Felafel said:
thanks.
okay so far, but how do you exactly do the rest of the demonstration later on?
The point is to find an interval I \ni 1/3 such that 1 \notin I and f(I) \subset I (ie, if x \in I then f(x) \in I also). The interval (0,1) will do nicely.
Because then if, given a_1 \in \mathbb{R}, there exists a finite N such that a_N \in I, then a_n \in I for all n \geq N. In other words the sequence is trapped there, so to determine whether it converges I only have to look at what happens if a_1 \in I (because if I don't start here or at a special case then I'm just inserting terms at the beginning of the sequence and that cannot affect convergence, which is about what happens eventually).
But first I have to check what other choices of a_1 lead to sequences that get to that interval in a finite number of steps.
Clearly if a_1 = 1 then a_n = 1 for all n, so that doesn't (and the limit will be 1).
If a_1 \leq 0 then 0 < a_2 = 1/(4 - 3a_1) \leq 1/4 so that a_2 \in I.
If a_1 > 4/3 then a_2 < 0 so from the above I know a_3 \in I.
The problem comes if 1 < a_1 < 4/3. Here x < 1/(4 - 3x) so initially the sequence will be increasing. But that means that for some finite n, a_n \geq 4/3. If a_n > 4/3 then from the above I know that a_{n+2} \in I.
If a_n = 4/3 then the sequence blows up. To find the a_1 where the sequence hits 4/3, I started at 4/3 and worked backwards, which gave me the sequence
<br />
b_1 = 4/3,\quad b_{n+1} = \frac{4b_n-1}{3b_n}<br />
(You can check that if y = 1/(4 -3x) then x = (4y-1)/(3y).)
So if a_1 \neq 1 and there is no a_n such that a_n = 4/3 then the sequence is in I after a finite number of steps.
So now I can check what happens if a_1 \in I. In general, it's possible that for different choices of a_1 the sequence might tend to a limit (the possible such limits in this case are 1/3 and 1, which I can't exclude because I is open), or tend to a periodic cycle, or might be chaotic. But in this case I can exclude every possibility other than tending to 1/3:
You can also show that for x \in (0,1),
<br />
|f(x) - 1/3| < |x - 1/3|<br />
which means that |a_n - 1/3| is monotonic decreasing here, and so a_n tends to a limit which must be 1/3.