TITLE: Finite collection of Lipschitz functions.
QUESTION [5 upvotes]: Let $I=\{1,...,n\}$, $\{f_i\}_{i\in I},f_i:\mathbb{R}\rightarrow\mathbb{R},\forall i$, $a>0$. Assume:
$$|f_i(x)-f_i(y)|\leq a|x-y|,\forall x,y\in \mathbb{R},i\in I.$$
Put: $$f(z)=\max_{i\in I}f_i(z),\forall z\in \mathbb{R}$$
Can we have that: $$|f(x)-f(y)|\leq a|x-y|,\forall x,y\in \mathbb{R}.$$

REPLY [9 votes]: Yes, $f$ is Lipschitz continuous with Lipschitz constant $a$ too.
If $f(x) = f(y)$, the inequality $\lvert f(x) - f(y)\rvert \leqslant a\lvert x-y\rvert$ is trivially true. If $f(x) \neq f(y)$, choose the naming so that $f(y) > f(x)$. Let $i,j$ so that $f(x) = f_i(x)$ and $f(y) = f_j(y)$. Then $f_j(x) \leqslant f_i(x)$ and hence
$$0 < f(y) - f(x) = f_j(y) - f_i(x) \leqslant f_j(y) - f_j(x) \leqslant a\lvert y-x\rvert$$
by the assumption on $f_j$.