# Convergence to fixed point

Question:

Let $$f:[0,1] \rightarrow [0,1]$$ be a continuous function that is strictly increasing on $$[0,a)$$ and strictly decreasing on $$(a,1]$$. Moreover, $$f(a) \leq a$$.

Consider the discrete map where

$$x_{n+1} = f(x_n)$$

Show that this map has at least one fixed point in $$[0,f(a)]$$, and that each initial point in $$[0,1]$$ converges to one of them (not necessarily the same one).

Attempt:

Consider the function $$g(x) = f(x)-x$$

The range of $$f$$ is $$[0,1]$$, so in particular $$f(0) \geq 0$$. Also, it is given that $$f(a) \leq a$$. It follows that

$$g(0) = f(0) - 0 = f(0) \geq 0 \qquad \qquad g(a) = f(a)-a \leq 0$$

So by the Intermediate Value Theorem, there exists $$c \in [0,a]$$ such that $$g(c) = 0$$, i.e. that $$f(c) = c$$, which means that $$c$$ is a fixed point of $$f$$.

Suppose that $$c \in (f(a),a]$$ (i.e. $$f(a)). Note that $$f$$ is strictly increasing on this interval, so $$f(c) \leq f(a)$$. But also, $$f(c) = c>f(a)$$. This gives a contradiction unless $$c = a = f(a)$$.

Thus, there exists a fixed point in $$[0,f(a)]$$.

As for the second part, I'm not so sure how to go about it.

I tried using the Banach Fixed Point Theorem (i.e. the Contraction Mapping Theorem), but it just can't be applied directly because (intuitively, I think that) $$f$$ is a contraction only in a neighborhood of one of the fixed points.

Any hints would be much appreciated. Thanks!

p.s. Also, I'm having trouble understanding why there cannot be a $$2$$-cycle (for in that case, there will be points that converge to this $$2$$-cycle which is not a fixed point).

You already noticed that $$f$$ maps $$[0,f(a)]$$ into itself. Any iteration sequence that starts inside this interval stays inside that interval. If $$x_1\le x_0$$ the iteration sequence is monotonically decreasing, if $$x_1\ge x_0$$ it is monotonically increasing. In any case the sequence is bounded, thus convergent, and the limit must be a fixed point.
Any sequence with $$x_0\in[0,a]$$ will converge to a fixed point. If $$x_0\in(a,1]$$, then $$f(a)\ge x_1=f(x_0)\ge f(1)$$ so that from then on the sequence converges monotonically to a fixed point. There is no possibility for a periodic cycle.
• Why is it that $x_1 \leq x_0$ implies the sequence is monotonically decreasing? – glowstonetrees May 4 '19 at 13:35
• Because then $x_2=f(x_1)\le f(x_0)=x_1\le x_0$ and so on. The first observation ensures that the sequence does not leave the region where $f$ is monotonically increasing. – Lutz Lehmann May 4 '19 at 14:06