Timeline for Transform Riesz basis $\{e^{i \lambda_n t}\}_{n\in\mathbb Z}$ to an orthogonal basis
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2015 at 18:41 | vote | accept | Mark | ||
| Sep 21, 2015 at 8:00 | answer | added | user75485 | timeline score: 1 | |
| Sep 19, 2015 at 8:11 | comment | added | Mark | @Josep, what is $\hat f(n)$? Is there a relation between $T$ and $A$? Yes, I don't know if these are useful but they represent an opportunity for me to learn new things, therefore I thank you. But perhaps we are writing too many comments; I'd like to read your answer, if you like :-) | |
| Sep 16, 2015 at 13:11 | comment | added | user75485 | In the conditions of Kadec's result , you have $A=(I-S)^{-1}=\sum_{m=0}^\infty S^m,$ where $S(f)=\sum \hat f(n) (e^{inx}-e^{i\lambda_n x})$ verifies $\Vert S\Vert<1$ ( I don't know if this is useful). | |
| Sep 15, 2015 at 19:16 | comment | added | Mark | @Josep, thank you. This is an equivalent formulation where (I think I understand) $T$ has the same function of $A$. I'd like to calculate $T$ in the case of $\{e^{i \lambda_n t}\}_{n\in\mathbb Z}$ and $\{e^{i n t}\}_{n\in\mathbb Z}$, although I do not know how. | |
| Sep 15, 2015 at 9:47 | comment | added | user75485 | If $T$ is a topological isomorphism that takes a Riesz basis $\{x_n\}$ to a orthonormal system $\{e_n\}$, ( $ T(x_n)=e_n$ ), then $(x,y)'=(Tx,Ty)$ is an equivalent inner product and $\{x_n\}$ is orthonormal with respect to this product. | |
| Sep 14, 2015 at 19:29 | comment | added | Mark | @user64494, ok. Maybe my question is less trivial than I thought at first. | |
| Sep 14, 2015 at 19:28 | comment | added | Mark | @LSpice, I think that the existence of $(\cdot, \cdot)'$ is automatic, but I could not prove it, because I'm not an expert in this topic. | |
| Sep 13, 2015 at 20:10 | comment | added | user64494 | Making use of Maple package OrthogonalExpansions, I tried a math experiment and obtained huge expressions (too long to be stated here). | |
| Sep 13, 2015 at 20:00 | comment | added | LSpice | Is the existence of $(\cdot, \cdot)'$ automatic—that is, does a norm equivalent to one coming from an inner product always come from an inner product—or is the existence of such a product in this case part of the statement? | |
| Sep 13, 2015 at 8:03 | history | asked | Mark | CC BY-SA 3.0 |