Consider the following fragment from Takesaki's book "Theory of operator algebra I":

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In this proof, we encounter expressions like $$x = \int_{-\|x\|}^{\|x\|}\lambda de(\lambda).$$

Can someone explain me how to understand this integral or give me an 'accessible' reference where I can read more about these kinds of integrals? Is this related to the "Borel functional calculus"?

  • 1
    $\begingroup$ A very good reference on the matter is Schmüdgen's book "Unbounded Self-adjoint Operators on Hilbert Space" $\endgroup$ Dec 30, 2021 at 19:48

1 Answer 1


This is a result of the spectral theorem for self adjoint, bounded operators on a Hilbert space. The theorem which can be found in Reed and Simons for example provides a way of constructing an operator out of an integral over the spectrum of the operator with respect to a special kind of measure called a spectral measure.
In the case of self adjoint $x$ we know $\sigma(x)\subset \mathbb{R}$ and moreover, the spectral radius is equal to $\|x\|$.

The $de(\lambda)$ is the spectral measure defined on $\sigma(x)$ and it allows you to integrate continuous functions or even measurable functions with respect to this measure to define the analogue of the given function on the operator $x$. For example, if $f(\lambda)=e^\lambda$ defined on the spectrum of $x$, then the spectral theorem says that $$f(x)=\int_{\sigma(x)}f(\lambda)de(\lambda)$$ So you can give meaning to the exponential of an operator (in the case of bounded operators this machinery is not necessary since you can represent it as a power series in the operator norm). In the case where $f(\lambda)=\lambda$ i.e the $id_{\sigma(x)}$ then the theorem yields $$x=\int_{\sigma(x)}\lambda de(\lambda)$$ But remember if $x$ is self adjoint then the spectral radius is equal to $\|x\|$ and $\sigma(x)\subset \mathbb{R}$ we have $\sigma(x)\subset [-\|x\|,\|x\|]$ (if $\sigma(x)\subsetneq [-\|x\|,\|x\|]$ you can extend the integration to the endpoints by declaring that the spectral measures are exactly zero in the compliment) and hence $$x=\int_{-\|x\|}^{\|x\|}\lambda de(\lambda)$$

  • $\begingroup$ Thanks for your answer! I think $\sigma(x)\subseteq [-\|x\|,\|x\|]$ can be strict (think for example about matrices where the LHS is finite). $\endgroup$
    – Andromeda
    Dec 30, 2021 at 18:36
  • $\begingroup$ How would you interpret $e_n$ in Takesaki's proof? As the integral of the indicator function $\chi_{\{\lambda: |\lambda|\ge n^{-1}\}\cap \sigma(x)}$? $\endgroup$
    – Andromeda
    Dec 30, 2021 at 18:40
  • $\begingroup$ @Andromeda For your first comment I agree and I will fix the statement. For the second, yes you are correct; projections can be obtained by integrating against the characteristic function of any Borel measurable set. Remember, the measures $de(\lambda)$ are operator valued $\endgroup$ Dec 30, 2021 at 19:35
  • $\begingroup$ Thanks! One final question: Given a normal element $u\in B(H)$, is $\int f(\lambda)de(\lambda)$ just fancy notation for the Borel functional calculus $f(u)$? $\endgroup$
    – Andromeda
    Dec 30, 2021 at 19:46
  • $\begingroup$ @Andromeda I believe they are equal, no notation but now the spectrum is not confined to $\mathbb{R}$ but is some compact subset of $\mathbb{C}$ and the same principles apply. The road to proving the spectral theorem is difficult and takes some getting used to. If you have the time, Rudin's Functional Analysis is beautiful but it takes some work. $\endgroup$ Dec 30, 2021 at 19:49

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