In my functional analysis class I was asked to check if these operators are unitary equivalent: multiplication operators in $L^2\left[-\pi,\pi\right]$ corresponding to functions $\sin t$ and $\dfrac t \pi$.
I know that unitary equivalent operators share the same norm and spectrum. And these operators have the same norm $\|A_1\|=\|A_2\|=1$ and spectrum $\sigma\left(A_1\right)=\sigma\left(A_2\right)=\left[-1,1\right]$. But it does not imply unitary equivalence. So what can I do next?
Thanks
I was asked to provide additional context. I'm familiar with the definition of cyclic vector, more over in my lecture notes there's
Theorem If $A=A^*$ has a cyclic vector than it is unitary equivalent to $A_t(x) (t)= tx(t) $ in $L^2(\sigma(A)) $ for some borel measure.
As a task was left the following statement: operators, multiplying by an argument are unitary equivalent iff corresponding measures are equivalent.
In my case theorem holds for both operators, but I have no idea how to check if measures equivalent.
The operator $A_2$ admits a cyclic vector $f(x)\equiv 1.$ This follows from the Weierstrass theorem as $A_2^n1=t^n/\pi^n,$ and the polynomials are dense in $L^2(-\pi,\pi).$ On the other hand the operator $A_1$ does not admit a cyclic vector. It is more convenient to consider the unitary equivalent operator $\tilde{A}_1$ acting by $$(\tilde{A}_1f)(t)=\cos t\, f(t)$$ The operator $\tilde{A}_1$ does not admit a cyclic vector. Indeed, for $g\neq 0$ let $$h(t)=\begin{cases} \overline{g(-t)} & -\pi <t<0\\ -\overline{g(-t)} & 0<t<\pi \end{cases}$$ Then $$\langle \tilde{A}_1^ng,h\rangle = \int\limits_{-\pi}^0(\cos t)^ng(t){g(-t)}\,dt-\int\limits_0^\pi(\cos t)^ng(t){g(-t)}\,dt=0$$ Thus $g$ is not a cyclic vector.
Summarizing, the operators $A_1$ and $A_2$ are not unitarily equivalent.
Remark We could as well stay with the operator $A_1,$ but the definition of $h$ would be less transparent. We would have to define $h$ separately on $(-\pi,-\pi/2)\cup (0,\pi/2)$ and on $(-\pi/2,0)\cup (\pi/2,\pi).$