= Solution
Let $k_0=\lceil(n+1)/2\rceil$, $\theta=\pi/(n+1)$, and define the high-frequency endpoint parameters
$$
a=1-\cos(k_0\theta),\qquad b=1+\cos\theta.
$$
They obey $1\leq a\leq b<2$ (with equality of endpoints for a single high mode). Since $|1-\omega t|$ is convex in $t$ and the endpoint modes are included, the exact high-frequency attenuation is
$$
\mu_\omega=\max\{|1-\omega a|,|1-\omega b|\}.
$$
The minimax choice balances the endpoint values with opposite signs. If they were not balanced, moving $\omega$ toward their balancing point would lower the larger endpoint value. Thus
$$
\boxed{\omega_*^{(n)}=\frac2{a+b},\qquad
\mu_*^{(n)}=\frac{b-a}{a+b}.}
$$
For odd $n$, $a=1$, giving $\omega_*=2/(2+\cos\theta)$ and $\mu_*=\cos\theta/(2+\cos\theta)$. For even $n$, $a=1+\sin(\theta/2)$, giving
$$
\omega_*^{(n)}=\frac2{2+\cos\theta+\sin(\theta/2)},\qquad
\mu_*^{(n)}=\frac{\cos\theta-\sin(\theta/2)}{2+\cos\theta+\sin(\theta/2)}.
$$
These give the <exact finite-grid high-frequency Jacobi smoothing optimum>, including complete annihilation when there is only one high-frequency mode. Their grid-independent limits are \b[$\boxed{\omega_*=2/3,\ \mu_*=1/3}$]. Choosing $2/3$ damps every high mode by at most $1/3$ on every grid, while its low-mode factor remains $1-O(n^{-2})$. Thus high frequencies decay geometrically much faster. The familiar $2/3,1/3$ pair is the optimal uniform-in-grid smoother; it need not be the exact optimum on a particular finite sine-mode grid.
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