Chebyshev polynomial domination lemma

If $f$ has degree at most $n$ and $|f|<1$ on $[-1,1]$, alternation at the extrema of $T_n$ implies $|f(t)|<|T_n(t)|$ outside that interval.

For $0<a<b$, the displayed polynomial minimizes ${\max }_{t\in [a,b]}|p(t)|$ among real polynomials of degree at most $k$ with $p(0)=1$. Its $k+1$ alternating extremal values prove the assertion: a strictly smaller competitor would make the difference change sign in each consecutive extremal interval and also vanish at zero outside $[a,b]$, giving too many roots. The normalizing factor makes either the monic or the usual Chebyshev polynomial convention give the same ratio. For $\kappa =b/a$ and $q=(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$, its maximum is $2q^k/(1+q^{2k})$. This yields an error bound for polynomial acceleration of Richardson iteration.

On real polynomials of degree at most $n\ge 1$ with the uniform norm on $[-1,1]$, set $x_j=\cos (j\pi /n)$ and let ${\ell }_j$ be the Lagrange interpolation basis. If $u>1$, the denominator of ${\ell }_j(u)$ has exactly $j$ negative factors while its numerator is positive. Therefore $|{\ell }_j(u)|=(-1{)}^j{\ell }_j(u)$. Since $T_n(x_j)=(-1{)}^j$, interpolation gives ${\sum }_j|{\ell }_j(u)|=T_n(u)$. Hence $|P(u)|\le \Vert P{\Vert }_{\infty }{T}_n(u)$, and $P=T_n$ attains equality. Reflection and Chebyshev parity give the case $u<-1$. Dividing the interpolation weights by $T_n(u)$ yields an exact signed evaluation representation of norm one on $n+1$ nodes. Constants give the immediate degree-zero case.