Is the range of a self-adjoint operator stable by its exponential? 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?
 A: 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.
A: 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 $e^{-A}$. 
