Fourier transform of gradient

Question

I encountered in a physics book the Fourier transform $F$ of the gradient of a function $g$ smooth with compact support on $\mathbb R^3$. Up to some multiplicative constants:

$F(\nabla g)(k)=k.F(g)(k) $

The book claims that this is proved by integration by parts, again up to some multiplicative constants:

$\int exp(-ik.x).\nabla g(x)dx= k\int exp(-ik.x).g(x)dx$

I understand the logic but I can't make sense of an expression like $\int_{\mathbb R^3} \nabla g(x)dx$ and can't see how the equality above holds. I tried to relate this to the divergence theorem but without luck, because $\nabla g(x)$ is a vector after all.

What is the definition of the expression $\int_{\mathbb R^3} \nabla g(x)dx$ and why is the equality above true?

Answer 1

The integral is interpreted coordinate-wise; $$\left(\int \mathbf v\right)_j =\int v_j$$

Now looking at the result given, you have $$\int f(k,x) \nabla_j g(x) = \int \nabla_j (fg)-g\nabla_j f$$ Applying the divergence theorem to the first term (it's a divergence of $fg \mathbf n$ for a particularly simple $\mathbf n$) makes it go away.

Answer 2

It's common practice to interpret the integral of a vector to be the vector of integrals of coordinates. This recipe actually has a name when the vectors come from infinite-dimensional spaces: the Bochner integral. Put simply, let $f : X \rightarrow V$ be a "measurable" map from a measure space $(X,\mu)$ to some topological vector space $V$ (in applications $V$ is assumed Banach or Frechet).

Then you have to write some theorem down that makes sense of $\int_X f(x) d \mu(x)$ for $f$ nonnegative. This is usually stated in the following way: there exists a unique element $\int_X f(x) d\mu(x)$ of $V$ such that for each $l \in V^*$, $$ l\left(\int_X f(x) d\mu(x) \right) = \int_X (l \circ f)(x) d \mu(x) $$ This integral is additive, homothetic, and obeys dominated convergence and many other classical results. The definition I give here works perfectly well for finite-dimensional vector spaces, by the way.

Think of the above as a generalization of "coordinatewise integral".