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.


1 Answer 1


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!

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    $\begingroup$ You are right. But Hall does claim that $\rho\ne0$ a.e. $\endgroup$ Apr 16, 2020 at 19:09
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    $\begingroup$ I have no much time now, but I'll try to check the proof later. $\endgroup$ Apr 16, 2020 at 19:15
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    $\begingroup$ 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). $\endgroup$ Apr 16, 2020 at 19:16
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    $\begingroup$ 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. $\endgroup$ Apr 17, 2020 at 1:16
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    $\begingroup$ Yes, and yes. $ $ $\endgroup$ Apr 17, 2020 at 1:41

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