Show that $B_n(f)^2 \le B_n(f^2)$ where $B_n(f)$ is Bernstein's polynomials. HINT : Expand $B_n((f-a)^2)$
I really do not know how to solve it only a hint.
according to HINT $$B_n((f-a)^2)=B_n(f^2-2af)+a^2$$
and then how to derive given inequality? please give me a hint!
Since the basis polynomials $P^n_k(x)=\binom{n}{k}x^k(1-x)^{n-k}$ are $\ge0$ on $[0,1]$, $f\ge0$ implies $B_n(f)\ge0$ on $[0,1]$. So $B_n((f-a)^2)=B_n(f^2)-2aB_n(f)+a^2\ge0$ on $[0,1]$ for all $a\in\mathbb{R}$. Now put $a=B_n(f)$.