Showing $Q_n = \frac{T_n(x)}{2^{n-1}}$ is the monic polynomial of least norm

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.

• I don't understand; you give a whole sequence of polynomials (one for each $n$). How then can 'it' be the monic polynomial of smallest norm? What is 'it' here? – Servaes Apr 16 at 22:28
• Can you prove $Q_n$ is a monic polynomial? Also, are you sure which norm is being used? – Somos Apr 16 at 23:23
• @Somos I'm sure about the inner product, not the norm, it's just a guess. About the first question, I don't know how, because for $n=2$ the thing looks at least squared to me – Guerlando OCs Apr 17 at 0:02
• The wikipedia article on Chebyshev polynomials states "The Chebyshev polynomials Tn are polynomials with the largest possible leading coefficient, but subject to the condition that their absolute value on the interval [−1,1] is bounded by 1. They are also the extremal polynomials for many other properties.[2" see also the "Minimal $\infty$-norm" section. – Somos Apr 17 at 1:59
• @GuerlandoOCs: The word "uniform" means that the norm to be minimized is $\| f \| = \sup _{x \in [-1,1]} |f(x)|$. A proof sketch has already been suggested by Somos, no need for bounty here. – Alex M. Apr 19 at 7:12

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, $$\|P_n(x)\|^2 = \sum\limits_{k=0}^{n} c_k^2\int\limits_{-1}^1\dfrac{Q_k^2(x)}{\sqrt{1-x^2}}\,dx+\sum\limits_{1\le i $$= \|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).$$

• First, on the very last line it should be $Q_n$ instead of $Q_k$. Second, you are showing the minimality of the norm of $Q_n$ in the $L^2$-norm, when in reality you were supposed to show it in the uniform convergence norm (check the question once more, carefully, and then my comment under it). The OP, not understanding a thing, has rushed to accept the answer... – Alex M. Apr 20 at 19:17
• @AlexM.Thank you for a serious attention, you are right. As I feel the situation, OP did not need in the detalized sollution, only in the plan. At the same time, your comment is actual even in this case. – Yuri Negometyanov Apr 20 at 19:37
• Why can you pass the $||.||$ into the sum? – Guerlando OCs Apr 21 at 21:36
• @GuerlandoOCs Thank you for the comment. Detalized. – Yuri Negometyanov Apr 22 at 5:34
• Can you explain what you mean by $d( \arccos x)$? – Guerlando OCs Apr 24 at 16:43