Is there a generic criterion for a function to be the gradient of another function?

Question

I am interested to know whether there is a generic criterion in which a function $f:\mathbb{R}^n \to \mathbb{R}^n$ satisfies $f(x) = \nabla d(x)$ for some function $d:\mathbb{R}^n \to \mathbb{R}$.

Many thanks!

Answer 1

Provided your function is defined on all of $\mathbb{R}^n$, then you want to know if $f$ is exact. Since the cohomology of $\mathbb{R}^n$ is trivial, this happens exactly when $f$ is closed. That is, whenever $\mathrm{d} \! f = 0$. Writing this condition down properly requires your function be treated as a differential form, which is typically not much to ask for.

Broadly though, the condition you're looking for is (for all $i$ and $j$)

$$\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i}.$$

You can think of this as being an "obvious" necessary condition, since if $f_i = \frac{\partial f}{\partial x_i}$, as it would be if $\langle f_i \rangle = \nabla f$, then this condition is asking that $\frac{\partial^2 f}{\partial x_i \partial x_j} = \frac{\partial^2 f}{\partial x_j \partial x_i}$, which is true provided $f$ is smooth enough.

In $\mathbb{R}^3$, this is the familiar theorem that $f$ has a potential exactly when $\text{curl} f = 0$ (cf. this answer).

If your function is not defined everywhere, then this condition might fail. That is, you can have a function where $\mathrm{d} \! f = 0$ but it doesn't have a potential! In this case, the geometry of $U \subset \mathbb{R}^n$ where $f$ is defined becomes important, and indeed the cohomology tells you when the theorem is still true, and gives you more precise information about how the theorem fails when it does.

As mentioned in the below comment, this also relies on each of the $f_i$ being differentiable. But in the absence of such a condition, I don't know of any results.

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I hope this helps ^_^