you don't need a basis to write $V^*\otimes V\to{\rm End}(V)$. define $\phi\otimes v\mapsto \phi(-)v$ for pure tensors, check it makes sense to extend linearly. proving it's an isomorphism might require picking a basis though
@RandomVariable I rarely see you joining in a conversation that has little to do with mathematics problems :P so I guess it's an expected consequence :P just jokin
@RandomVariable nyway .. it seems galactus is back at I&S :-) I hope the forum becomes active again :)
@r9m He hasn't posted anything on the site this year.
@DanielFischer I ended up convincing myself (perhaps wrongly) that the answer I posted the other day was just Parseval's theorem in disguise. So I deleted it since there is already an answer that uses Parseval's theorem.
My ignorance re: infinite-dimensional manifolds is showing lately. I have no idea what carries over, what doesn't. I guess it's time for a crash course.
@RandomVariable Probably. You lost some reputation by the deletion, and the notifications wait until your reputation rises above the level you had before the deletion. Could be that they come back before that if you close your browser window, but maybe not.
@MikeMiller Take any non-complemented closed subspace and look at the canonical projection $E \to E/F$. Doesn't split. And I don't think it has a (differentiable) [local] section.
@DanielFischer: If it had a differentiable section, that should provide a section on the level of tangent spaces; since the map is linear the differential should be exactly the same as the original map.
Does anyone know if the following identity holds: Suppose $F$ is the CDF of some RV. So $F(x) = \Pr(X < x)$. I am reading a paper, where they come across the inverse of the survival function $[1 - F(x)]^{-1}$ and reduce it to $F^{-1}(1 - x)$
Does such an identity hold and if so, can someone point me to a quick proof of it
@DanielFischer: In my case, the map has Fredholm derivative, so it's automatically complemented. I wonder if the correct statement is just that the "kernel of the derivative can be given a continuously chosen complement" implies that there's a smooth section.
I'm trying to prove that for every $\varepsilon>0$ there exists $N>0$ such that $\left|\arctan(x)-\dfrac{\pi}{2}\right|<\varepsilon$ whenever $x>N$. I thought we could let $N=\tan\left(\dfrac{\pi}{2}-\varepsilon\right)$ but it's not defined for every $\varepsilon>0$. What $N$ should I use?
