Contraction and $\max$ function

Question

$f: \Bbb R \mapsto \Bbb R$

$g: \Bbb R \mapsto \Bbb R$

$h: \Bbb R \mapsto \Bbb R$

$h:=\max\{f(x), g(x)\}$

Is $h$ a contraction on $ \Bbb R$ if $f$ and $g$ are both so?

First attempts of mine -although they are not very fruitful- is the following: since every contraction mapping is Lipschitz cont's and hence uniformly cont's, and I just proved that $h$ is uniformly cont's if $f$ and $g$ are so, $h$ could be a contraction if $f$ and $g$ are so.

Answer 1

Assume that $f$ and $g$ are both $L$-Lipschitz continuous, i.e. $$ |f(x) - f(y)| \leq L |x-y|, \quad |g(x) - g(y)| \leq L |x-y|, \quad \forall x,y, $$ and let us prove that $h$ is $L$-Lipschitz continuous too. (The required result follows with $L < 1$.)

Let $x,y\in\mathbb{R}$. To fix the ideas, suppose that $h(x) = f(x)$ (the case $h(x) = g(x)$ is similar). Since $h(y) \geq f(y)$, we have that $$ h(x) - h(y) = f(x) - h(y) \leq f(x) - f(y) \leq L |x-y|. $$ A similar argument shows that $h(y) - h(x) \leq L |x-y|$, so that $|h(x) - h(y)| \leq L|x-y|$.

Answer 2

Let $f: \mathbb{R}\rightarrow \mathbb{R}$ and $g: \mathbb{R}\rightarrow \mathbb{R}$ be contractions. Now, the proof given by Rigel assumes that they are both L-Lipschitz although question does not indicate as such. For all $x,y\in \mathbb{R}$. We have, $$|f(x)-f(y)|\leq M_1 |x-y|$$ $$|g(x)-g(y)|\leq M_2 |x-y|$$ for some $M_1, M_2 \in (0,1)$. Set $M=max \{M_1,M_2\}$. Now, assume without loss of generality that $h(x)=f(x)$. We have two cases:

1. If h(y)=f(y), then we clearly have the result as $f$ is a contraction.
2. For the case of h(y)=g(y), consider the subcase $f(x)\geq g(y)$. Then, $$|h(x)-h(y)|=|f(x)-g(y)|=f(x)-g(y)\leq f(x)-f(y) \; (since \; g(y)\geq f(y))$$ $$\leq |f(x)-f(y)|\leq M_1|x-y|\leq M|x-y|$$ Showing that $h$ is a contraction. Proof for the other sub-case is similar.