Let $T$ be the operator $Tf = xf$ on $L^2(-1,1)$. I am trying to figure out if any of $x, \mathrm{sgn}\ x$, or $x\mathrm{sgn}\ x$ is a cyclic vector.
I believe that $x$ is not, since the span of $\{x, x^2, x^3, \dots\}$ is the set of all polynomials which vanish at zero. I think we can invoke Stone-Weirestrass and say that since it doesn't separate points it isn't dense in $L^2(-1,1)$, however SW requires that it contains a non-zero constant function, which it does not. So I am a bit stuck on this. Similarly, I think $x$ sgn $x$ is not cyclic since the span doesn't separate points.
I think that $f =$ sgn $x$ is cyclic since the span of $\{A^nf\}_n$ separates points.
The issue here is that none of these sets contain a non-zero constant function
The function $g$ is a cyclic vector of $T$ iff $$\int\limits_{-1}^1x^{n}g(x)f(x)\,dx=0,\ n\ge 1\implies f=0$$ Equivalently $$\int\limits_{-1}^1x^{n}\,xg(x)f(x)\,dx=0,\ n\ge 0\implies f=0\quad (*)$$ Assume that all the integrals in $(*)$ vanish. Since the polynomials are dense in $L^2(-1,1)$ we get $xg(x)f(x)=0\,{\rm a.e.}$ In all three cases $xg(x)\neq 0$ for $x\neq 0.$ Therefore $f(x)=0\,{\rm a.e.}$ Hence all three functions are cyclic vectors.