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Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and there exists a $A \in [0,1[$ so that $|f'(t)| \leq A$ for every $t \in \mathbb{R}$. Then $f$ has a unique fixed point.
How does one prove the existence of this fixed point? I succeeded in proving its unicity, however that's not much worth if you can't verify its existence.
And what about it if $|f'(t)|<1$ for every $t \in \mathbb{R}$? Does $f$ then still have a fixed point?
For the first, uniqueness is straightforward, all you need to do is show existence.
Suppose $f(0) >0$. Note that for $t >0$ $f(t) = f(0) + f'(\xi) t \le f(0)+At$, and hence $f(t) -t \le f(0) + (A-1)t$. Since $A-1 <0$, there is some $t'$ such that $f(t')-t' < 0$ and the intermediate value theorem shows that there is some $t^*$ such that $f(t^*) = t^*$.
If $f(0) <0$ apply the previous reasoning to $-f$.
For the second part, try $f(x) = ({1 \over 2}+{1 \over \pi} \arctan x) x +{1 \over \pi}$, then $f'(x) = { \arctan x \over \pi }+{ x\over \pi(1+x^2)} + {1 \over 2} $ and $f(x) > x$ for all $x$.
Note that $f''(x) = {2 \over \pi (1+x^2)^2} >0$ and $\lim_{x \to \infty} f'(x) = 1$, hence $f'(x) \in (0,1)$ for all $x$.
Since for any $x, y\in \mathbb R$, $$|f(x)-f(y)|=|f'(\xi)||x-y|\le A|x-y|,$$ now we can apply Banach Fixed Point Theorem to conclude that $f$ has a unique fixed point..
