# Possible significant error in proof of the spectral theorem, Brian C Hall, Quantum Theory for Mathematicians

Note/Edit: Read the below paragraphs for context. I think I found a counterexample, though I don't have the energy to work through it now. Suppose $$\mathcal{H} = L^2([0,1],m) \oplus L^2([0,1],m)$$ where $$m$$ denotes Lebesgue measure. Consider the "position" operator $$X$$ given by $$Xf(x) = xf(x)$$ and consider the operator $$A$$ given by $$Af = 1_{[1/2,1]}f$$, each operator is a bounded self-adjoint operator on $$L^2([0,1],m)$$. Then define the bounded self-adjoint operator $$X \oplus A$$ on $$\mathcal{H}$$ in the obvious way: $$X \oplus A (f,g) = (Xf,Ag)$$. You can look at Hall's construction of the $$(W_j, \psi_j)$$, but you can convince yourself that we could take $$W_1 = L^2([0,1],m) \times \{0\}$$ and have $$\psi_1 = (1,0)$$ (standard techniques using density of polynomials will then convince you that the closed span of the $$(X \oplus A)^n \psi_1$$ is $$W_1$$). Then note that $$1 \in \sigma(X \oplus A|_{W_1})$$, $$1_{\{1\}}(X \oplus A) \neq 0$$ (as $$(0, 1_{[1/2,1]} g)$$ is an eigenvector of $$X \oplus A$$ with eigenvalue $$1$$ for all $$g \in L^2([1/2,1],m)$$), hence $$\mu(\{1\}) \neq 0$$ but $$\mu_{\psi_1}(\{1\}) = 0$$, which contradicts surjectivity as I noted in the 3rd to last paragraph. Let me know if you think this works or doesn't. If if does then I guess there's a serious problem with the proof in the book.

So I am in the middle of reading through B.C. Hall's proof of the spectral theorem for bounded self-adjoint operators on separable Hilbert spaces, so this question is a little hard to state given the amount of context.

Let $$A$$ a self-adjoint bounded operator on a separable Hilbert space $$\mathcal{H}$$. Suppose $$\{W_j, \psi_j\}$$ is a (possibly finite) sequence of pairwise orthogonal subspaces of $$\mathcal{H}$$ s.t. they are invariant under $$A$$, for fixed $$j$$ the span of $$A^n \psi_j$$ is dense in $$W_j$$, and $$\mathcal{H} = \bigoplus_j W_j$$ (the orthogonal direct sum). Let $$A_j = A|_{W_j}$$. Let $$\mu^A$$, $$\mu^{A_j}$$ denote the projection valued measure on the Borel $$\sigma$$-algebra on $$\sigma(A), \sigma(A_j)$$ resp. (where $$\sigma(B)$$ is the spectrum of $$B$$). Let $$\mu_{\psi_j}$$ be the positive measure given by $$\mu_{\psi_j}(E) = (\psi_j, \mu^{A_j}(E) \psi_j)$$. Let $$\mu$$ be a positive measure on $$\mathcal{B}_{\sigma(A)}$$ s.t. $$\mu(E) =0 \iff \mu^A(E) = 0$$. Then note that $$\mu^{A_j}(E) = 1_E(A_j) = 1_E(A|_{W_j}) = 1_E(A)|_{W_j}$$ (where we are referring to the functional calculus induced by the respective operators, and the last equality follows from a lemma). Thus $$\mu(E) = 0$$ implies $$0=\mu^A(E)= 1_E(A)$$, hence $$\mu^{A_j}(E) = 1_E(A_j) = 1_E(A)|_{W_j} = 0$$ which implies $$\mu_{\psi_j}(E) = 0$$. So $$\mu_{\psi_j}$$ is absolutely continuous w/r/t $$\mu$$ on $$\mathcal{B}_{\sigma(A_j)}$$. Let $$\rho_j$$ denote its Radon-Nikodym derivative.

Finally we get to the part I'm stuck at. The author now claims "one can easily see that" the map $$f \mapsto \rho_j^{1/2}f$$ is unitary from $$L^2(\sigma(A_j), \mu_{\psi_j}) \to L^2(\sigma(A_j), \mu)$$. It is easy to verify it is norm preserving but why is it surjective?

Note that we only showed that $$\mu_{\psi_j}$$ was absolutely continuous w/r/t $$\mu$$. I claim that we need the converse to hold. Suppose we had the converse. Then $$\rho_j$$ is a.e. (both $$\mu$$ and $$\mu_{\psi_j}$$ since these agree on null sets) to a function that is everywhere nonzero, so we can WLOG suppose that $$\rho_j$$ is everywhere nonzero. Then we note that $$\int_{\sigma(A_j)} |f \rho_j^{-1/2}|^2 d\mu_{\psi_j} = \int_{\sigma(A_j)} |f|^2 \rho_j^{-1} \rho_j d\mu = \int_{\sigma(A_j)} |f|^2 d\mu$$. So if $$f \in L^2(\sigma(A_j), \mu)$$, then $$f\rho_j^{-1/2} \in L^2(\sigma(A_j), \mu_{\psi_j})$$. So the map is onto. On the other hand if we have that the converse doesn't hold, by considering a set that is $$\mu_{\psi_j}$$ null but not $$\mu$$ null ($$\mu$$ can be taken to be a finite measure, we can WLOG suppose this set has finite measure), and considering the indicator function on this set, we can see that no function in $$L^2(\sigma(A_j), \mu_{\psi_j})$$ will map onto it.

So why does this converse hold (to me it doesn't seem like it actually does so it feels like there is a near fatal error in this proof)? I tried constructing examples, but they got confusing fast.

Sorry for the wall of text, the question is really deep into a long proof but I'm totally stuck and any help would be greatly appreciated.

You don't need any converse, if $$\rho\ne0$$ a.e. $$\mu_\psi$$. Let me drop the $$j$$ to type less. If $$f\in L^2(\sigma(A),\mu)$$, all you need for surjectivity is that $$\rho^{-1/2}f\in L^2(\sigma(A),\mu_\psi)$$. And that works: you have that, for any measurable nonnegative $$g$$, $$\int_E g\,d\mu_\psi=\int_E g\,\rho\,d\mu.$$ Thus $$\int_{\sigma(A)}|\rho^{-1/2}\,f|^2\,d\mu_\psi=\int_{\sigma(A)}\rho^{-1}\,|f|^2\,\rho\,d\mu=\int_{\sigma(A)}|f|^2\,d\mu<\infty.$$ So $$U$$ is surjective: $$f=U(\rho^{-1/2}f)$$.

More details about Hall's proof: As written, the proof doesn't work. But it's more about the implementation than the idea. As written, the proof seems to work only when $$H$$ is separable, but that's an assumption over the whole book, so not an issue.

In Lemma 8.12, one should be more careful about the choice of $$\mu$$. Namely, instead of taking $$\mu$$ as described in its proof, one can take $$\tag1 \mu=\sum_k\tfrac1{k^2}\,\mu_{\psi_k}.$$ Here, for the nontrivial part of the proof, if $$\mu_{\psi_k}(E)=0$$ fr all $$k$$, we have $$\mu^A(E)\psi_k=0$$, and then for any $$f$$ $$0=f\mu^A(E)\psi_k=\mu^A(E)\,f\psi_k.$$As $$\psi_k$$ is cyclic, we get that $$\mu^A(E)=0$$ each $$W_k$$, and so $$\mu^A(E)=0$$.

The other important observation is this: a point $$\lambda\in \sigma(A)$$ is an eigenvalue for $$A$$ if and only if $$\mu(\{\lambda\})>0$$. That is, the eigenvalues of $$A$$ are precisely the atoms of $$\mu$$. Indeed, if $$\lambda$$ is an eigenvalue for $$A$$, then there exists $$j$$ such that $$\lambda$$ is an eigenvalue for $$j$$. As $$A_j$$ goes to the multiplication operator in $$L^2(\sigma(A_j),\mu_{\psi_j})$$, the eigenvalue equation says that there exists $$g$$ with $$\|g\|_2=1$$ $$tg(t)=\lambda g(t),\ \ \ \ t\in \sigma(A_j)\ \text{ a.e.}$$ This forces $$g(t)=0$$ a.e. on $$\sigma(A_j)\setminus\{\lambda\}$$. Thus $$1=\int_{\{\lambda\}} |g(t)|^2\,d\mu_{\psi_k}=|g(\lambda)|^2\,\mu_{\psi_k}(\{\lambda\})$$ and so $$\{\lambda\}$$ is an atom for $$\mu_{\psi_k}$$.

Now, whenever there is an eigenvalue $$\lambda\in\sigma(A)$$, we can fix an orthonormal basis $$\{\phi_r\}$$ of eigevectors for it, and we can decompose the eigenspace as $$\bigoplus_r \mathbb C\,\phi_r$$, each invariant for $$A$$ with cyclic vector. What this means is that in the original decomposition in subspaces $$W_j$$ we may have all eigenvalues accounted for. In other words, each $$\mu_{\psi_j}$$ will be either a Dirac delta or a diffuse measure. So, after renaming and changing coefficients, we may write $$(1)$$ as $$\tag2 \mu=\sum_k\tfrac1{k^2}\delta_{\lambda_k} + \sum_k\tfrac1{k^2}\,\mu_{\psi_k},$$ where each of the sums could be finite, and the $$\mu_{\psi_k}$$ are all diffuse.

Now suppose that $$A_j=\{\lambda_k\}$$. Then the Radon-Nikodym derivative is $$c\,1_{\lambda_k}$$ for an appropriate $$c>0$$. Indeed, $$\int_{\sigma(A_j)}f\,1_{\lambda_k}\,d\mu=f(\lambda_k)\,\mu(\{\lambda_k\})=f(\lambda_k)\,\sum_{j:\ \lambda_j=\lambda_k}\mu_{\psi_j}(\{\lambda_k\})=f(\lambda_k)\,\sum_{j:\ \lambda_j=\lambda_k}\tfrac1{j^2}.$$ In particular you get that $$\mu_{\psi_k}$$ and $$\mu$$ are mutually absolutely continuous on $$A_j$$.

When $$A_j$$ does not contain any eigenvalue, I know how to salvage the proof, but not with elementary arguments. I'll write them anyway. The result, stated in our context, is that if $$\mu_{\psi_k}$$ is diffuse, then $$L^2(\sigma(A_j),\mu_{\psi_k})$$ is isomorphic to $$L^2([0,1],m)$$ via a unitary that preserves the multiplication operator. Because both $$\mu$$ and $$\mu_{\psi_k}$$ are diffuse on $$A_j$$, we can map from one to the other via $$L^2([0,1],m)$$. The proof I know of this is about masas in von Neumann algebras. Namely, if $$\mathcal A\subset B(H)$$ is a diffuse masa, then there exists a unitary $$U:H\to L^2[0,1]$$ such that $$U\mathcal A U^*=L^\infty[0,1]$$ acting as multiplication operators (usually this result is quoted as "well-known" in von Neumann algebra literature, so written proofs do not abound; the one I know is Lemma 2.3.6 in Sinclair-Smith's Finite von Neumann Algebras and Masas). At this stage, I'm not sure if all the above can be done without the Spectral Theorem!

• You are right. But Hall does claim that $\rho\ne0$ a.e. – Martin Argerami Apr 16 '20 at 19:09
• I have no much time now, but I'll try to check the proof later. – Martin Argerami Apr 16 '20 at 19:15
• But if you are choosing your measures, it is not clear to me that you can produce those measures withing Hall's proof (which I only saw superficially, so I cannot be sure). – Martin Argerami Apr 16 '20 at 19:16
• Ok, so I wrote some ideas. As stated in the book, and unless I'm missing something, I think the problem you found is essential. – Martin Argerami Apr 17 '20 at 1:16
• Yes, and yes.  – Martin Argerami Apr 17 '20 at 1:41