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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 :->

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    $\begingroup$ 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
    Commented May 15, 2016 at 16:48
  • $\begingroup$ the paper arxiv.org/abs/1309.0164 can be interest too $\endgroup$
    – poupy
    Commented May 15, 2016 at 16:52
  • $\begingroup$ For normal closed operators there is. $\endgroup$
    – Robert Israel
    Commented 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 $\endgroup$
    – Liviu Nicolaescu
    Commented May 15, 2016 at 20:46

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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.

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