Sobolev Freud polynomials Mohamed BOUALI 5 1 0 2 Abstract n a J Weinvestigatetheuniformasymptoticof someSobolevorthogo- 1 nal polynomials. Three term recurrence relation is given, moreover ] wegivearecurrencerelationbetweentheso-calledSobolevorthogo- A nalpolynomialsandFreudorthogonalpolynomials. C . h t a 1 Introduction m [ During the past few years, orthogonal polynomials with respect to an in- 1 v ner product involving derivatives (so-called Sobolev orthogonal polyno- 0 0 mials) have been the object of increasing number of works (see, for in- 5 6 stance [1], [5], [6], [4], [7], [8]). Recurrence relations, asymptotics, alge- 0 ff braic, di erentiation properties and zeros for various families of polyno- . 2 0 mials have been studied. In this paper we study a connection between a 5 particular case of non-standard orthogonal polynomials. 1 : v For λ1,λ2 0, we defined the inner product ≥ i X r f ,g = f (x)g(x)e x4dx+λ f (0)(x)g(0)+λ f (0)(x)g (0), a S − 1 2 ′ ′ h i ZR We denote also by the norm associate to the inner product .,. . Let S S ||·|| h i Q be the sequence of orthogonal polynomial with respect to .,. . We n S h i denoted k = Q 2 = Q ,Q . n || n||S h n nis LetP bethesequenceofmonicpolynomialsorthogonalwithrespectto nb + the inner product f ,g = ∞f (x)g(x)e x4dx : They have been consid- F − h i Z ered by Nevai [14,15]. These−p∞olynomials satisfy a three-term recurrence 1 relation xP (x)=P (x)+c P (x), n n+1 n n 1 − with initial conditions P (x) = 1 and P (x) = x; where the parameters c 0 1 n satisfy a non-linear recurrence relation (see [4]) n=4c (c +c +c ), n 1, n n+1 n n 1 − ≥ Γ Γ with c = 0 and c = (3/4)/ (1/4). Moreover the polynomial P satisfies 0 1 n therecurrence relation P (x)=nP (x)+d P (x), n 3 n′ n 1 n n 3 − − ≥ + where d = 4k /k , with k = P 2 = ∞ P (x) 2e x4dx. One can see n n n−3 n || n||F Z n − (cid:16) (cid:17) from the three-term recurrence relation th−a∞t k =c k , (1.1) n n n 1 − and d =4c c c . n n n 1 n 2 − − Lemma1.1 1. ForallpolynomialsP,Q, xmP,Q = P,xmQ = xmP,Q = P,xmQ , S S F F h i h i h i h i wherem 1 ifλ =0, and m 2 if λ >0. 2 2 ≥ ≥ 2. For a polynomial Q, we denote Q(x) =Q( x). For all polynomials P,Q, − wehave Q,P = Q,P . h iS h iS e 3. Q ( x)=e( 1)nQ (x).e n n − − Proof. Theproof of the first item iseasy. 2) Using the symmetry of the Freud inner product Q,P = Q,P and F F h i h i thefact that P (0)= P (0) ′ − ′ e e e Q,P = Q,P +λ Q(0)P(0) λ Q (0)P (0) S F 1 2 ′ ′ h i h i − e = Qe,P F +λ1Q(0)P(0) λ2Q′(0)P′(0) h i − . = Q,P +λ Q(0)P(0)+λ Q (0)P (0) e F 1 2 ′ ′ h i = Q,P e S e e h i e 2 3) From the first step we have by orthogonality Q ,P = 0, for all poly- n S h i nomials P with deg(P) n 1. ≤ − e ffi HenceQ (x)=α Q (x),equalingtheleadingcoe cientweobtainα = n n n n ( 1)n. − f 2 Case λ = 0 2 Proposition 2.1 ThepolynomialsP and Q are relatedby n n xP (x)=Q (x)+a Q (x), n 1 n n+1 n n 1 − ≥ xQ (x)=P (x)+b P (x), n 1 n n+1 n n 1 − ≥ k k n n Q (x)=1, Q (x)=x, wherea = , b = ,. 0 1 n n kn 1 kbn 1 − − Proof. Since, let write b n xP (x)=Q (x)+ α Q (x), (2.2) n n+1 k k X k=0 By orthogonality one gets + Q 2α = ∞P (x)Q (x)xe x4dx, || k||S k Z n k − −∞ Since for k n 2, byorthogonality theintegral vanishes, moreover using ≤ − the symmetry of the inner product, one as P ( x)=( 1)nP (x), Q ( x)=( 1)kQ (x), n n k k − − − − hence α =0, anda =α . n n n 1 − + k α = ∞P (x)xQ (x)e x4dx=k , n 1 n 1 n n 1 − n − − Z − −∞ b where we used xQ (x) = P (x)+... The second statement can be proved n 1 n − by asame argument. Proposition 2.2 3 c c n+2 n+1 1. +a =c +c . n n+1 n a n+2 2. a b =c c n+1 n n+1 n a b 1 n n 3. lim = lim = . n √n n √n 2√3 →∞ →∞ Proof. 1) Since xP (x)=Q +a Q (x), hence n n+1 n n 1 − xP 2 = Q 2 +a2 Q 2, || n||S || n+1||S n|| n−1||S moreover k Q 2 = n, || n−1||S an and xP 2 = xP 2, from the tree-term recurrence relation onegets || n||S || n||F xP 2 = P 2 +c2 P 2 =(c +c )k , || n||F || n+1||F n|| n−1||F n+1 n n thus k n+2 +a k =(c +c )k , n n n+1 n n a n+2 hence c c n+2 n+1 +a =c +c . n n+1 n a n+2 k k n+1 n 2) We saw that a = , b = , and k =c k ,hence n+1 n n n n 1 kn kbn 1 − − b k n+1 a b = =c c . n+1 n n+1 n k n 1 − c c c +c 3) Let λ =a √n, σ = n+2 n+1, δ = n+1 n. Then we obtain, n n n n+1 n √n+1 σ √n 1 n + − λ =δ , (2.3) n 1 n λn+1 √n+1 − c 1 n Using the factthat lim = , (see for instance [7], [10]) onegets, n √n 2√3 →∞ 1 1 lim σ = , lim δ = . n n n + 12 n + √3 → ∞ → ∞ 4 Moreover since λ 0, and n ≥ n+2 λ δ , n ≤r n n+1 thus the sequence λ is bounded. Let ℓ be the limit of a subsequence. It n follows from equation (2.3), 1 1 +ℓ = , 12ℓ √3 and the unique solution is ℓ = 1 . Hence the unique limit of a sub- 2√3 sequence of λ is ℓ = 1 , then the bounded sequence λ converges to n 2√3 n ℓ = 1 . 2√3 b The same hold for n from therelation a b =c c . √n n+1 n n+1 n Theorem2.3 Theasymptoticbehavior xϕ 4 3x lim Qn(√4nx) = √412 (cid:16)q4 (cid:17) , n→∞ Pn(√4nx) 1+ϕ2 4 3x 4 q (cid:16) (cid:17) holduniformly on compact subset of C [ √44/3,√44/3], where \ − ϕ(x)= x+√x2 1, with √x2 1> 0 for x > 1, i.e., the conformal mapping of − − C [ 1,1] ontotheexterior of theclosedunit disk. \ − Proof. We saw that xQ (x)=P (x)+b P (x). n n+1 n n 1 − It is well-known (see [16]) that from the three-term recurrence relation of non normalizing Freud polynomials xS (x)=α S (x)+α S (x), n n+1 n+1 n n 1 − we can obtain asymptotic properties of the orthonormal polynomials S : n α 1 n Indeed, as lim = .We deduce (see [11]) n √4n √412 →∞ Sn 1 √4nx 1 lim − = , (cid:16) (cid:17) n→∞ Sn √4nx ϕ 4 3x 4 (cid:16) (cid:17) q (cid:16) (cid:17) 5 uniformly on compact subsets of C [ √44/3, √44/3]. Then, for the monic \ − Freud polynomial P one gets n Pn 1 √4nx √412 lim √4n − = , (cid:16) (cid:17) n→∞ Pn √4nx ϕ 4 3x 4 (cid:16) (cid:17) q (cid:16) (cid:17) uniformly on compact subsets of C [ √44/3, √44/3]. \ − Since Q (√4nx) 1 P (√4nx) b √4nP (x) n n+1 n n 1 = + − . Pn(√4nx) x(cid:18)√4nPn(√4nx) √n Pn(x) (cid:19) Asngoestoinfinity,onegetsoneverycompactsubsetsofC [ √44/3, √44/3], \ − lim Qn(√4nx) = 1 ϕ(q4 34x) + 1 √412 . n→∞ Pn √4nx x(cid:18) √412 √12ϕ( 4 3)(cid:19) 4 (cid:16) (cid:17) q Asimple computation gives 2 1+ ϕ 4 3x Q (√4nx) 1 4 lim n = (cid:18) (cid:16)q (cid:17)(cid:19) . n→∞ Pn √4nx √412 xϕ 4 3x 4 (cid:16) (cid:17) q (cid:16) (cid:17) Theorem2.4 The polynomials Q have all their zeros real and simple. For n n 3 thepositivezeros of Q interlacewith thoseof P . n n ≥ Proof. We distinguish two cases: the even and the odd one, respectively. ff Theproofs are similar with slight di erences. Even case: Let x ,k =1,...,m, be the positive zeros of P in increas- 2m,k 2m ing order, that is, x <...<x . First, we need to study the sign of the 2m,1 2m,m integrals + P (x) I = ∞Q (x) 2m e x4dx, m 2, k =1,...,m, 2m,k Z 2m x2 x2 − ≥ −∞ − 2m,k We have + m 1 m 1 + I = ∞Q (x) − (x2 x2 )e x4dx= − b ∞Q (x)x2re x4dx=b (k)λ Q (0), 2m,k Z 2m − 2m,j − rZ 2m − 0 1 2m −∞ j=Y1,j,k Xr=0 −∞ 6 m 1 m − whereb (k)=( 1)m 1 x2 ,andQ (0)= a Q (0)=( 1)m a , 0 − − j=Y1,j,k 2m,j 2m − 2m 2m−2 − Yk=2 2k (a >0), moreover sign(b )=( 1)m 1, hence 2k 0 − − sign(I )= 1. (2.4) 2m,k − On the other hand, using Gaussian quadrature in all the zeros of P and 2m ff taking into account the symmetry of the polynomials Q , the Christo el 2m numbers (see, for example, [7, p140]) 1 µ = , i =1,..,m, 2m,i 2m 1P2(x ) j=0− j 2m,i P together with thefact m P (x ) 2′m 2m,k = x2 x2 , 2x 2m,k − 2m,j 2m,k j=Y1,j,k(cid:16) (cid:17) we get P (x ) 2′m,k 2m,k I =µ Q (x ) , 2m,k 2m,k 2m 2m,k 2x 2m,k andfrom (2.4) we deduce sign(Q )= sign(P ). 2m − 2′m,k Since P (x) has opposite sign in two consecutive zeros of P (x), we de- 2′m 2m duce that it also occurs for Q (x), and therefore Q (x) has one zero in 2m 2m each interval (x ,x ), k =1,...,m 1 (and from the symmetry it has 2m,k 2m+1,k − one zero in each interval ( x , x ), k = 1,...,m 1. Thus Q (x) 2m+1,k 2m,k 2m − − − has at least 2m 2 real and simple zeros interlacing with those of P (x). 2m − Finally, as P (x )>0 then Q (x )<0 and since Q (x) is monic 2′m 2m,2m 2m 2m,2m 2m we deduce the existence of one zero of Q (x) in (x ,+ ) and another 2m 2m,m ∞ zero in ( , x ), which complete theresult for theeven case. 2m,m −∞ − Odd case: Let m 2, 0 < x < ... < x be the positive simple 2m+1,1 2m+1,2m ≥ zeros of P , since P (0)=Q (0)=0, let define theintegral 2m+1 2m+1 2m+1 + P (x) I = ∞Q (x) 2m+1,k e x4dx 2m+1,k Z 2m+1 x2(x2 x2 ) − −∞ − 2m+1,k 7 hence + Q (x) m I = ∞ 2m+1 (x2 x2 )e x4dx, 2m+1,k Z x − 2m,j − −∞ j=Y1,j,k I =m−1b (k) +∞Q2m+1(x)x2re x4dx, 2m+1,k r − Z x Xr=0 −∞ Since for 1 r m 1, ≤ ≤ − + Q (x) ∞ 2m+1 x2re x4dx= Q ,x2r 1 λ (Q x2r 1) , − 2m+1 − S 1 2m+1 − x=0 Z x h i − | −∞ hence byorthogonality one getsfor 1 r m 1 ≤ ≤ − + Q (x) ∞ 2m+1 x2re x4dx=0. − Z x −∞ Thus + Q (x) I =b (k) ∞ 2m+1 e x4dx, (2.5) 2m+1,k 0 − Z x −∞ since xP (x)=Q (x)+a Q (x), 2m 2m+1 2m 2m 1 − hence + + + Q (x) Q (x) ∞P2m(x)e−x4dx= ∞ 2m+1 e−x4dx+a2m ∞ 2m−1 e−x4dx Z Z x Z x −∞ −∞ −∞ thus byorthogonality +∞P (x)e x4dx=0, and 2m − R−∞ + Q (x) + Q (x) ∞ 2m+1 e−x4dx= a2m ∞ 2m−1 e−x4dx Z x − Z x −∞ −∞ and + Q (x) 5 m ∞ 2m+1 e x4dx=2Γ( )( 1)m a . (2.6) − 2k Z x 4 − Y −∞ k=1 Moreover m b (k)=( 1)m 1 x2 , (2.7) 0 − − 2m+1,j j=Y1,j,k from equations (2.5), (2.6) and(2.7)one gets sign(I )= 1. 2m+1,k − Therest of the proof is as in the even case. 8 , 3 Case λ 0 2 Proposition 3.1 For all n 1, ≥ xP (x)=Q (x)+a Q (x), 2n 1 2n n 2n 2 − − x2P (x)=Q (x)+b Q (x)+α Q (x), n n+2 n n n n 2 − x2Q (x)=P (x)+σ P (x)+δ P (x). n n+2 n n n n 2 − k k x2P ,Q with Q0(x) = 1, Q1(x) = x, where, an = k22nn−21, αn = knn2, bn = hhQnn,QnniiSS, δn = − − k k n n b b , σ =b . n n k k bn 2 bn − Proof. The proof isas in proposition 1.1. Proposition 3.2 For all n 1, ≥ c c 2n+1 2n 1. +a =c +c . n 2n 2n 1 an+1 − c c c c c c 2. c c +c c +(c +c )2 = n+4 n+3 n+2 n+1 +b2 n+2 n+1 +α . n+2 n+1 n n 1 n+1 n n n − αn+4 αn+2 n 3. σ = +c b . n n 2 n 2 4cn − − − σ n 1 n+2 4. c c +b σ +α = + . n+2 n+1 n n n b 2 4 n+2 δ b n n 5. σ = . n c c n n 1 − a 1 b 1 α 1 n n n 6. lim = , lim = , lim = . n √2n 2√3 n √n √3 n n 12 →∞ →∞ →∞ σ 1 δ 1 k n n n 7. lim = , lim = , lim =1. n→∞√n √3 n→∞ n 12 n→∞bkn Proof. 1) The first relation can beproved as the case λ =0. 2 2)Fromthesecondrelation intheprevious proposition andorthogonality one gets x2P 2 =k +b2k +α2k . || n||S n+2 n n n n−2 b b b 9 k Using the factthat k =c k , andk = n+2. Onegets n n n 1 n α − n+2 x2P 2 = cn+4cn+3cn+2bcn+1 +b2cn+2cn+1 +α k (3.8) || n||S α n α n n (cid:16) n+4 n+2 (cid:17) Since from the three term recurrence relation xP (x)=P (x)+c P (x), n n+1 n n 1 − andorthogonality we have xP 2 = xP 2 =k +c2k , || n||S || n||F n+1 n n−1 and x2P 2 = x2P 2 = xP +xc P 2 || n||S || n||F || n+1 n n−1||F = xP +c2 xP 2 +2c xP ,xP || n+1||F n|| n−1||F nh n+1 n−1iF =k +c2 k +c2k +c2c2 k +2c k (3.9) n+2 n+1 n n n n n 1 n 2 n n+1 − − =(c c +c2 +c2+c c +2c c )k n+2 n+1 n+1 n n n 1 n n+1 n − = c c +(c +c )2+c c k . n+2 n+1 n+1 n n n 1 n − (cid:16) (cid:17) From equation (3.8), (3.9)one getsthe desired result. 3) By orthogonality one gets σ k = x2Q ,P , n n n n F h i since bydefinition of the Sobolev inner product we have x2Q ,P = x2Q ,P = Q ,x2P n n F n n S n n S h i h i h i = x2P b Q α Q ,x2P , n 2 n 2 n 2 n 2 n 4 n S h − − − − − − − i = x2P ,x2P b Q ,x2P α Q ,x2P n 2 n S n 2 n 2 n S n 2 n 4 n S h − i − − h − i − − h − i since + x2P ,x2P = ∞x4P (x)P (x)e x4dx n 2 n S n 2 n − h − i Z − 1−∞+ 1 + = ∞P (x)P (x)e x4dx+ ∞xP (x)P (x)e x4dx 4Z n−2 n − 4Z n′−2 n − 1 −+∞ −∞ + ∞xP (x)P (x)e x4dx, 4Z n−2 n′ − −∞ byorthogonality the first and thesecond integralvanished. Moreover + + ∞xP (x)P (x)e x4dx= ∞xP (x) nP (x)+d P (x) e x4dx n 2 n′ − n 2 n 1 n n 3 − Z − Z − − − (cid:16) (cid:17) −∞ −∞ =nk +d k . n 1 n n 2 − − 10