The Chebyschev polynomials are denoted by $T_n(x) = \cos (n\arccos x)$
and are orthogonal in relation to $\langle f,g\rangle =
\int_{-1}^1\frac{f(x)g(x)}{\sqrt{1-x^2}}$Show that $T_n(x) = \frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{2}$
and that $Q_n = \frac{T_n(x)}{2^{n-1}}$ is the monic polynomial of
smallest norm uniform in $[-1,1]$
I've found that $T_n(x) = \frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{2}$ here https://math.stackexchange.com/a/2800290/166180
Now I'm trying to prove it's the polynomial of smallest norm. I think the norm should be derived from $\sqrt{\langle f, f\rangle^2}$, so I should analyze
$$||Q_n||=\sqrt{\int_{-1}^1\frac{\left(\frac{T_n(x)}{2^{n-1}}\right)^2}{\sqrt{1-x^2}}dx}$$
but it looks too ugly and I don't know where to begin with.
Best Answer
HINT
Let us consider Chebyshev Polynomial of the First Kind.
Firstly, recurrence relation $$T_{n+1}(x) = 2xT_n(x)-T_{n-1}(x),\tag1$$ where $$T_0(x)=1,\quad T_1(x) = x,\tag2$$ provides that the polynomials $Q_n(x)$ are monic, if $n>0$.
So should be $$Q_0=1.$$
Secondly, one can prove the orthogonality in the form of $$\int\limits_{-1}^1\dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}}\,dx = \int\limits_{-1}^1 \cos(m\arccos x)\cos(n\arccos x)\,d(arccos x)$$ $$=\int\limits_{-\large\frac\pi2}^{\large\frac\pi2}\cos mt\cos nt \,dt =\dfrac12\int\limits_{-\large\frac\pi2}^{\large\frac\pi2}\left(\cos(m+n)t+\cos (m-n)t\right) \,dt$$ $$=\begin{cases} \pi,\quad\text{if}\quad m=n=0\\ \frac12,\quad\text{if}\quad m=n\not =0\\ 0,\quad\text{if}\quad m\not=n.\\ \end{cases}$$ Thus, polynomials $\{Q_n(x)\}$ also are orthogonal.
On the other hand, the arbitrary $n$-th order monic $P_n(x)$ allows decomposition in the form of $$P_n(x) = \sum\limits_{k=0}^{n-1} c_k Q_k(n),\quad\text{where}\quad c_n=1.$$ Therefore, $$\|P_n(x)\|^2 = \sum\limits_{k=0}^{n} c_k^2\int\limits_{-1}^1\dfrac{Q_k^2(x)}{\sqrt{1-x^2}}\,dx+\sum\limits_{1\le i<j\le n}2^{i+j-1}c_ic_j \int\limits_{-1}^1\dfrac{T_i(x)T_j(x)}{\sqrt{1-x^2}}\,dx$$ $$ = \|Q_n(x)\|^2 + \sum\limits_{k=0}^{n-1} c_k^2\|Q_k(x)\|^2 > \|Q_n(x)\|^2, \quad\text{if}\quad P_n(x)\not = Q_n(x).$$