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I was looking on at the Convolution page on wikipedia and saw that it stated that we can define the convolution of two Borel measures of bounded variation on $\mathbb{R}^d$, $\mu$ and $\nu$, to be $\int_{\mathbb{R}^d} f(x) d(\mu\ast \nu)(x) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x+y) d\mu(x) d\nu(y),$ and that we have the result $||\mu \ast \nu || \leq ||\mu|| ||\nu||.$

I'm not sure what the second statement means though. Is there a natural norm for the space of Borel measures with bounded variation?

2 Answers 2

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DJC's answer is of course to the point. I'm adding a short sketch of a proof of the inequality $\|\mu \ast \nu\| \leq \|\mu\| \,\|\nu\|$ since you asked about it in a comment. In that comment you mentioned $\sigma$-finiteness. Note that bounded variation implies that all measures involved are in fact finite.

One of the most convenient ways of writing the total variation norms is as

$\|\mu\| = \sup_{|f| \leq 1}{\;\left|\int f\,d\mu\right|}$

where it doesn't matter much what kinds of measurable functions $f$ you allow in the supremum (continuous; simple; Borel; smooth; compactly supported or not). I'll take Borel here.

Given this, the inequality is then clear: By Fubini we have for every Borel function $f$ with $|f|\leq 1$ that the function $y \longmapsto \left|\int f(x+y)\,d\mu(x)\right|$ is Borel and bounded by $\|\mu\|$ hence the fact that $\|\nu\| = \| \,|\nu|\,\|$ gives $\left|\int f \, d(\mu \ast \nu) \right| = \left|\iint f(x+y)\,d\mu(x)\,d\nu(y)\right| \leq \int\left|\int f(x+y)\,d\mu(x)\right|\,d|\nu|(y) \leq \|\mu\| \,\|\nu\|.$ The norm of $\|\mu \ast \nu\|$ is by definition the supremum over the left hand side over $|f| \leq 1$.

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    It's rather easy. See e.g. Kechris, *[Classical descriptive set theory](http://books.google.com/books?id=pPv9KCEkklsC&pg=PA107)* section 17.C, theorems 17.10 and 17.11. page 107 (link should go directly to that page on Google Books). It's also in most of the serious texts on measure theory.2011-08-25
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Yes. Generally, $\| \mu\|$ is defined as

$ \| \mu \| = |\mu|(\mathbb{R}^d), $

where $|\mu |$ is the total variation.

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    That definition should be well-defined for every signed(not necessarily being Borel) measure right? Also, using that definition, does the convolution inequality then come from using applying Fubini's Theorem (like the way it would if the two measures were absolutely continuous with respect to Lebesgue measure)? You would also need the $\sigma$-finite property in that case then right?2011-08-24