Can the Chebyshev polynomials be constructed from the extremal property?

Question

It is a known fact that the Chebyshev polynomials, defined as $T_n := \cos(n \arccos x)$, have the following extremal property:

Theorem: Of all monic degree $n$ polynomials, $\frac1{2^{n-1}} T_n$ has the minimal $\infty$-norm on $[-1, 1]$, i.e. $\frac1{2^{n-1}} T_n$ minimises $\|p\|_\infty :=\max_{x \in[-1,1]} |p(x)|$.

I was wondering if it is possible to derive the Chebyshev polynomials from this property. I found the following question (from 14 years ago) but I'm struggling to understand the accepted answer and was wondering if someone could clarify.

The accepted answer looks at the equivalent problem of finding the polynomial $p \in P_n$ with $\|p\|_\infty = 1$ with the largest possible leading coefficient. It is not difficult to argue that this polynomial should oscillate between $\pm 1$ $n+1$ times and also take on the values $\pm 1$ at the endpoints, so there should exist $x_1, \dotsc, x_{n-1} \in (-1, 1)$ such that $|p(-1)| = |p(x_1)| = \dotsc = |p(x_{n-1})| = |p(1)| = 1$ and $p'(x_1) = \dotsc = p'(x_{n-1}) = 0$. Therefore, $1 - p^2$ should be divisible by $(1-x^2) p'(x)$. I get the answer up to this point, but then they claim that at this point the substitution $x = \cos t$, $p = \cos f$ is "very natural" and that the Chebyshev polynomials can be invented from there, and that is the part I don't get.

Could anyone either clarify this step or give a different explanation for how to derive the Chebyshev polynomials from the extremal property?

Answer 1

The argument in the answer you refer to actually shows that (using the notation from that answer) $t_2, \ldots, t_n$ are extrema of $p$ with value $\pm 1$ and thus double roots of $1-p^2$. Hence $(1-x^2)p'^2$ still divides $1-p^2$, and these polynomials have the same degree, so they are multiples of one another, and then we find the constant by comparing the leading terms. It follows that $n^2(1-p^2)=(1-x^2)p'^2$ or $$ \frac{p'}{\sqrt{1-p^2}} = \pm\frac{n}{\sqrt{1-x^2}} . $$ If we take the $+$ sign, this has the desired solution $p=\cos (n\arccos x)$ (and other solutions, which are not polynomials). More precisely, we solve on an interval $(t_j,t_{j+1})$ and continue holomorphically.

We can also try the $-$ sign. We then obtain solutions such as $p(\cos\theta)=\sin n\theta$, which have some typographical similarity to the $T_n$ but are not polynomials.