= Solution
Take $n\geq1$, as intended by the stated normalization. The <Chebyshev polynomial> recurrence $T_{r+1}=2xT_r-T_{r-1}$, with $T_0=1$ and $T_1=x$, shows by <mathematical induction> that $T_n$ has <polynomial degree> $n$ and leading coefficient $2^{n-1}$. Consequently $f_n=2^{1-n}T_n$ is a <monic polynomial>. At the ordered <Chebyshev extreme points> $t_j=-\cos(j\pi/n)$,
$$
f_n(t_j)=2^{1-n}(-1)^{n+j},\qquad0\leq j\leq n,
$$
and $\|f_n\|_\infty=2^{1-n}$.
Suppose a <monic polynomial> $q$ of the same <polynomial degree> had $\|q\|_\infty<2^{1-n}$. The difference $f_n-q$ has degree at most $n-1$ because its leading coefficients cancel. Its signs at the $n+1$ ordered extreme points strictly alternate. The <intermediate value theorem> supplies a distinct <root of a polynomial> between each consecutive pair, giving at least $n$ roots. This contradicts its degree bound. Hence the <monic Chebyshev extremal polynomial> satisfies
$$
\boxed{\|f_n\|_\infty=2^{1-n}\leq\|q\|_\infty.}
$$
At $n=0$ the printed formula instead gives $f_0=2$, which is not monic; the degree-zero minimizer is the constant one. This endpoint case requires the separate normalization $f_0=1$.
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