# How to smoothly extend a function?

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 $f$).

• 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 $U$. – mollyerin Jan 26 '15 at 16:02
• @mollyerin Thank you for your comment. I had a new idea and edited my question. – user174981 Jan 27 '15 at 0:22
• 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 $U$. – mollyerin Jan 27 '15 at 2:50
• @mollyerin I don't understand how to use the partition of unity, assuming I have one subordinate to $\overline{V}^c$ and $U$... – user174981 Jan 27 '15 at 2:55
• 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.) – mollyerin Jan 27 '15 at 2:57

## 1 Answer

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 $F(v) = f \varphi(v) = f \varphi (v) + f \cdot 0 = f \varphi (v) + f \cdot \psi (v) = f$.

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