Skip to main content
6 events
when toggle format what by license comment
Sep 14, 2012 at 7:45 comment added Suvrit Thanks Robert for this result; I did not know this theorem, but it seems that I might be able to use it. :-)
Sep 14, 2012 at 3:10 comment added Robert Israel $g$ is non-expansive (in the usual metric on $\mathbb R$) iff $g(x) = g(0) + \int_0^x u(t)\ dt$ where $u$ is a measurable function with $|u| \le 1$ almost everywhere.
Sep 13, 2012 at 7:57 comment added Suvrit @Robert: True! I had thought of doing a variable change, but ultimately, I am searching for sufficient conditions that allow one to easily test nonexpansivity for at least a class of functions. And a variable change merely pushes the burden of testing into a different metric.... :-)
Sep 13, 2012 at 6:57 answer added Dirk timeline score: 1
Sep 13, 2012 at 0:01 comment added Robert Israel If you write $g(x) = \log h(e^x)$, so $g$ is a continuous function from $\mathbb R$ to itself, your condition is equivalent to $g$ being non-expansive in the usual metric on $\mathbb R$.
Sep 12, 2012 at 20:49 history asked Suvrit CC BY-SA 3.0