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Let the polynomial $p(x)= a_0 + a_1 x + . . . + a_n x^n$ have coefficients satisfying the relation $$\sum_{i=1}^{n} a_i^{2} = 1$$

Prove that $\int_{0}^{1} |p(x)| dx \leq \frac{\pi}{2} $.

I don't have any idea to prove this inequality, is there any reference to study about integrating polynomial ?

user326307
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    I think that the condition should be $\sum_{i=0}^{n} a_i^{2} = 1$, otherwise the inequality does not hold. – Robert Z Sep 29 '17 at 09:00

1 Answers1

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For $x\in [0,1)$, by Cauchy-Schwarz inequality, $$|p(x)|^2\leq \sum_{i=0}^{n} a_i^{2}\cdot \sum_{i=0}^{n} x^{2i}\leq \frac{1}{1-x^2}.$$ Hence $$\int_{0}^{1} |p(x)| dx \leq \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} dx=\frac{\pi}{2}.$$

P.S. I am assuming that $\sum_{i=0}^{n} a_i^{2} \leq 1$. If we have the weaker condition $\sum_{i=1}^{n} a_i^{2} = 1$ the inequality does not hold. Take for example $p(x)=2$.

Robert Z
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