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?
Best Answer
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 $\overline{f}(it)=e^t$.