Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a smooth ( real) manifold, if $ p\in M$ and $f\in C^{\infty}(U)$ ($U$ is an open subset of $M$), the symbol $[ f]_p$ indicates the smooth germ of $f$ at $p$ . Consider the following set $$S=\{[f]_p\; :\; f\in C^{\infty}(M)\}$$

If $C^{\infty}_p(M)$ is the algebra of all smooth germs at $p$, clearly $S\subseteq C^{\infty}_p(M)$, but is it true or not that $S=C^{\infty}_p(M)$ ? Does exist a smooth function defined over an open subset of $M$ that doesn't coincide around $p$ with an element of $C^{\infty}(M)$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

If $U$ is an open neighborhood of $p$ and $f\in C^\infty(U)$, then there is a function $g\in C^\infty(M)$ such that $f = g$ in some neighborhood of $p$. To prove this, one constructs a smooth function $\chi\colon M\to [0,1]$ such that $\chi$ is supported in a compact subset of $U$ and $\chi\equiv 1$ in a small neighborhood of $p$ (this $\chi$ is called a "bump" function). You can then set $g = \chi f$ on $U$ and $g\equiv 0$ outside of $U$. This function $g$ is smooth, and agrees with $f$ on a neighborhood of $p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.