I have been wondering how to extend a result of (Yamamoto 1971)¹. The part my advisor and I are very interested is the following: given $p(z) \in \mathbb{C}[z]$ a polynomial and an (unbounded, close, densely defined) operator $A: \mathsf{Dom}(A) \subseteq H \to H$ on a Hilbert space, we have

$$\mathsf{Ker}(p(A)) = \bigoplus_{\lambda \in \mathsf{Zeros}(p)} \mathsf{Ker}((\lambda - A)^{\nu(\lambda)}).$$

Making sense of the poles

One may exploit this decomposition for more general rational functions $r(z) = \frac{p(z)}{q(z)} \in \mathbb{C}(z)$ with $\mathsf{deg}(p) < \mathsf{deg}(q)$. In order to do this, it is required to invert $q(A)$, so we require $\mathsf{Poles}(r) = \mathsf{Zeros}(q)$ to be disjoint from the spectrum of $A$, $\sigma(A)$. Note that a partial fraction decomposition²³ allows us to write $r(A)$ as a sum of resolvent operators $R(\lambda, A)$, which are bounded.

A step further into a more general statement can be taken by allowing $\mathsf{deg}(p) \geq \mathsf{deg}(q)$. In this case, a simple application of Euclidean division and partial fraction decomposition allows us to split the now unbounded operator $r(A)$ into a holomorphic part $h(A)$ and a singular part $s(A)$, where $s(A)$ is a bounded operator.

In order to use the result above, we need to consider $\mu \in \mathbb{C}$ and compare the (generalized) eigenspace associated with $r(A) - \mu$ with the direct sum of the (generalized) eigenspaces of $A - \lambda$, where $\lambda$ corresponds to a zero of $r(z) - \mu$. What is interesting is that, in addition to the pole condition above, we require conditions on the zeros of the holomorphic part of $r(z)-\mu$, that is, $\mathsf{Zeros}(h - \mu) \cap \sigma(A) = \emptyset$. This is required to be able to invert $h(A) - \mu$ using the fundamental theorem of algebra, when dealing with some regularity issues concerning the domain of $A$.

More zeros and more problems

What if we consider a more general holomorphic part? One idea would be to express $h(z)$ in its factorized form, using Weierstrass theorem. If $\mathsf{Zeros}(h) = \{ a_n : n \in \mathbb{N} \} \cup \{0\}$ with no accumulation points in $\mathbb{C}$, then there is an entire function $g$ such that

$$h(z) = z^m \exp(g(z)) \prod_{n \in \mathbb{N}} E_{n-1} \left(\frac{z}{a_n}\right),$$

where $E_n(z) = (1-z) \exp\left(\sum_{j=1}^n \frac{z^j}{j}\right)$ is an elementary factor, verifying

$$\left\lvert 1 - E_n(z) \right\rvert \leq |z|^{n+1},$$

for $|z| \leq 1$. Then we face the issue of defining $A$, its powers, and its exponential.

Taking a step back

One way to tackle this is to topologize the domain of our operator. Define

$$\mathsf{Dom}(A^{n+1}) = \left\{x \in \mathsf{Dom}(A) : A x \in \mathsf{Dom}(A^n) \right\},$$

for $n \in \mathbb{N}$, endowed with a stronger graph norm $\|x\|_n = \sum_{j=1}^n \|A^j x\|$. Provided all of these spaces are Banach, we are tempted to take the projective limit of these spaces, yielding a Fréchet space, where the underlying set is given by

$$\mathsf{Dom}(A^{\infty}) = \bigcap_{n\in\mathbb{N}} \mathsf{Dom}(A^n).$$

This could allow us to start defining $\exp(A)$ more naturally, yet with some great effort and constraints ahead.

Analytic semigroups

In a more general setting, Hille-Yosida and continuous semigroup theory comes to our aid. A semigroup behaves like an exponential function!

Let $\boldsymbol{\omega} \in (0, \pi)$, and define the open sector

$$\Sigma(\boldsymbol{\omega}) = \left\{ z \in \mathbb{C}\setminus \{0\} : |\mathsf{arg}(z)| < \boldsymbol{\omega} \right\},$$

where the argument is taken in $(-\pi, \pi)$. The analytical version of Hille-Yosida states that, given a densely defined closed operator $A$, the following assertions are equivalent:

• $A$ generates a bounded analytic semigroup on $\Sigma(\eta)$ for some $\eta \in \left(0, \frac{1}{2}\pi\right)$;
• there exists $\theta \in \left(\frac{1}{2}\pi, \pi\right)$ such that $\Sigma(\theta) \subset \rho(A)$ and $$\sup_{\lambda \in \Sigma(\theta)} \left\| \lambda R(\lambda, A) \right\| < + \infty.$$

Moreover, the suprema of such $\eta$ by $\boldsymbol{\omega}_\mathsf{Holo}$ and the suprema of such $\theta$ by $\boldsymbol{\omega}_\mathsf{Res}$, we have

$$\boldsymbol{\omega}_\mathsf{Res} = \frac{1}{2}\pi + \boldsymbol{\omega}_\mathsf{Holo}.$$

In this case, the inverse Laplace transform provides a concrete representation of the associated semigroup

$$\mathbb{E}(t)x = \frac{1}{2 \pi \mathsf{i}} \int_\Gamma \exp(\lambda t) R(\lambda, A) x \mathrm{d} \lambda,$$

where $\Gamma = \partial \left(\Sigma(\tilde{\theta}) \setminus B_0\right)$ is the upwards-oriented boundary, for any $\tilde{\theta} \in \left( \frac{1}{2}\pi, \boldsymbol{\omega}_\mathsf{Res}\right)$ and $B_0$ is a closed ball around the origin.

Now, note that

$$E_n\left(\frac{z}{a}\right) = (1-\frac{z}{a}) \prod_{j=1}^n \exp\left(\frac{z^j}{j a^j}\right),$$

thus it is plausible to define

$$E_n(A) = (1 - \frac{1}{a} A) \prod_{j=1}^n \mathbb{E}_j\left(\frac{1}{j a^j}\right),$$

where each $\mathbb{E}_j$ is generated by $A^j$. This also relies on all the operators $A^j$ being closed and densely defined.

I wonder if this may yield a more explicit representation for a general $h$, though Dunford-Taylor calculus already yields an answer to it! Maybe understanding the Laplace transformation may help with this.

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Footnotes

1. Yamamoto, Tetsuro. “A note on the spectral mapping theorem.” SIAM Journal on Mathematical Analysis 2, no. 1 (1971): 49-51. DOI ↩

2. Lang, Serge. Algebra. Vol. 211. Springer Science & Business Media, 2012. DOI ↩

3. Conway, John B. Functions of one complex variable II. Vol. 159. Springer Science & Business Media, 2012. DOI ↩