I have shown easily that a function is uniformly continuous if it is globally Lipschitz, but I am not sure how to approach the case when we only consider locally Lipschitz functions.
my attempt: I fixed an epsilon, and chose a delta (to be determined in terms of epsilon later). Then fixed two points with a distance less than delta. I then argued that each of these points has a neighborhood for which the function is Lipschitz (by definition) and tried fiddling around with the definitions and the triangle inequality with no luck.
Any hints or direction would be appreciated.
What is the domain of the function? I believe that if the function is $f:K\to \mathbb{R}$, where $K$ is compact, then the statement is true.
Because $f$ is locally Lipschitz, for each point $x\in K$, there exists an open neighborhood around $x$, say $U_x$, and a constant $L_x$ such that $\forall x_1,x_2\in U_x:|f(x_1)-f(x_2)|\leq L_x|x_1-x_2|$.
Then, $\{U_x:x\in K\}$ is an open covering of $K$, and because $K$ is compact, finitely many such sets (and corresponding Lipschitz constants) cover $K$, then you can choose the largest constant...