Dropping Conditions on the Stone-Weierstrass Theorem

It’s a good exercise, in general, to take well-known theorems and see what happens. Can you find counterexamples? What is salvageable? As a C*-algebraist, one of my favorite theorems from baby Rudin to do this with is the Stone-Weierstrass theorem.

First, we recall the theorem as it’s presented in Rudin.

Theorem. Let K be a compact metric space. Let A be a self-adjoint subalgebra of C(K), the algebra of complex-valued continuous functions from K. If

  1. A vanishes nowhere (i.e. for every point x\in K, there exists a function f \in A such that f(x) \neq 0) and
  2. A separates points on K (i.e. for every x,y\in K such that x\neq y, there exists a function f\in A such that f(x) \neq f(y)),

then A is uniformly dense in C(K).

It’s well-known that K can be replaced by a compact Hausdorff space, but since, baby Rudin doesn’t deal with abstract topological spaces, it’s stated for metric spaces. The proof first establishes the theorem for boolean subalgebras (in the real case), and then one realizes that closed algebras (with the usual addition and multiplication) are closed boolean subalgebras since the absolute value of a function can be used to recover the max and min operations.

The first generalization is the fact that if one replaces K with a locally compact space and the algebra C(K) with C_0(K) of continuous functions vanishing at infinity, then a similar theorem holds. This is clear from the fact that the subalgebra C_c(K) of continuous functions with compact support is dense in C_0(K).

Especially the way I wrote the theorem lends itself to the following exercise:

Exercise. What does the closure of A correspond to when condition (1) is dropped? What about condition (2)?

Before proceeding, I’d like to give a spoiler alert. I found this exercise to be fun, even though it probably should have been obvious from the start. Before proceeding, it might be more fun to think about the question for a little bit, though those of you smarter than myself may find the exercise trivial.

Now if we drop the first condition, then A vanishes somewhere. If we denote Z = \{ x\in K\mid f(x) = 0 \text{ for all} f\in A \}, then we can consider C_0(K\setminus Z) a self-adjoint subalgebra of C(K) by extending by 0 on Z. With this identification in mind, we see that A is a subalgebra of C_0(K\setminus Z) that satisfies conditions (1) (for the subspace) and (2). So A is dense in C_0(K\setminus Z). Since Z is closed, A corresponds to dense subalgebras of C_0(U) where U is an open subset of K.

Dropping the second condition means that A can’t separate some points. So we define an equivalence relation on K by x \sim y if f(x) = f(y) for all f\in A. Now, I’m not sure if K/\sim is a compact metric space, but if we use the topological version of the theorem, we can so that the quotient space is a compact Hausdorff space by using the fact that the quotient is Hausdorff if and only if the relation \sim as a subspace of K\times K is closed in the product topology. Then we see that A is a subalgebra of C(K/\sim) satisfying conditions (1) and (2). So A corresponds to dense subalgebras of C(Q), where Q is a quotient space of K.

Of course, I was being a little sneaky here by not listing self-adjointness as a condition. But it seems clear that we can’t say much about non-self-adjoint subalgebras of C(K). In particular, if we take the classic example with K being the unit disk and A is the closed subalgebra generated by the constant function 1 and the map z\mapsto z, then we get the disk algebra, a proper subalgebra of C(K), where all the functions are holomorphic in the interior of the disk.

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About minimalrho
Unemployed guy with a PhD in math.

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