As part of an exercise, my textbook (Analysis on Manifolds) first gives an alternate definition as to what it means for a function $f$ to be differentiable at a point.

enter image description here

It then asks us to extend $f$ to $h$ using the method described below.

enter image description here

I question why one must use a partition of unity to make this extension. I would define a new function, albeit not a tame one, as follows:

$$g(x): A \to \mathbb{R}^{n}$$

$$g(x) = g_{x}(x)$$

where $$A = \cup U_{x}$$ where $U_{x}$ is the open set given in the definition in part $a$ over which some $C^{r}$ function $g_{x}$ exists so that $g_{x} = f$ on $U_{x}$

I sense that this function is not easy to work with. How is a function defined with a partition of unity any better?

If it helps anyone, here is the entire problem statement:

enter image description here

EDIT I sense it would be good to compare my naive definition above with what I presume is a representation using a partition of unity.

Let the open sets $U_{x}$ for each $x \in S$ be put into a collection $\{U_{1}, U_{2}, \dots \}$ (can we do this if $U_{i}$ are uncountable)? Let their union be $A$, and make a $C^{\infty}$ partition of unity $\{\phi_{i}\}$ on $A$ dominated by this collection. Now create our extension:

$$H(x) = \sum_{i=1}^{\infty}\phi_{i}(x)g_{i}(x) = \sum_{i=1}^{\infty}h(x) $$

where $g_{i}(x)$ is the $C^{r}$ function agreeing with $f$ on $U_{i}(x)$ and $h(x)$ is a proven (this was part a of the exercise) well-defined function.By the local finiteness condition, this sum has only finitely many non-zero terms, so the finite summands $\phi_{i}(x)g(x)$, each $C^{r}$, are also $C^{r}$ when summed together.

  • $\begingroup$ How do you know that the map $x \to g_x(x)$ is even continuous? $\endgroup$ – guest Mar 5 '17 at 5:00
  • 1
    $\begingroup$ How do you know $x \mapsto g_x(x)$ is well-defined? If $z \in U_x$ and $z \in U_y$, you would need $g_x(z) = g_y(z)$. $\endgroup$ – user14972 Mar 5 '17 at 5:16
  • $\begingroup$ @guest I see. Thanks $\endgroup$ – Muno Mar 5 '17 at 5:33
  • $\begingroup$ Hurkyl's anwer has it all! It is the well-definedness issue. You must have a way of passing from $g_x$ on one neighborhood to $g_y$ on a different one in a $C^r$ smooth manner. To be able to do this, you take a weighted average of them. And this weights come from the partition of unity. $\endgroup$ – Behnam Esmayli Mar 5 '17 at 6:04

As you seem to have observed yourself, the problem is that your function $g$ is not "tame". There's no reason to expect it to be continuous, let alone differentiable: you're using a different function $g_x$ to define its value at each different point $x$, and it would be a huge coincidence if somehow they all managed to come together to give a continuous function. On the other hand, as you note, the function constructed with the partition of unity is guaranteed to be $C^r$, because in a neighborhood of each point you are just taking a finite sum of $C^r$ functions. If you just wanted some extension of $f$ to a function on $A$ and didn't care about it being a nice function, there is a much easier way to do that: for instance, you could just define your function to be $0$ at every point of $A\setminus S$.


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.