Let $n \in \mathbb{N}^*$, $W$ be a monic polynomial of degree $n$, $Q = \frac { 1 } { 2 ^ { n - 1 } } T _ { n } - W$, and for all $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. In this question, we prove $$\sup _ { x \in [ - 1,1 ] } | W ( x ) | \geqslant \frac { 1 } { 2 ^ { n - 1 } }$$ by contradiction.
If we assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | < \frac { 1 } { 2 ^ { n - 1 } }$, show that, for all $k \in \llbracket 0 , n - 1 \rrbracket , Q \left( z _ { k } \right) Q \left( z _ { k + 1 } \right) < 0$.
Deduce a contradiction and conclude.
Let $n \in \mathbb{N}^*$, $W$ be a monic polynomial of degree $n$, $Q = \frac { 1 } { 2 ^ { n - 1 } } T _ { n } - W$, and for all $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. In this question, we prove
$$\sup _ { x \in [ - 1,1 ] } | W ( x ) | \geqslant \frac { 1 } { 2 ^ { n - 1 } }$$
by contradiction.
\begin{itemize}
\item If we assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | < \frac { 1 } { 2 ^ { n - 1 } }$, show that, for all $k \in \llbracket 0 , n - 1 \rrbracket , Q \left( z _ { k } \right) Q \left( z _ { k + 1 } \right) < 0$.
\item Deduce a contradiction and conclude.
\end{itemize}