I am trying to prove the following:
Let $p$ be a polynomial of degree n and let $I=[0,1]$ and $E\subset I$ a measurable set of non-zero measure, i.e., $\mu(E)\neq 0$. Then, $$\sup_{x\in I}|p(x)|\leq \Big(\frac{4}{\mu(E)}\Big)^n \sup_{x\in E}|p(x)|$$
This is a small part of the proof of Lemma 1 in Kovrijkine, "Some results related to the Logivenko-Sereda Theorem", Proc. Amer. Math. Soc, 2001. The article says it is a consequence of the Remez Inequality, but I cannot see the connection. I easily believe that the factor in the RHS comes from a rescaling but I am having difficulties working the details. Can you help me or give a good reference to look it up?
Thank you!
Michela
We will rescale our interval in order to apply the Remez inequality. We need to find $s$ such that after the map $f(x)= (2+s)x-1$ the set $E$ will have measure at least $2$.
We can write $2\le\mu (f(E)) = (2+s)\mu(E)$, hence we can take $s= \frac{2}{\mu(E)}-2$.
Rescale $p$ by its supremum on $E$: $$q(x)=\frac{p(x)}{\sup_{E}|p(x)|}.$$
With this value of $s$ we can write the estimate $$\sup_{[0,1]}|q(x)|=\sup_{f([0,1])}|q(f(x)|\le T_n(s+1) = T_n\left(\frac{2}{\mu(E)}-1\right).$$ Now we need to write an estimation on Chebyshev polynomials $T_n$:
$$T_n\left(\frac{2}{\mu(E)}-1\right) = \frac{\left(\frac{2}{\mu(E)}-1+\sqrt{\left(\frac{2}{\mu(E)}-1\right)^2-1}\right)^n+\left(\frac{2}{\mu(E)}-1-\sqrt{\left(\frac{2}{\mu(E)}-1\right)^2-1}\right)^n}{2} \\\le \left(\frac{2}{\mu(E)}-1+\sqrt{\left(\frac{2}{\mu(E)}-1\right)^2-1}\right)^n\\\le \left(\frac{2}{\mu(E)}-1+\sqrt{\left(\frac{2}{\mu(E)}-1\right)^2}\right)^n\le \left(\frac{4}{\mu(E)}-2\right)^n\\\le \left(\frac{4}{\mu(E)}\right)^n$$