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

A character of an algebra is a nonzero multiplicative complex-linear map $\chi :A\to \mathbb{C}$. A maximal proper ideal is a proper ideal contained in no larger proper ideal. For a character, $\chi (1)=1$ and $\chi (a)\in \sigma (a)$: otherwise $a-\chi (a)1$ would be invertible although its character value is zero. The spectral bound gives $|\chi (a)|\le \Vert a\Vert$, so the character is continuous. It is onto, and its kernel is a closed maximal ideal because $A/\ker \chi \cong \mathbb{C}$.

Conversely let $M$ be maximal. Its closure is an ideal and remains proper: otherwise an element of $M$ would lie within distance one of the unit, hence would be invertible by the Neumann series, forcing $1\in M$. Maximality then gives $\overline{M}=M$. The Banach algebra $A/M$ is a field. For any $b$ in this quotient choose $\lambda \in \sigma (b)$, using nonemptiness of the complex spectrum. In a field only zero is noninvertible, so $b-\lambda 1=0$. Hence every quotient element is a scalar multiple of the unit. The quotient map therefore determines a character ${\chi }_M$, uniquely normalized by ${\chi }_M(1)=1$. These constructions are inverse:For the convolution algebra, write $e_j$ for the sequence supported at $j$ with value one. Then $e_j{e}_k=e_{j+k}$, $e_0=1$, and $e_1{e}_{-1}=1$. If $z=\chi (e_1)$, continuity and $\Vert e_{\pm 1}{\Vert }_1=1$ imply $|z|\le 1$ and $|z^{-1}|\le 1$. Thus $|z|=1$, so uniquely $z=e^{it}$ for $0\le t<2\pi$, and $\chi (e_j)=e^{ijt}$ for every integer $j$. Absolute convergence in ${\ell }^1$ now givesConversely each displayed map is bounded, nonzero, and multiplicative, since the absolutely convergent double sum can be rearranged along $k+l=j$. Thus these are all characters, and all maximal proper ideals areThey are distinct because the character is uniquely recovered from its kernel and its value on the unit.

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

For a bounded linear functional on a unital C*-algebra, the state criterion isEquivalently it is positive and takes the unit to one.

In the Gelfand–Naimark–Segal construction, set $N_x=\{b:x(b^{\ast }b)=0\}$. This is a left ideal. Give $A/N_x$ the inner product $⟨[b],[c]⟩=x(c^{\ast }b)$, linear in the first variable, and complete it to a Hilbert space $H_x$. Left multiplication gives the bounded *-representation ${\pi }_x(a)[b]=[ab]$. The vector $\xi =[1_A]$ is cyclic and satisfies $x(a)=⟨{\pi }_x(a)\xi ,\xi ⟩$.

The Hermitian matrix $\rho$ has eigenvalues ${\lambda }_{\pm }=1/2\pm |\alpha |$ and trace one. In an orthonormal eigenbasis $v_{+},v_{-}$,If $|\alpha |\le 1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha |>1/2$, evaluating on the positive rank-one projection onto the negative-eigenvalue eigenspace gives a negative value, so it is not a state. ThereforeTo obtain the representation dimension, cyclicity of trace givesThus $A/N_{x_{\rho }}$ identifies isometrically with the matrix space $\{b{\rho }^{1/2}:b\in M_2(\mathbb{C})\}$, with left multiplication as the representation. If $r=\mathrm{rank}\rho$, this space consists, after a unitary change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. Rank is one at $|\alpha |=1/2$ and two at $|\alpha |<1/2$, proving