Minimal deviation property of Chebyshev polynomials

Remarks

Proof idea: the polynomial $\tilde T_n = 2^{1-n} T_n$ is monic and equals $(-1)^k 2^{1-n}$ at the $n + 1$ points $\cos(k\pi/n)$, $k = 0, \dots, n$. If $\max \abs{p} < 2^{1-n}$, then $\tilde T_n - p$ has degree at most $n - 1$ and alternates in sign at these $n + 1$ points. So it has $n$ zeros, hence it is $0$, a contradiction. The equality case follows from a refinement of the same counting, or from the uniqueness of best uniform approximation.

Example: for $n = 2$, $x^2$ has maximum $1$ on $[-1, 1]$, while $\tilde T_2 = x^2 - 1/2$ has maximum $1/2$, the smallest possible value.

Equivalently, $x^n - \tilde T_n$ is the best uniform approximation of $x^n$ on $[-1, 1]$ by polynomials of degree at most $n - 1$, with error $2^{1-n}$. On an interval $[a, b]$, an affine change of variable gives the minimum $2 \left(\frac{b - a}{4}\right)^n$.

Going deeper

For a student

This is the first and model case of Chebyshev's theory of best uniform approximation. It explains why Chebyshev nodes are good for interpolation: the node polynomial $\prod (x - x_k)$ appears in the interpolation error, and the zeros of $T_n$ make its maximum as small as possible. It also gives the minimax error of $x^n$ and lower bounds for polynomial approximation of smooth functions.