Chebyshev polynomials increase more quickly than any other polynomial outside $[-1,1]$

Recently, I need this result for another question. Since I cannot find a easy reference of this, I have proved it myself. I will extract the part relevant to this question and presented here.

Let $f(x)$ be any polynomial with degree at most $n$ and $|f(x)| \le 1$ over $[-1,1]$.

Let $x_k = \cos\frac{k\pi}{n}$ for $0 \le k \le n$, we have $1 = x_0 > x_1 > \ldots > x_n = -1$.

For any $\epsilon > 0$, consider the polynomial

$$g(x) = (1+\epsilon)T_{n}(x) - f(x)$$

Since $T_n(x_k) = (-1)^k$, we find: $\displaystyle\quad g(x_k) \begin{cases} > 0, & k \text{ even }\\ < 0, & k \text{ odd }\end{cases}$

This implies $g(x)$ has at least $n$ roots $y_1,\ldots,y_n$ with the $k^{th}$ root $y_k$ falls inside the $k^{th}$ interval $(x_k,x_{k-1})$. Since $g(x)$ is a non-zero polynomial with degree at most $n$. These are all the roots of $g(x)$. We can factorize $g(x)$ as $A\prod_{k=1}^n (x - y_k)$ for some constant $A$.

Notice $g(x_0) > 0$ and $x_0 > y_1,\ldots, y_k$. we get $A > 0$. So for all $t \ge x_0 = 1$, we have

$$(1+\epsilon)T_n(t) - f(t) = A\prod_{k=1}^n(t - y_k) > 0 \implies (1 + \epsilon)T_n(t) > f(t)$$ Apply a similar argument to $-f(x)$, we get $(1 + \epsilon)T_n(t) > -f(t)$

Combine these, we find $(1 + \epsilon)T_n(t) > |f(t)|$.

Since this is true for all $\epsilon > 0$, we can deduce $T_n(t) \ge |f(t)|$.

The case for $t \le -1$ is similar. At the end, we have following bound over $\mathbb{R}$.

$$|f(x)| \le \max( 1, |T_n(x)| )$$