Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 / B

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

Use the source's notation for the vanishing-at-zero ideal $I=\{f\in C[0,1]:f(0)=0\}$. It is the kernel of bounded evaluation at zero, hence closed. It is stable under pointwise multiplication and complex conjugation, so inherits the C-star identity and completeness. If it had an identity $e$, multiplication against $f(t)=t$ would give $e(t)=1$ for $t>0$, contradicting continuity and $e(0)=0$. Thus $I$ is a non-unital C-star subalgebra. This explicit meaning of the notation matters because the usual notation $C_0$ for a compact space has a different convention.

Define the unitization map by $\Phi (f,\lambda )=f+\lambda 1$. Its inverse is $h\mapsto (h-h(0),h(0))$. The specified multiplication and involution are preserved by direct substitution. The pullback norm $\Vert (f,\lambda )\Vert =\Vert f+\lambda 1{\Vert }_{\infty }$ is a C-star norm, so equals the unique unitization norm supplied in the question. Therefore $\Phi$ is a unital isometric C-star isomorphism onto $C[0,1]$.

For the approximation put $h(s)=a(\sqrt{s})$ on $[0,1]$, so $h$ is continuous and $h(0)=0$. The Stone-Weierstrass theorem supplies a polynomial $p(s)$ with $\Vert h-p{\Vert }_{\infty }<\epsilon /2$. Then $b(t)=p(t^2)-p(0)$ is a polynomial in the positive even powers of $t$, with zero constant term, andThis gives exactly the required finite form.

The continuous functional calculus for a Hermitian element of a C-star algebra identifies its generated unital C-star algebra with continuous functions on its real spectrum, preserves the involution and norm, and sends functions to the corresponding spectral values by spectral mapping. Consequently an element is a square of a Hermitian element of a C-star algebra exactly when it is a Hermitian element of a C-star algebra with nonnegative spectrum: the reverse direction uses the continuous square root on that spectrum.

Nonnegative scalar multiples preserve this property directly. For positive $a,b$ put $C=\Vert a\Vert +\Vert b\Vert$. Functional calculus gives $\Vert \Vert a\Vert 1-a\Vert \le \Vert a\Vert$ and similarly for $b$, so $\Vert C1-(a+b)\Vert \le C$. The sum is a Hermitian element of a C-star algebra, and spectral mapping now places its spectrum in $[0,2C]$. It therefore has a positive square root in a C-star algebra. If $C=0$ both elements are zero. Thus $A_{+}$ is a cone. It is pointed because an element positive with its negative has spectrum contained in $\{0\}$ and hence is zero by the self-adjoint norm formula. It is closed as well: for a converging sequence of Positive elements of a C-star algebra use a common norm bound $C$ and pass to the limit in $\Vert C1-a_n\Vert \le C$.

For the two-by-two matrix the exact criterion isA positive matrix is Hermitian with nonnegative eigenvalues; evaluating its quadratic form at the coordinate vectors gives the diagonal conditions, and its determinant, the eigenvalue product, gives the final inequality. Conversely if $a_{11}>0$, completing the square writes the quadratic form as $a_{11}|z_1+(a_{12}/a_{11})z_2{|}^2+(a_{22}-|a_{12}{|}^2/a_{11})|z_2{|}^2$, which is nonnegative. If $a_{11}=0$, the determinant inequality forces $a_{12}=0$ and the conclusion is immediate. The spectral theorem then gives nonnegative eigenvalues and their Hermitian square root, proving positivity in the question's sense.

A state on a C-star algebra is a positive linear functional of norm one. A cyclic representation of a C-star algebra is a representation of a C-star algebra $\pi :A\to B(H)$ with a vector $\xi$ whose orbit $\pi (A)\xi$ has dense linear span in $H$. On $c$, pointwise multiplication and coordinatewise complex conjugation preserve convergence; the constant sequence one is the unit. The supremum norm is submultiplicative and obeys $\Vert {\alpha }^{\ast }\alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }^2$. Completeness is given, so $c$ is a commutative unital C-star algebra.

If $F=\varphi (\beta )$ is a state, testing its positivity on the coordinate projections gives ${\beta }_{j+1}\ge 0$. To recover the independent coefficient at infinity, let $q_N$ have its first $N$ coordinates zero and all later ones one. This is a positive element of $c$, and $F(q_N)={\beta }_1+{\sum }_{j>N}{\beta }_{j+1}\ge 0$. Passing to the limit gives ${\beta }_1\ge 0$. By the given isometry, $\Vert F\Vert =\Vert \beta {\Vert }_1={\sum }_j{\beta }_j$, which is one. Conversely these nonnegative unit-sum coefficients make $F({\alpha }^{\ast }\alpha )$ a sum of nonnegative weighted squares, and the isometry gives norm one. Thus the stated coefficient conditions are necessary and sufficient.

For the specified state, all other coefficients are zero because they are nonnegative and the first three already sum to one. Its GNS null ideal isEvaluation at infinity and the first two coordinates identifies $c/N_x$ with ${\mathbb{C}}^3$, with inner product $\frac{1}{2}{u}_{\infty }\overline{v_{\infty }}+\frac{1}{4}{u}_1\overline{v_1}+\frac{1}{4}{u}_2\overline{v_2}$. Left multiplication induces the diagonal representation; the class of the unit is $(1,1,1)$. In the equivalent standard inner product the resulting GNS triple isIts coefficient is exactly $\frac{1}{2}{\alpha }_{\infty }+\frac{1}{4}{\alpha }_1+\frac{1}{4}{\alpha }_2=F(\alpha )$. The three evaluations can be assigned arbitrary independent values by a convergent sequence, and all three entries of ${\xi }_x$ are nonzero. Therefore its orbit spans ${\mathbb{C}}^3$, verifying cyclicity and the required state identity directly.