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).


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. $f(a)<c \leq a$). 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.

  • $\begingroup$ Why is it that $x_1 \leq x_0$ implies the sequence is monotonically decreasing? $\endgroup$ – glowstonetrees May 4 '19 at 13:35
  • 1
    $\begingroup$ 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. $\endgroup$ – Lutz Lehmann May 4 '19 at 14:06

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