Let $T_n\in P_n[-1,1]$ the n-th Chebyshev polynomial. Show that the lowest-norm monic polynomial in $P_n[a,b]$ is of the form $$\frac{(b-a)^n}{2^n}\frac{1}{2^n}T_n\left(\frac{2}{b-a}x-\frac{b+a}{b-a}\right)$$
I know that for any given $n\geq 1$, among the polynomials of degree $n$ with leading coefficient $1$ is $$f(x)=\frac{1}{2^{n-1}}T_n(x)$$ is the one of which the maximal absolute value on the interval $[−1, 1]$ is minimal. I think I must transform this interval into the $[a,b]$ interval but I'm not sure if it's the right way.
Any suggestions would be great!