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.


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 $f$.

  • $\begingroup$ your $f$ need not be smooth $\endgroup$ – Ignorant Mathematician Apr 20 '19 at 15:57
  • $\begingroup$ $M$ is smooth, but $g$ might not be. $\endgroup$ – Selene Apr 20 '19 at 15:59
  • $\begingroup$ @XIAODAQU Yes, you are right, but how to find $f$ now ? $\endgroup$ – onurcanbektas Apr 20 '19 at 16:24
  • $\begingroup$ Do it for a small co-ordinate chart with compact closure at first. Then use a partition of unity to glue them together. $\endgroup$ – Ignorant Mathematician Apr 20 '19 at 16:27
  • $\begingroup$ @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 $x$. $\endgroup$ – 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 $|f-g|<m/2$ 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.

  • $\begingroup$ Looks like an overkill. $\endgroup$ – Ignorant Mathematician Apr 21 '19 at 1:00
  • $\begingroup$ It was the first thing I thought of so may very well be so. Do you know a more elementary/slick way to do it? @IgnorantMathematician $\endgroup$ – Tim The Enchanter Apr 21 '19 at 6:54

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