Timeline for Existence of topology on the space of continuous functions
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 16, 2015 at 17:31 | comment | added | YCor | Assume we have such a topology with (2) replaced with the slightly stronger: the function $f\mapsto (f(0),f(1),\int_{[0,1]}f)$ is continuous. Since addition by 1 is a homeomorphism, the set of continuous functions valued in $[0,2]$ should be closed. Then let $(f_n)$ be zero on $[1/n,1]$, and affine on $[0,1/n]$ with $f_n(0)=1$. By compactness, let $f$ be an accumulation point of $(f_n)$. By the conditions, we have $f(0)=1$ and $\int_{[0,1]}f=0$, and by what's just written, $f\ge 0$. Since $f$ is continuous, we get a contradiction. So there is no such topology. | |
| Jul 16, 2015 at 17:09 | history | edited | CodeGolf | CC BY-SA 3.0 |
added 44 characters in body
|
| Jul 16, 2015 at 17:04 | history | edited | CodeGolf | CC BY-SA 3.0 |
added 44 characters in body
|
| Jul 16, 2015 at 15:29 | comment | added | CodeGolf | @BillJohnson I don't really need neither local convexity, nor the complicity, just a topological space. | |
| Jul 16, 2015 at 15:26 | comment | added | CodeGolf | @BillJohnson Thanks for replying. Indeed, I may choose that metric defined by $d(f,g):=|f(0)-g(0)|+|f(1)-g(1)|+|\int_0^1f(t)dt-\int_0^1g(t)dt|$. But how can we show the unit ball is compact? | |
| Jul 16, 2015 at 15:25 | history | edited | CodeGolf | CC BY-SA 3.0 |
deleted 25 characters in body
|
| Jul 16, 2015 at 15:25 | comment | added | CodeGolf | @YCor Thanks for pointing out that. I mean compact and I have modified. | |
| Jul 16, 2015 at 14:57 | comment | added | Bill Johnson | If you want a Hausdorff locally convex topology, the answer is no, because it would make $C[0,1]$ into a dual Banach space, and $C[0,1]$ is not even isomorphic to a dual Banach space. | |
| Jul 16, 2015 at 14:55 | comment | added | Bill Johnson | Sure. Take the weak topology generated by the three functionals ptwise evaluation at 0, ptwise evaluation at 1, and integration. | |
| Jul 16, 2015 at 14:52 | comment | added | YCor | I'd improved the formatting, and added the condition that the topology is a TVS topology, please fix if I'm mistaken (but otherwise I can't figure out how "weakly compact" makes sense). | |
| Jul 16, 2015 at 14:42 | history | edited | YCor | CC BY-SA 3.0 |
Corrected many typos and added "topological vector space topology" since otherwise "weakly compact" is senseless
|
| Jul 16, 2015 at 14:32 | history | asked | CodeGolf | CC BY-SA 3.0 |