Timeline for Closed sets in the space of Fourier transforms $\mathcal{F}L^{1}$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2014 at 13:32 | vote | accept | Inquisitive | ||
| Apr 3, 2014 at 13:31 | comment | added | Inquisitive | @JH; all right; thanks for every thing; | |
| Apr 3, 2014 at 13:26 | comment | added | Johannes Hahn | Who is "we" ? Fourier transform is defined for tempered distributions (including $L^1$) and an automorphism of the space of tempered distributions. | |
| Apr 2, 2014 at 12:01 | comment | added | Inquisitive | @JH;Thanks; I am bit confused; If that is the case, then, Why do we required, both $f$ and $\mathcal{F}f$ are in $L^{1}$ ? (In the hypothesis of Fourier inversion formula) | |
| Apr 2, 2014 at 11:50 | comment | added | Johannes Hahn | Yes it is true for all $f\in L^1$. What do you mean by "if $\mathcal{F}f=\hat{f}$ ? $\hat{\cdot}$ and $\mathcal{F}$ are two notations for the same thing, the fourier transform... | |
| Apr 2, 2014 at 11:48 | comment | added | Inquisitive | @JH; thanks; Is $\mathcal{F^{2}}f(x)=f(-x)$ true for any $f\in L^{1}$ ? (I know, this is true, if $f,\text{and} \ \mathcal{F}f= \hat{f} \ \text{both are } \in L^{1}$, by inversion formula) | |
| Apr 2, 2014 at 11:09 | comment | added | Johannes Hahn | @DivyangBhimani: Because $\mathcal{F}^2(f(x))=f(-x)$ (up to constants depending on your definition of $\mathcal{F}$) | |
| Apr 2, 2014 at 4:25 | comment | added | Inquisitive | @JH;I think, It is clear to me that, $\mathcal{F}:\mathcal{F}L^{1}\to L^{1}$ is an isometry; but why is it onto map ? (I think, this is needed to show, two spaces are isometric by definition;); Thanks a lot; | |
| Mar 31, 2014 at 16:47 | comment | added | Johannes Hahn | @DivyangBhimani: Yes, you're missing that $L^1$ is complete and $(0,1)$ isn't. Completeness is obviously invariant w.r.t. to isometries that is: If $X_1, X_2$ are isometric metric spaces then $X_1$ is complete iff $X_2$ is. In your case $\mathcal{F}: \mathcal{F}L^1\to L^1$ is by your definition of the norm on $\mathcal{F}L^1$ an isometry. | |
| Mar 31, 2014 at 16:26 | comment | added | Inquisitive | @JH;thanks; but I am sorry, I could not follow: see, we consider $\mathbb R$ with usual metric, namely modulus, $|\cdot|$, which is complete; I think, $d:(0, 1)\to \mathbb R $ such that, $x\mapsto |x|$; preserves distance; but (0, 1) is not a complete metric space; Or am I missing something ? | |
| Mar 31, 2014 at 15:30 | history | answered | Johannes Hahn | CC BY-SA 3.0 |