It is easy to prove that if the functions $f_i : \mathbb{R}^n \to \mathbb R$ are Lipschitz with a uniform Lipschitz constant, then $\sup f_i$ is also Lipschitz with the same constant, provided that the supremum is finite everywhere.
I am curious if such a property extends to locally Lipschitz functions.
Suppose that $f_i$ is locally Lipschitz, and suppose that $\sup f_i$ is finite everywhere. Can I conclude that $\sup f_i$ is locally Lipschitz?
Best Answer
NO. Suppose $f$ is Lipschitz in $|x| >\epsilon$ for each $\epsilon$ but not Lipschitz on $\mathbb R$. Let $g_n$ be a smooth function which is $0$ on $(-1/n,1/n)$, $1$ for $|x| >2/n$ taking values between $0$ and $1$. Then $f(x)g_n(x)$ gives a counter-example since the supremum is $f(x)$ which is not Lipschitz.