$F(L_2([0,1],\mu))$ is dense in $L_2([0,1],\mu)$, where $F$ is the multiplication operator

Question

Consider $A=L_2([0,1],\mu)$ where $\mu$ is the Lebesgue measure. Let $F:A\rightarrow A$ be linear given by $Ff(x)=xf(x)$ for $f\in A$ and $x\in[0,1]$.

I want to show that $F(A)$ (the range of $F$) is a proper dense subspace of $A$. I've managed to show that $F(A)$ is a proper subspace of $A$, but I'm having trouble showing it's dense.

Answer 1

A space $S$ of a Hilbert space $\mathcal H$ is dense if its orthogonal is reduced to $\{0\}$ (because $\left(S^\perp\right)^\perp$ is the closure of $S$).

Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $f\in A$, $$\int_{[0,1]}xf(x)g(x)\mathrm d\lambda(x)=0;$$ in particular, choosing $f\colon x\mapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.

Answer 2

If you are lazy (just like me), then you can use the identity $\ker F=(\operatorname{ran}F^\ast)^\perp$. Clearly $F=F^\ast$ and $\ker F=0$, so $(\mathrm{ran}F)^\perp=\{0\}$.

If you are slightly less lazy, note that $f\in C_c((0,1))$ implies $(x\mapsto f(x)/x)\in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.