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...

Question

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?.

Answer 1

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?

Answer 2

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 $g(x)<g(0)+Kx = g(0)-K|x|$.

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 $g(x)=0$.