# Showing inequality in integrating polynomials

Let the polynomial $|P(x)| = a_0 + a_1x + \dots + a_nx^n$ have coefficients satisfying the relation $$\sum_{i=0}^{n} a_i^2 = 1.$$

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

Show that this inequality continues to hold if $\pi/2$ is replaced by $\pi/\sqrt{6}.$

I was able to solve the first part using $P(x) \leq \sqrt{1 + x^2 \dots + x^{2n}}$ by the Cauchy inequality and the showing that the original integral is less than or equal to $$\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} = \frac{\pi}{2}.$$

I'm not sure how to do the second part. The hint is to use the Schwarz inequality for integrals.

• Perhaps you should try to prove that $\left(\int_0^1P(x)dx\right)^2\le\sum_{n=1}^\infty\frac1{n^2}$ – ajotatxe Jul 27 '15 at 12:49
• I think I thought about this. Wouldn't I still need the modulus signs inside integral since it is larger than the modulus of the integral? – user110503 Jul 27 '15 at 13:13

$$\int_{0}^{1}|P(x)|\,dx \leq \sum_{i=0}^{n}\int_{0}^{1}|a_i|x^i\,dx = \sum_{i=0}^{n}\frac{|a_i|}{i+1}\color{red}{\leq}\sqrt{\sum_{k=1}^{n+1}\frac{1}{k^2}}\leq\sqrt{\zeta(2)}=\frac{\pi}{\sqrt{6}}$$ where $\color{red}{\leq}$ follows from the Cauchy-Schwarz inequality.