# Exponential of unbounded normal operator

One can define the exponential of bounded operators on some Banach space in different ways. For example, the "Continuous functional calculus" theorem is an important result to define the exponential of an bounded normal operators on Hilbert spaces.

For unbounded operators, almost all cases that I found is about the definition of the exponential of self-adjoint operators $A^* = A$ or operators satisfying $A^* = -A.$

In some references, ex. Kato. Perturbation theory of linear operators P. 281, the so-called Dunfer-Taylor integral $$F(A) = \frac{1}{2i\pi }\int_{\Gamma} f(\xi)(A-\xi)^{-1}d\xi,$$ is used to define the square root of some unbounded operators. Here $\Gamma$ is an appropriate path depending on the geometry of the spectrum of $A.$

My question is: Can we define the exponential of unbounded operators in the following settings?: Let $H$ be a complex Hilbert space and and $N : D(N) \subset H \to H$ be unbounded normal operator with Domain $D(N)$ (might be dense in $H$). and spectrum $$\sigma(N) \subset \{\lambda \in \mathbb{C}| {\rm Re}(\lambda) \leq 0\}.$$

If we think to use the above integral formula by integrating on a path that runs from $+\infty$ to $+\infty$ in the resolvent set of $N$, we might define the operator $$E(N) := \frac{1}{2i\pi }\int_{\Gamma} \exp(-\xi)(N-\xi)^{-1}d\xi,$$ The above integral is convergent which means that $E(N)$ is a bounded operator and in some sense should represent the operator $\exp(-N)$ which is a priori unbounded ! Then the above formula fails to define $\exp(N).$

I'd be grateful for any useful information.

• Do you have a specific question? – amsmath Nov 1 '17 at 12:52
• the question is in the Bold-paragraph. – A. PI Nov 1 '17 at 12:57
• If $N$ is normal, you can define $f(N)$ in a meaningful way for any measurable function, namely $f(N) = \int f\,dE$, where $E$ is the spectral measure of $N$. You have $\operatorname{dom}f(N) = \{x\in H : f\in L^2(\mu_x)\}$, where $\mu_x$ is the measure $\|E(\cdot)x\|^2$. – amsmath Nov 1 '17 at 13:01
• the properties of the exponential here are still satisfied ? is $e^N$ normal ? $e^N$ commute with $N$? $\sigma(e^N) = \{e^\lambda, \lambda \in \sigma(N)\}?$ – A. PI Nov 1 '17 at 13:07
• Don't understand the first question. Otherwise: Yes, yes, and no. I think you just have $\sigma(e^N)\subset\{e^\lambda : \lambda\in\sigma(N)\}$, in general. – amsmath Nov 1 '17 at 17:37

## 1 Answer

The spectral theorem for a (densely-defined) normal operator $N$ on a complex Hilbert space $H$ allows you to write $$Nx = \int_{\sigma(N)} \lambda dE(\lambda)x,$$ where the domain $\mathcal{D}(N)$ is characterized as the set of all $x\in H$ for which $$\int_{\sigma(N)}\lambda^2 d\|E(\lambda)x\|^2 < \infty.$$ This precisely characterization of the domain is what makes everything work out so nice.

If $\sigma(N)\subseteq\{ z : \Re z \le 0 \}$, then the operator $e^{N}$ is a bounded normal operator defined by $$e^{N}x = \int_{\sigma(N)}e^{\lambda}dE(\lambda)x.$$ And $e^{-N}$ is defined on the range of $e^{N}$ by $e^{-N}y=\int_{\sigma(N)}e^{-\lambda}dE(\lambda)y$, which is automatically dense in $H$. These operators are inverses of each other. The first is bounded, while the second may be unbounded, and $$e^{-N}e^{N}=I,\;\;\; and\;\; e^{N}e^{-N}x=x,\; x\in \mathcal{D}(e^{N}).$$

• Thank you for this answer. If $N$ is densely defined on $H,$ is it necessary that $(D(N), \|\cdot\|_N)$ is a Banach space for $\|x\|_N = \|x\|+ \|Nx\|$ ? – A. PI Nov 1 '17 at 22:14
• One more question; If in general two unbounded normal operators $N,M$ commute, can we prove, in view of the above definition of $e^N$ that $M, e^N$ commute ? – A. PI Nov 1 '17 at 22:15
• @A.MONNET : An unbounded normal operator should be closed, which means that the graph of $N$ is a Banach space under the graph norm. As for your second question, how are you defining that $N,M$ commute? Are you defining this on a common domain, etc.? Domain issues are always tricky. – DisintegratingByParts Nov 2 '17 at 0:48
• In fact, I want to prove that for $N, M$ unbounded normal operators having the same domain $D(N)$ and spectrums in the left half plane, we have $\|N e^{tM} x\| \leq \|x\|_N$ for all $x\in D(N)$ and $t\geq 0.$ Providing that $e^{tM}(D(N)) \subset D(N).$ – A. PI Nov 2 '17 at 12:30
• As for $N,M,$ I assume that $D(NM)=D(MN)=:D$ and $NM=MN$ on $D.$ – A. PI Nov 3 '17 at 10:56