Has anyone stated a fixed point theorem based on a condition of the derivative?

I've come up with the following one, and have some questions about it.

If $f(x)$ is a continuous function from $\mathbb{R} \rightarrow \mathbb{R}$ and $g(x)$ is the infinite composition of $f$ on $x$, that is: $g(x) = f(f(...f(x)))$, then from the chain rule,

$g'(x)= f'(g(x)) f'(g(x)) f'(g(x)) ... = \lim_{n\rightarrow \infty} \left[ f'(g(x)) \right]^n$

If $\vert f'(x) \vert < 1$ for all $x$ then the infinite product reduces to zero, so that $g'(x)=0$ for all $x$. If a fixed point for which $f(x_0)=x_0$ exists, then $g(x)=x_0$ for all $x$.

Does that make sense? But what about situations where there is more than 1 fixed point?

If valid, has this result, or a variation of it been stated somewhere? How does it compare to other fixed-point theorems?

Can one prove that $g(x)$ is continuous?


It seems like the theorem you are stating boils down to the following:

Theorem: If $f:\mathbb R\rightarrow\mathbb R$ is a continuously differentiable function such that $f'(x)<1$ for all $x$, then if $f$ has a fixed point $x_0$, it is true that $\lim_{n\rightarrow\infty}f^n(x)=x_0$ for all $x_0\in \mathbb R$

where the notation $f^n(x)$ refers to the composition $f(f(\ldots f(x)\ldots ))$ where $f$ is applied $n$ times. Your reasoning goes along the right intuition, but it doesn't quite hold up to rigorous scrutiny for two reasons: first, the definition of $g$ assumes the conclusion that this thing converges, and second, you are exchanging a derivative with a pointwise limit, which is not generally allowed.

Instead, we can use some lemmas to achieve the goal rigorously. Firstly, the derivative is a somewhat finnicky concept - instead of saying "the derivative is bounded", there's a much more robust idea that is almost unavoidable in proving your theorem. This definition captures the idea that the amount that the function's value may change is some fixed proportion of how much the function's input changed:

Definition: A function $f:\mathbb R\rightarrow\mathbb R$ is $c$-Lipschitz if $|f(x)-f(y)| \leq c|x-y|$ for all $x,y\in\mathbb R$.

and we will also talk about this concept on subsets of $\mathbb R$ as well (intervals, in particular). Lipschitz is a stronger condition than continuity, and one can prove the following lemma (though we won't here):

Lemma: If $f:\mathbb R\rightarrow\mathbb R$ is continuous and differentiable almost everywhere and $|f'(x)| \leq c$ for all $x$, then $f$ is $c$-Lipschitz.

While this machinery suffices to show your theorem, we can use it to state an even stronger theorem: the Banach contraction mapping theorem which, states, essentially:

Theorem: Let $f$ be a $c$-Lipschitz function for $c<1$ mapping an appropriately nice space to itself. Then $f$ has a unique fixed point $x_0$ in that space such that $\lim_{n\rightarrow\infty}f^n(x)=x_0$ for all $x$.

For our purposes, closed intervals of $\mathbb R$ are suitably nice, whereas other sets, like the rational numbers or bounded open intervals, are not nice - and we just note as an aside that this theorem is far more general than just commenting about the real numbers. The proof of this very directly uses a chain of reasoning like the following to say that the sequence $x,f(x),f^2(x),\ldots$ must converge because the distance between consecutive terms shrinks quickly. In particular, note the following consequences of being $c$-Lipschitz: $$|f^2(x) - f(x)| \leq c|f(x) -x|$$ $$|f^3(x) - f^2(x) | \leq c |f^2(x) - f(x)| \leq c^2 |f(x) -x |$$ $$|f^{n+2}(x) - f^{n+1}(x)| \leq c^n |f(x) -x|$$ and then we can note that, since the distances between consecutive terms are shrinking exponentially fast, this sequence had better converge - which is basically what "nice space" means here.

Your theorem follows by noting that if $f$ has a fixed point $x_0$, then for any $\ell>0$ we can apply the contraction mapping theorem to $f$ restricted to the interval $[x_0-\ell,x_0+\ell]$. Note that, by the extreme value theorem, since $f$ is continuously differentiable, there is some $z$ in this interval maximizing $f'(z)$ - but $f'(z) < 1$. Set $c=f'(z)$. On this interval, $f$ is $c$-Lipschitz. Moreover, since $x_0$ is a fixed point, the Lipschitz condition implies that no point can get further from $x_0$ under application of $f$ - so this interval maps into itself. Then, we can apply the theorem to see that everything maps to $x_0$.

Note that this is basically the same as your reasoning, but made rigorous through a different lens that behaves better in limits: the idea that the $n^{th}$ iterate of $f$ has derivative something like the $n^{th}$ power of $f'$ is exactly the intuition that goes here to basically note that if $f$ is $c$-Lipschitz, then $f^n$ is $c^n$-Lipschitz - so has to flatten out to a constant function in the limit.

One note is that, while there are functions like $$f(x) = \begin{cases}x + 1/x & \text{if }x \geq 1 \\ 2 & \text{if }x\leq 1 \end{cases}$$ that have $|f'(x)| < 1$ everywhere, but lack a fixed point, if you strengthen the condition to say that there is some universal $c<1$ such that for all $x$ we have $|f'(x)| \leq c$, then the existence of a fixed point is guaranteed!

  • 1
    $\begingroup$ Thank you for the thorough answer Milo! $\endgroup$ – user3433489 Sep 29 '19 at 13:51

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