As in the title, Royden proves a fixed-point theorem as a prelude to the Kakutani's fixed point theorem:
$\pi:G\to GL(E)$ is a representation of the compact group $G$ on $E$, a Banach space. Suppose that $g\mapsto \pi(g)x$ is continuous for each $x\in E$, where we give $E$ the norm topology. Further, suppose that there is a non-empty convex, weak$^*$- compact $K^*\subseteq E^*$ such that $\pi^*(g)(K^*)\subseteq K^*.$ Then the conclusion is that there is a functional $\psi\in K^*$ such that $\pi^*(g)(\psi)=\psi.$ (Note that $\pi^*(g)(\psi)=\psi\circ \pi(g^{-1})).$
The proof proceeds along the lines of Krein-Milman, and is fairly straightforward, but there is an estimate that I can't verify. Having chosen a particular $x_0\in E$ (with a property that, for the purposes of my question, is not necessary to state), we have that
$1).\ K^*$ is bounded and $p$ is weak$^*$- continuous, where
$2).\ $ the functional $p(\psi)=\sup\{\psi(\pi(g)x_0):g\in G\}$ for $\psi\in K^*.\ p$ is positive homogeneous and subadditive. Consequently, the balls
$3).\ B_0(\eta,r)=\{\psi\in K^*: p(\eta -\psi)<r\}$ are open in the weak$^*$- topology on $K^*.$ So,
$4).\ d=\sup \{p(\psi-\varphi):\psi,\varphi\in K^*\}$ is properly defined. Now, compactness gives us $\{\psi_1,\cdots,\psi_n\}$ such that $K^*=\bigcup B(\psi_k,d/2).$ Put
$5).\ \psi^*=\frac{\psi_1+\cdots + \psi_n}{n}\in K^*$ Then, if $\psi\in K^*,$ we must have $\psi\in B(\psi_{k_0},d/2)$ for some $1\le k_0\le n$ so
$6).\ p(\psi-\psi^*)\le p(\psi-\psi_{k_0})+p(\psi^*-\psi_{k_0})\le \frac{d}{2}+\frac{n}{n-1}d\color{red}{<d}$, where the red indicates what Royden has. It's obviously a typo. I guess he means $\frac{d}{2}+\frac{n}{n-1}\frac{d}{2}<d$ but when I do the calculation I get $\frac{d}{2}+p(\frac{\psi_1+\cdots + \psi_n}{n}-\psi_{k_0})=\frac{d}{2}+\frac{1}{n}p(\psi_1+\cdots + \psi_n-n\psi_{k_0})<\frac{d}{2}+\frac{n-1}{n}d$ which is what Royden has. How can we patch this up?
The step sketched is indeed incorrect. A correct version could read:
$$p(\psi - \psi^*) = p\left(\frac1n\sum_k (\psi-\psi_k)\right)≤ \frac1np(\psi - \psi_{k_0})+ \frac1n \sum_{k\neq k_0}p(\psi -\psi_k)≤\frac1n \frac d2+\frac{n-1}nd <d$$