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The exposition I'm talking about can be found here (page 6): http://www.math.ucla.edu/~tao/preprints/Expository/prime.dvi

Essentialy, Tao proves the prime number theorem in the elementary way, although in disguised form. I have been looking at this for a while now and I still have some problems understanding the notation

The definition of convolution of measures is the following, let $\mu_1, \mu_2$ be two measures, then their convolution $\mu_1 * \mu_2$ is defined as

$\int f(x) (\mu_1 * \mu_2)(x)=\iint f(x+y) \mu_1(x) \mu_2(y)$

I can see that if you take an arithmetical functions $f(n)$ that you can define a measure $M_f(x)=\sum_n \frac{1}{n} f(n) \delta(x - \log n)$, (where $\delta$ denotes the Dirac Delta measure). Also if $f(n),g(n)$ are arithmetical functions then the Dirichlet convolution is related to the convolution of measures since

$M_f M_g=\sum_n \sum_m \frac{1}{nm} f(n) g(m) \delta (x - \log n) \delta(x- \log m)$ and now substituing $r=nm$

$=\sum_r \frac{1}{r} \delta(x- \log r) \sum_{d|r} f(d) g(r/d)$

$=\sum_r \frac{1}{r} \delta(x- \log r) (f\#g)(r) $

Where $\#$ denotes Dirichlet convolution. What I dont really grasp is mostly his notation, what does he mean with

$ H(x) \mathrm{d}x * H(x) \mathrm{d}x =x H(x) \mathrm{d}x$

(H(x) is simply $\mathbb{1}_{[0,\infty]})$.

EDIT: Specifically, I'm trying to prove that the map

$\phi: A \to C \\f \mapsto M_f$

from the algebra of arithmetical functions with Dirichlet convolution and addition to the algebra of measures with convolution of measures and addition is an algebra homomorphism. Why does the fact that $M_f M_g=M_{f \#g}$ imply that $\int u(x) (M_f * M_g)(x) = \int u(x) M_{f \# g}(x)$

Also his asymptotic notation I dont really understand, but it is most likely the same problem, what does he mean with this?

$M_\mu * H(x) \mathrm{d}x=O(1)H(x) \mathrm{d} x$

Does this mean that

$\int \mathbb{1}_{[0,x]} (M_\mu * H(x) \mathrm{d} x)=\mathrm{O}(x)$ ?

I have taken undergraduate measure theory, but I have not seen this type of notation before (we used Measures, Integrals and Martingales by Rene L. Schilling).

As a follow up question, I wonder why he used $\delta(x- \log n)$ instead of $\delta(x-n)$, i.e. why he uses the log integers instead of the integers, I feel like this might be a convergence problem, but it isn't really explained in his exposition.

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1 Answer 1

up vote 9 down vote accepted
  • Firstly, the only "multiplication" of measures that makes sense is convolution, and when you are writing $M_f M_g$, you should be writing $M_f *M_g,$ or, at least, understanding that this is what is meant. Then the fact that $\int u(x) M_f *M_g = \int u(x) M_{f\# g} $ follows from the definition of convolution and the computation that $M_f * M_g = M_{f \# g}$ made earlier.

  • Secondly, the formula $H(x) dx * H(x) dx = x H(x) dx$ means exactly what it says; each of $H(x) dx$ and $x H(x) dx$ is a measure, and this is a formula that says that the convolution of the former with itself equals the latter. To check it, just use the definition of convolution, and make a simple change of variable in the resulting double integral.

  • A general formula (that you should check) is that $\delta(x + a) * \delta(x + b) = \delta (x + a+ b).$ This is why the $\log$'s appear; because $\log m + \log n = \log m n,$ and thus one gets a relationship between convolution (which is essentially about the additive structure of the real numbers) and Dirichlet convolution (which is about the multiplicative structure of the natural numbers).

  • $M_{\mu} * H(x) dx = O(1) H(x) dx$ means that the LHS is equal to $f(x) dx$ for some function $f(x)$ which is uniformly bounded by some multiple of $H(x)$. Other instances of asymptotic notation have similar meanings.

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Thanks, this makes alot of sense. –  user45878 Jul 16 at 15:40

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