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?.
Best Answer
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?