136 P. ERDös We ca ow put the followig questio : Let (2) _1<xi')<...<x()<_1, 1<_m<oo be a triagular matrix. What is the ecessary ad sufficiet coditio

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1 ON THE BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS By P. ERDŐS i Budapest, Hugary Let -1<xI<x2<...<x <_1 be poits i ( - 1, + 1). A well kow theorem of Faber [5] states that there always is a polyomial of degree - 1 for which but IP - 1(xj)I<l, l<i< max I P,,-(.x) I > c log. -1<x<1 Throughout this paper P,(x) will deote a polymoial of degree ; c, c 1 c 2,". will deote positive absolute costats ot ecessarily the same if they occur at differet places. T other words : for o choice of the poits X 1,"',X ca we deduce from the boudedess of I P _ 1(x) 1 <_ 15 I, the boudedess of I P,,-,(x) I i the whole iterval ( - 1, + 1). Berstei [1] asked himself the questio whether oe ca deduce the boudedess of I P,(x) I i ( - 1, + 1) if we kow that I P (x) I < 1 for i > (1 + c) values of x. His aswer was affirmative. I fact he showed that if I P (x("'~) < 1 for all roots of the i h I Tchebicheff polyomial T,,(x) where m > (1 + c), the (1) max j P,,(x) I < A(c) _ 1 < x < 1 where A(c) depeds oly o c. Zygmud [7] proved that (1) holds if T (x) is replaced by P,(x) the m` h Legedre polyomial. 135

2 136 P. ERDös We ca ow put the followig questio : Let (2) _1<xi')<...<x()<_1, 1<_m<oo be a triagular matrix. What is the ecessary ad sufficiet coditio o the matrix that if for m > (1 + c). P,(xim) ) I < 1, 1 < i < ) the (1) holds. A priori it is ot obvious that a reasoable ecessary ad sufficiet coditio ca be formulated, but we will fid such a coditio which is ot too complicated. Put COS O,(m)= xim), 0 < 0<?C Let 0 <_ a < # < ad deote by N(a, f) the umber of O(j' ) satisfyig a < 0i"' ) < P. Let a < Di m) <... < O í') < N be the 0's i (a, /f), for each ri we defie a subsequece 0,á,' ;... 0 ;( m) of these 0's where i, = i ad if i,, i r _ 1 have already bee defied the 0 0m) is the smallest 0,( '), i r _, < l 5 j with W' )- O, _, > thus the distace 112 betwee ay two 0(,"0 is > /i ad ay other 0,x < 0;"' ) )<f3 is at distace < 0 /m from at least oe of the 0,, ~1 <= r <= k. Put N,) (a, f3) = k = k( 0) Now we formulate Theorem 1. Let x(j ` ) satisfy (2), ad assume that (3) I P (x ;") I S 1, 1 _< i <_ r, ) > (l + c) holds. The the ecessary ad sufficiet coditio that (3) should imply (1)

3 BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS is that there should be a h > 0 idepedet of m so that for every a< satisfyig - a,,,) -* 00 (4) N ;~(af li,,,) -(I +0(1)) (Í~, - a.). Coditio (4) meas that every iterval large compared to I cotais m asymptotically at least as may poits 0 < as T(x). o two of which are "too close", Before we give the fairly difficult proof T would like to call Theorem I proved 20 years ago [3]. attetio to a Theorem 2. Let x ;satisfy (2). The ecessary ad sufficiet coditio that to every cotiuoils fuctio f(x), - 1 <_ x < 1 ad to every c > 0 there should exists a sequece of polyomials P (x), < m(1 + c), such that P(xi 'I)-f(x~"'~), 1 ~ 1 ~ m ad (5) P (x)-f(x) uiformly i (- 1, + 1) as - oo, is that Jim if m. mi (0j(0(,+,,), -0 " ~)>0 ad that i.f 7(a - a ) O)o the (6) fl,) ( 1 + 0(1)) I (fl. - am)" Coditio (6) meas that every iterval (i 0) which is large compared 1 - cotais asymptotically at most as may x i 's as T,,,(x). The classical orthogoal polyomials as is well ko%v satisfy both (5) ad (6), ad also (4), thus our Theorem 1 cotais the results of Berstei ad Zygmud as special cases. I [3] the proof of Theorem 2 was oly outlied. The proof of Theorem 2 is i fact similar to the proof of Theorem 1. It ca be show that Theorem 2 is substatially equivalet to the followig to

4 1 3 8 P. ERDŐS Theorem 3. Let x~"'~ satisfy (2). The ecessary ad sufficiet coditio that there should exist to every c > 0 ad A(c) so that to every < 1, 1 < r < i, I < 1 < oo there should exist a polyomial P (x), Ir < ( I + c)m satisfyig P,,(xi "'» = y mt, I P (x)1 < A(c), - 1 < x <_ 1 is that (5) ad (6) should be satisfzed. Theorem 3 (ad therefore Theorem 2 too) is clearly related to Theorem 1. 1 this paper we do ot further discuss Theorem 2 ad 3. Now we prove Theorem 1. First we show that (4) is sufficiet, i other words if (4) holds the for every c > 0 (3) implies (1). 'To show this it will clearly suffice to prove the followig. Theorem 4. Let q > 0, c > 0 be arbitrary give umbers, e = r(q, c) is s cietly small ad B = B(r) is give. The there is a A = A(ij, c, e, B) so that if - I < x i <... < x," < 1, COS O i = x i, r = 1,, m is a sequece for which for every 0 <_ a < /i <= zz satisfyig /i-a > - B z N;"/i) > (1 - c) 177 (/1- a). The if P (x), < 1 + c is ay polyomial satisfyig (9) I P (xi) I< 1, 1< i _< m. We have P (x)i < A(q,c,c,B) for - 1 <_ x < 1.

5 BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS To prove Theorem 4 we will eed two Lemmas : Lemma l. Let z, <,... < z be the roots of the "' TchebicheT polyomial T (x). Let t be a fixed iteger ad - I < y, <... < y < l. where y j = z r for 1 <_ i <_ a ad u + t < i _<_. Assume further that for a fixed > 0. (10) arc cos arc cos y, s > r1/, 1 < s < t. The for a absolute costat C = Q1, t) we have where Ilk (x)i <C(rl,0, -1<x<1, 1<_Ic<it (Q((x co(x) = 11 (x - A) k = 1 are the fudametal polyomials of the Lagrage iterpolatio formula belogig to the yk. Deote by Lk(x) the fudametal polyomials of the Lagrage iterpolatio belogig to the zk. It is well kow that [4] (11) ILk(x)I<4, -1<x<-1, 1<-k<ai. Lemma 1 follows from (11) by a simple computatio by comparig the factors of Ik (x) ad L,(x) term by term ad by usig (10). We leave the simple details to the reader. Before we state Lemma 2 (which will be the most difficult part of the paper) we itroduce the followig otatios : Let P (x) = IZ (x - x i), cos B o = x o is a arbitrary poit i ( - 1, + 1). I(a) deotes the iterval {cos o, cos(d o + a)} ad I( - a t3) the iterval {cos (0, - g), cos (0, + fl)}* N,,(a) respectively N ( - a, f3) deotes the umber of the x, i I(a) respectively i I( - a, /3). Lemma 2. To every t, ad c i there is a t z = t z (t,, c,) so that if > o (t i, t z, c,) ad for every t i < t < t z

6 1 4 0 P. ERDŐS (12) N ( t ) > ( 1 + c l ) t/rc ad N ( - t) > ( 1 + c 1 ) t/ the (13) I P (xo) I< 2 max I Pjx) I. 15x51 I other words qualitatively speakig if P (x) has much more roots i every large eighborhood of x o tha the 11 th Tchebicheff polyomial the I P (x o )I is much smaller tha the absolute maximum of I P (x)i i ( - 1, + 1). (13) would hold with a arbitrary c z istead of z i but the t z (t,,c,) has to be replaced by t z (t 1,c l,c 2 ). Oe further remark : I (12) we oly cosider those itervals for which 0<0,+ t < 7T. To prove Lemma 2 we replace our P (x) by a ew polyomial Q,(x). Outside of I ( - t? tz), all the roots of P (x) are also roots of Q (x). Q (x) has the further roots (14) 2i - 1 t 2 2i. - I Cos _ < i _< j, = N,, ad cos ( ?r), a Our Q (x) has ow roots. By (14), i the iterval 1<i< t2 ). I ( _ 2jz -1,~ 2J 1-1 the roots of Q (x) are cogruet to those of T~(x) ad by the well kow theorem of M. Riesz [6] Q (x) must assume its absolute maximum i ( - 1, + 1) outside the iterval (15). By (12) 1( iterval (15). t2 ) is iside the

7 Assume ow that ( 1 b) I Q (zo) I= max I Q (x) - 15X<_I By what has bee just said we ca assume that z o is outside the iterval (15). Now we prove BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS (17) I Q,,(xo) IP(xo) I i 2I Q(z0) I P,,(ZO) I. Assume that (17) has already bee proved. By (16) we have Q (-z,) I >_ I Q,(x,) I, thus from (17) P"(xo) I < 2 P(zo) I max I Pa,(x) -15X51 which proves (13) ad thus Lemma 2 is proved. Thus to complete our proof we oly have to prove (17). The proof of (17) is quite simple i priciple ad to avoid simple ad routie computatio we will ot give all the details. Without loss of geerality we ca assume that lies to the right of the iterval (15). Deote by x, <_... < x j, the roots of P (x) i I (it ) ad by y i <... < y ;, the roots of Q (x) i I (2ll 1 ). X, >_ >_ x', are the roots of P (x) i I( -? ad yi < - < yj z those of z o Q (x) i I ( - Put 2j 2-1 ll Q,,(xo) P,.(--o) P,,(xo) Q,,( z o) where 7" ~7 (xo - yi)(z0 - xi) 7''~ _ ~` (x0- A(zo- xi) (19) i-1. (x0 - xi)(zo - ~' i) ' i=1 ( x 0 - x i) ( z o - Yi) 112 Now it immediately follows from (12) ad the defiitio of the x's ad y's t t that for every x i ad x i ot i I

8 142 P. ERDÖS (20) xo - ti `> ad xo - Y i > 1 + S, b = S(c i ). X 0 - x j x o - xí Also sice z o is to the right of (15) we have for every x ; ot i I \ tl) (21) z 0 - x i > z 0 - y i. From (12), (19), (20) ad (21) we obtai by a simple computatio that for 12 > t 2 (tl, c i ) ad > 11 0 (1 2, t i, CO (22) H, > 2 sice for sufficietly large t 2 = t 2 (t l, c,) ad > 11 0 (1 2, t i, c i ) the cotributio to 1-I, of the x, ad y, correspodig to the x i i I t, which do trot satisfy (20) ca be igored. Similarly we see that for t 2 = t i (t i, c i ) ad > 0 (t2, t i, CO (23) > 1 sice for the xí ad yí ot i I - i tt ) we have (24) (xo- Yí)(zo- XI) > 1. (xo - xí) (zo - Yí) (24) ca be deduced by a simple geometric (or aalytic) argumet from (20) ad z o > x 0. The x, i I (t ) ca agai be igored for sufficietly large t 2. (18), (22) ad (23) prove (17) ad hece the proof of Lemma 2 is complete. It is a ope questio if (13) remais true if istead of (12) we assume oly t t that for every t i < t < t 2 N - -, - > (1 + c i )2t/. Now we are ready to prove Theorem 4. If Theorem 4 would be false the there would be a fixed c, s, ri ad B so that for every D there would be arbitrarily

9 BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS large values of m for which there is a sequece - 1 <_ x 1 <... < x, <_ 1 satisfyig (25) arc cos x,,, - are cos x, > il/m ad for every x ad fl satisfyig (7), (8) is satisfied. Fially this sequece would be such that there would exist a P (x), < 1 I c satisfyig (9) ad (26) max I P (x) I = D. 1<x<_1 From these assumptios we have to derive a cotradictio for sufficietly large D. Assume that P (x o ) I = D, - 1 _< x o <_ 1 (i.e. I P (x) I assumes its absolute maximum i ( - 1, + 1) at x o ). Put cos 0, = x o, ad let B < t < T where T is sufficietly large ad will be determied later i). By (25) ad (8) (sice (25) holds the rl i (8) ca be left out) (T is idepedet of (27) N,,, (~t ) > (1 - a) N, (-~ deotes the umber of the x, i {cos 0 0, cos 0+ m t)~. O the other rat had T (x) has at most (28) t 11 7r m + 2 < (1-2 ) N, (~ ) roots i I G t) if e = s(c) is sufficietly small. m Deote ow by y 1 <... < y v _<_ 1 the roots of T (x) outside T T T T I ( - -, - ad our x i i I - -, -. By (25) ad (27), N = + 0(1) m m to I where the error term 0(1) depeds oly o T. Deote by SN _ 1 (x) the polyomial of degree at most N - 1 which coicides with P (x) o the x ; i I ( - t, t ad is 0 o the other y's. By the Lagrage iterpolatio formula m m

10 1 4 4 P. PFDös _ ( 29 ) SN- 1(x) - P ;,(xi) 1k(x), ik(x) - COW w'(3'k) (x W(x), = 11 (x.yk), N -.Yk) k = 1 t where i -'k rus over the yk (i.e. the x k) i I (-l12 t, l71. By (25) Lemma 1 ca be applied for the 'k(x) of (29) ad we obtai for every - 1 < x < 1 I SN-1(x) I < C(il, T) E' I P,,-1(xi) I By (25) the umber of summads i E' is less tha 2T hece by (9) q (3 0) SN- 1(x) < 2 C(, T), -1 < x <_ 1. Choose ow D = 6T Q1, T) ad put (31) R, (x) = P (x) - S,-,(x), i = max(, N - 1) = + O(t). R,(x) vaishes at the N,,, ( _ T T) X'S i I ( - T Tl. Thus by (31) i J7 (27) ad (28), (12) (of Lemma 2) is satisfied by R,,(x) with B = t 1, T= t z ad 1 1-c/2 = 1 + e 1. At x o we have by (26), (29), (31) ad the choice of D (32) I R,,,(x o ) I > 2 max I R (x) 1<_x<1 but this cotradicts Lemma 2 for sufficietly large T. Hece the proof of Theorem 4 is complete ad we showed that (4) is a sufficiet coditio that (3) should imply (1). To complete the proof of Theorem 1 we have to prove the ecessity of (4). I other words we shall show that if (4) is ot satisfied the (3) does ot imply (1). To show this it will suffice to show that the coditios of Theorem 4 are best possible. I other words we shall prove Theorem 5. Let A be a arbitrary positive umber, ri = q(a) is suf-

11 BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS ficietly small, e > 0 is fixed ad 6 < e/2 is arbitrary. The there is a B = B(A,q,e,5) so that if - 1 < x i <... < x, < 1, m > m o (A, q, e, b, B) is ay sequece satisfyig for some 0 <_ a < a + B < 7r m (33) N ') (a, x + B ) < B(1 - r). m The there is a polyomial P (x), < m(1-6) satisfyig (34) I P (x ;) I< 1, max I P (x) I> A. To make the idea of the proof more itelligible we first assume istead of (33) the stroger coditio (35) N,,, (a, a + B) < B(1 - e) 111 ad deduce the existece of a polyomial satisfyig (34) from (35). It will the be easy to modify our argumet to show that (34) follows from (33) too. First we defie a auxiliary polyomial Q (x). All the x i i (a, a + B are m roots of Q (x) (the iterval cos fl < x < cos y will be deoted (/3, y)). I 01 + t - s ( i 10, 0 ad i 7r, a + BWi ~) all the roots of T,.(, -,, ),(x) are roots of Q (x). By (33) ad 6 < s we obtai that the degree of Q (x) is less tha m(1 - (S). Thus by the theorem of M. Riesz [6] Q (x) assumes its absolute B(r; - S) maximum i ( - 1, + 1) i ( a + B(e, a + m B (1 - e 10-5, say at tom x o = cos 0 0. Our polyomial P (x) is obtaied from Q (x) as follows : All the commo roots of Q,,(x) ad T, x -a»( ) ~ I a, a,.~~~ + B lom a) are moved to cosa ad

12 1 4 6 P. ERDŐS all the commo roots of Q(x) ad T,,,, (, a),(x) i (a + B (1-6 E10 )' a + rb ) are moved to cos (a + B i Thus B(e - b) 10 + O(1) roots are moved away from x o i both directios. We ow show that P (x) satisfies (34). First of all P (x,) = 0 for all the x, i (a, a + B/m), thus to complete our proof it will suffice to show that (36) I P(xo) I /I P(zo) I> A, I P (z 0 ) I= max I P (x) I where i (36) the maximum is take over the x i (-1, +1) which are ot i a, a + B-1 (sice (36) clearly implies I P (xo) I > A max P (x,) which is m (34)). By assumptio we have I Q (x o) I >_ I Q (zo) I, hece (36) will follow if we ca prove (37) Q (xo) > A I Q zo) The proof of (37) is almost idetical with the proof of (17) ad ca be left to the reader. This completes the proof of 'Theorem 5 if we assume (35). To complete our proof we ow have to show that a polyomial P (x), < m(1-6) exists satisfyig (34) if we oly assume (33) (istead of (35)). Choose rl = 2/1rA. By (33) ad the defiitio of N ;,';t (see the itroductio) there is a subsequece xi,, x,,, r < B(l - e) of the x,'s i (a, a + B i satisfyig 2 arccosx,. +l - arccosx, > rrat ~ A so that for every x i (a, a + B ) there is a x ij satisfyig (38) Iarecosx -arccosx,.i < ' 2 Ai

13 BOUNDEDNESS AND UNBOUNDEDNESS OF POLYNOMIALS 1 47 I view of our previous costructio there is a polyomial P (x), < m(1-6) satisfyig (39) max I P,,(x) I = A, P (x i.) = 0, j = 1,..., r -1<_x<_1 ad (40) max I P (x) I < 1 where i (40) the maximum is take over the - 1 <_ x <_ 1 ot i (a, a + B I A well kow theorem of Berstei [2] states that if f(o) is a trigoometric polyomial of degree satisfyig max IL(0) I = 1 the max If', (O) I <, 0<8<2 0<B<_2 (thus from this theorem of Berstei we easily obtai from (38) ad (39) that for every x i a, )c + B) I (41) I P,'(x ) I < 1. (39), (40) ad (41) prove (34) ad hece the proof of Theorem 5 is complete, but this also fiishes the proof of Theorem 1. Fially we state without proof Theorem 6. To every A however large there is a e > 0 so that if > a (A,e), m = [(1 + e)], the for every - 1 < x l <... < x, < 1 there is a P,(x) satisfyig P (x,) 1< I i = 1,.., m ad max I P(x) I >A. -1<x<1 We do ot give the proof of Theorem 6. REFERENCES 1. S. Berstei, Sur ue classe de formules d'iterpolatio, Bill. Acad. Sci. U.S.S.R. (7) vol. 4 (1931), , see also Y. Marcikievicz ad A. Zygmud, Mea values of trigoometric polyomials, Fud. Math. 28 (1937), see p S. Berstei, Belg. Mem p. 19.

14 1 4 8 P. ERDÖS 3. P. Erdős, O some covergece properties of the iterpolatio polyomials, Aals of Math. 44 (1943), P. Erdős ad G. Grümvald, Note o a elemetary problem of iterpolatio, Bull. Amer. Math. Soc. 44 (1938), The slightly weaker iequality of Fejér would have served our purpose just as well (Math. Aale 106 (1932), G. Faber, Über die iterpolatorische Darstellug stetiger Fuktioe, Jahresbericht der Deutsche Math. Ver. 23 (1914), M. Riesz, Eie trigoometriche Iterpolatiosformel etc. Jahresbericht der Deutsche Math. er. 23 (1914), A. Zygmud, A property of the zeros of Legedre polyomials, Tras. Amer. Math. Soc. 54 (1943), THE HUNGARIAN ACADEMY OF SCIENCES BUDAPEST, HUNGARY AND ISRAEL INSTITUTE OF TECHNOLOGY HAIFA, ISRAEL (Received Jauary 2, 1967)