Timeline for An example of non-invertible operator $F$ such that $P_nF$ is invertible on $\operatorname{Im}P_n$ or proving that It is impossible
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 28, 2023 at 18:39 | comment | added | TorteDeline | Yeah. Few seconds ago I proved it by checking kernel. Thank you very much for your help! | |
| Nov 28, 2023 at 18:36 | comment | added | Jochen Glueck | @TorteDeline: " Why from: "$F$ is bounded below" follow that $F$ is also injective." You will certainly figure that out yourself if you simply have a look at the definitions. I feel obliged to point out that, if you are interested in questions like the one that you asked in the original post, it is highly recommendable to acquaint yourself with basic functional analysis. | |
| Nov 28, 2023 at 17:12 | comment | added | TorteDeline | Yeah, I thoght a little bit and It is true. Looks like "our invertibility" equals injectivity. But if so I have one (hope last) question. Why from: " $F$ is bounded below" follow that $F$ is also injective. Is there theorem or it can be easily showed? | |
| Nov 28, 2023 at 16:53 | comment | added | TorteDeline | Hmm. So if my defenition of "invertebility" is injectivity than there are no such counterexample. Because you prove that $F$ is always injective (according to my conditions). Am I right? If so, I will check my defenition, maybe I missed something. But If "our invertebility" equals injectivity so I think it is a key. I will think about that, thank you | |
| Nov 28, 2023 at 16:41 | comment | added | Jochen Glueck | @TorteDeline: Ok, but checking the proof of (1) shows that the same argument also works if $P_n F P_n$ is only assumed to be bijective from $\operatorname{Im}P_n$ to $\operatorname{Im}P_nFP_n$. | |
| Nov 28, 2023 at 16:33 | comment | added | TorteDeline | Yeah, sorry about this. We have another terminology probably, I am very sorry about that. In such case as I mentioned, do you have ideas of example? (maybe it is easier than it was before or harder, don't know) | |
| Nov 28, 2023 at 16:31 | comment | added | TorteDeline | Okay. So lets take $Im P_n$ to $Im P_nFP_n$, as you already mentioned. | |
| Nov 28, 2023 at 16:30 | comment | added | Jochen Glueck | Apart from this, your definition of invertibility simply means injectivity (which is of course equivalent to having trivial kernel). But calling injectivity "invertibility" in the setting of Banach spaces is really strange terminology because it does not imply that the inverse map is continuous. | |
| Nov 28, 2023 at 16:28 | comment | added | Jochen Glueck | @TorteDeline: Well, if the $P_nFP_n$ are bijective from $\operatorname{Im}P_n$ to $\operatorname{Im}P_n$, then there is no counterexample; see the partial result (1) in my answer. | |
| Nov 28, 2023 at 9:53 | comment | added | TorteDeline | About your question about: $P_nFP_n$. To be honest: You can choose the easist case. (what is more easy for you to come up example). So, do you have any ideas how to modify your example? And again, thank you very much for involving!! | |
| Nov 28, 2023 at 9:46 | comment | added | TorteDeline | And of course: operator is invertible when it has invert operator. And it is not invertible when such invert operator doesn't exist | |
| Nov 28, 2023 at 9:44 | comment | added | TorteDeline | To get rid of misunderstanding. This defenition of invertible operator I use: The invert operator to operator $A$ is an operator that assigns to each $y$ from the set of values $ImA $of operator $A$ a single element $x$ from the domain of a function $D(A)$ of operator $A$, which is a solution to the equation $Ax=y$. | |
| Nov 28, 2023 at 8:03 | comment | added | Jochen Glueck | @TorteDeline: If you define invertibility as "bijective onto its range", then what about your assumption that $P_nFP_n$ is "invertible on $\operatorname{Im} P_n$? Do you indeed assume that it is bijective from $\operatorname{Im} P_n$ to $\operatorname{Im} P_n$? Or only that it is bijective from $\operatorname{Im} P_n$ to $\operatorname{Im} P_n FP_n$? | |
| Nov 28, 2023 at 7:47 | comment | added | TorteDeline | @JochenGluek If all that I wrote upper is right: do you have any ideas how to modify your example? Need such $F$ that is non-invertible on Im$F$ and satisfies all another conditions. Thank you very much for your involvement! | |
| Nov 28, 2023 at 7:44 | comment | added | TorteDeline | @JochenGluek Probably I understand: In your case $F$ is non-invertible on X. But $F$ is invertible on $Im F$. (if I'm not right please correct me) But in my case It is still consider "invertible". | |
| Nov 28, 2023 at 7:26 | comment | added | TorteDeline | Addition: the theorem I’m talking about only works when $F$ is linear, but we are considering such a case, so it still works | |
| Nov 28, 2023 at 7:21 | comment | added | TorteDeline | Or maybe Im wrong and this $F$ has another kernel | |
| Nov 28, 2023 at 7:21 | comment | added | TorteDeline | This link. But I understand what you mean: "The inverse of $f$ exists if and only if $f$ is bijective". Yeah, hope so. But it the same moment: if $Ker(F)=\{0\}$ it means that $F$ is invertible. I don't know how to link this two points in this case | |
| Nov 28, 2023 at 7:10 | comment | added | Jochen Glueck | @TorteDeline: If you don't use "invertible" as a synonym for "bijective", could you please specify what definition of the notion you use? | |
| Nov 28, 2023 at 6:47 | comment | added | TorteDeline | But $F$ is invertible. For example: because $Ker(F) = \{0\}$. Or I am wrong? | |
| Nov 27, 2023 at 0:36 | history | edited | Jochen Glueck | CC BY-SA 4.0 |
added 279 characters in body
|
| Nov 27, 2023 at 0:18 | history | answered | Jochen Glueck | CC BY-SA 4.0 |