I learned that there is a holomorphic functional calculus for closed operators: If $T$ is a closed operator on a Hilbert space, and $f$ is a function that is holomorphic on some open subset $\Omega$ of $\mathbb{C} \cup \{\infty\}$ that contains $\infty$ if $T$ is unbounded, then I can form $f(T)$.

Does there also exist a continuous or measurable functional calculus, for a non-self-adjoint closed operator?

If not, what are the problems?

P.S.: The problem is that if I understand the results correctly, $\Omega$ has to contain $\infty$ if $T$ is unbounded, so I can never form $f(T)$ for an entire function, unless $f$ is a polynomial entire function. I would, however, very much like to take $\cos(T)$, for example :->

  • 2
    $\begingroup$ it depends on what you need from the functional calculus. In general, one has problems for $f(x)=|x|$ as there are many candidates for $f(T)$, for example $(T^*T)^{1/2}$ ot $(TT^*)^{1/2}$. $\endgroup$
    – poupy
    May 15, 2016 at 16:48
  • $\begingroup$ the paper arxiv.org/abs/1309.0164 can be interest too $\endgroup$
    – poupy
    May 15, 2016 at 16:52
  • $\begingroup$ For normal closed operators there is. $\endgroup$ May 15, 2016 at 19:55
  • $\begingroup$ Try Kato's book Perturbation theory of linear operators or the classic 3 volume by Dunford and Schwartz Linear Operators $\endgroup$ May 15, 2016 at 20:46

1 Answer 1


If your operator $T$ is additionally sectorial, then many of your questions are addressed in the Monograph

Haase, Markus The functional calculus for sectorial operators. Operator Theory: Advances and Applications, 169. Birkhäuser Verlag, Basel, 2006. xiv+392 pp. ISBN: 978-3-7643-7697-0; 3-7643-7697-X

especially cosine functions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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