2
$\begingroup$

Let $T$ be a distribution, $f$ be smooth. It is known that

$T(f(x - \cdot) ) \in C ^ \infty$ .

On the other hand, for $\phi$ a test function

$( T \ast f ) (\phi) = T( f( - \cdot ) \ast \phi ) = T( \int f( x - \cdot ) \phi( \cdot ) )$,

because the convolution is still a distribution. However, with regard to the first result we expect

$ T( \int f( x - \cdot ) \phi( \cdot ) ) = \int T ( f( x - \cdot ) \phi( \cdot ) )$.

This intuitively makes sense, as $T$ is a distribution in the $x$-variable, whereas we integrate over the unnamed second variable.

Nevertheless, I don't know a proof of the last equality, and I am not familiar enough with the topologies of the smooth functions, test functions and distribution, to use approximative arguments. Can you help?

Thanks.

  • 1
    (1) Your notation is a bit messy. The left-hand side of the last displayed formula should not have a $*$. The last term on the right hand side of the second displayed formula is incorrect: it should have $\cdot - x$ instead of what you wrote as an argument to $f$. (2) Convolution of a distribution with merely a smooth function is not always well-defined. Generally you require $f$ to be of compact support.2010-11-16
  • 1
    (3) If $T$ were representable by a locally integrable function, then your identity (when correctly phrased) follows from Fubini's theorem. In the general case, the RHS of your expression needs some interpretation, as $T$ is technically a linear functional on $\mathcal{D}(\mathbb{R}^d)$, whereas you are trying to act it on $\mathcal{D}(\mathbb{R}^{2d})$. So technically the $T$ on the RHS should be the pull-back distribution of $T$ by the projection map $\mathbb{R}^{2d}\to\mathbb{R}^d$. After that there should be a simple argument, which I'll get back to you later.2010-11-16
  • 0
    It just occurred to me that you need to consider tensor products of distributions to make the argument precise. See, e.g., chapter 5 of volume 1 of Lars Hormander's "Analysis of Linear Partial Differential Operators".2010-11-16

1 Answers 1

2

I'll do it for when $f(x)$ has compact support; the general Schwartz function case is a modification of this. I use the fact that the Schwatrz functions are sequentially dense in the distributions, meaning that there is a sequence of Schwartz functions $g_n$ such that $\int g_n(x) \alpha(x)$ converges to $T(\alpha)$ as $n$ goes to infinity for any Schwartz function $\alpha(x)$.

If you replace $T$ by $g_n(x)$, then your identity holds by Fubini's theorem. So one takes limits now as $n$ goes to infinity. The left hand side converges to $T \int f(x - \circ)\phi(\circ) d\circ$ immediately.

For the right-hand side, for each $x$ one has that $g_n(f(x - \circ) \phi(\circ))$ converges to $T(f(x - \circ) \phi(\circ))$. Since the $g_n(f(x - \circ) \phi(\circ))$ will be uniformly bounded (the construction of the $g_n$ is quite explicit and readily implies this) and since $f(x - \circ) \phi(\circ)$ is the zero distribution for $x$ outside a compact set, the dominated convergence theorem gives that the right hand side converges to $\int T (f(x - \circ) \phi(\circ)) dx$ as needed.