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 Thank you for your comment. I had a new idea and edited my question.

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 I don't understand how to use the partition of unity, assuming I have one subordinate to $\overline{V}^c$ and $U$...

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 That was my previous idea but the problem with it is that $\varphi_U f$ is not smooth.

$\varphi_U f$ is certainly smooth; it's smooth on $U$ (since it's the product of two functions there); but $\varphi_U$ is equal to zero outside $U$, which forces it to be smooth everywhere.

Hello

hello!

The latex doesn't render in chat... too bad.

Yeah I really dislike that

I am trying to smoothly extend a smooth map $f: U \to \mathbb R$ where $U \subset M$ and $M$ is a smooth manifold.

Right. So ideally you'd like a function $\phi$ which is smooth and zero outside $U$, and also the constant 1 on a neighborhood of $C$

$\phi$ is like a bump function

If you have such a $\phi$ you can just take $\phi f$, which is smooth on all of $M$

I think even weaker than that: constant one on a neighborhood of $C$ would be good enough.

Or does it have to be zero outside $U$?

It had better go to zero outside $U$, otherwise what does $\phi f$ mean outside $U$?

Oops, good point :-)

But then $f \varphi_U$ is still not smooth because the support of $\varphi_U$ is contained in $U$.

We need to consider a partition of unity with respect to $V$ and $C^c$.

So why isn't $f \varphi_U$ smooth?

(I believe that it is smooth: It's smooth on $U$ and it's smooth on the complement of the support of $\varphi_U$.)

I cannot prove that it is not smooth but in general the product of smooth functions isn't necessarily smooth. Or is it?

The product of smooth functions is certainly smooth (on a domain where they're both defined)

For the same reason that the product of differentiable functions is differentiable

Good point: we can just apply the product rule of differentiation...

Beat me to it :-)

:)

Then the only problem we have is that $\varphi_U f$ does not necessarily equal $f$ on some neighborhood of $C$.

That's why you want to ensure that $\varphi_U$ is equal to 1 on a neighborhood of $C$

Right. But if we just use the existence of a partition of unity to obtain $\varphi_U$ then this is not necessarily true.

So suppose your partition of unity is subordinate to $U$ and $\overline{V}^c$

Then I claim you can show that $\varphi_U = 1$ on $V$.

Then on $U$, $\varphi_{\overline{V}^c} = 0$ therefore on $U$, $\varphi_U f + \varphi_{\overline{V}^c} = (\varphi_U + \varphi_{\overline{V}^c})f = f$?

This is true if you replace "on $U$" by "on $V$" everywhere it occurs

Oops, right, $V$ is contained in $U$.

Awesome. Thank you for your help!

You should write it up as an answer to the question and accept

You're welcome

I was just about to write that you should write it as an answer so that I can accept it and up vote.

You should get some reputation for helping me.

But if you'd rather not that's fine.

My instinct is that it works better if you do it :D

Good, I will do it.