Pulling out a distribution ? $u\circ f$?

Question

In an attempt to generalize solutions with discontinuities to a PDE, we want to consider distributions.

We have a continuous function $g\in C(Y)$ and a smooth function $f:\mathbb{R}^n\rightarrow Y,\ f$ is $C^\infty$ , we want to generalize the classical continuous function $g\circ f: \mathbb{R}^n\rightarrow \mathbb{R}$ to a distribution.

We are able to generalize $g$ as a distribution $u\in \mathscr{D}'(Y)$ but we want a distribution $U\in \mathscr{D}'(\mathbb{R}^n)$ that generalize $g\circ f$ , by pulling back : $U= u\circ f$ .

I don't quite understand how am I allowed to do that ? Here is what I tried :

If $\phi$ is a test function on $\mathbb{R}^n$ then we can map it to the test function $T(\phi)$ on $Y$ that is defined by $$T(\phi) (y)= \int_{f^{-1}\big(\{y\}\big)}\phi(x)dx \quad, \text{ for all } y\in Y$$

Here $T$ is a well defined operator. Is this a well known construction ?
Then we can consider $U(\phi)= u(T(\phi))= u\circ T(\phi)$ .

But does $U$ generelize $g\circ f$ ?

Attendum :
The space $Y$ is the real line $\mathbb{R}$ and my $f$ is not even injective, $f: \mathbb{R}^n\rightarrow \mathbb{R}$ is rather a projection and has the form $f(x)= x\cdot v$ where $v$ is a fixed vector of $\mathbb{R}^n$ .

Answer 1

There is a simple formula for the case where the distribution $g$ is of finite order and $f$ is a diffeomorphism. The first condition means that $f$ has the form $D^nG$ (distributional derivative) for a continuous $G$ and integer $n$. The required distribution is then $\left(\frac 1{f‘}D\right)^n (G\circ f)$. This suffices for most practical situations, e.g., where $g$ is the Dirac distribution, and can be generalised to more general ones using standard localisation methods („recollement des morceaux“).

Answer 2

As a matter of strategy, try to define the object that you are looking for, take note of where your approach fails and correct it in these places.

If $u \in \mathcal D'(Y)$ and $f : X \to Y$ is smooth between smooth manifolds, you would like to somehow obtain $u \circ f \in \mathcal D' (X)$. Let $\varphi \in \mathcal D (X)$ be a test function. No matter how you defined $u \circ f$, you would clearly want it to have the reasonable property that $\langle u \circ f, \varphi \rangle = \langle u, (\varphi \circ f^{-1}) \, |\det (\mathrm d f)^{-1}| \rangle$, because you want to recover the usual change of variable formula when $u$ is an integrable function. This shows that $f$ must be a diffeomorphism, otherwise there is nothing left for you to do. Next, you encounter another problem: while it is true that $(\varphi \circ f^{-1}) \, |\det (\mathrm d f)^{-1}|$ is smooth on $Y$, nobody guarantees that is has compact support. You can fix this in only two ways:

• either require $u$ to be a compactly supported distribution (because the test functions for these need not be compactly-supported);
• or require $f$ to be a proper map.

In any case, the requirement that $f$ be a diffeomorphism massively restricts who $Y$ can be. This shows that the pull-back is not a very interesting operation on distributions.