How does $\int (u*v)dx = \int u dx \int v dx $ follow from $ (u*v)*w = u*(v*w) $ by taking $w = 1$?

Question

In Hormander's first book on PDEs he states on page 17 that

$$ (u*v)*w = u*(v*w) $$ if all except one of the continuous functions $u,v,w$ $\in C(\mathbb{R})$ have compact support. He then says

taking $w=1$ we find that

$$ \int (u*v)dx = \int u dx \int v dx \quad \quad (*) $$ when $u$ and $v$ have compact support.

I am already familiar with this identity, my question is how this follows from taking $w = 1$ in the previous statement. If I take $w=1$, for the left hand side I have

$$ ((u*v)*1)(x) = \int (u*v)(x-y)dy $$ Then integrating this I get

$$ \begin{align} \int ((u*v)*1)(x) dx & = \int \int (u*v)(x-y)dy dx \\ & = \int \bigg( \int (u*v)(x-y) dx \bigg) dy \\ & = \int \bigg( \int (u*v)(x) dx \bigg) dy \\ & = \bigg( \int (u*v)(x) dx \bigg) \int dy \\ & = \infty. \end{align} $$

So I don't see how setting $w = 1$ is going to lead to the expression $(*)$? Is the book incorrect or have I made a mistake?

Answer 1

You don't want to integrate again. Each side looks like a function of one variable, but this function is actually constant because the integral is translation-invariant. In particular, $$ (v * w)(x) = \int v(x-y) 1 \, dy = \int v(z) \, dz, $$ so "${}*w$" is just the integral over $\mathbb{R}$. This applies equally well to $(u * v) * w$, and convolution with any constant does the same by linearity, so for any $y$, $$ \int u * v \, dx = ((u * v) * 1)(y) \\ = (u * (v * 1))(y) = \left(u * \int v \, dx \right)(y) = (u * 1)(y) \int v \, dx = \int u \, dx \int v \, dx $$

Answer 2

\begin{align} (u*v)*1 &= \int (u*v)(x) \, dx\\ &= \int \int u(y)v(x-y) \, dy \, dx\\ &= \int u(y) \int v(x-y) \, dx \, dy \quad \text{[Fubini]}\\ &= \int u(y) \int v(x) \, dx \, dy \quad \text{[Change of variables $\;x-y \mapsto x$]}\\ &= \int u(y) \, dy \int v(x) \, dx \end{align}