 Solution

Chebyshev's equal-ripple criterion says that a polynomial $q$ of degree at most $m$ is a best uniform approximation to a continuous function $F$ if and only if the error $F-q$ attains alternating extrema of maximum magnitude at $m+2$ ordered points.

Since $T_n$ has leading coefficient $2^{n-1}$, the polynomial $2^{1-n}{T}_n$ is monic and alternates between $\pm 2^{1-n}$ at $t_j=\cos (j\pi /n)$. Therefore

At each $t_j$, both $T_n-f$ and $T_n+f$ have sign $(-1{)}^j$, because $|f(t_j)|<1$. Each has a root in every interval between consecutive extrema, hence all its at most $n$ roots lie in $(-1,1)$. For $t>1$ both polynomials remain positive, giving $|f(t)|<T_n(t)$. For $t<-1$ both have sign $(-1{)}^n$, giving $|f(t)|<|T_n(t)|$. On $[-1,1]$ the hypothesis gives $|f|<1$. Thus