 1 ] (31) {t a m} hence, using Favard inequality, it follows e m (f j ;w ) 6 C 1 m [ f w m; 1 + fw m; 1 ]; (32) i.e., the theorem is proved for r = 1.Now let r 1.Let P M P m such that (f P M )w 1 6 CE M (f) w ;1.With the notation (P M ) j = P M (1 j ), we have By (32) e m (f j ;w ) = e m (f j P M ;w ) 6 e m ((f P M ) j ;w ) + e m (P M j ;w ) = :B 1 + B 2 : (33) B 1 6 C 1 m [ (f P M ) w m; 1 + (f P M )w m; 1 ]: Using a result in [8, p.310], we get (f P M ) w m; 1 6 C[ m (f P M )w m; 1 + E M (f ) wm; ;1; whence by (11) we have B 1 6 C f(r) w m; 1 ( m) r : (34)

208 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 Now consider B 2.We have B 2 = e m (P M j ;w ) 6 The rst member is dominated by P M (x) w (x)dx6ce Am x a m Moreover, whence k j+1 k j+1 m;k (w ) P M (x k ) P M (x) j (x) w (x)dx + k j+1 m;k (w ) P M (x k ) 6C k j+1 k j+1 m;k (w ) P M (x k ) : P M (x) w (x)dx 6 Ce Am fw 1 : (35) w (x k )x k P M (x k ) ; xk x k 1 P M (x)w (x)dx+ Since x k 6 m 1=6 for k j + 1, we have m;k (w ) P M (x k ) 6 P M (x) w (x)dx + C k j+1 [x j;x j] k j+1 xk x k P M (x) w (x)dx : x k 1 ( ) 1 1=3 P M (x) w (x)dx: m [x j;x j] The rst integral can be estimated as before.for the second one rst we use (10) and then Bernstein inequality (see [13]) to obtain whence k j+1 m;k (w ) P M (x k ) 6 Ce A m fw 1 B 2 6 Ce A m fw 1 : Theorem 2.1 follows combining (36), (34) with (33) and (29). Proof of Theorem 2.2. In order to prove the theorem we need the following lemma proved in [4]. (36) Lemma 3.1. Assume f be a function such that f (r 1) is absolutelycontinuous in [ a; a] \{0}, 0 a 1, and f (r 1) L p [ a;a] for 1 6 p 6, where (x)= x ; 0. Let t a. If + 1 p r; then we have f L p ( t;t) 6 Ct r [ f (r) L p ( a;a) + f L p ( a;a)] (37)

while if G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 209 + 1 p 6 r with + 1 p j; j =1; 2;:::;r; and in addition, f (r s 1) AC( a; a), where s=[+1=p], then there exist a polynomial p P r s 1 such that [f p] L p ( t;t) 6 Ct r [ f (r) L p ( a;a) + f L p ( a;a)]: (38) Finally, if +1=p=r, then (37) holds with t r log t 1 in place of t r and if +1=p=j; j =1;:::;r 1, then (38) holds with t r log t 1 in place of t r. Here C is a positive constant independent of f and t. First we prove (21).Dene the polynomial Q := P r ][ 1 (x) interpolating f at the points {0} { i } r ][ 1 i=1 where i are arbitrary points in [x 1 =2;x 1 =2].By the proof of Lemma 3.1 [4], we have that for functions f D 2, the following estimate holds: as [f Q]w L 1 [ x 1;x 1] 6 C ( m) [ fw r 1 + f (r) w 1 ]: (39) Now we construct a suitable combination of the function f and the polynomial Q.Let F j dened F j = f j (1 2 )+Q 2 ; 2 (x)= ( 2(x1 x ) where f j was introduced in (13).By denition Q(x); x 6 x 1 2 ; f(x)(1 2 (x)) + Q(x) 2 (x); x 1 =2 6 x 6 x 1 ; F j (x)= f(x); x 1 6 x 6 x j ; f(x)(1 j (x)); x j 6 x 6 x j+1 ; 0; x x j+1 : We observe that F j Wr 1.By f(x)w (x)dx = [f(x) F j (x)]w (x)dx + and using the Gaussian rule w.r.t. w we have f(x)w (x)dx f(x k ) k (w )= where e m (F j ; w )= 06 k 6j F j (x)w (x)dx k= x 1 ) ; (40) F j (x)w (x)dx [f(x) F j (x)]w (x)dx + e m (F j ; w ); (41) F j (x k ) k (w ):

210 G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 We observe that the quadrature sum in (41) is the same which was introduced in (19).By denition of F j it follows f(x) F j (x) w (x)dx = f(x) Q(x) w (x)dx x 6x 1=2 + f(x) Q(x) 2 (x)w (x)dx x 1=26 x 6x 1 + f(x) j (x)w (x)dx x j6 x 6x j+1 + f(x) w (x)dx x x j+1 =:I 1 + I 2 + I 3 + I 4 : (42) I 1 has been estimated in (39). I 3 ;I 4 can be estimated as in the proof of the Theorem 2.1 (see (31) and (32)). Let us estimate I 2.Since 0 6 2 6 1, using (39), I 2 6 C ( m) [ fw r 1 + f (r) w 1 ]: (43) Therefore we obtain f(x) F j (x) w (x)dx6 C ( m) r [ f(r) w 1 + fw 1 ]+Ce Am fw 1 : (44) Now we estimate e m (F j ; w ).Since F j Wr 1,by(16) we obtain (r) F j w m; 1 e m (F j ;w ) 6 C ( : (45) m) r From denition of F j it follows F (r) j (x)=0 if x [ x 1 =2;x 1 =2] and for x x j+1.moreover, being w m (x) w (x) for x x 1, we have F (r) j w m; 1 6 C F (r) j w 1 : For x 1 =2 6 x 6 x 1 F (r) j = f (r) + r k=0 (r k) 2 (f Q) (k) : Since x 1 =2 1= m, it follows (r k) 2 (x) 6 (max x (r k) (x) )=x 1 =2 6 C( m) r k.so we have r F (r) j w L 1 [x 1=2;x 1] 6 f (r) w 1 + C ( m) r k w () (f Q) (k) x 1 1 ; 2 x 1: Using inequality (see [6, p.191]), k=0 (b a) k G (k) [a;b] 6 M(r) G +(b a) r G (r) [a;b] ; 0 j r; (46)

G. Mastroianni, D. Occorsio / Journal of Computational and Applied Mathematics 169 (2004) 197 212 211 which holds for any function G C (r) ([a; b]), we have r F (r) j L 1 [x 1=2;x 1] 6 f (r) w 1 + C w ()[( m) r (f Q) L 1 [x 1=2;x 1] + (f Q) (r) L 1 [x 1=2;x 1]]; k=0 6 C [ f (r) w 1 +( ] m) r (f Q)w L 1 [x 1=2;x 1] and using (38), we obtain F (r) j L 1 [x 1=2;x 1] 6 C[ fw 1 + f (r) w 1 ]: (47) For x j 6 x 6 x j+1 r F (r) j = f (r) (r k) + j f (k) : k=0 By similar arguments used in the previous estimate, we get F (r) j L 1 [x j;x j+1] 6 C[ fw 1 + f (r) w 1 ]: (48) Eq.(21) follows combining (47 48) with (45), (44) and (41).The proof of (20) follows a similar scheme, only the polynomial Q is identically zero. Acknowledgements We are grateful to the referee, whose careful reading of the paper strongly improved the original manuscript. eferences [1] G.Criscuolo, G.Mastroianni, Fourier and Lagrange operators in some weighted Sobolev-type spaces, Acta Sci. Math.(Szeged) 60 (1995) 131 166. [2] P.J.Davis, P.abinowitz, Methods of Numerical Integration, Academic Press, New York, 1975. [3] M.C.De Bonis, B.Della Vecchia, G.Mastroianni, Approximation of the Hilbert transform on the real semiaxis using Laguerre zeros, J.Comput.Appl.Math.140 (2002) 209 229. [4] M.C.De Bonis, G.Mastroianni, M.G.usso, Polynomial approximation with special doubling weights, Acta Sci. Math.(Szeged) 69 (2003) 159 184. [5] B.Della Vecchia, G.Mastroianni, Gaussian rules on unbounded intervals, J.Complexity 19 (3) (2003) 247 258. [6] Z.Ditzian, V.Totik, Moduli of Smoothness, in: Springer Series in Computational Mathematics, Springer, Berlin, 1987. [7] G.Freud, Orthogonal Polynomials, Pergamon Press, New York, 1971. [8] O.Kis, Lagrange interpolation with nodes at the roots of Sonin Markov polynomials (ussian), Acta Math.Acad. Sci.Hungar.23 (1972) 389 417. [9] N.X. Ky, A contribution to the problem of weighted polynomial approximation of the derivative of a function by the derivative of its approximating polynomial, Acta Sci.Math.Hungar.10 (1975) 309 316. [10] G.Mastroianni, Generalized Christoel functions and error of positive quadrature rules, Numer.Algorithms 10 (1 2) (1995) 113 126. [11] G.Mastroianni, G.Monegato, Truncated Gauss Laguerre quadrature rules, in: D.Trigiante (Ed.), Advances in the Theory of Computational Mathematics, ecent Trends in Numerical Analysis, Vol.3, Nova Science, 2000, pp.213 222.

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