Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 4

By the spectral radius formula, $r(T)<1$. ChooseThe formula gives a constant $C$ such that $\Vert T^n\Vert \le C{r}^n$ for every $n$. Since $f$ is holomorphic on the unit disc, the Cauchy estimate on the circle of radius $s$ gives $|a_n|\le M_s{s}^{-n}$. ThereforeThus ${\sum }_{n\ge 0}{a}_n{T}^n$ converges absolutely in operator norm.

If $T$ is normal, then $S=T-I$ is normal and $\sigma (S)=\{0\}$. The C-star identity and normality imply $\Vert S^n\Vert =\Vert S{\Vert }^n$. The spectral radius formula consequently givesso $T=I$.

Normality is essential. On ${\ell }^2$, let $N$ send $e_2$ to $e_1$ and every other basis vector to zero. Then $N\ne 0$ but $N^2=0$, sowhile $I+N\ne I$.

The Commutative Gelfand--Naimark theorem says that every commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C(\Delta (A))$, where $\Delta (A)$ is its compact character space and the map is the Gelfand transform

Indeed, maximal ideals give enough characters to identify $\sigma (a)$ with the range of $\hat{a}$. The C-star identity and the spectral radius formula giveso $\Gamma$ is isometric and has closed range. Characters send $a^{\ast }$ to $\overline{\chi (a)}$, so the range is self-conjugate; it contains constants and separates distinct characters. The complex Stone-Weierstrass theorem makes the range dense in $C(\Delta (A))$, and closedness makes it all of that algebra. This proves the theorem.

An element $x$ of a C-star algebra is positive when it is self-adjoint and $\sigma (x)\subseteq [0,\infty )$. Consider the commutative C-star subalgebra $C^{\ast }(1,x)$. Under its Gelfand–Naimark isomorphism, $x$ becomes a nonnegative continuous function $\hat{x}$. The function $\sqrt{\hat{x}}$ is continuous and nonnegative, so its inverse image $y$ is positive and satisfies