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Let $A$ be a bounded normal operator in a Hilbert space. Then we know there exists a continuous functional calculus defining $f(A)$ for $f\in C(\sigma(A))$ in a reasonable way. But there is also the Riesz functional calculus for functions $f$ that are holomorphic in a neighborhood of $\sigma(A)$. We have $$ f(A) = \frac 1 {2\pi i}\int_C f(z)(z-A)^{-1}\,dz, $$ where $C$ is, e.g., a circle around the spectrum of $A$.

My question: If $f$ is now only continuous on $\sigma(A)$, it makes sense to define the operator $$ T = \frac 1 {2\pi i}\int_C f(z)(z-A)^{-1}\,dz. $$ But does $T$ coincide with $f(A)$, defined through the continuous functional calculus? More specific: Do we have (in the case $f(z) = \overline z$) $$ A^* = \frac 1 {2\pi i}\int_C \overline z(z-A)^{-1}\,dz\,? $$ Thanks in advance for answers.

Friedrich Philipp
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  • $f(z) = \overline{z}$ ? which is not holomorphic at all ? and you wrote $f$ holomorphic in a neighborhood of $\sigma(A)$ you meant holomorphic on $U$ a simply connected open containing $\sigma(A)$ ? – reuns Apr 29 '16 at 02:32
  • no apparently no need for $U$ to be connected as we allow $C$ to be any "Jordan curve" https://en.wikipedia.org/wiki/Holomorphic_functional_calculus – reuns Apr 29 '16 at 02:39
  • and for expressing $T^$ you should express your contour integral explicitely : parametrize $C$ by $\gamma(t), t \in (a,b)$, change of variable $z = \gamma(t)$ , $dz = \gamma'(t) dt$ hence $T = \int_C f(z) (z-A)^{-1} dz = \int_a^b f(\gamma(t)) (\gamma(t)-A)^{-1} \gamma'(t) dt$ and $T^ = \int_a^b \overline{f(\gamma(t))} (\overline{\gamma(t)}-A^)^{-1} \overline{\gamma'(t)} dt = \int_a^b g(\overline{\gamma(t)}) (\overline{\gamma(t)}-A^)^{-1} \overline{\gamma'(t)} dt = \int_{\overline{C}} g(z) (z-A^*)^{-1} dz$ with $g(z) = \overline{f(\overline{z})}$ which is holomorphic on $\overline{C}$ – reuns Apr 29 '16 at 02:44
  • hence with $f(z) = z$ you get $g(z) = f(z)$ and $T^* = \int_{\overline{C}} z (z-A^*)^{-1} dz$ – reuns Apr 29 '16 at 02:47
  • But I have $f(z) = \overline z$ and thus $g(z) = \overline z$. Hence $T^* = -\frac 1 {2\pi i}\int_{\overline C}\overline z(z-A^*)^{-1},dz$. The question is: Is this $A$? – Friedrich Philipp Apr 29 '16 at 03:07
  • so you read nothing of what I wrote ? do you understand that $f(z) = \overline{z}$ is not holomorphic and hence that the holomorphic functional calculus doesn't apply ? – reuns Apr 29 '16 at 03:32
  • @user1952009 Of course it does not apply. However, I asked whether $\frac 1 {2\pi i}\int_C \overline z(z-A)^{-1},dz = A^*$ if $A$ is normal. Do you have an answer? I have the impression you don't. – Friedrich Philipp Apr 29 '16 at 03:56
  • you don't understand ... if $h$ is not holomorphic then $\int_C h(z) dz$ depends on the contour ... (this is why we never use this notation for non holomorphic functions) and if you mean $\int_a^b \overline{\gamma(t)} (\gamma(t)-A)^{-1} \gamma'(t) dt$ then for particular cases of $\gamma$ (for example in the case of a circular contour) you can reduce it to an holomorphic contour integral and compute it, but it has no reason to simplify to $A^*$... so the point is that it is useless, you have no reason to be interested in those kind of things. – reuns Apr 29 '16 at 04:37
  • and first of all, you have to prove the equivalent of the residue theorem for the resolvent formalism i.e. prove what happens when computing $\int_C (z-A)^{-1} dz$ with $C$ enclosing only one eigenvalue of $A$ https://en.wikipedia.org/wiki/Resolvent_formalism – reuns Apr 29 '16 at 04:39
  • Ok, I see your point. The thing depends highly on the contour. Thanks for pointing me to this. – Friedrich Philipp Apr 29 '16 at 05:25

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