Let $M$ be a smooth manifold ($dim\ge 1$). Let $f:M\to\mathbb{R}$ be a positive continuous function. Prove there is a smooth map $g\in C^{\infty}(M)$ such that $0<g<f$.
I knew this would involve a partition of unity. I took a locally finite cover of $M$ by precompact open sets ${\bf U}=\{U_i\}_{i\in I}$. Since $f$ is continuous I know $f$ attains a minimum on each ${\rm cl}(U_i)$, say at $x_i$, and that this minimum is still positive. So I defined maps $g_i: U_i\to\mathbb{R}$ given as a constant maps: $g_i(x)=\epsilon_i$, where $\epsilon_i$ is chosen to be $0<\epsilon_i<f(x_i)$. But each $x\in M$ is contained in only finitely many $U_i$, so I defined $h(x)=\min{\{g_i(x)\;|\; x\in U_i\}}$. Now I take a partition of unity $\{\psi_i:M\to\mathbb{R}\}_{i\in I}$ subordinate to ${\bf U}$. My thinking is to smoothly stitch together these constant maps... but I'm not sure what to do from here. Should I define $g(x)=\sum_{i\in I}{\psi_i(x)\,h(x)}$? I know each of these sums will be finite because $supp(\psi_i)\subseteq U_i$, but does this function actually do what I want it to? I thought at first it would, but now it just looks like I'm multiplying by $1$ in a really fancy way...
Thanks for your help