The problem is stated as follows: Assume that $f:\mathbb{R} \rightarrow \mathbb{R}$ is differentiable. (a) If there is an L < 1 such that for each $x \in \mathbb{R}$ we have $f '(x) < L$, prove that there exists a unique point x such that $f(x) =x$. (b) Show by example that (a) fails if $L=1$.
I find this problem interesting because, since the derivative has no lower bound, it is not a contraction mapping problem nor is the function necessarily Lipschitz continuous. I think I understand the essence of the problem in that since the function grows at a rate strictly slower than 1 (and a rate not approaching 1), then the graph of the function must necessarily intersect the graph of $y = x$. Only, I have no idea how to prove this mathematically. Any help with a starting point for proof of this theorem would be greatly appreciated. I tried analyzing the function $h(x) = f(x) - x$ and even $h(x)=f(x) - (x + f(0))$, but couldn't really see how this lead to a solution.
Uniqueness holds because if $f(x) = x$ and $f(y) = y$, then $\frac{f(x)-f(y)}{x-y} = 1$, which by MVT must equal to $f'(\xi)$ for some $\xi$ between $x$ and $y$, clearly impossible by the conditions.
For existence, if $g(x) = f(x) - x$ then $g(x) -g(y) < (L-1) (x-y)$ for every $x>y$. We will prove that $g$ has a real root.
Let $g(0) > 0$. Then, set $y=0$ to get $g(x) - g(0) < (L-1)x$, or $g(x) < g(0) + (L-1)x$. By taking $x \to \infty$, the RHS goes to $-\infty$, since $L-1 < 0$. Thus, $g(x) \to -\infty$ as $x \to \infty$. Conclude.
Let $g(0) < 0$. Then set $x = 0$ to get $g(0) - g(y) < (L-1)(-y)$ , so $g(y) > g(0) +y(L-1)$. As $y \to -\infty$, the RHS goes to $+\infty$, so $g(y) \to \infty$ as $y\to -\infty$. Conclude again.
If $g(0) = 0$, we are done anyway.
Simply take $f(x) = x+1$ for a counterexample if $L$ isn't allowed to be below $1$.