Closed convex hull of unitaries

Question

If a C*-algebra ${\cal U}$ contains a non-unitary isometry $S$, show that $$\|S-A\|>\frac{1}{2n}$$ for every $A=\sum_{i=1}^n \lambda_iU_i$ which is the convex combination of $n$ unitaries.

Thanks in advance.

Answer 1

Represent $\mathcal U$ as a sub-algebra of $\mathcal B(H)$, the bounded linear operators on some Hilbert space $H$.

As $s$ is non-unitary, $s(H)^\perp\neq \{0\}$ in $H$. Choose an index $1\leq j\leq n$ such that $\lambda_j\geq \frac 1 n$ and a unit vector $\xi\in H$ such that $u_j(\xi)\in s(H)^\perp$ (this is possible since $u$ is surjective).

Then we have

\begin{align*}\|a(\xi)-s(\xi)\|^2&=\|a(\xi)\|^2+\|s(\xi)\|^2-2\mathrm{Re}\langle a(\xi),s(\xi)\rangle\\ &=2(1-\lambda_j\mathrm{Re}\langle u_j(\xi),s(\xi)\rangle-\sum_{i\neq j}\lambda_i\mathrm{Re}\langle u_i(\xi),s(\xi)\rangle)\\ &\geq 2(1-\sum_{i\neq j}\lambda_i\|u_i(\xi)\|\|s(\xi)\|)\\ &=2\lambda_j\\ &\geq \frac 2 n.\end{align*} Of course, an intrinsic argument (without referring to Gelfand-Naimark) would be desirable.