Solve for $|p(x)| \leq 1$.

Question

Let $p(x)=ax$$^2+bx+c$ such that $|p(x)|\leq 1$ for $|x|\leq 1.$

Show that $|cx^2+bx+a|\leq 2$ and $|2ax+b|\leq 4$ for $|x|\leq 1$.

Answer 1

We have that $$p(0)=c,\; p(1)=a+b+c,\; p(-1)=a-b+c.$$ Therefore $$c=p(0),\; b=\frac{p(1)-p(-1)}{2},\;a=\frac{p(1)+p(-1)}{2}-p(0).$$ Now use these relations in $cx^2+bx+a$ and $2ax+b$ and consider that $|p(0)|\leq 1$, $|p(1)|\leq 1$, $|p(-1)|\leq 1$.

For example, $$|2ax+b|=\left|(p(1)+p(-1)-2p(0))x+\frac{p(1)-p(-1)}{2}\right|\\ \leq |p(1)|\left|x+\frac{1}{2}\right|+|p(-1)|\left|x-\frac{1}{2}\right|+2|p(0)||x|\\ \leq \left|x+\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|+2|x|.$$

Can you take it from here?