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?

  • $\begingroup$ 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. $\endgroup$ – Guillaume Garrigos Feb 18 '16 at 15:06

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 $e^{-A}$.

  • $\begingroup$ 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. $\endgroup$ – PhoemueX Feb 18 '16 at 20:33

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