# Functional Calculus of closed operators

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 :->

• 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 $(TT^*)^{1/2}$. – poupy May 15 '16 at 16:48
• the paper arxiv.org/abs/1309.0164 can be interest too – poupy May 15 '16 at 16:52
• For normal closed operators there is. – Robert Israel May 15 '16 at 19:55
• Try Kato's book Perturbation theory of linear operators or the classic 3 volume by Dunford and Schwartz Linear Operators – Liviu Nicolaescu May 15 '16 at 20:46

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.