Some coefficient estimates for polynomials on the unit interval

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1 Some coefficient estimates for polynomials on the unit interval M. A. Qazi and Q.I. Rahman Abstract In this paper we present some inequalities about the moduli of the coefficients of polynomials of the form fx) := n ν=0 a νx ν, where a 0,..., a n C. They can be seen as generalizations, refinements or analogues of the famous inequality of P. L. Chebyshev, according to which a n n if n ν=0 a νx ν 1 for x AMS Classification: 6C05, 6C10, 30A1, 30D15, 4A05, 4C05. Key words: Polynomials, Inequality, Weighted L p norm. 1. Chebyshev Polynomials T n Using de Moivre s theorem, i.e. the formula cos nθ + i sin nθ = cos θ + i sin θ) n, we readily see that cos nθ can be expressed as a polynomial of degree n in cos θ. The polynomial T n such that 1) T n cos θ) cos nθ is called the Chebyshev polynomial of the first kind of degree n. Formula 1) implies that T n x) 1 for x 1 and that T n cos νπ ) = ) ν ν = 0, 1,..., n). n From 1), it also follows that the zeros of T n are cos ν 1)π n ν = 1,..., n). 1

2 M.A. Qazi and Q.I. Rahman Clearly, they all lie in the open interval, 1). Writing 1) in the form we see that e iθ + e iθ ) T n einθ + e inθ ) z + z τz) := z n T n zn + 1 θ R) is a polynomial of degree not exceeding n, which vanishes at each point of the unit circle. Since a polynomial of degree at most n cannot vanish at more than n points without being identically zero, we conclude that ) ) z + z T n zn + z n z 0). Let T n z) := n ν=0 t n,ν z ν be the Maclaurin expansion of T n. Then, comparing the coefficients of z n on the two sides of ) we find that 3) t n,n = n, i.e. the leading coefficient in the Maclaurin expansion of T n is n. It is useful to know that 4) T n z) := n n/ k=0 k n k 1)! ) k! n k)! z)n k.. The polynomial inequality of Chebyshev The polynomials {T n } are known to have several remarkable extremal properties. They play an important role in the Theory of Approximation and related areas of mathematics. Let P m denote the class of all algebraic polynomials of degree at most m. For any function Φ C[a, b], let E n [Φ] = inf { max Φx) P x) }. P P n x [a,b] A polynomial P P n is said to be a polynomial of best approximation to Φ if E n [Φ] = max a x b Φx) P x).

3 Coefficient estimates for polynomials 3 By a classical result of P. L. Chebyshev, a polynomial P P n is a polynomial of best approximation to Φ if there exist n + points such that a x 0 < < x n+1 b ±) ν [Φx ν ) P x ν )] = max a x b Φx) P x) ν = 0,..., n + 1), i.e. if Φx) P x) takes on the values ± max a x b Φx) P x) in alternating succession at the points x 0,..., x n+1. This alternation theorem shows, in particular, that P is unique. For a historical perspective of the alternation theorem, we refer the reader to an interesting article of Butzer and Jongemans []. As an example, let us find the best uniform approximation of the function Φx) := x n on [, 1] by polynomials in P n. Clearly, the polynomial of best approximation must be real. In view of 3), we can write x n = T nx) n + a polynomial in P n. Since T n x) takes the value +1 and at n + 1 points in [, 1] and n + 1 can be written as n)+, the alternation theorem shows that the desired polynomial of best approximation to x n, in P n, is the polynomial x n 1 n T nx). Equivalently, the monic polynomial P x) = x n +a n x n + +a 0 of best approximation to the function Φx) 0 on the interval [, 1] is the Chebyshev polynomial 1 n T n x). The maximum deviation of this polynomial from zero is 1 n. The above discussion may be summarized as follows. Theorem A. Let T n x) := n ν=0 t n,ν x ν be the Chebyshev polynomial of the first kind of degree n. Then, for any polynomial fx) := n ν=0 a ν x ν of degree at most n, not identically zero, we have 5) a n max x 1 fx) t n,n max x 1 T n x) = n.

4 4 M.A. Qazi and Q.I. Rahman 3. Upper bound for a n and the distribution of the zeros of f In 5), equality holds if and only if fz) e iγ T n z), γ R. It was shown by Schur [16] that if the polynomial fz) := n ν=0 a ν z ν vanishes at c = 1 or c = and fx) 1 for x 1, then 6) a n cos π 4n) n n, where equality holds if and only if fz) := e iγ T n z cos π 4n c π ) sin 4n γ R). In fact, the following more general result see [9, Theorem 1]) holds. Theorem B. Let fz) := n ν=0 a ν z ν be a polynomial of degree n such that fc) = 0 for some c belonging to [, cos π/n)) cos π/n), 1]. Furthermore, let fx) 1 at the points ξ ν = ξ ν,c := c 1+ c ) cos νπ/n)+c sign c) cos π/n) c cos π/4n) Then 7) ) 1 + cos π/n) n a n n. 1 + c 1 ν n). The upper bound for a n given in 7) is attained for the polynomial ) 1 + cos π/n))z c + sign c) cos π/n) e iγ T n γ R), 1 + c which satisfies the conditions of Theorem B. The above mentioned result of Schur has also been generalized [1, Theorem ] as follows. Theorem C. Inequality 6) holds for any polynomial fz) := n ν=0 a ν z ν of degree n with real coefficients which has at most n 1 distinct zeros in the open interval, 1). It was conjectured by K. Mahler and stated by him in some of his lectures see [0, p. 15 ]) that if f is as above, then { π exp log f e iθ) } 8) dθ n. π π

5 Coefficient estimates for polynomials 5 It is well-known [18, p. 84d] that 1 π π π log e iθ ζ dθ = log + ζ := Hence, if z 1,..., z n are the zeros of f then 1 π π π log f e iθ) dθ = log an + { 0 for ζ 1 log ζ for ζ 1. n log + z ν = log a n + log z ν, ν=1 z ν >1 where the product on the right is to be replaced by 1 if z ν 1 for 1 ν n. Thus 8) is equivalent to the inequality 9) z ν a n n. z ν >1 Turán [0] settled the conjecture of Mahler by proving it in the form 9). Inequality 9) gives a quantitative improvement on 5) only if f has some zeros outside the unit circle. However, as mentioned above, inequality 5) is strict unless fz) e iγ T n z), γ R, and so certainly if some of the zeros of f lie in C\[ cos π/n), cos π/n)]. Theorem B deals with the case where f has a zero in [, cos π/n)) cos π/n), 1]. What if f has some zeros in C\[, 1]? An answer to this question is to be found in Corollary 1. We shall deduce it from the following theorem, which is a refined version of a result from [11, Theorem 1]. Theorem 1. Let ψζ) := m µ=0 γ µ ζ µ be a polynomial of degree m such that γ 0 γ m > 0 and ψ e lπi/m) 1 for l = 0, 1,..., m 1. Furthermore, let ζ 1,..., ζ J be amongst the zeros of ψ, where 1 J m. Then 10) γ 0 Jj=1 ζ j + γ m J ζ j 1. j=1 The following modified form of a famous inequality of C. Visser [1] plays an important role in the proof of Theorem 1. Lemma 1. Let φζ) := m µ=0 c µ ζ µ be a polynomial of degree m such that c 0 c m > 0. Furthermore, let φ e lπi/m) 1 for l = 0, 1,..., m 1. Then c 0 + c m 1.

6 6 M.A. Qazi and Q.I. Rahman Proof. Let ω := e lπi/m, where l {1,..., m 1}. Then ω m = 1 and because ω 1, we have m µ=0 e lπi/m) µ = m µ=0 ω µ = 1 ωm 1 ω = 0 l = 1,..., m 1). Hence and so m l=0 m l=0 e lπi/m) µ = m φ e lπi/m) m = l=0 l=0 m c 0 + e µπi/m) l = 0 µ = 1,..., m 1), µ=1 Since c 0 c m > 0, this implies that m c µ l=0 e lπi/m) µ + m l=0 c m = m c 0 + c m ). c 0 + c m = c 0 + c m 1 m φ e lπi/m) 1. m l=0 Proof of Theorem 1. Let φζ) := ψζ) Then φζ) := m µ=0 c µ ζ µ, where and so Furthermore, J j=1 ζ j ζ 1 ζ ζ j. c 0 = γ 0 Jj=1 ζ j and c m = γ m J c 0 c m = γ 0 Jj=1 ζ j γ m J j=1 j=1 ζ j, ζ j = γ 0 γ m > 0. φ e lπi/m) 1 for l = 0, 1,..., m 1 since ζ j ζ 1 = 1 ζ = 1). ζ ζ j Thus, Lemma 1 applies and so 10) holds.

7 Coefficient estimates for polynomials 7 We are ready to state and prove our refinement of Chebyshev s inequality, which takes into account each and every zero that does not belong to [, 1]. Corollary 1. 11) Let fz) := a n nν=1 z z ν ), a n 0, and suppose that cos f kπ ) 1 k = 0, 1,..., n). n Furthermore, for any z ν [, 1] let R ν denote the sum of the semi-axes of the ellipse whose foci lie at ±1 and which passes through z ν ; otherwise let R ν = 1. Then ) 1 1) R 1 R n + a n n. R 1 R n Proof. Let Then ψζ) := n ν=0 γ ν ζ ν, where ) ζ + ζ ψζ) := ζ n f. 13) γ 0 = γ n = n a n. In particular, γ 0 γ n = n a n > 0. Furthermore, since )) lπi exp ψ = n cos f lπ ) 1 l = 0, 1,..., n 1) n cos lπ n l)π = cos l = n + 1,..., n 1). n n Thus, ψ satisfies the conditions of Theorem 1 with m = n. For any given R > 1, let C R denote the circle ζ = R e it, 0 t π. Also, let E R be the ellipse whose foci lie at ±1 and whose semi-axes are R + R and R R It may be noted that as ζ goes around the circle C R, the point z := ζ + ζ )/ describes the ellipse E R once counterclockwise, starting at the point R + R )/. Hence, corresponding to a zero z ν of f, not belonging to [, 1], there is a number ζ ν lying outside the closed unit disc such that. 14) 1 ζ ν + ζ ν ) = z ν.

8 8 M.A. Qazi and Q.I. Rahman In fact, ζ ν is equal to R ν, the sum of the semi-axes of the ellipse whose foci lie at ±1 and which passes through z ν. Note that 1 ζ + ζ ) = z ν has a double root at ζ = 1 if z ν = 1 and a double root at ζ = if z ν =. For any z ν, 1) there are two different values of ζ ν, both of modulus 1, that satisfy 14). So, we take R ν = ζ ν = 1 if z ν is a zero of f that lies somewhere in [, 1]. Amongst the zeros of ψ we can therefore count n zeros ζ 1,..., ζ n with ζ ν = R ν 1 for ν = 1,..., n. Applying Theorem 1 with m = n, J = n and taking note of 13), by which γ 0 γ n > 0, we see that 1) holds. Remark 1. In [11, Corollary 4], inequality 1) was obtained under the assumption that fx) 1 for x 1. Some other interesting results appear in [7]. The result, stated above as Corollary 1, first appeared in [9], but the proof presented here is different. It is easily proved by induction that if R 1,..., R n are as in Corollary 1, then n Rν + 1 nν=1 Rν + 1. ν=1 This is clearly equivalent to the inequality 15) n ν=1 R ν + R ν nν=1 R ν + n ν=1 R ν, which may be combined with 1) to obtain the following result. Corollary. 16) Since Under the conditions of Corollary 1, we have n R ν + Rν ) a n n. ν=1 ζ ν + ζ ν z ν = R ν + Rν, we see that 1) is considerably stronger than 9). In addition, the hypothesis in Corollary 1 is quite a bit weaker. Instead of supposing fx) to be bounded by 1 on [, 1] we only require this restriction on fx) to be satisfied at the extrema of T n.

9 Coefficient estimates for polynomials 9 The following result is an immediate consequence of Corollary 1. Corollary 3. For any R > 1, let E R be the ellipse whose foci lie at ±1 and the sum of whose semi-axes is R. Furthermore, let fz) := a n nν=1 z z ν ) satisfy 11), and have at most n l zeros inside E R. Then 17) R l + R l ) a n n. Corollary 3 says in particular that if fz) 0 inside E R, then 18) R n + R n ) a n n. This estimate is sharp. In order to see this, let us consider the polynomial f R z) := R n + R n T n z) + i Rn R n ). Clearly, f R x) R n + R n 1 + R n R n ) = 1 x 1). Using ) we find that R n + R n f R ζ + ζ ) = ζn + ζ n + i Rn R n, which is easily seen to vanish for ζ = ζ ν = R e 4ν)πi/n ν = 1,..., n). Hence, f R z) = 0 at each of the n points z = z ν = 1 { R e 4ν)πi/n + R e 4ν)πi/n} ν = 1,..., n). Since f R is a polynomial of degree n, it has no other zeros. Setting A := R + R the points z ν can also be written as and B := R R, z ν = A cos 4ν 1)π n + i B sin 4ν 1)π n ν = 1,..., n).

10 10 M.A. Qazi and Q.I. Rahman Thus, f R has all its zeros on the ellipse E R. Since f R z) is of the form f R z) = n n R n + R n zn + a ν z ν, we conclude that the bound for a n, given by 18), cannot be improved even if 11) is replaced by the stronger condition: fx) 1 for x 1. ν=0 4. Relevance of the points { cos kπ n Corollary 1 says that in 5) we may replace max x 1 fx) by the maximum of fx) at the extrema of the extremal) polynomial T n. One might wonder about the reason behind it. The following result see [3, Theorems 4 and 5]) provides an explanation. Theorem D. Let t 0 < < t n be an arbitrary set of n + 1 real numbers. Setting P z) := n ν=0 z t ν ) let t 0,k < < t n k,k be the zeros of P k), the k th derivative of P. In addition, let y 0,..., y n be any set of n + 1 non-negative numbers, not all zero, and denote by π n the unique polynomial of degree at most n such that π n t ν ) = ) n ν y ν for ν = 0, 1,..., n. Then, for any real polynomial f of degree at most n, other than fz) ±π n z), such that ft ν ) y ν for ν = 0, 1,..., n, we have 19) and 0) fz) < π n z) if z satisfies z t 0 + t n } ) t n t 0, z {t 0, t n }, f k) z) < π k) n z) k = 1,..., n) z t 0,k + t n k, k t n k, k t 0,k, z {t 0,k, t n k, k }. Remark. The case k = n of 0) says that if π n z) := n ν=0 π n,ν z ν, then a n < π n,n unless fz) ±π n z). This is a good deal more than what Theorem A says. In the important case where t ν = cosνπ/n), inequality 19) goes back to S. Bernstein [1] and to Erdős [4]. One may consult [14, Chapter 1] for related results. The following application of 19) gives a meaningful refinement of 5) provided that n is odd.

11 Coefficient estimates for polynomials 11 Theorem. For any odd integer n, let t 0 = < < t n = 1 be a set of n + 1 real numbers such that t ν = t n ν for ν = 0, 1,..., n. Also, let y 0,..., y n be a set of n+1 non-negative numbers, not all zero, satisfying the condition y ν = y n ν for ν = 0, 1,..., n. Finally, let π n z) := n ν=0 π n,ν z ν be the unique polynomial of degree at most n for which π n t ν ) = ) n ν y ν for ν = 0, 1,..., n. Then for any real polynomial fz) := n ν=0 a ν z ν of degree at most n, other than fz) ±π n z), satisfying ft ν ) y ν for ν = 0, 1,..., n, we have 1) a n + a 0 < π n,n. Proof. Because of the restrictions on {t ν } and {y ν }, the polynomial π n must be odd, and so π n,0 = 0. Without loss of generality we may assume a n to be positive. Applying 19) with t 0 := and t n := 1 we see that fz) < π n z) if z 1, z ±1. Hence, the polynomial π n z) fz) has all its zeros in E := {z C : z < 1} {, 1}. We claim that the zeros cannot all lie at ±1. In fact, if they did then π n x) fx) would be either strictly positive or strictly negative on, 1). If it was strictly positive then in particular we would have y n ft n ) = π n t n ) ft n ) > 0, which is not possible since ft n ) y n by hypothesis. Similarly, if π n x) fx) was strictly negative on, 1) then for x = t n we would obtain ft n ) > y n, which is a contradiction since ft n ) is supposed to be bounded above by y n. Thus, the polynomial n π n z) fz) = π n,n a n )z n + π n,n ν a n ν ) z n ν a 0, which has has all its zeros in E, has at least one in {z C : z < 1}. Since E {z C : z 1} this implies that ν=1 a 0 < π n,n a n, and so 1) holds. 5. Bounds for the other coefficients Generalizing Theorem A, Wladimir Markoff [8]; see also [5]) found sharp bounds for all the Maclaurin coefficients of a polynomial f of degree at most n in terms of its maximum modulus on [, 1].

12 1 M.A. Qazi and Q.I. Rahman Theorem E. Let fx) := n ν=0 a ν x ν be a polynomial of degree at most n such that fx) 1 for x 1. Then, a n µ is bounded above by the modulus of the corresponding coefficient of T n for µ = 0,..., n/, and a n µ is bounded above by the modulus of the corresponding coefficient of T n for µ = 0,..., n 1)/. Erdős [4, p. 1176] remarked that if the coefficients of fz) := n ν=0 a ν z ν are all real, then n ν=0 a ν is maximal for fz) ±T n z). According to him, Szegő proved that a k + a k+1 is maximal for fz) ±T n z), which is stronger than the observation made by Erdős himself. Unfortunately, Szegő never published his proof. Erdős learned about the result orally see [4, p. 1176]) from Szegő. It would be nice to know Szegő s proof of his result. A few years ago see [3, Theorem 3]) we proved the following result, which covers that of Szegő. Theorem F. Let x 0 < < x n be n+1 real numbers, and let y 0,..., y n be a sequence of n+1 non-negative numbers, where we suppose that x ν = x n ν and y ν = y n ν for ν = 0,..., n, and that n ν=0 y ν > 0. In addition, let F x) := n/ µ=0 A n µ x n µ be the unique polynomial of degree n such that F x ν ) = ) n ν y ν for ν = 0,..., n. Furthermore, let fx) := n ν=0 a ν x ν be a real polynomial of degree at most n, whose modulus does not exceed that of F at the points x 0 < < x n, that is fx ν ) y ν = F x ν ) ν = 0,..., n). Then, ) a n k + a n k A n k ) n 1 k = 0,...,. Theorem F says, in particular, that if T n x) := n/ ν=0 t n,ν x ν is the Chebyshev polynomial of the first kind of degree n then, for any real polynomial fx) := n ν=0 a ν x ν of degree n satisfying 11), we have ) n 3) a n k + a n k t n,n k k = 0,...,. Not only T n but all the ultraspherical polynomials P n λ) have the properties the polynomial F of Theorem F is required to have for ) to hold. This observation suggests an inequality, more general than 3), involving the coefficients of P n λ) for any λ > / see Corollary 4). In order to present

13 Coefficient estimates for polynomials 13 such a generalization of 3) we need to recall some basic facts about the polynomials P n λ). This will come in handy in connection with certain extensions see Theorem 3 and Corollary 5) of the well-known L analogues of Chebyshev s inequality, which we also intend to discuss in this paper. 6. The polynomials P λ) n and an extension of 3) For λ 1 λ), 0) 0, ), the ultraspherical polynomials P n [17], see 4.7.1) on p. 81 and ) on p. 85) by are given 4) P λ) n x) = Γλ + 1 ) Γλ) = n/ k=0 Γn + λ) Γn + λ + 1 ) P λ n 1,λ 1 ) x) ) k Γn k + λ) Γλ) Γk + 1) Γn k + 1) x)n k. Thus P n λ) is a multiple of the Jacobi polynomial P λ 1,λ 1 ) n. The multiplicative factor Γλ + 1 ) Γn + λ) Γλ) Γn + λ + 1 ), which depends on λ and n, vanishes for λ = 0. It is known [17, p. 8], see 4.7.8)) that lim λ 0 λ P n λ) x) = n T nx) n = 1,,...), where T n is the Chebyshev polynomial of the first kind of degree n. Hence, we may define 5) 6) P 0) n x) := n T nx) n = 1,,...). It is known [17, p. 8], see )) that if λ 1, 0) 0, ), then 1 x ) λ 1 P λ) n x) P m λ) x) dx = In view of 5), the corresponding formula for λ = 0 is { 0 if m n, 1 λ π Γn+λ) {Γλ)} n+λ) Γn+1) if m = n.

14 14 M.A. Qazi and Q.I. Rahman 6 ) P 0) n x) P 0) m x) We find it convenient to define P n λ) Γλ) 7) x) := { dx 0 if m n, = 1 x π if m = n 1. n n+λ) Γn+1) 1 λ π Γn+λ) P n λ) x) if λ > 1, λ 0, n π P n 0) x) if λ = 0. Because of 6) and 6 ), the polynomials {P n λ) }, λ > / are orthonormal in the sense that 8) 1 x ) λ 1 P λ) n x) P m λ) x) dx = { 0 if m n, 1 if m = n. For any given λ > / and any n, let x n,1 λ) <... < x n,n λ) be the extrema of P n λ) in, 1). Then, Theorem F, applied with and x 0 :=, x ν := x n,ν λ) for ν = 1,..., n 1, x n := 1 y ν := gives us the following result. Corollary 4. 9) P n λ) x ν ) ν = 0,..., n), For any given λ > / and any integer n, let P λ) n x) := n ν=0 p λ) n,ν x ν be the ultraspherical polynomial of degree n, as defined in 7). Also let x 1 := x n,1 λ),..., x n := x n,n λ) be the extrema of P n λ) in, 1), arranged in increasing order. Furthermore, let fx) := n ν=0 a ν x ν be a real polynomial of degree at most n such that fx ν ) P n λ) 30) x ν ) ν = 0,..., n), where x 0 := and x n := 1. Then, ) a n k + a n k p λ) n 1 31) n,n k k = 0,...,.

15 Coefficient estimates for polynomials A refined L analogue of Chebyshev s inequality In Corollary 4 we require fx) to be dominated by P n λ) x) at the points x 0, x 1,..., x n. How large can a n k + a n k be if fx) is not necessarily bounded by P n λ) x) at any specific points of [, 1] but in some average sense, like wx) fx) dx wx) dx λ) wx) P n x) dx wx) dx for some appropriate weight function w? We are able to shed some light on this question in the case where wx) := 1 x ) λ 1, the weight whose relevance is clear from 8). As in 9), let p λ) n,n denote the leading coefficient in the Maclaurin ex- x). Then, from 7), 4), 5) and 4), it follows that Γn + λ + 1)Γn + λ) = n 1 λ λ > 1 ), πγn + 1)Γn + λ) pansion of P λ) n 3) p λ) n,n where the formula proves to be in order for λ = 0 because P 0) n := /n)t n. Theorem 3. For any given λ > / and any integer n, let p λ) n,n denote the coefficient of x n in the Maclaurin expansion of the ultraspherical polynomial P n λ) x), as defined in 7). Furthermore, let fx) := nν=0 a ν x ν be a polynomial of degree n with coefficients in C such that 33) Then, 34) 1 x ) λ 1 fx) dx a n + 4 n + λ)n + λ 1) nn + λ 1) 1 x ) λ 1 P n λ) x) dx = 1. { a n p λ) n,n }. The inequality is best possible in the sense that the coefficient of a n in 34) cannot be replaced by any number larger than 4 n + λ)n + λ 1) nn + λ 1) It is well-known that the sharp upper bound for each of the two terms a n and 4 n + λ)n + λ 1) nn + λ 1). a n,

16 16 M.A. Qazi and Q.I. Rahman { appearing on the left-hand side of 34) is p λ) }. n,n So apparently, inequality 34) is not to be sneezed at. Nevertheless, in view of 31), it would be interesting to know the largest admissible value of ε for the inequality a n + ε a n p λ) n,n to be true for any polynomial fx) := n ν=0 a ν x ν satisfying the hypothesis of Theorem 3. The following corollary of Theorem 3 gives the sharp upper bound for a n + ε a n in terms of ε > 0. The upper bound is larger than p λ) n,n for any ε > 0. Corollary 5. For any given λ > / and any integer n, let p λ) n,n denote the coefficient of x n in the Maclaurin expansion of the ultraspherical polynomial P n λ) x), as defined in 7). Furthermore, let fx) := nν=0 a ν x ν be a polynomial of degree n, with coefficients in C, satisfying 33). Then, 35) a n + ε a n 1 + ε 4 nn + λ 1) n + λ)n + λ 1) pλ) n,n ε > 0). The inequality is sharp. It cannot be improved even if the coefficients of f are all real. Proof of Theorem 3. such that For any given λ > / there exist constants b ν fx) = n ν=0 b ν P λ) ν x). Since P ν λ) is odd or even according as n is odd or even, respectively, we note that p λ) n,n = 0. Because of this fact, comparing the coefficients in the two expansions of fx) and taking 3) into account, we see that 36) a n = p λ) n,n b n = n Γn + λ + 1)Γn + λ) 1 λ πγn + 1)Γn + λ) b n and 37) a n = p λ) n,n b n = n Γn + λ)γn + λ 1) 1 λ πγn)γn + λ 1) b n.

17 Coefficient estimates for polynomials 17 The orthonormality of {P λ) 0, P λ) 1, P λ),...}, given in 8), implies that n 38) 1 x ) λ 1 fx) dx = b ν b n + b n, where, for n 3, the inequality becomes an equality if and only if b 0,..., b n are all zero, that is, if and only if fx) := b n P n λ) x) + b n P λ) n x). From 33) and 38) it follows that ν=0 39) b n + b n 1. 4 By 37), we have n + λ)n + λ 1) nn + λ 1) which, in conjunction with 36), shows that a n + 4 n + λ)n + λ 1) nn + λ 1) a n = n Γn + λ + 1)Γn + λ) 1 λ π Γn + 1)Γn + λ) b n, { a n = b n + b n ) This, in view of 39), gives us the desired inequality 34). p λ) n,n }. In order to see that 34) is sharp let us consider, for any t [0, 1], the polynomial 40) fx) := t e iα P λ) n x) + 1 t e iβ P λ) x) α R, β R). n Then 1 x ) λ 1 fx) dx = t 1 x ) λ 1 P λ) n x) dx + 1 t ) 1 x ) λ 1 P λ) n x) dx = 1 by 8). Considering the Maclaurin expansion fx) := n ν=0 a ν x ν of the polynomial defined in 40), we see that a n = t e iα p λ) n,n and a n = 1 t e iβ p λ) n,n.

18 18 M.A. Qazi and Q.I. Rahman Hence a n n + λ)n + λ 1) + 4 a n nn + λ 1) { } = t p λ) n,n + 1 t n + λ)n + λ 1) { ) 4 p λ) } n,n nn + λ 1) { } { } { } = t p λ) n,n + 1 t ) p λ) n,n = p λ) n,n. The above calculations also show that 34) would fail if its left-hand side was replaced by a n + c a n with any n + λ)n + λ 1) c = c n, λ) > 4. nn + λ 1) Proof of Corollary 5. From 34) it follows that a n p λ) and then a n 1 nn + λ 1) n + λ)n + λ 1)) Hence a n + ε a n ϕ a n ), where ϕu) := u + ε nn + λ 1) n + λ)n + λ 1)) n,n { p λ) } n,n an. { p λ) } n,n u for 0 u p λ) n,n. The function ϕ has only one critical point u = p nn+λ) n+λ)n+λ) ε λ) n,n in 0, p λ) n,n ) and there it has a local maximum. Simple calculations then lead us to the estimate for a n + ε a n that is given in 35). To see that 35) cannot be improved even if fx) is real for all real x, let δ := pλ) n,n p λ) n,n = n + λ + 1)n + λ) n n + λ 1) and, for any ε > 0, consider the real polynomial fx ; ε) := δ ε + δ P λ) ε n x) + ε + δ P λ) n x) = n ν=0 a ν x ν.

19 Coefficient estimates for polynomials 19 Then, and Clearly, and a n = a n + ε a n = a n = δ ε + δ pλ) n,n ε ε + δ pλ) n,n = ε δ ε + δ pλ) n,n. 1 x ) λ 1 fx ; ε) dx = δ δ + ε + ε δ + ε = 1 δ ε + δ pλ) n,n + ε δ ε + δ pλ) n,n = 1 + ε δ pλ) n,n. This shows that 35) becomes an equality for the polynomial fx ; ε). 8. The L p mean of f with Chebyshev weight Let Φz) := m µ=0 c µ z µ be a polynomial of degree m 1. In addition, let Az) := z m + 1. It was shown by one of us [10, Theorem ] that for any p [1, ), we have 41) c 0 + c m 1 π 1 π π A eiθ ) p ) 1/p π dθ π If fx) := n ν=0 a ν x ν is a polynomial of degree n then ) z + z Φz) := z n n f = c ν z ν is a polynomial of degree n, where c 0 = c n = 1 n a n, and 41) may be applied with m = n to obtain a n 1 π n π 0 cos nθ p dθ) 1/p 1 π π 0 ν=0 π Φ e iθ) ) p 1/p dθ. fcos θ) p dθ ) 1/p 4) = 1 π n π 0 T nx) p dx 1 x ) 1/p 1 π fx) p ) dx 1/p 1 x

20 0 M.A. Qazi and Q.I. Rahman for any p [1, ). Note that this result is a direct consequence of 41) and that the restriction imposed on p comes solely from the restriction on p in that inequality. Hence, 4) should hold for any p for which 41) may turn out to be true. In [13, Theorem 3], it was shown that 41) holds not only for p [1, ) but also for p 0, 1). Hence, so does 4). Thus, the following result holds. Other references for this result are [6, p. 183] and [15, p. 10]. Theorem G. For any polynomial fx) := n ν=0 a ν x ν of degree n and any p 0, ), we have 43) a n n π Γ p + 1) 1/p ) ) 1 Γ p+1 Γ 1 π 0 fx) p The inequality is sharp for all p 0, ). It may be noted that dx π = dθ = π 1 x and that 1 π fx) p Hence, 43) is a generalization of 5). dx 1 x ) 1/p. ) dx 1/p max fx) as x. 1 x x 1 In looking for a generalization of the case λ = 0) of Theorem 3 in the spirit of Theorem G we did not find anything of interest for p, ) or for p 0, 1]. We do have the following result for values of p 1, ]. Theorem 4. For any p 1, ] let p := p p 1. Then, for any polynomial fx) := n ν=0 a ν x ν of degree n with coefficients in C and any p 1, ], we have 44) a n p + a n p ) 1/p n 1/p 1 π fx) p dx 1 x ) 1/p. Proof of Theorem 4. As the first step, we write fx) = n ν=0 b ν T ν x). Then, clearly 45) a n = n b n and a n = n b n.

21 Coefficient estimates for polynomials 1 Let where 46) n gθ) := fcos θ) = b ν cos νθ = n c ν e iνθ, ν=0 ν= n c 0 = b 0 and c ν = c ν = 1 b ν ν = 1,..., n). By a well-known result due to Hausdorff Young see [, p. 101]), if g belongs to L p 0, π) for some p 1, ] and c k := 1 π π then with p := p/p 1), we have k= 0 gθ) e ikθ dθ k = 0, ±1, ±,...), 1/p c k p 1 π Hence, taking 46) into account, we obtain π 0 gθ) p dθ) 1/p. { 1 1/p b n p + b n p ) ) 1/p bn p ) bn p } 1/p = + n ) 1/p c ν p ν= n π 1/p 1 gθ) dθ) p = π 0 π 0 1 ) = fx) p dx 1/p. π 1 x In view of 45), this is equivalent to 44). π ) 1/p gθ) p dθ 9. Polynomials satisfying fx) e x dx 1 In this last section we wish to present the analogue of 34) for polynomials fx) := n ν=0 a ν x ν of degree n for which fx) e x dx 1. For this we need to recall certain facts about Hermite polynomials [19, p. 5]. There is a unique polynomial n ν=0 a ν x ν of degree n, with prescribed a n 0, that satisfies the differential equation y x y + n y = 0.

22 M.A. Qazi and Q.I. Rahman The one whose leading term is n x n is denoted by H n and is called the Hermite polynomial of degree n. Hermite polynomials of odd degree are odd and those of even degree are even. The first four Hermite polynomials are H 0 = 1, H 1 x) = x, H x) = 4x, H 3 x) = 8x 3 1x, and for n = 4, 5, 6,... the Maclaurin expansion of H n is H n x) = x) n nn 1) 1! x) n + nn 1)n )n 3)! The Hermite polynomials are orthogonal, with { 0 if m n, H n x) H m x) e x dx = n n! π if m = n. Hence, the polynomials x) n 4. 47) H nx) := 1 n n! π 1/4 H nx) n = 0, 1,,...), are orthonormal in the sense that 48) H nx) H mx) e x dx = { 0 if m n, 1 if m = n. Let Hmx) := m µ=0 h m,µ x µ be the Maclaurin expansion of Hmx), and note that h m,m = 1 m 49) π 1/4 m = 0, 1,,...). m! Now, we are ready to prove the following analogue of Theorem 3. Theorem 5. Let fx) := n ν=0 a ν x ν be a polynomial of degree n with coefficients in C such that 50) fx) e x dx 1. Then, 51) a n + n a n n n! π. The inequality is best possible in the sense that the coefficient of a n in 51) cannot be replaced by any number larger than /n.

23 Coefficient estimates for polynomials 3 Here it may be mentioned that the sharp upper bound for each of the two terms a n and n a n, appearing on the left-hand side of 51) is also n /n! π). Proof of Theorem 5. Let fx) = n m=0 β m H mx) be the Hermite Fourier expansion of f in terms of H 0,..., H n. Then 5) n n a ν x ν m β m h m,µx µ. ν=0 m=0 µ=0 Because of the fact that h n,n = 0, when we compare the coefficients of xn and of x n on the two sides of 5), and take 49) into account, we obtain 53) and 54) a n = h n,n β n = 1 n π 1/4 n! β n a n = h n,n β n = 1 n π 1/4 n 1)! β n. Formula 48), applied in conjunction with condition 50), which f satisfies, implies that 55) β n + β n 1. Using 53) and 54) in 55), we obtain a n + n a n = β n + β n ) n n! π n n! π, which proves 51). It is easily checked that 51) becomes an equality for any polynomial of the form 56) f t x) := t e iα H nx) + 1 t e iβ H nx) α R, β R), where t can be any number in [0, 1]. For any t in [0, 1), the polynomial f t appearing in 56) shows that the coefficient of a n in 51) cannot be replaced by any number larger than /n.

24 4 M.A. Qazi and Q.I. Rahman By Schwarz s inequality, a n + ε a n 1 + ε n a n + n a n, and so Theorem 5 readily implies the following result Corollary 6. Let fx) := n ν=0 a ν x ν be a polynomial of degree n, with coefficients in C, satisfying 50). Furthermore, let h n,n denote the coefficient of x n in the Maclaurin expansion of the polynomial Hnx) defined in 47). Then a n + ε a n 1 + ε n 1 57) n π 1/4 n!. The example δ ε fx) := ε + δ H nx) + ε + δ H nx), δ := n shows that inequality 57) is sharp, and that it cannot be improved even if the coefficients of f are all real. References [1] S. N. Bernstein, Sur une propriété des polynômes, Comm. Soc. Math. Kharkow Sér ), 1 6. [] P. Butzer, F. Jongemans, P. L. Chebyshev ), J. Approx. Theory ), [3] D. P. Dryanov, M. A. Qazi and Q. I. Rahman, Certain extremal problems for polynomials, Proc. Amer. Math. Soc ), [4] P. Erdős, Some remarks on polynomials, Bull. Amer. Math. Soc ), [5] O. D. Kellogg, On bounded polynomials in several variables, Math. Z ), [6] A. Kroó and E.B. Saff, On polynomials of minimal L q - deviation, 0 < q < 1, J. London Math. Soc. ) ), [7] Xin Li, A Note on Chebyshev s Inequality. In Approximation Theory N.K. Govil, R.N. Mohapatra, Z. Nashed, A. Sharma, J. Szabados, Eds.), pp , Marcel Dekker, Inc., New York, 1998.

25 Coefficient estimates for polynomials 5 [8] W. Markoff, Über Polynome, die in einem gegebenen Intervalle möglichst wenig von Null abweichen, Math. Ann ), [9] M. A. Qazi and Q. I. Rahman, On a polynomial inequality of P. L. Chebyshev, Archives of Inequalities and Applications 1 003), [10] Q. I. Rahman, Inequalities concerning polynomials and trigonometric polynomials, J. Math. Anal. Appl ), [11] Q. I. Rahman and G. Schmeisser, Location of the zeros of polynomials with a prescribed norm, Trans. Amer. Math. Soc ), [1] Q. I. Rahman and G. Schmeisser, Inequalities for polynomials on the unit interval, Trans. Amer. Math. Soc ), [13] Q. I. Rahman, G. Schmeisser, L p inequalities for polynomials, J. Approx. Theory ), 6 3. [14] Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, London Math. Society Monographs New Series) No. 6, Clarendon Press, Oxford, 00. [15] T. J. Rivlin, Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory nd edn., John Wiley & Sons, Inc., New York, [16] I. Schur, Über das Maximum des absoluten Betrages eines Polynoms in einem gegebenen Intervall, Math. Z ), [17] G. Szegő, Orthogonal polynomials, 3rd edn), Colloquium Publications, 3, American Mathematical Society, Providence, R.I., [18] E. C. Titchmarsh, The theory of functions, nd ed., Oxford Univ. Press, Oxford, [19] J. Todd, Introduction to the constructive theory of functions, Academic Press, Inc., New York, Birkhaüser Verlag Basel, [0] P. Turán, On an inequality of Ceby sev, Annales Universitatis Scientiarum Budapestinensis de Rolando Eőtvős Nominale, Sectio Mathematica ), [1] C. Visser, A simple proof of certain inequalities concerning polynomials, Indagationes Mathematicae ),

26 6 M.A. Qazi and Q.I. Rahman [] A. Zygmund, Trigonometric series, Vol. II, nd edn). Cambridge University Press, New York, 1959 Revised, 1968). Department of Mathematics Tuskegee University Tuskegee, Alabama U.S.A. Département de Mathématiques et de Statistique Université de Montréal Montréal Québec) H3C 3J7 Canada