If we have a set of functions $f_i$ such that every $f_i$ has a common Lipschitz constant $M$, is it true that the supremum of these functions at some $x$ is also Lipschitz? Assume that we can uniformly bound all the functions by $K$.
The first part of this answer suggests it is true, so I tried to do a little triangle inequality manipulation with this, similar to how one might prove that $fg$ or $f + g$ are Lipschitz assuming $f, g$ Lipschitz and bounded. However, I had trouble with finding the common term. What are some other approaches to this proof?
Best Answer
Let $f(x) := \sup\{f_i(x) : i\in I\}$. Let $x,y$ be arbitrary. Then for every $\epsilon>0$ there exists $i\in I$ such that $f(x)-\epsilon\le f_i(x)$. Hence \begin{align*} f(x)-f(y) &\le f_i(x)-f(y) + \epsilon\le f_i(x)-f_i(y) + \epsilon\\ &\le |f_i(x)-f_i(y)| + \epsilon \le M|x-y| + \epsilon. \end{align*} Since $\epsilon>0$ was arbitrary, it follows that $f(x)-f(y)\le M|x-y|$. Interchanging the roles of $x$ and $y$ shows that also $f(y)-f(x)\le M|x-y|$, hence $|f(x)-f(y)|\le M|x-y|$.