# Converse of Stone-Weierstrass Theorem for Complex Continuous Functions

Stone-Weierstrass Theorem for complex continuous functions says:

Let $$K$$ be a compact Hausdorff space and $$\mathcal{A} \subseteq C(K, \mathbb{C})$$ be a subalgebra. If $$\mathcal{A}$$ separates points and is closed under conjugation, then $$\mathcal{A}$$ is dense in $$C(K, \mathbb{C})$$.

I am trying to figure out whether the converse holds, at least in case $$K$$ is a metric space. It is obvious that $$\mathcal{A}$$ separates points if $$\mathcal{A}$$ is dense in $$C(K, \mathbb{C})$$, but it seems difficult to prove or disprove that $$\mathcal{A}$$ is closed under conjugation if $$\mathcal{A}$$ is dense. Is there any proof or counterexample for this?

• As another classic example, if $K$ is any proper closed subset of the unit circle, the algebra of polynomials is clearly not closed under conjugation as $\bar z =1/z$ is not a polynomial, but it is still dense in $C(K)$ as one can show (either by Runge or a Runge inspired proof, as $\bar z=1/z$ can be extended to an analytic function on a neighborhood of $K$ so, is approximable by polynomials on $K$ and then Feijer's theorem from Fourier series theory does the rest, or by the Riesz brothers theorem about analytic measures and Hahn Banach) Commented Jul 18, 2022 at 6:13

Note that if a subalgebra $$\mathcal{A}$$ is dense, then so is any $$\mathcal{B}$$ containing $$\mathcal{A}$$. But there's no reason that an arbitrary subalgebra $$\mathcal{B}$$ containing $$\mathcal{A}$$ would have to be closed under conjugation. For instance, if you adjoin one new function to $$\mathcal{A}$$, then there's no reason the subalgebra it generates should contain the conjugate of that new function.
Now, actually proving the conjugate is not in the generated subalgebra in a specific example will take some cleverness. Here's one way to do it. Let $$K=[0,1]$$ and let $$\mathcal{A}$$ be the algebra of polynomial functions on $$[0,1]$$. Then by Stone-Weierstrass, $$\mathcal{A}$$ is dense in $$C([0,1],\mathbb{C})$$. Now let $$\mathcal{B}$$ be the algebra generated by $$\mathcal{A}$$ together with the function $$f(x)=e^{ix}$$. I claim that the function $$\overline{f}(x)=e^{-ix}$$ is not in $$\mathcal{B}$$. To prove this, note that every element of $$\mathcal{B}$$ extends uniquely to a holomorphic function on all of $$\mathbb{C}$$. For any $$g\in\mathcal{B}$$, this holomorphic extension has the property that $$|g(it)|$$ is bounded by a polynomial in $$t$$ as $$t\in\mathbb{R}$$ goes to $$+\infty$$ (since this is true of $$f$$ and of any polynomial). However, this is not true of the holomorphic extension of $$\overline{f}$$, since $$\overline{f}(it)=e^t$$.