Timeline for Is this operator invertible?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2019 at 21:52 | comment | added | S. Maths | @JochenGlueck for example if $G$ is defined just on "constant functions" will be defined on $X$ but won't be invertible I guess. Maybe an operator in terms of $(Gh)(t_0)$, for some $t_0$, would be. | |
| Jul 24, 2019 at 21:25 | comment | added | Jochen Glueck | @S.Maths: Well, to find an explicit formula for all powers of $B$ - even for the case where $T$ is a semigroup and $A$ is constant - would imply to have an explicit formula (whatever "explicit" means) for all terms that occur in the Dyson-Phillips series for perturbed semigroups. I think this is much more than what we can expect. Concerning your second question again, I still have difficulties to follow; what does it mean to "deduce" an operator? How should this invertible operator on $\mathcal{L}(X)$ that you seek be related to $G$? | |
| Jul 24, 2019 at 20:55 | comment | added | S. Maths | @Jochen, Thanks. I calculated $B^2$ in terms of $B$ but I don't know if this help. For the last question, I mean $G$ is invertible in $\mathcal{L}(C(0,\tau;X))$, can we deduce an invertible operator in $\mathcal{L}(X)$. | |
| Jul 24, 2019 at 19:54 | comment | added | Jochen Glueck | @S.Maths: We can compute the inverse of $G = \operatorname{id} - B$ by means of the von Neumann series, but I think in most cases it will not be possible to explicitly compute the powers of $B$ that occur in this series, even if $T$ is a semigroup. Concerning your second question, I'm not sure what you mean by "a like invertible operator on $X$". | |
| Jul 23, 2019 at 0:38 | comment | added | S. Maths | @Jochen can we explicit the inverse in case of semigroups? and can we get a like invertible operator on $X$. Thanks in advance. | |
| Apr 5, 2019 at 16:33 | comment | added | Nik Weaver | Very nice argument! | |
| Apr 5, 2019 at 16:19 | history | answered | Jochen Glueck | CC BY-SA 4.0 |