# Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If there is an $L < 1$ such that for any $x\in \mathbb{R}$ we have...

Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be differentiable.

If there is an $$L < 1$$ such that for any $$x\in \mathbb{R}$$ we have $$f'(x) < L$$, then there exists a unique $$x$$ such that $$f(x) = x$$

Attempt: Let $$P(x) = f(x) - x$$. we have $$P'(x) = f'(x) - 1 < L - 1 < 0$$. thus $$P$$ is strictly decreasing. this proves that if there is a point $$x$$ such that $$P(x) = 0$$, then it must be unique. And that's pretty much all I got.

If I could prove that $$P$$ is positive for some numbers and negative for others, then by the IVT, $$P$$ would have a zero, but after trying for a while, I think it's not the way to go because the hypothesis doesn't provide much information. So I tried proof by contradiction, but I couldn't find any contradiction. I have a feeling that there's something i'm missing, maybe a theorem?.

• Its an application of the Banch fixed-point theorem. Note that $f'(x)<L<1$ implies the Lipschitz condition from the theorem by using the mean value theorem. Sep 30, 2019 at 0:09
• @F. Conrad: No, because Banach's fixed-point theorem requires $|f'(x)|<L$; we may have $f'(x)$ very negative. Sep 30, 2019 at 0:11
• Oh, you are right! I was completely missing out on that detail. Sep 30, 2019 at 0:27
• Just to clarify the statement of the question, I assume "for any $x \in \Bbb R~f'(x) \lt L$" means that this condition holds for all $x \in \Bbb R$. It could be read to mean as long as there is any $x \in \Bbb R$ satisfying the condition. Sep 30, 2019 at 1:47

You're on the right track. Here's a lemma you might find useful:

Suppose $$\phi'(x)\leq-\alpha$$ for all $$x$$. Then: \begin{align*} x>0&\Rightarrow\phi(x)\leq \phi(0)-\alpha x \\ x<0&\Rightarrow\phi(x)\geq \phi(0)-\alpha x \end{align*}

This is proven by integration from $$0$$ to $$x$$.

Now note that if $$-\alpha<0$$, then $$x\mapsto-\alpha x$$ tends to $$\pm\infty$$ as $$x\to\mp\infty$$. Can you show that $$P$$ does likewise?

Let $$g(x)=x-f(x)$$. Then $$g'(x)=1-f'(x)>1-L$$, so putting $$K=1-L$$, we have $$K>0$$ such that $$g'(x)>K$$ for all $$x$$. And we want to show that there exists a unique $$x$$ such that $$g(x)=0$$.

Uniqueness is a simple consequence of the Mean Value Theorem.

And for existence, we have $$g(x)\to +\infty$$ as $$x\to +\infty$$ and $$g(x)\to-\infty$$ as $$x\to-\infty$$, because if $$x>0$$ then $$g(x)>g(0)+Kx$$, and if $$x<0$$ then $$g(x).

So we can certainly find $$u$$ and $$v$$ such that $$g(u)<0$$ and $$g(v)>0$$. Then by the IVT there exists $$x\in(u,v)$$ such that $$g(x)=0$$.