Given a smooth map $g: M\to R^+$, show that there is a smooth function $f:M → R^+$ such that $f(x) < g(x)$ for all $x ∈ M$.

Question

I'm trying to prove the following proposition.

Let M be a smooth manifold and let $g:M→R^+$ be a continuous function. Show that there is a smooth function $f:M→ R^+$ such that $f(x) < g(x)$ for all $x ∈ M$. (Hint: Use partition of unity on an open cover of M by coordinate balls.)

I've only come up with $f(x) := g(x)/2 $.However, considering the given hint, I'm guessing that does not work, and I don't know why.

Edit:

As it is pointed out, since $g$ is not smooth, $f$ does not have to be as well, so the question is now, how to find $f$.

Answer 1

As mentioned in the comments, the issue is now constructing such an $f$ on a neighbourhood of a point $x$. Consider a fundamental neighborhood $U_0$ around $x$ and restrict the local diffeomorphism $\phi$ to an open $U$ containing $x$ such that $\bar U \subset U_0$ and is compact. Now with the local coordinate system $(x_1,x_2\dots x_n)$ induced by $\phi$ consider the subalgebra of $C(\bar U,\mathbb R)$ generated by polynomials. It contains the constant functions and separates points. Thus, by Stone-Weierstrass we can approximate $g$ uniformly by such polynomials, which are in fact smooth.

Now as $\bar U$ is compact, $g$ has a minimum, say $m>0$. We can find $f\in C(\bar U,\mathbb R)$ such that $|f-g|<m/2$ on $\bar U$. Then we have that $f$ maps into $\mathbb R^+$ and furthermore $f/2 < g/2 + m/2 \le g$. Thus we have constructed the desired $f$ on $\bar U$. The rest follows by using a partition of unity.