In my applied math research group, we are studying and going over functional analysis results from papers and theses from our institution to generalize their results and apply them in our discrete dynamics in quantum chemistry and coding theory research. Right now, we are dealing with self-adjoint operators in the context of the spectral theorem's many forms. One form, the multiplication operator form, says

Let $A$ be a bounded, self-adjoint operator on a separable Hilbert space $H$. Then, there exist measures $\{\mu_n\}_{n=1}^{N}$ (where $N$ is a natural number or infinity) on $\sigma(A)$ and a unitary operator $$ U : H \rightarrow \bigoplus_{n=1}^N L^2(\mathbb{R},d\mu_n) $$ and we have $$ (UAU^{-1}\psi)_n(\lambda) = \lambda \psi_n(\lambda) $$ where we write an element $ \psi \in \bigoplus_{n=1}^N L^2(\mathbb{R},d\mu_n) $ is written as an N-tuple $(\psi_1(\lambda),\psi_2(\lambda),...,\psi_N(\lambda))$. If we do not insist on the function we multiply by to be $f(\lambda)=\lambda$, we have that $A$ is unitarily equivalent to the multiplication operator $ M_F$ on $L^2(M,d\mu) $ which multiplies by the function $F(\lambda)$. That is the background we are using.

Here is my problem. We consider the operator $A=L+R$ on $\ell^2(\mathbb{Z})$ which is the sum of the left and right shift operators on square-summable sequences indexed by all integers. I know, via simple Fourier series, that $A$ is unitarily equivalent to $M_{2\cos(x)}$ on $L^2([0,2\pi),dx/2\pi)$. Here is what I am missing and need. We take $(Bf)(x)=xf(x) \; \; \text{on} \; \; L^2([-2,2],dx)$. The old Russian thesis I am working on says that $A$ is unitarily equivalent to $ B \oplus B $ on $L^2([-2,2],dx)=H_1 \oplus H_2$, and it defers the proof of this to the appendix which, as luck would have it, seems to be missing (I checked all databases I know of it is not there, it is quite old so maybe not digitized and uploaded to any database). This result is really important to my research in dynamics in quantum chemistry and coding theory where $A$ shows up a lot, and having the decomposition of $L^2([-2,2],dx)$ as stated above into two disjoint invariant subspaces of $B$ and actually finding the unitary $U$ that takes $ A $ to $ B \oplus B $ (or vice versa) can really boost my research, but I am just now entering functional analysis as a user/researcher whose background is mainly in applied math. I was hoping that someone here can help me find $U$ such that $UAU^*=B \oplus B$. Unfortunately, I am not so proficient in functional analysis and spectral theory to do anything that is not mostly intuitive. I would also love to share more about the context of our research if people are curious. I thank all helpers.

  • 1
    $\begingroup$ There really isn't much functional analysis involved here, it's basically calculus: split $(0,2\pi)$ into two pieces $I_j$ in such a way that $2\cos x$ is injective on each, and then set up unitary maps $L^2(I_j)\to L^2(-2,2)$ by a change of variable, by making $2\cos x$ the new variable. $\endgroup$ Oct 16, 2020 at 14:37

1 Answer 1


If you think about it, then on $[0,2\pi)$, the function $2\cos(t)$ takes all the values in $[-2,2]$, with multiplicity $2$ (except for $\pm 2$ which have multiplicity $1$). So the claim seems very plausible. The exercise now is to pick the correct unitary, which is basically a "change of variables" problem.

The unitary you want (assuming my rusty Calculus is okay) is $$ U : L^2([0,2\pi), dt/2\pi) \rightarrow L^2([-2,2],dx) \oplus L^2([-2,2],dx); \quad \xi\mapsto (f,g), $$ where \begin{align*} f(2\cos(t)) &= \xi(t) / \sqrt{4\pi\sin(t)} \quad (0\leq t\leq \pi), \\ g(2\cos(t)) &= \xi(t) / \sqrt{-4\pi\sin(t)} \quad (\pi < t < 2\pi) \end{align*} Notice that $g(\pm 2)$ is not defined, but we are in an $L^2$ space, so it's okay to not define a function at some isolated points. Similarly, where $\sin(t)=0$ the functions are not defined.

Similarly let $U(\eta) = (h,k)$. Then \begin{align*} (U^*(B\oplus B)U\xi|\eta) = \int_{-2}^2 x f(x) \overline{h(x)} \ dx + \int_{-2}^2 x g(x) \overline{k(x)} \ dx \end{align*} Change variables to see that this equals \begin{align*} \int_{\pi}^0 2\cos(t) f(2\cos(t)) \overline{h(2\cos(t))} (-2)\sin(t) \ dt \\ + \int_\pi^{2\pi} 2\cos(t) g(2\cos(t)) \overline{k(2\cos(t))} (-2)\sin(t) \ dt. \end{align*} Now substitute in, and see why the strange $\sin$ parts occurred before, \begin{align*} \int_0^\pi 2\cos(t) \frac{\xi(t)}{\sqrt{4\pi\sin(t)}} \frac{\overline{\eta(t)}}{\sqrt{4\pi\sin(t)}} 2\sin(t) \ dt \\ + \int_\pi^{2\pi} 2\cos(t) \frac{\xi(t)}{\sqrt{-4\pi\sin(t)}} \frac{\overline{\eta(t)}}{\sqrt{-4\pi\sin(t)}} (-2)\sin(t) \ dt \end{align*} This cancels down to $$ \int_0^{2\pi} 2\cos(t) \xi(t) \overline{\eta(t)} \ \frac{dt}{2\pi}, $$ as we want.

Edit: Here I have written $(\cdot|\cdot)$ for the inner-product. (A sort of mix between maths and physics notation; but I often write papers which need to consider both bilinear, and sesquilinear, pairings, and it's nice to have a notational difference between these). $\xi,\eta$ are members of $L^2([0,2\pi), dt/2\pi)$ (so functions, or equivalence classes thereof).

A similar calculation shows that $U$ is a unitary. More accurately, repeat the calculation without the operator $B\oplus B$ to show that $U^*U=1$. To see that $UU^*=1$, equivalently, $U$ has dense range, one could argue that "this is clear"; or you could compute $U^*$ (use change of variable again) which has a similar form.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.