Every $L^2$ function is the divergence of a $L^2$ vector field

Question

As in the title, I am struggling with the following statement:

For any $f\in L^2(\mathbb R^n)$ there exists a vector field $F\in L^2(\mathbb R^n, \mathbb R^n) $ such that $f=div F$.

Apparently this should follow from the Hodge decomposition...

Any help appreciated!

Answer 1

As was pointed out in another answer, this is not true in general. A simple example is the function

$$f(x) = \begin{cases} 1,&\text{if } -1 \leq x \leq 1\\ 0,&\text{otherwise.} \end{cases}$$

All antiderivatives of this function are not $L^2$.

However, there are some simple conditions one can place on $f$ to make the result true. To see how this may be done, let's try to construct $F$. One way to do this is to solve Poisson's equation

$$\Delta u = f \ \text{ in } \mathbb{R}^n$$

and then set $F=\nabla u$. The solution $u$ satisfies $\hat{u}(k) = \hat{f}(k)/|k|^2$ and we need $u\in H^1(\mathbb{R}^n)$ to ensure $F=\nabla u \in L^2(\mathbb{R}^n)$. Hence, a sufficient condition for the existence of $F$ is

$$\frac{(1 + |k|^2)^{1/2}}{|k|^2} \hat{f}(k) \in L^2(\mathbb{R}^n).$$

This is optimal for any proof passing through the Poisson problem, but may not be optimal in general.

Answer 2

I think the statement is false.

If the statement were true it shall also be true in the case of $n = 1$. Take a function $f \in L^2$, then if $\partial_x \, \varphi = f$ that will imply that, up to constants, $\hat\varphi(\xi) \xi = \hat{f}(\xi)$. If you choose $f$ so that its Fourier transform is locally constant around $0$ (something you can always do by Plancherel's Theorem) that will imply that $\varphi(\xi)$ is comparable to $\xi^{-1}$ around the origin and therefore not in $L^2$.