Unitarily equivalent multiplication operators

Question

Let $A$ be the operator given by $Ax(t)=\sin(t)\,x(t)$ in $L_2[0,2\pi]$, and $B$ the operator given by $Bx(t)=\sin(t)\,x(t)$ in $L_2[-2\pi,2\pi]$.

Are these operators unitarily equivalent?

Answer 1

Edit: when I answered I had forgotten about multiplicity theory, so I'm updating the answer. I have left my old answer at the end.

First, it is convenient to move both operators to a common space. The unitaries $$ V:L^2[0,2\pi]\to L^2[0,1],\qquad (Vf)(t)=\sqrt{2\pi}\,f(2\pi t) $$ and $$ W:L^2[-2\pi,2\pi]\to L^2[0,1],\qquad (Wf)(t)=2\sqrt\pi\,f(4\pi t-2\pi) $$ satisfy $$ V^*M_{\sin t} V=M_{\sin2\pi t},\qquad W^*M_{\sin t}W=M_{\sin4\pi t}. $$ So the question reduces to whether $M_{\sin2\pi t}$ and $M_{\sin4\pi t}$ are unitarily equivalent in $B({L^2[0,1]})$. Now we want to use multiplicity theory (see, for instance, section IX.10 in Conway's A Course in Functional Analysis).

If we call our two functions $s_1$ and $s_2$, it is enough if we show that $$ M_{\frac12\,(s_1+1)}=\frac12\,(M_{s_1}+I)\ \text{ is unitarily equivalent to } \frac12\,(M_{s_2}+I)=M_{\frac12\,(s_2+1)}. $$ The gain we obtain from this is that we may assume without loss of generality that $$s_1([0,1])=s_2([0,1])=[0,1].$$

Write $m$ for the Lebesgue measure on $[0,1]$. It is not hard to show, using multiplicity theory, that on $L^2[0,1]$ the multiplication operator $M_g$ is unitarily equivalent to the canonical multiplication operator $M_z$ (i.e., $(M_zf)(t)=tf(t)$) if and only if $m\circ g^{-1}$ is mutually absolutely continuous with $m$. In other words, $g^{-1}(E)$ is a nullset if and only if $E$ is a nullset. A sufficient condition for $g$ to satisfy this is that $g$ is surjective, differentiable, and $\{g'=0\}$ is a nullset. For any two such functions we have $M_{g_1}\simeq M_{g_2}$. In particular, $M_{s_1}\simeq M_{s_2}$, and so the two multiplication operators are unitarily equivalent.

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(old answer)

I cannot answer about unitary equivalence, but it looks like I can about approximate unitary equivalence. In this case, the answer is that $A$ and $B$ are indeed approximately unitarily equivalent. More generally:

Theorem. Let $f\in L^\infty(X,\mu)$ and $g\in L^\infty(Y,\nu)$, where $L^2(X,\mu)$ and $L^2(Y,\nu)$ are separable. The following statements are equivalent:

• $M_f$ and $M_g$ are approximately unitarily equivalent;

• $\overline{\operatorname{ran}f}=\overline{\operatorname{ran}g}$ and, for any closed set $I\subset\mathbb C$, $$\tag1\operatorname{Mult}(\{f^{-1}(I)\})=\operatorname{Mult}(\{g^{-1}(I)\}),$$ where for any measurable set $E$ we define $\operatorname{Mult}(E)=0$ if $E$ is a nullset, and $$\operatorname{Mult}(E)=\max\{m:\ \exists \{E_j\}_{j=1}^m,\ \text{ partition, }\bigcup_jE_j=E\}$$ if $E$ has positive measure. Here partition means "pairwise disjoint, each with positive measure".

For the operators $A$ and $B$ in the question, the conditions are trivially satisfied, so $A$ and $B$ are approximately unitarily equivalent. Any function with range $[-1,1]$ and no constant sections will also do (as the multiplicities will be zero or infinite): for instance $A$ is approximately unitarily equivalent to $M_h$, where $h\in L^\infty[0,1]$ is $h(t)=2t-1$.

Proof. If $M_f$ and $M_g$ are approximately unitarily equivalent, then it is well known (see for instance Theorem II.4.4 in Davidson's C$^*$-Algebras by Example) that $M_f$ and $M_g$ have the same essential spectrum, and that the eigenvalues not in the essential spectrum have the same (finite) multiplicity. The spectrum of $M_f$ is the closure of the range of $f$; eigenvalues $\lambda $ with finite multiplicity $m$ correspond to sets with $m$ atoms.

So, if the interval $I$ has nontrivial intersection with $ \sigma_{\rm ess}(M_f)=\sigma_{\rm ess}(M_g)$, the spectral projection $P_{M_f}(I)$ corresponding to the interval $I$ is infinite; using my very first answer in MSE, this means that the projection of multiplication by $1_{f^{-1}(I)}$ is infinite, so $\operatorname{Mult}(f^{-1}(I))=\infty=\operatorname{Mult}(g^{-1}(I))$. If $I$ has no intersection with the essential spectra, then each of its elements is either not in the spectrum, or an eigenvalue with finite multiplicity; there can only be finitely many of these, because otherwise we would have accumulation points and these would be in the essential spectrum. So $f^{-1}(I)$ is a finite union of sets each with finite multiplicity, and then $\operatorname{Mult}(f^{-1})$ is the sums of these multiplicities; same with $g$.

Now for the converse, suppose that $\overline{\operatorname{ran}f}=\overline{\operatorname{ran}g}$ and $\operatorname{Mult}(\{f^{-1}(I)\})=\operatorname{Mult}(\{g^{-1}(I)\})$ for all closed $I\in\mathbb C$. Fix $n$. Choose $\lambda_1,\ldots,\lambda_m\in\overline{\operatorname{ran}f}$ such that $$ \left\|f-\sum_j\lambda_j\,1_{E_j}\right\|<\frac1n,\ \ \ \ \left\|g-\sum_j\lambda_j\,1_{F_j}\right\|<\frac1n, $$ where $E_j=\{\lambda_{j-1}<f\leq\lambda_j\}$, $F_j=\{\lambda_{j-1}<g\leq\lambda_j\}$. Write $f_n=\sum_j\lambda_j\,1_{E_j}$, $g_n=\sum_j\lambda_j\,1_{F_j}$. By hypothesis, each pair $E_j,F_j$ have the same multiplicities; this allows us (using separability to guarantee that infinite sets have the same cardinality) to construct orthonormal bases $\{e_{kj}\}_k$ and $\{h_{kj}\}_k$ of $L^2(E_j)$ and $L^2(F_j)$ respectively, with the same cardinalities. We then define a unitary $V_n:L^2(X)\to L^2(Y)$ by $V_n:e_{kj}\longmapsto h_{kj}$. Now \begin{align} M_{g_n}V_n\sum_{k,j}\alpha_{kj}\,e_{kj} &=g_n\,\sum_{k,j}\alpha_{kj}\,h_{kj} =\sum_{k,j}\alpha_{kj}\,\lambda_j\,h_{kj}\\[0.3cm] &=V_n\,\sum_{k,j}\alpha_{kj}\,\lambda_j\,e_{kj} =V_nM_{f_n}\,\sum_{k,j}\alpha_{kj}\,e_{kj}. \end{align} So $M_{g_n}V_n=V_nM_{f_n}$ and, as $V_n$ is a unitary, $M_{g_n}=V_nM_{f_n}V_n^*$. Finally, \begin{align} \|M_g-V_nM_fV_n^*\|&\leq\|M_g-M_{g_n}\|+\|M_{g_n}-V_nM_{f_n}V_n^*\|+\|V_nM_{f_n}V_n^*-V_nM_fV_n^*\|\\[0.3cm] &=\|g-g_n\|+\|f_n-f\|\leq\frac2n. \end{align} Thus $M_g=\lim_n V_nM_fV_n^*$.