 Solution

Map $[a,b]$ to $[-1,1]$ by $z(r)=(2r-a-b)/(b-a)$. Since $a>0$, the normalization point $z(0)=-(a+b)/(b-a)$ lies outside that interval. The proposed Chebyshev polynomial ratio has degree $k$ and equals one at zero. On the spectral interval its maximum absolute value isThe Chebyshev minimax polynomial with exterior normalization minimizes this maximum among degree-at-most-$k$ polynomials with $p(0)=1$. One may quote the minimax property. Its reason is the alternating $k+1$ extremal values: a strictly better competitor would force their difference to have at least $k$ roots in $[a,b]$ and the additional root at zero outside it, impossible for degree at most $k$.

The paper uses a monic normalization of $T_k$ for $k\ge 1$, rather than the usual $T_k(1)=1$ convention. The common factor cancels in the ratio. With $\kappa =b/a>1$ and $q=(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$, the hyperbolic-cosine formula gives the error factor $\boxed{2q^k/(1+q^{2k})\le 2q^k}$. This tends rapidly to zero when the spectral interval is well-conditioned. If $a=b$, use $p(r)=1-r/a$ for exact one-step termination instead of the displayed interval map.