Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 / B

For the real lattice version of the Stone-Weierstrass theorem, let $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. For any $x,y\in \Omega$, separation of points and constants give $h_{xy}\in L$ matching $f$ at both $x$ and $y$: take an affine combination of a separating function, and for $x=y$ take a constant. Fix $x$. The open sets where $h_{xy}>f-\epsilon$ cover $\Omega$ as $y$ varies. Compactness gives finitely many such functions; their maximum $g_x$ belongs to the lattice $L$, satisfies $g_x>f-\epsilon$ everywhere, and has $g_x(x)=f(x)$. The open sets where $g_x<f+\epsilon$ now cover $\Omega$ as $x$ varies. Choose finitely many and take their minimum $h\in L$. Then $f-\epsilon <h<f+\epsilon$ everywhere. Thus $L$ is uniformly dense in the real continuous functions.

For the complex *-algebra, its real subalgebra $A_{\mathbb{R}}$ of Hermitian C-star elements contains the real and imaginary parts of every member. It separates points, because at least one of those parts distinguishes any pair distinguished by a complex function. Its uniform closure is a unital real subalgebra and is a lattice by the allowed result. The real theorem makes this closure all of $C_{\mathbb{R}}(\Omega )$. Approximating the real and imaginary parts separately proves $A$ is uniformly dense in $C(\Omega )$.

On $[0,1]$, the algebra of polynomials in $t^2$ is unital, closed under conjugation, and separates points because $t\mapsto t^2$ is injective there. Choose $p(t^2)$ within $\epsilon /3$ of $a(t)$, and correct its endpoints byThis is still a polynomial in $t^2$ and has exactly the requested endpoint values. The correction has absolute value at most $\epsilon /3$, because it is a convex combination of two errors of that size. Therefore

The spectrum of an element is ${\sigma }_A(a)=\{\lambda \in \mathbb{C}:a-\lambda 1_A\text{ is not invertible in }A\}$. A Normal element of a C-star algebra satisfies $a^{\ast }a=a{a}^{\ast }$.

For a Hermitian C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. For normal $a$, commutation givesAll powers remain normal, so induction yields $\Vert a^{2^k}\Vert =\Vert a{\Vert }^{2^k}$. Taking the spectral-radius limit along these powers givesLet $\phi :A\to B$ be a C-star homomorphism, without assuming continuity or preservation of the ambient unit. Put $p=\phi (1_A)$, a projection in a C-star algebra. If $p=0$, then $\phi =0$. Otherwise its image lies in the unital corner $pBp$ and it is unital as a map into that corner. An inverse in $A$ maps to an inverse in the corner, so ${\sigma }_{pBp}(\phi (b))\subseteq {\sigma }_A(b)$. Apply this to the normal element $b=a^{\ast }a$ and use the C-star identity in both algebras:Thus every C-star homomorphism is norm non-increasing, including nonunital maps.

A Unitary element of a C-star algebra satisfies $u^{\ast }u=u{u}^{\ast }=1$. A projection in a C-star algebra satisfies $p=p^{\ast }=p^2$.

If $q=u^{\ast }u$ is a projection in a C-star algebra, the C-star identity gives $\Vert u(1-q){\Vert }^2=\Vert (1-q)q(1-q)\Vert =0$. Hence $u=uq=u{u}^{\ast }u$. Conversely $u{u}^{\ast }u=u$ implies $(u^{\ast }u{)}^2=u^{\ast }u$, and $u^{\ast }u$ is a Hermitian C-star element, so it is a projection in a C-star algebra. Taking adjoints of the triple-product equality gives the corresponding equality for $u^{\ast }$, provingFor nonzero $u$, let $p=u{u}^{\ast }$ and $q=u^{\ast }u$. These are projections, $u=pu=uq$, and $\Vert u\Vert =1$. The rectangular corner $pAq$ is closed because it is the range of the bounded idempotent map $a\mapsto paq$.

Define $\Theta :pAq\to qAq$ by $\Theta (a)=u^{\ast }a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $u{u}^{\ast }a=pa=a$ and $u^{\ast }ub=qb=b$ verify both compositions. Both maps are contractive because $\Vert u\Vert =\Vert u^{\ast }\Vert =1$, so $\Theta$ is an isometry. For the proposed multiplication and involution,Thus the operations correspond exactly to the ordinary operations in the closed C-star algebra $qAq$. This proves associativity, closure, all involution identities and the C*-norm identity at once. Moreover $\Theta (u)=q$, so the unit of $A_u$ is $u$. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^{\ast }a$ is an onto isometric *-isomorphism from $A_u$ to $A$.

First suppose $x$ is a positive linear functional, without assuming boundedness. Every Hermitian C-star element $h$ is a difference of positive elements, since $\Vert h\Vert 1\pm h\ge 0$. Thus $x(h)$ is real. For a Hermitian C-star element $h$ of norm at most one, $-1\le h\le 1$ gives $|x(h)|\le x(1)$.

For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The real part of $\zeta a$ is a Hermitian C-star element and has norm at most $\Vert a\Vert$. Since $x$ is real on Hermitian C-star elements, it respects real parts, givingThis proves boundedness and $\Vert x\Vert \le x(1)$. Evaluation at the unit gives the reverse inequality, so positivity implies $\Vert x\Vert =x(1)$.

Conversely let $x$ be bounded with $x(1)=\Vert x\Vert =c\ge 0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a Hermitian C-star element $h$ and real $t$, the C-star identity givesWriting $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\le t^2\Vert h{\Vert }^2$. Taking $t\to 0$ through positive and negative values forces $b=0$. Hence $x$ is real on Hermitian C-star elements. If $0\le a\le 1$, then $1-a$ has norm at most one, so $x(1-a)\le 1$ and $x(a)\ge 0$. Scaling proves nonnegativity on every positive element, in particular every $b^{\ast }b$. Therefore