MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to make sense of some operators that come up on Buchholz and Summers' work on warped convolutions (two works on arxiv: 2008 and 2011).

There, they work on a Hilbert space $H$ and on the bounded operators algebra $B(H)$ using some operator valued integrals similar to

$\int_\mathbf{R} A(x)\,dE(x)\;$ and $\;\int_\mathbf{R} dE(x)\,A(x)$,

where $E$ is the spectral resolution of a self-adjoint operator and $A$ is a $B(H)$ valued (norm-continuous) function. I don't know how one defines that.

I don't even know how that one above is defined, since both the measure and the function are operator-valued kinda. What I read about is that you can integrate vector valued functions with respect to a scalar valued measure (Bochner integral or Pettis integral), or scalar valued functions with respect to a spectral resolution (projection valued measure - spectral theorem).

If anyone knows how to define it and/or standard references for it, I would be thankful!

Some further remarks. If this helps, the authors also states that if $E$ is the spectral resolution of the self-adjoint operator $P$, then $P\,dE(p) = p\,dE(p)$, and if $B$ is a compact subset of $\mathbf{R}$ and $F$ is a finite-rank projection in $H$, then by spectral calculus $\int_B A(x)\,F\,dE(x)\;$ and $\;\int_B dE(x)\,F\,A(x)$ are well defined. I can make sense of the second type of integrals, since those will be of finite rank, but not the first type.

Thank you.

share|cite|improve this question
I'm not an operator theory expert, so the following may be way off base, but it seems to me that you could just copy the definition of the Riemann-Stieltjes integral in the scalar-valued world, replace the all the scalars there (values of the integrand and differences between values of the cumulative measure function) by operators, and replace (in the limiting process, as the partitions get finer) the absolute-value distance between scalars with the norm distance between operators. (I trust that, if this is wrong, an expert will soon come along and tell me, and then I'll delete this.) – Andreas Blass Jan 6 '13 at 21:16
Also on stackexchange: – Martin Jan 6 '13 at 22:08
@Vahid: your comment seems to be a bit lower level than the question itself... – András Bátkai Jan 7 '13 at 0:02
@Andreas (not András): your definition is reasonable, but as noted in my answer, convergence is dangerous. That is because the "difference between values of the cumulative measure function" is a projection, so its norm can be 1 no matter how fine the partition gets. If you're integrating a scalar function, that is not so bad because for a given partition these projections are orthogonal to each other, but if you're integrating an operator function they can get pushed onto each other. – Nik Weaver Jan 7 '13 at 19:25
thank you all for your comments! andreas and nik, I think the Riemann sums can be used with the strong operator topology given that the spectral measure $E$ has compact support and the function $A$ has uniform continuity, with maybe some more technical regularity issues. I'm trying to check this, but then again, it seems that it will take quite some time... – Yul Otani Jan 10 '13 at 19:06
up vote 4 down vote accepted

Well, I would make sense of expressions like this by inserting $|e_n\rangle\langle e_n|$ between the operator and the measure and then summing over $n$.

Say we want to give a meaning to $A = \int A(x)d\mu$. If we know how to integrate scalar-valued functions against a spectral measure then we can define $$\langle v|A = \sum_n \int \langle v|A(x)|e_n\rangle\langle e_n|d\mu = \sum_n \langle e_n|\int \langle v|A(x)|e_n\rangle d\mu$$ for any $v \in H$, and if we know how to integrate vector-valued functions against scalar measures then we can define $$A|w\rangle = \sum_n \int A(x)|e_n\rangle\langle e_n|d\mu|w\rangle$$ for any $w \in H$. So take your pick, say what the operator is by saying either what happens when you apply it to a vector or what happens when you apply a covector to it.

Do these sums converge? No, not in general, even assuming $A(x)$ is a bounded operator-valued function and $\mu$ is the spectral measure coming from a compact self-adjoint operator. For example, consider the spectral measure coming from the operator of multiplication by $1/n$ on $l^2({\bf N})$. This is a spectral measure defined on $X = \{1/n: n \in {\bf N}\}$ with $\mu(\{1/n\}) = |e_n\rangle\langle e_n|$.` Define an operator-valued function on $X$ by setting $A(1/n) = |e_1\rangle\langle e_n|$. Then the integral $\int A(x)d\mu$ becomes a sum $\sum_n |e_1\rangle\langle e_n|e_n\rangle\langle e_n| = \sum_n |e_1\rangle\langle e_n|$ which does not converge.

Edit: I just looked at their 2011 paper and I found that they have some discussion about how to make their integrals rigorous in Section 2.2. So I think the full answer is that integrals of this form don't always make sense, but there is a heuristic and one can prove convergence in some special cases.

share|cite|improve this answer
@Nik, sorry for taking so much time, but thank you for the answer. I'll make remarks here too, as I have been glancing at that paper for some time now. They seem to use the ansatz $\int A(x)dE(x)=\int A(x)dE(y)\delta(x-y)$, and taking $e^{izy}dE(y) = U(z)$ (the generated unitary group), start working the definition for sufficiently regular functions, with some oscillatory integral techniques, mostly with strong operator topology. Rieffel calls those "pseudodifferential operator" techniques. Your definition and example does make sense, of course, but I still have to check how they fit together. – Yul Otani Jan 10 '13 at 19:14
I see. Yes, I remember that Rieffel built up some technology for doing oscillatory integrals in his monograph Deformation Quantization for Actions of ${\bf R}^d$. – Nik Weaver Jan 10 '13 at 19:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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