Let $M$ be a $n$-dimensional smooth manifold and $U \subseteq M$ be a open subset such that there is a diffeomorphism $\phi: U \to B$ where $B$ is the $n$-dimensional open unit ball.
Let $f: \mathbb R^n \to \mathbb R$ be a nonnegative smooth function such that $f^{-1}((0,+\infty))=B$. In particular, $f$ and its derivatives of all orders are $0$ in $\mathbb R^n \backslash B$. Define a function $g: M \to \mathbb R$ such that $g(x) = f\circ \phi(x)$ for $x\in U$ and $g(x) = 0$ for $x\not\in U$.
Question: Is $g$ a smooth function? I can prove $g$ is continuous. Also, $g|_U$ is smooth by definition. I believe $g$ should also be smooth, but I don't know how to show it.
I'm interested in the question, because in Lee's book Introduction to Smooth Manifolds, in the proof of the existence of partition of unity, he defined a function similar to $g$ except $U$ is chosen to be a regular coordinate ball, is this case it is easy to show $g$ is smooth. I wonder whether the condition that the coordinate ball $U$ is regular is only technical.
Let $\tilde f:(-\infty,1) \to [0, 1)$ be a smooth function such that $\tilde f|_{(-\infty,-1]} = \{0\} $, $\tilde f|_{(-1,1)}: (-1,1) \to (0,1)$ is diffeomorphism and $\tilde f$ is strictly increasing on $(-1, 1)$. Let $\phi: (0,1) \to (-1,1)$ be the diffeomorphism $(\tilde f|_{(-1,1)})^{-1}$, then $\tilde f\circ \phi (x) = x$ for $x\in(0,1)$.
Let $\tilde g: (-\infty,1) \to \mathbb R$ be the function that $\tilde g(x) = \tilde f\circ \phi(x)$ for $x \in (0,1)$ and $\tilde g(x) = 0$ otherwise. Then the one sided derivatives of $\tilde g$ at $0$ are different and thus $\tilde g$ is not differentiable at $0$.
To construct a counterexample, we only need to let $M = (-\infty, 1), \ B= (-1,1),\ U = (0,1)$ and $f(x) = \tilde f(x)$ for $x < 0$. Then $\tilde g(x) = g(x)$ for $x < \phi^{-1}(0)$, and $g$ is not differentiable at $0$.