I have a sequence $x_0 = 0$, $x_{n+1} = f(x_n)$ and I know that:
- $f(x)$ is stricty convex on $[0,1]$
- $f(x)$ is increasing on $[0,1]$
- $f(1) = 1$
if need be, I can further assume that
- $f(x)$ is continuous on $(0,1)$
I want to show that if $f'(1) \leq 1$, then $x_n \to 1$ as $n \to \infty$. Any ideas on how to approach this?
Using 1.-3. I can show that $1>x_{n+1}>x_n$ for all $n$. Hence I know that the sequence must converge. It is also clear that the limit is a fixed point of $f(x)$. However, how can I be sure that $x = 1$ is the only fixed point of $f$ on $[0,1]$?
Suppose $f$ has some other fixed point $x' < 1$, so $f(x') = x'$.
This will cause problems. Intuitively, the fact that $f$ is strictly convex and $f'(1) \le 1$ means $f$ certainly cannot dip below the line $y = x$. Because of the strictness, it also can't touch it again.
More formally: by strict convexity, for all $x$ with $x' < x < 1$, we have $f(x) < x$.
Now pick any $t \in (x', 1)$. Let $m$ be the gradient of the line segment joining $(t, f(t))$ and $(1, 1)$. Then $m > 1$. We'll show this violates convexity by using the fact that $f(1 - h) \approx 1 - f'(1)h$ for small $h$. In particular, for $h$ sufficiently small, we must have $f(1 - h) > 1 - mh$, since $m > f'(1)$. This contradicts strict convexity on the interval $[t, 1]$, which says that $f(1 - h) < 1 - mh$. So there is no fixed point besides $1$!