2
$\begingroup$

Here is what I am trying to do:

Let $X$ be a paracompact smooth manifold.

Let $C$ be closed, $U$ open and $C\subset U \subset X$ and $f$ is a smooth map on $U$. I want to show that then there exists an open $V$ and smooth $F$ on $X$ with $F\mid_V = f\mid_V$ and $C\subset V \subset U$.

To this end I would like to smoothly extend $f$ to $X$.

Please could someone show me how one can extend a smooth map smoothly?

Thoughts:

Because $X$ is paracompact a partition of unity subordinate to $U, C^c$ exists. Let's call the maps in this partition of unity $\varphi_U$ and $\psi$.

I suspect the answer will involve convolution of $f$ with $\varphi_U$ but I have been unable to prove that this is smooth or that it extends $f$ (in fact, it may not extend $f$).

$\endgroup$
  • $\begingroup$ This $F$ is definitely not smooth in general. For instance, suppose $f$ has support $K$ contained in $U$; then $F$ is zero on $U \setminus K$, but $1$ on $M \setminus U$, so your $F$ won't even be continuous at the "boundary" of $U$. $\endgroup$ – mollyerin Jan 26 '15 at 16:02
  • $\begingroup$ @mollyerin Thank you for your comment. I had a new idea and edited my question. $\endgroup$ – user174981 Jan 27 '15 at 0:22
  • $\begingroup$ I'm not convinced that you can even make sense of the notion of "convolution of $f$ with $\varphi_U$" on a general smooth manifold. (Convolution requires that the underlying manifold be a group.) Here's a suggestion: show that there exists an open set $V$ containing $C$ whose closure $\overline{V}$ is contained in $U$. Consider a partition of unity subordinate to $\overline{V}^c$ and $U$. $\endgroup$ – mollyerin Jan 27 '15 at 2:50
  • $\begingroup$ @mollyerin I don't understand how to use the partition of unity, assuming I have one subordinate to $\overline{V}^c$ and $U$... $\endgroup$ – user174981 Jan 27 '15 at 2:55
  • $\begingroup$ Look at $\varphi_U f + \varphi_{\overline{V}^c}$. (Or even just $\varphi_U f$. The "partition of unity" part isn't the important part.) $\endgroup$ – mollyerin Jan 27 '15 at 2:57
1
$\begingroup$

Let $U$ be the open set containing the closed set $C$: $C \subset U$.

By assuption $X$ is paracompact and Hausdorff. This implies that $X$ is normal which means that closed sets can be separated.

Let $V,V'$ be separating sets for $C$ and $U^c$. Then $C \subset V \subset \overline{V}\subset U$. Note that $U$ and $\overline{V}^c$ cover $X$. Since $X$ is paracompact there exists a partition of unity subordinate to it. Let us denote the maps in the partition by $\varphi: U \to \mathbb R$ and $\psi: \overline{V}^c \to \mathbb R$.

Define $F = f \varphi + \psi$. Note that $F$ is smooth on $X$.

On $V$ we have $F(v) = f \varphi(v) = f \varphi (v) + f \cdot 0 = f \varphi (v) + f \cdot \psi (v) = f $.

$\endgroup$
  • $\begingroup$ Looks good to me. $\endgroup$ – mollyerin Jan 27 '15 at 3:56
  • $\begingroup$ What is $f\varphi + \psi$? Is it multiplication or composition in $f\varphi$? $\endgroup$ – user152715 Sep 12 '18 at 5:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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