Given a Banach space $V$ and a bounded linear operator $A:V\to V$, the operator $e^A$ is bounded and invertible. When $V$ is finite dimensional, every invertible operator is of the form $e^B$ (one can reduce the problem to Jordan blocks, and then to operators of the form $I+N$ with $N$ nilpotent, which follows from the power series for $\log$).

What happens in the infinite dimensional case? What if we restrict to seperable Hilbert spaces? Is there a nice description of the operators in the image of the operator exponential?

  • $\begingroup$ This is a very nice question. $\endgroup$ – Tomasz Kania Oct 10 '14 at 18:40

In finite dimensional complex spaces, you can find a branch of complex log function that doesn't intersect any of the eigenvalues. And that offers one explanation of why everything works so nicely. You can see it by using Cauchy's integral formula $$ \log(A) = \frac{1}{2\pi i}\oint_{C} \frac{1}{\lambda I-A}\log(\lambda)\,d\lambda. $$ You can prove that this works if $C$ is a positively-oriented simple closed rectifiable curve enclosing the eigenvalues of $A$ in its interior, and if $\log(\lambda)$ is a branch of the log that is holomorphic on a neighborhood of $C$ and its interior. (Please note the similarity of the above formula and Cauchy's integral formula.)

Now suppose that $X$ is a Banach space and $A$ is a continuous linear operator on $X$ (all complex matrices are continuous in the Euclidean norm topology.) The generalization of eigenvalue is the spectrum $\sigma(A)$ consisting of all $\lambda$ for which there is no continuous operator $R(\lambda)$ such that $(\lambda I-A)R(\lambda)=I=R(\lambda)(\lambda I-A)$. In other words, $R(\lambda)=1/(\lambda I-A)$ does not exist for $\lambda$ in the spectrum. It can be shown that $\sigma(A)$ is a compact subset of $\mathbb{C}$, and $R(\lambda)=(\lambda I-A)^{-1}$ is a holomorphic operator-valued function of $\lambda$ for $\lambda \notin\sigma(A)$. The set $\sigma(A)$ can be an arbitrary compact set in the plane; it's not just a discrete bunch of points as for finite-dimensional spaces. Even if you're working on a separable Hilbert space, the spectrum of $A$ can be any compact subset of $\mathbb{C}$.

So, even though $A$ may be invertible (which guarantees that the spectrum is a finite distance from $0$), you may not be able to get a branch of the logarithm that won't pass through the spectrum. For example, the spectrum of an invertible $A$ may be an annulus $0 < r_{\mbox{inner}} \le |\lambda| \le r_{\mbox{outer}}$, and this technique won't work. But, if you can get $C$ as described in the first paragraph, then the same integral will do what you want. This condition is not absolutely necessary, but it's practically necessary. There are specialized conditions you can put on $R(\lambda)$ that will allow the integral to exist for a dense subset of $X$, and you can eventually bootstrap to a bounded operator $\log(A)$, but that discussion is typically long and tedious. This still gives you a large class of operators that are in the range of the exponential, but definitely not everything.

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  • $\begingroup$ +1. Thank you, this is a very nice answer. I will likely accept it, but I want to give a little more time for other people to add different perspectives. Can you give a reference for the fact that the spectrum can be an arbitrary compact set (ideally an online reference)? $\endgroup$ – Aaron Oct 10 '14 at 23:10
  • $\begingroup$ @Aaron : math.stackexchange.com/questions/683196/… . I'll add: if the spectrum is closed, non-empty and bounded, then the operator is bounded. $\endgroup$ – DisintegratingByParts Oct 11 '14 at 0:27

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