Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 / B / 5

An element of a C*-algebra is a Normal element of a C-star algebra if

An element is a Positive element of a C-star algebra if it is self-adjoint and its spectrum lies in $[0,\infty )$, equivalently if $a=b^{\ast }b$ for some $b$.

For a normal $a$, the continuous functional calculus is the unique unital isometric *-isomorphism $C(\sigma (a))\to C^{\ast }(1,a)$ sending the coordinate function $z$ to $a$. It respects sums, products, conjugation and the unit, satisfies $\Vert f(a)\Vert ={\max }_{\sigma (a)}|f|$, and has $\sigma (f(a))=f(\sigma (a))$. For self-adjoint $a$ it preserves the order of real-valued functions.

The piecewise definition of $g_n$ is consistent for $n\ge 2$. At $n=1$ the printed branches conflict, even assigning both zero and one at $t=1/2$. All the requested limit and face arguments use only $n\ge 2$; discarding this first index resolves that harmless initial-index defect without altering any limit.

For $n\ge 2$, interpolate linearly from $(1/(2n),0)$ to $(1/n,g(1/n))$, and from $(1-1/n,g(1-1/n))$ to $(1-1/(2n),1)$. On the middle interval the error is zero. On the left both $g$ and $g_n$ lie between zero and $g(1/n)$; on the right both lie between $g(1-1/n)$ and one. Thusby continuity at the endpoints.

Put $\delta =1/(2n)$. To prove the bounds on $h_n$, it suffices to show $0\le t-\delta g_n(t)\le 1-\delta$. For $t\le \delta$, $g_n=0$; for $t\ge 1-\delta$, $g_n=1$. On the remaining interval, $\delta \le t\le 1-\delta$ and $0\le g_n\le 1$, so both inequalities follow. Consequently $g_n$ and $h_n$ are positive contractions.

For the face assertion, write $r=\Vert a\Vert$. Since $a\in [0,1_A]$ and $0\notin F$, $0<r\le 1$. If $r<1$, then $a=r(a/r)+(1-r)0$, with both endpoints in $[0,1_A]$. The face property would force $0\in F$, a contradiction. Thus $\Vert a\Vert =1$. Its spectrum contains one by positivity and the norm formula, and contains zero because it is noninvertible.

Applying functional calculus to the scalar identity givesSince $0<\delta <1$ and $a\in F$, the face property gives $g_n(a)\in F$. Isometry of functional calculus gives $g_n(a)\to g(a)$ in norm; closedness of $F$ therefore proves