Let $X$ be a Radon measure on a second countable locally compact space $X$. Let $C_c(X)$ be the space of continuous functions with compact support.
Is
$$F=\mathrm{span}_{\mathbb{C}}\{|f|^2 :f \in C_c(X)\}$$
dense in $L^1(X)$ ?
My feeling is that it is true. Indeed, any continuous positive function $g$ with compact support can be written as $g=|\sqrt{g}|^2$. Moreover, $C_c(X)$ is (?) the span of positive continuous functions with compact support. Hence, $F=C_c(X)$.
Is it true?
Best Answer
Well, we do know that $C_c(X)$ is dense in $L^1(X)$, so it's enough to check if $F$ is dense in $C_c(X)$.
Take $f\in C_c(X)$. This takes complex values in general, so we can write it as $f=f_1+if_2$ where $f_1,f_2$ are real-valued. It is not hard to see that $f_1,f_2\in C_c(X)$ as well. Now we can write $f_1=f_1^+-f_1^-$ and likewise $f_2=f_2^+-f_2^-$ and there are all positive functions with compact support. In other words, yes, $C_c(X)$ is indeed the span of positive functions with compact support. It suffices thus to approximate such a function through $F$.
Let $f\in C_c(X)_+$. Then we can write $f=|\sqrt{f}|^2$ so if we show that $\sqrt{f}\in C_c(X)$, we are done. But indeed, if $\sqrt{f}(x)\neq0$, then $f(x)\neq0$, so $\text{supp}(\sqrt{f})\subset\text{supp}(f)$ and closed subsets of compact sets are compact, thus $\sqrt{f}$ has compact support.