 Solution

Take $n\ge 1$, as intended by the stated normalization. The Chebyshev polynomial recurrence $T_{r+1}=2x{T}_r-T_{r-1}$, 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 $t_j=-\cos (j\pi /n)$,and $\Vert f_n{\Vert }_{\infty }=2^{1-n}$.

Suppose a monic polynomial $q$ of the same polynomial degree had $\Vert q{\Vert }_{\infty }<2^{1-n}$. 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 satisfiesAt $n=0$ 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 $f_0=1$.