0
$\begingroup$

Once I met a notation of "spectral decomposition function" (for a self-adjoint operator). No definition was given. Could someone give me a clue what can that be, cause I can't find this exact phrase anywhere. Thank you!

Improve this question
$\endgroup$
3
  • $\begingroup$ Context would help here. Where was it? $\endgroup$
    – András Bátkai
    Commented May 26, 2013 at 17:49
  • $\begingroup$ University professor's lectures on operator algebras. $\endgroup$
    – haijo
    Commented May 26, 2013 at 18:01
  • $\begingroup$ As you see from Robert Israels guesses, it is still unclear. Why don't you ask the professor? $\endgroup$
    – András Bátkai
    Commented May 27, 2013 at 7:24

1 Answer 1

Reset to default
2
$\begingroup$

My guess is that this is referring to the resolution of the identity corresponding to the operator, which is sometimes called the "spectral decomposition" (e.g. in Rudin, Functional Analysis) or "spectral function" (e.g. in Akhiezer and Glazman, Theory of Linear Operators in Hilbert Space).

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .