# 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 $x ∈ M$.

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 $$f$$.

• your $f$ need not be smooth – Ignorant Mathematician Apr 20 '19 at 15:57
• $M$ is smooth, but $g$ might not be. – Selene Apr 20 '19 at 15:59
• @XIAODAQU Yes, you are right, but how to find $f$ now ? – onurcanbektas Apr 20 '19 at 16:24
• Do it for a small co-ordinate chart with compact closure at first. Then use a partition of unity to glue them together. – Ignorant Mathematician Apr 20 '19 at 16:27
• @IgnorantMathematician If I could do that, hint clearly says how to extend that to the whole manifold already; but the problem is, how to construct $f$ for a neighbourhood of $x$. – onurcanbektas Apr 20 '19 at 16:29

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 $$|f-g| on $$\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.