= 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 $\pm2^{1-n}$ at $t_j=\cos(j\pi/n)$. Therefore
$$
\boxed{q_*(t)=t^n-2^{1-n}T_n(t),\qquad
\min_{\deg q<n}\lVert t^n-q\rVert_\infty=2^{1-n}.}
$$
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
$$
\boxed{|f(t)|<\max\{1,|T_n(t)|\}\quad(t\in\mathbb R).}
$$
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