 Solution

We construct the finite evaluation representation explicitly by Lagrange interpolation. This supplies all the polynomial facts needed for the bound.

Define the Chebyshev polynomial by $T_0=1,T_1=x$ andInduction shows that $T_m$ has degree $m$; the cosine addition identity also proves $T_m(\cos \theta )=\cos (m\theta )$. Therefore $\Vert T_n{\Vert }_{[-1,1]}=1$, and $T_n(-x)=(-1{)}^n{T}_n(x)$ follows from the recurrence. The case $n=0$ is immediate, since both $P$ and $T_0$ are constant.

For $n\ge 1$, choose the descending extremal nodesso $T_n(x_j)=(-1{)}^j$. Define the fundamental interpolation polynomialsThey satisfy ${\ell }_j(x_k)={\delta }_{jk}$. Consequentlyfor every polynomial of degree at most $n$: the difference has degree at most $n$ and vanishes at the $n+1$ distinct nodes, so it is zero. The root count used here follows by successively factoring $(t-x_j)$ from a polynomial having such a root; a nonzero degree-$n$ polynomial cannot have more than $n$ distinct roots. Thus the interpolation identity has been proved.

If $u>1$, every numerator factor in ${\ell }_j(u)$ is positive, while exactly $j$ denominator factors are negative. Hence $\mathrm{sgn}{\ell }_j(u)=(-1{)}^j$. Applying interpolation to $T_n$ itself givesPutting $M={\sup }_{[-1,1]}|P|$ in the interpolation formula yieldsFor $u<-1$, apply the proved positive-side inequality to $Q(t)=P(-t)$ at $-u>1$, and use the parity identity. The result isThis is sharp, with equality for $P=T_n$ or its scalar multiples.

In functional terms, the evaluation map $E_u:P\mapsto P(u)$ on the normed polynomial space has $\Vert E_u\Vert =|T_n(u)|$: the inequality gives the upper bound and $T_n$ attains it. For $u>1$, its normalized version has the exact representationThus Chebyshev interpolation represents external evaluation using $n+1$ nodes. The necessary finite representation and its norm are derived directly, so no convex-hull theorem is being invoked without proof.