Does $|f^\prime|<1$ imply that $\forall_{x,y}|f(x)-f(y)|<|x-y|$?

Question

I have a task that I think reduces to proving that $f$ is a contraction mapping. We know that $\forall_x|f^\prime(x)|<1$. Therefore if I could prove that $|f^\prime|<1\implies|f(x)-f(y)|<|x-y|$ then I think the task would be solved.

I feel this property does hold and that it is also somewhat obvious. Unfortunately, saying that something is obvious is obviously not a valid proof.

I have feeling proving this belongs to an elementary course on analysis, but it is somehow surprising how much have I forgotten from this course... How to prove this property and does it even hold?

Answer 1

Suppose that $|f'(x)|<1 ~\forall x \in (a,b).$

SIMPLER PROOF (thanks to BigbearZzz)

Consider $y<x\in [a, b]$.

By the mean value theorem, there exists a number $c \in (y,x)$:

$$f'(c) = \frac{f(x)-f(y)}{x-y} \Rightarrow \\ f(x)-f(y) = (x-y)f'(c) \Rightarrow\\ |f(x)-f(y)| = |x-y||f'(c)|.$$

Since $|f'(c)| <1 $, then:

$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| \Rightarrow \\ |f(x)-f(y)| < |x-y|.$$

OLD PROOF

Consider $y<x\in [a, b]$. Moreover, consider the function $g(x) = x$ (continuous and differentiable everywhere).

By the Cauchy's mean value theorem, there exists a number $c \in (y,x)$:

$$\left(f(x)-f(y)\right)g'(c) = \left(g(x)-g(y)\right)f'(c) \Rightarrow\\ \left(f(x)-f(y)\right) = \left(x-y\right)f'(c) \Rightarrow \\ |f(x)-f(y)| = |x-y||f'(c)|.$$

Since $|f'(c)| <1 $, then:

$$|f(x)-f(y)| = |x-y||f'(c)| < |x-y| \Rightarrow \\ |f(x)-f(y)| < |x-y|.$$

Answer 2

I am making a guess as to what your final problem is. Draw a careful graph of $$ \frac{3x + \sqrt{1+x^2 \;}}{4} $$ It has no fixpoint.