This is the exercise 13 of Analysis I of Amann and Escher, page 349.
Define the normalized Tchebyshev polynomials by $\tilde T_n:=T_n/2^{n-1}$ and $\tilde T_0:=T_0$. Let $\mathcal P_n\subset\Bbb R[X]$ for $n\in\Bbb N_{\ge 0}$ such that if $p\in\mathcal P_n$ then $\deg(p)=n$ and $[x^n]p(x)=1$. Prove that the $\tilde T_n$ is the best approximation of zero in $\mathcal P_n$ in $[-1,1]$, that is $$\|\tilde T_n\|_\infty\le \|p\|_\infty,\quad p\in\mathcal P_n$$
This exercise is killing me. Previous known facts:
$T_n(x)=\cos(n\arccos x)=\sum_{k\ge 0}\binom{n}{2k}x^{n-2k}(x^2-1)^k$.
$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$.
The local extrema of $T_n$ are attained at $y_k:=\cos\frac{k\pi}{n}$ for $k=0,1,\ldots,n$, i.e. $T_n(y_k)=(-1)^k$.
The zeros of $T_n$ are $x_k:=\cos\frac{(2k-1)\pi}{2n}$ for $k=1,2,\ldots,n$.
If we define $\tilde T_n(x):=\sum_{k=0}^n a_kx^k$ then $a_n=1$.
From above we have the identity $\tilde T_n(x)=\prod_{k=1}^{n}(x-x_k)$.
From all of this we know that $\|\tilde T_n\|_\infty=1/2^{n-1}$. But I dont know exactly where start or what to do to prove the statement of the exercise.
I tried to start proving that for each $p\in\mathcal P_n$ with some zero it must exists some $x_0\in[-1,1]$ such that $|p(x_0)|=1/2^{n-1}$ (but now Im not sure that this deduction was correct or useful).
I tried induction over $n$ for polynomials in $p_n\in\mathcal P_n$, writing something like $p_{n+1}=xp_n+c$, but I dont get something useful.
The exercise comes after some theory about Taylor theorem and some polynomial approximation based in Newton/Lagrange polynomials (there is no theory in the book, by now, related to integration). But trying to relate the exercise to the theory was fruitless.
Some hint (or solution) will be appreciated.
UPDATE: I think I get the way to do, but the proof below is not (by now) totally correct.
We knows that $\|\tilde T_n\|_\infty=1/2^{n-1}$, so the statement to prove must be re-written as
$$\frac1{2^{n-1}}\le \|p\|_\infty,\quad \forall p\in\mathcal P_n$$
Now define $p_n:=p$ for any $p\in\mathcal P_n$, i.e. the degree of $p_n$ is $n$. I define too $p_n(0):=c_n$. Then if $|c_n|\ge\frac1{2^{n-1}}$ we are done.
Then suppose that $|c_n|<\frac1{2^{n-1}}$. Now we prove by induction:
Base case: $\|p_1\|_\infty=\|x+c_1\|\ge 1$ for $x\in[-1,1]$.
Induction hypothesis: assume that $\|p_n\|_\infty\ge 1/2^{n-1}$.
Induction step: first observe that $p_{n+1}=xp_n+c_{n+1}$ for some $p_n$, then we want to prove that $$\|p_{n+1}\|_\infty=\|xp_n+c_{n+1}\|_\infty\ge1/2^n$$
Then
$$\|xp_n\|+|c_{n+1}|\ge\|xp_n+c_{n+1}\|_\infty\ge\big|\|xp_n\|_\infty-|c_{n+1}|\big|\ge\ldots \text{(working around)}$$