Show that hermitian element $h=\sum p_n/3^n$ generates $ C_0(\Omega)$

Question

Let $\Omega$ be a locally compact Hausdorff space, and suppose that the C*-algebra $C_0(\Omega)$ is generated by a sequence of projections $(p_n)_{n=1}^{\infty}$.

Show that the hermitian element $h=\sum_{n=1}^\infty \frac{p_n}{3^n}$ generates $C_0(\Omega)$.

My attempt: firstly, I show that $h\in C_0(\Omega)$. It means that $h$ is continuous and for every $\epsilon >0$, the set $\{x\in \Omega ; |h(x)|\geq \epsilon\}$ is compact. Suppose $p_np_m=0$ for every $n,m\in \Bbb N$ and $n\neq m$.

Let $x_m\to x$ , there is $n_0\in \Bbb N$ such that $x\in p_{n_0}$. Also there is a subsequence $\{y_m\}$ of $\{x_m\}$ such that $\{y_m\}\subset p_{n_0}(\Omega)$. Then $$|h(y_m) - h(x)| \leq \sum_{n=1}^\infty\frac{|p_n(y_m)-p_n(x)|}{3^n}=0$$ So $h$ is continuous. To show for every $\epsilon >0$, the set $\{x\in \Omega ; |h(x)|\geq \epsilon\}$ is compact. I know that for every $\epsilon>0$, there is $n_0$ such that $\sum_{n=n_0+1}^\infty\frac{1}{3^n} <\epsilon$. Thus for $x\in p_1+...+p_{n_0}(\Omega)$, $|h(x)|\geq \epsilon$. Clearly $Im(p_1+...+P_{n_0})$ is closed, but I can not show that it's compact.

Now I show that $h$ generates $C_0(\Omega)$. Let $f\in C_0(\Omega)$. there is a sequence of polynomials $\{q_m\}\subset C_0(\Omega)$ such that $f=\lim q_m$. Also every polynomial has a representation of projections $\{p_n\}$ as $q_m = \sum \lambda_n p_n$. Clearly for $n$, there is $t_n$ such that $\lambda_n=t_n/3^n$ . So $h$ generates $C_0(\Omega)$.

Please check my attempt and give me a hint to show that $h\in C_0(\Omega)$. Thanks in advance

Answer 1

The space $C_0$ is a Banach space. This implies that absolutely convergent sequences are convergent, i.e. if $$\sum_n \Vert \frac{p_n}{3^n} \Vert_\infty$$ is finite, then $\sum_n \frac{p_n}{3^n} \in C_0$. But projections have norm at most one, which means $$\sum_n \Vert \frac{p_n}{3^n} \Vert \leq \sum_n 3^{-n} = \frac{1}{1-1/3} < \infty.$$

This simplifies your argument considerably, but assumes that you know already that $C_0$ is complete.

Also, you do not have to assume $p_n p_m = 0$ for $n \neq m$! You just introduced this assumption without justification in your proof!

For the "generating" part of the proof, it is not at all clear what you mean with a sequence of "polynomials" $(q_m)_m$ in $C_0 (\Omega)$, because $\Omega$ is an arbitrary LCH space, so that it is not clear what you mean by a polynomial.

I think the approach to be taken here is that since each $p_n$ is a projection, you have ${\rm range}(p_n) \subset \{0,1\}$. Now, for each $n \in \Bbb{N}$, the point $3^{-n}$ has positive distance from

$$ \bigcup_{m \neq n} \{3^{-m}\}. $$

This should allow you to find a continuous function $f_n : \Bbb{R} \to \Bbb{R}$ such that $f_n (3^{-n}) = 1$ and $f_n (3^{-m}) = 0$ for $m \neq n$.

This will imply (why?) that

$$ f_n (h) = f_n\big(\sum_m \frac{p_m}{3^m}\big) = \frac{p_n}{3^n}. $$

Hence (why?), each $p_n$ is contained in the algebra generated by $h$.

Answer 2

By Stone-Weierstrass theorem, we only need to prove that $h$ separates points of $\Omega$ and $h$ vanishes nowhere.

If $h(x)=h(y)$, then since $p_n$ takes values in $\{0,1\}$, we get that $p_n(x)=p_n(y)$ for every $n\geqslant 1$. Since points of $\Omega$ can be realized as characters on $C_0(\Omega)$, which is generated by $(p_n)_{n=1}^\infty$, we must have $x=y$.

Meanwhile, if $h(x)=0$, then $p_n(x)=0$ for every $n\geqslant 1$, and the C*-algebra generated by $(p_n)_{n=1}^\infty$ must be contained in $C_0(\Omega\setminus\{x\})$, which leads to a contradiction.