| Abstract
| - Summary. The following problem was raised by P. Turán. Letϕ(x) ⩾ 0 for −1 ⩽x ⩽ 1 and consider the classPn,ϕ of all polynomials of degree n such that |pn(x)| ⩽ϕ(x) for −1 ⩽x ⩽ 1. How large can max-1 ⩽x ⩽ 1 |pn(k)(x)| be ifpn is an arbitrary polynomial belonging topn,ϕ? He pointed out two cases of special interest. These cases beingϕ(x) = (1 − x2)1/2,ϕ(x) = (1 − x2). The above problem has been extensively studied by Q. I. Rahman and his associates. The object of this paper is to consider the solution of the Turán problem in the weightedL2 norm. Main results of this paper may be summarized as follows. Theorem 1.Let pn+2 be any member of the set of those algebraic polynomials of degree n + 2 which have only real roots, all of them in the interval [−1, 1], and for which$$\left| {p_{n + 2} (x)} \right| \leqslant 1 - x^2 , - 1 \leqslant x \leqslant 1.$$ Then we have$$\begin{gathered} \int_{ - 1}^1 {(p'_{n + 2} (x))^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx} \leqslant \int_{ - 1}^1 {(f'_0 (x))^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx = \frac{\pi }{4}(n^2 + 4)} , \hfill \ \int_{ - 1}^1 {(p''_{n + 2} (x))^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx} \leqslant \int_{ - 1}^1 {(f''_0 (x))^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx} , \hfill \ end{gathered} $$ with equality iffpn + 2(x) = (1 − x2)Tn(x) = f0(x), Tn(x) = cosnθ, cosθ = x. Theorem 2. Let pn+2 be any real polynomial of degree n + 2 such that$$\left| {p_{n + 2} (x)} \right| \leqslant 1 - x^2 , - 1 \leqslant x \leqslant 1.$$ Then we have$$\int_{ - 1}^1 {\left| {p'''_{n + 2} (x)} \right|^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx} \leqslant \int_{ - 1}^1 {\left| {f'''_0 (x)} \right|^2 (1 - x^2 )^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-ulldelimiterspace} 2}} dx} ,$$ with equality iffpn+2(x) = f0(x), f0(x) = (1 −x2)Tn(x).
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