[proofplan] We first verify that $\tilde{T}_{n+1}$ is monic of degree $n+1$ with $\|\tilde{T}_{n+1}\|_\infty = 2^{-n}$ and that it attains $\pm 2^{-n}$ at $n+2$ alternating Chebyshev extrema. Then we prove minimality by contradiction: if a monic competitor $q$ had smaller sup-norm, the difference $\tilde{T}_{n+1} - q$ would be a polynomial of degree $\le n$ alternating in sign at $n+2$ points, hence having $n+1$ zeros, forcing it to be identically zero. [/proofplan] [step:Verify that $\tilde{T}_{n+1}$ is monic with sup-norm $2^{-n}$ and alternating extrema] Write $\tilde{T}_{n+1}(x) = 2^{-n}T_{n+1}(x)$. Since $T_{n+1}(x) = \cos((n+1)\arccos x)$, the polynomial $T_{n+1}$ has leading coefficient $2^n$ (verified from the recurrence $T_{k+1}(x) = 2x\,T_k(x) - T_{k-1}(x)$ with $T_0 = 1$, $T_1 = x$), so $\tilde{T}_{n+1}$ is monic of degree $n+1$. From $|T_{n+1}(x)| = |\cos((n+1)\arccos x)| \le 1$, we have $\|\tilde{T}_{n+1}\|_\infty = 2^{-n}$. Moreover, $\tilde{T}_{n+1}$ attains the values $\pm 2^{-n}$ at the $n+2$ Chebyshev extrema $x_k^* = \cos(k\pi/(n+1))$, $k = 0, \dots, n+1$, alternating in sign. [/step] [step:Prove minimality by contradiction using the intermediate value theorem] Suppose for contradiction that $q$ is a monic polynomial of degree $n+1$ with $\|q\|_\infty < 2^{-n}$. Then $r := \tilde{T}_{n+1} - q$ is a polynomial of degree $\le n$ (the leading monic terms cancel). At the $n+2$ Chebyshev extrema $x_0^*, \dots, x_{n+1}^*$: \begin{align*} r(x_k^*) &= \tilde{T}_{n+1}(x_k^*) - q(x_k^*). \end{align*} Since $|\tilde{T}_{n+1}(x_k^*)| = 2^{-n}$ and $|q(x_k^*)| < 2^{-n}$, the sign of $r(x_k^*)$ agrees with the sign of $\tilde{T}_{n+1}(x_k^*)$. Since $\tilde{T}_{n+1}$ alternates in sign at the $n+2$ extrema, $r$ also alternates in sign at these $n+2$ points. By the [intermediate value theorem](/theorems/629), $r$ has at least $n+1$ real zeros (one between each consecutive pair of extrema). But $\deg r \le n$, so $r = 0$, i.e., $q = \tilde{T}_{n+1}$, contradicting $\|q\|_\infty < 2^{-n}$. The zeros of $T_{n+1}$ are $x_k = \cos\!\left(\frac{2k+1}{2(n+1)}\pi\right)$ for $k = 0, \dots, n$, so the monic nodal polynomial $\omega(x) = \prod_{i=0}^n(x - x_i) = \tilde{T}_{n+1}(x)$ achieves the minimum $\|\omega\|_\infty = 2^{-n}$ precisely at these nodes. [/step]