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

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 ?

• 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 9:00

## 1 Answer

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 $p(x)=2$.

• A hint would have been a better answer to the question asked. – Paul Sep 29 '17 at 9:01
• @Paul Thanks for your comment. This time I could not resist ;-) – Robert Z Sep 29 '17 at 9:12
• how to find $\frac {1}{1-x^2}$ ? this is the first chapter of my undergraduate course, i'm sorry – user326307 Sep 29 '17 at 9:34
• @user326307 Note that $\sum_{i=0}^{n} x^{2i}\leq \sum_{i=0}^{\infty} x^{2i}=\frac {1}{1-x^2}$. See en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence – Robert Z Sep 29 '17 at 9:36