Skip to main content
10 events
when toggle format what by license comment
May 10, 2020 at 9:32 comment added Dirk Werner In your new rendering of the question you seem to have a family of mappings $f_a: (0,a) \to (0,a)$. What is the relation between two such maps?
May 9, 2020 at 22:15 history edited user918212 CC BY-SA 4.0
added 300 characters in body
May 9, 2020 at 21:57 comment added Dirk Werner I‘m sorry, but I still don‘t get it. The way you phrase it you have one interval and one function; i.e., $f:(0,a) \to (0,a)$. — Anyway, a remark worth keeping in mind is that a contraction on $(0,a)$ extends to a contraction on $[0,a]$.
May 9, 2020 at 20:16 history edited YCor CC BY-SA 4.0
edited tags; edited title
May 9, 2020 at 20:10 history edited user918212 CC BY-SA 4.0
deleted 92 characters in body
May 9, 2020 at 18:10 comment added user918212 @DirkWerner I edited the question which may help elucidate my problem.
May 9, 2020 at 18:09 history edited user918212 CC BY-SA 4.0
added 1331 characters in body
May 9, 2020 at 17:45 comment added user918212 @DirkWerner Yes, I would like to prove that statement. The reason I am distinguishing integer and non-integer values of $a$ is because the dynamics of the map over the interval are uniquely determined by the value of $a$, and it just so happens that when $a$ is an integer, this edge case occurs where the limiting behavior of the map is no longer well defined. As such, if I am seeking to prove the case when the behavior of the map is always well defined, I am not sure how to do this because it is only well defined over an open interval $(0, a)$ which is not complete.
May 9, 2020 at 17:33 comment added Dirk Werner Maybe I don‘t understand the problem. You have a contraction $f:(0,a) \to (0,a)$. You seem to assume that iteration of any $x\in (0,a)$ leads to a fixed point of $f$. But you say you want to prove exactly this. Plus I don‘t understand why you are making a difference between integer and non-integer $a$.
May 9, 2020 at 15:24 history asked user918212 CC BY-SA 4.0