Let $H$ be an Hilbert space, and $A \in L(H)$ be a bounded linear self-adjoint operator on $A$. We assume that $R(A)$, the range of $A$, is not closed.

Is it true or not that $R(A)$ is stable by $e^{-A}$? Meaning: if $y \in R(A)$ then $e^{-A} y \in R(A)$?

I make here two comments:

If $R(A)$ was closed, the answer would be simple, by writing $e^{-A}y$ as the limit of the sums $s_k:=\sum\limits_{k=0}^n \frac{(-A)^ky}{k!} \in R(A)$. But here the range is not closed so it could be that the limit "falls outside" the range.

On the other hand, $e^{-A}y$ is an infinite sum of the terms $\frac{(-A)^ky}{k!}$, which all lie in $R(A^k) \subset R(A^{k-1}) \subset ... \subset R(A)$. So here we are adding elements which go, somehow, deeper and deeper in $R(A)$. Could it be that this property save the game?

  • I noticed that my exponential might not be well-defined, so I assume additionally that $A$ is self-adjoint. Moreover, I am more looking for the stability by $exp(-A)$, which has a full domain, than $exp(A)$ so I changed that too. – Guillaume Garrigos Feb 18 '16 at 15:06
up vote 2 down vote accepted

This is true (for bounded $A$), for the simple reason that if $y=Ax$, then $e^{-A}y=e^{-A}Ax=Ae^{-A}x\in R(A)$ also.

Ok so I think I have a (partial) answer, by means of some spectral analysis.

Assuming that $A$ is self-adjoint, and that $H$ is separable, we can find a family $(\sigma_n)$ of decreasing nonnegative eigenvalues and $(u_n)$ of orthonormal eigenvectors such that for all $x \in H$, $Ax= \sum\limits_{n=0}^\infty \sigma_n \langle x,u_n \rangle u_n$. In particular, the range is characterized as $$R(A)= \{ x \in X \ | \ \frac{\vert \langle x , u_n \rangle \vert }{\sigma_n} \in \ell^2(\mathbb{N}) \}.$$ So if we take $y \in R(A)$, there exists some $x \in X$ such that $y=\sum\limits_{n=0}^\infty \sigma_n \langle x,u_n \rangle u_n$, and $e^{A}y$ writes as $\sum\limits_{n=0}^\infty e^{\sigma_n} \sigma_n \langle x,u_n \rangle u_n$. According to our characterisation of $R(A)$, $e^{A}y \in R(A)$ holds if and only if $e^{\sigma_n} \vert \langle x_n,u_n \rangle \vert \in \ell^2(\mathbb{N})$, which is true since $e^{\sigma_n}$ is contained in $]1,e^{\sigma_1}[$.

The above argument works also by replacing $e^A$ by $e^{-A}$.

  • What you write about $A$ is only true in general if $A$ is assumed to be compact. Noncompact selfadjoint operators might not have any eigenvalues. – PhoemueX Feb 18 '16 at 20:33

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.