Chebyshev constant

Remarks

The infimum is attained, and a monic polynomial $p$ of degree $n$ with $\norm{p}_K = t_n(K)$ is a Chebyshev polynomial of $K$. Since a product of monic polynomials is monic, $t_{m+n}(K) \le t_m(K) \, t_n(K)$. So the sequence $\log t_n(K)$ is subadditive, and Fekete's lemma shows that the limit exists and equals $\inf_n t_n(K)^{1/n}$.

Example: let $K$ be the unit circle. For a monic $p(z) = z^n + a_{n-1} z^{n-1} + \cdots + a_0$, Parseval's identity gives $\frac{1}{2\pi} \int_0^{2\pi} \abs{p(e^{i\theta})}^2 \, d\theta = 1 + \sum_k \abs{a_k}^2 \ge 1$. So $\norm{p}_K \ge 1$, with equality for $p = z^n$, and $t_n(K) = 1$ and $\tau(K) = 1$.

Non-example: the polynomial $2 z^n$ has maximum $2$ on the unit circle, but it is not monic, so it does not enter the infimum. Without the monic condition the infimum would be $0$.

If $K$ is a finite set with $m$ points, then $p(z) = \prod_{a \in K} (z - a)$ vanishes on $K$, so $t_n(K) = 0$ for $n \ge m$ and $\tau(K) = 0$.