Timeline for Equivalent metrics on Fréchet spaces and Lipschitz maps
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2011 at 14:09 | vote | accept | Angelo Lucia | ||
| Aug 8, 2011 at 19:34 | comment | added | Bill Johnson | Sure; zero. I is the identity operator. | |
| Aug 8, 2011 at 8:39 | comment | added | Angelo Lucia | ...but it still has a unique fixed point, right? | |
| Aug 8, 2011 at 7:52 | history | edited | Bill Johnson | CC BY-SA 3.0 |
Corrected the interchange of X and Y; added 1 characters in body
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| Aug 7, 2011 at 23:17 | comment | added | Bill Johnson |
As for having the same distance on $X$ and $Y$ when $X=Y$, that is not possible when e.g. $X$ is the countable product of lines. For this space $I/2$ is not a contraction under any compatible metric.
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| Aug 7, 2011 at 21:29 | comment | added | Angelo Lucia |
I meant $\sum 2^{-n} (\left\lVert x-y \right\rVert \wedge k)$ on $Y$.
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| Aug 7, 2011 at 20:12 | vote | accept | Angelo Lucia | ||
| Aug 7, 2011 at 21:25 | |||||
| Aug 7, 2011 at 20:03 | comment | added | Angelo Lucia |
I understand the idea, but I would say that if I take $ \sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1)$ on $X$, I need $\sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1/k)$ on $Y$. Anyway... what about maps from $X$ into himself? Are maps $f:X\toX$ satisfying $\left\lVert f(x) -f(y)\right\rVert \le k \left \lVert x-y \right \rVert$, with $k<1$, distance-contraction (and thus have a unique fixed point)? maybe this is a different question...
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| Aug 6, 2011 at 16:27 | history | answered | Bill Johnson | CC BY-SA 3.0 |