Smoothness on every compact $V\subset U$ implies smoothness on $U$?

Question

Quick question: while working out a problem I arrived at a conclusion that a certain function $u\in C^{\infty}(V)$ for all compact $V\subset U$ (where $U\subset \mathbb{R}^n$ is bounded). The conclusion I would want to arrive is that $u\in C^{\infty}(U)$. Does this necessarily follow out of what I have just shown or is there a subtlety I'm missing? Thanks.

Answer 1

Yes, it follows, differentiability isa local property and a closed ball is compact, every element $x$ is contained in a closed ball which is compact, so it is differentiable in the interior of that ball so it is differentiable at $x$.