Dirichlet density vs natural density

This is related to the edited form of this question. Suppose that I know that some set of primes has a certain Dirichlet density. What is the optimum statement one can make generally (for example, if the Dirichlet density is non-zero, is it true that the lower natural density is non-zero?

• Appears I had it backwards, I guess you know about this, if there is a natural density then the Dirichlet density exists and is the same, reverse maybe not. See answer at mathoverflow.net/questions/28047/… – Will Jagy Mar 11 '14 at 23:04
• @WillJagy I don't think this is true generally, but very likely true for some "special" properties (I think this is the content of Chambert-Loir comment on the original question) – Igor Rivin Mar 11 '14 at 23:05
• Just a quick note as I am in a hurry: the Dirichlet density equals the logarithmic density, so it is definitely a cruder measure than the natural density. Hope this helps to resolve your questions. – GH from MO Mar 12 '14 at 10:46
• @GH from MO. Thanks, that helps a lot. I didn't know that result. There is an elementary proof of it on math.stackexchange math.stackexchange.com/questions/444419/… Now I understand the better the wikipedia page on Dirichlet density, which cites the paper linked above for a proof that the set of primes beginning with 1 has no natural density but a Dirichlet density, while the article doesn't mention the Dirichlet density, just the logarithmic density. That said, the equality logarithmic / Dicichlet density should be mentioned in the wikipedia article. – Joël Mar 12 '14 at 14:24
• @Joël: Thanks for the link. It definitely proves that if the logarithmic density exists then so does the Dirichlet density (and they are equal). But it is also true that if the Dirichlet density exists then so does the logarithmic density (and they are equal). See Section III.1.3 of Tenenbaum: Introduction to analytic and probabilistic number theory. – GH from MO Mar 12 '14 at 20:20

Consider the set of integers $A$ which are in between $10^{n^2-n}$ and $10^{n^2}$ for some $n$. Then the upper natural density of $A$ is $1$, because among the $10^{n^2}$ first integers, at least $10^{n^2}-10^{n^2-n}$ are in $A$, so a proportion of $1-10^{-n}$ are in $A$. The lower density of $A$, on the other hand is $0$, for the number of integer up to $10^{n^2-n}$ of $A$ is at most $10^{(n-1)^2} = 10^{n^2-2n+1}$ so the proportion of those integers that are in $A$ is $10^{1-n}$.
Now let's compute the logarithmic density of $A$. Each interval $[10^{n^2-n},10^{n^2}]$ in $A$ contributes $\log(10^{n^2}) - \log(10^{n^2-n}) + O(1) = n \log 10 + O(1)$. So if $m$ is any number between $10^{n^2}$ and $10^{(n+1)^2}$, the number of elements of $A$ up to $m$ is $n^2 \log(10)/2 + O(n)$, and when divided by $\log m \sim n^2 \log(10)$, one gets $1/2 + O(1/n)$. Hence the logarithmic density of $A$ is $1/2$.
Of course, your question was about a set of primes, but then it suffices to replace $A$ by the set $A'$ of primes into $A$. But just applying the prime number theorem, we see that the computation of densities above are the same, and that proves that having a non-zero Dirichlet density says nothing about the lower or upper natural density.