Past exam of the Banach and C*-algebras course of the University of Oxford / 2011 / 1

A unital Banach star-algebra is a complex unital Banach algebra with a continuous conjugate-linear algebra involution satisfyingThe algebra norm is complete and submultiplicative, with the unit conventionally normalized to norm one. Many definitions additionally require the involution to be isometric, $\Vert a^{\ast }\Vert =\Vert a\Vert$; the function algebra here satisfies this stronger convention as well. A Banach star-algebra does not by definition require the extra C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

Here is the lattice proof of the Stone-Weierstrass theorem in its complex self-adjoint form. Let $B$ be the uniform closure of the given unital star-subalgebra in $C(\Omega )$, and let $B_{\mathbb{R}}$ be its real-valued part. Closure preserves multiplication and complex conjugation, so $B$ is still a unital star-algebra. Real and imaginary parts lie in $B_{\mathbb{R}}$, and separation by the complex algebra implies separation by its real-valued part.

For $h\in B_{\mathbb{R}}$, polynomial approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb{R}}$. For example, this polynomial approximation follows from the binomial series for the square root, whose coefficients give uniform convergence on the relevant closed interval. Thusalso lie in $B_{\mathbb{R}}$. It is therefore a lattice of functions.

To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $ϵ>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb{R}}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.

Fix $x$. The sets where $h_{xy}>f-ϵ$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb{R}}$ with $g_x>f-ϵ$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+ϵ$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb{R}}$ satisfyingThis is lattice approximation from two-point approximation. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb{R}}$ is closed, it contains every real continuous function. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega )}$.

The annulus test. For the closed annulus, the family holomorphic in the interior is not dense in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so locally uniform convergence of holomorphic functions makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega )$. It is proper because $z\mapsto \overline{z}$ is continuous but not holomorphic on the annulus interior.

There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)dz$ around its positively oriented boundary. The Cauchy integral theorem gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline{z})=2\pi i{r}^2\ne 0$. Since $|L(h)|\le 2\pi r\Vert h{\Vert }_{\infty }$, no uniform approximation of $\overline{z}$ by that family is possible.

The even-power family is dense on $[1,2]$. It is the algebra $A=\{p(t^2):p\in \mathbb{C}[s]\}$. It contains constants, is closed under complex conjugation, and separates points because the real function $t\mapsto t^2$ is injective on this positive interval. The complex Stone-Weierstrass theorem therefore applies.

Equivalently, the squaring map is a homeomorphism from $[1,2]$ onto $[1,4]$. Given $f\in C([1,2])$, approximate the continuous function $s\mapsto f(\sqrt{s})$ uniformly on $[1,4]$ by complex polynomials, using the Weierstrass approximation theorem on the real and imaginary parts. Substitution $s=t^2$ produces exactly the required even-power approximants. The positive interval is what lets even powers distinguish all its points.

The odd-power family is also dense on $[1,2]$. It is $B=tA$, where $A$ is the even-power family in part (i). Multiplication by $t$ is an invertible bounded linear map on $C([1,2])$, with inverse division by $t$, since $t\ge 1$.

For an explicit approximation, take any $f\in C([1,2])$ and $ϵ>0$. The quotient $f(t)/t$ is continuous, so part (i) gives an even polynomial $a$ with $\Vert a-f/t{\Vert }_{\infty }<ϵ/2$. Then $ta$ belongs to the odd-power family andAlthough this family is not a subalgebra and does not contain the constant function exactly, it can approximate every continuous function, including constants. This proves even and odd polynomial density on a positive interval without incorrectly applying the unital algebra theorem directly to the odd family.