= Solution
{c}
For $n\geq1$, set $p=2^{1-n}T_n$. It is monic by part (a), and its supremum <norm> on $[-1,1]$ is $E=2^{1-n}$. At the $n+1$ successive extreme points $x_j=\cos(j\pi/n)$ it alternates between $E$ and $-E$.
Suppose a monic degree-$n$ <polynomial> $q$ had $\|q\|_\infty<E$. At each $x_j$, the difference $p-q$ would have the sign of $p$. These signs alternate, so the <intermediate value theorem> forces a root in each of the $n$ successive gaps. But the leading monic terms cancel, making $\deg(p-q)\leq n-1$, a contradiction. Therefore
$$
\boxed{\|2^{1-n}T_n\|_\infty\leq\|q\|_\infty.}
$$
This is the monic <monic Chebyshev extremal polynomial>. The positive-degree restriction matters: at $n=0$ the printed scaling gives the constant two, while the only monic constant is one, so that version is false. The customary theorem uses $n\geq1$.
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