# Show that if $\lVert a_{pn}\rVert>0$ for primes $p$, then $\lVert a_n \rVert\geq0$

If we define

$$\lVert a_n\rVert = \lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_k$$

To be the mean of a sequence, and we let $$a_n$$ be a bounded sequence of integers where not only does $$\lVert a_n\rVert$$ converge but $$\lVert a_{pn}\rVert$$ converges for all prime numbers $$p$$, where

$$\lVert a_{pn}\rVert=\lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_{pn}$$

Is the mean of the elements that are multiples of $$p$$.

If $$\lVert a_{pn}\rVert>0$$ $$\forall$$ primes $$p$$, then it feels only natural that $$\lVert a_n \rVert>0$$ as well, but I simply cannot seem to prove it. I have tested this property on several sequences and it seems to hold, but if this theorem does not hold and someone out there has a counterexample that would be greatly appreciated as well.

I think that this theorem has some deep consequences about the distribution of primes since from this theorem one can reasonably easily deduce the PNT.

• In your definition of "$\lVert a_n\rVert = \lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_k$", with the RHS, if the limit exists, it would be a single value. However, on the LHS, you have $a_n$, which doesn't match unless you want all of your $a_i$ to be the same value. Commented Apr 24, 2020 at 18:58
• The LHS of my equation, $\lVert a_n \rVert$, is not meant to be dependent on $n$ and only includes $n$ since it is a sequence. Should I replace it with $\lVert a \rVert$ to make this more clear? Commented Apr 24, 2020 at 19:01

Say that a function $$\chi_q: \mathbf{Z} \rightarrow \{-1,0,1\}$$ is suitable if:

1. $$\chi_q(n)$$ only depends on $$n \pmod q$$.
2. $$\chi_q(0) = 1$$.
3. $$\displaystyle{\sum_{n=0}^{p-1} \chi_q(n) = 0}$$.

If $$\chi_q(n)$$ is suitable, then so is $$\chi_q(pn)$$ for any prime $$(p,q) = 1$$. The average of any suitable $$\chi_q$$ along an arithmetic progression with difference prime to $$q$$ is zero. Moreover, the partial sums of $$\chi_q(pn)$$ along any such progression are bounded in absolute value by $$q$$, since they are just sums over sequences which repeat every $$q$$ terms. Finally, $$\chi_q(qn) = 1$$ for any $$n$$.

There certainly exist many suitable functions, let us fix an explicit such function by the relations:

1. $$\chi_q(n)$$ only depends on $$n \pmod q$$.
2. If $$0 \le n \le p-2$$, then $$\chi_q(n) = (-1)^n$$.
3. If $$n = p-1$$, then $$\chi_q(n) = 0$$. That is, the values of $$\chi_q(n)$$ for $$n = 0, \ldots p-1$$ are $$\{1,-1,1,-1,1,-1,\ldots,1,-1,0\}$$

We now let $$b_{n,q}$$ denote the following sequence:

$$b_{n,q} = \begin{cases} \chi_q(n) & 2^{q} \| n, \\ 0 & \text{otherwise} \end{cases}$$

We have $$\|b_{n,q}\| = 0$$, since we are averaging $$\chi_q(n)$$ over the arithmetic progression $$2^q \pmod {2^{q+1}}$$. Similarly, $$\|b_{np,q}\| = 0$$, since we are averaging $$\chi_q(pn)$$ over the same progression. Finally, $$b_{nq,q} = 1$$ whenever it is non-zero, so $$\|b_{nq,q}\| = \frac{1}{2^{q+1}}.$$ We now let $$c_n = \sum_{q > 2} b_{n,q}$$ Note that each $$n$$ is divisible by a unique power of $$2$$, and so $$c_n \in \{-1,0,1\}$$. We claim that $$\|c_n\| = \|c_{2n}\| = 0$$, and $$\|c_{qn}\| = 2^{-q-1}$$ for any odd prime $$q$$. This would follow immediately from the identities $$\|c_{n}\| = \sum \|b_{n,q}\|, \quad \|c_{np}\| = \sum \|b_{np,q}\|,$$ but we need to be slightly careful as the sums are not absolutely convergent. Still, it is relatively easy. Consider the case of $$\|c_n\|$$. If we sum up to $$X$$, the only contributions we see are coming from $$b_{n,q}$$ with primes $$q$$ such that $$2^q < X$$, or $$q < \log(X)$$ (in base $$2$$ but I don't want to bother writing subscripts). The partial sums for each $$b_{n,q}$$ are always bounded by $$q$$, and thus the partial sums of the $$c_n$$ up to $$X$$ are bounded by $$\sum_{q < \log(X)} q < \sum_{q < \log(X)} \log(X) < (\log(X))^2 = o(X).$$ In particular, the average of the $$a_n$$ is certainly tending to zero. The other cases are similar. Finally, to get the contribution at $$2$$ to work out, we can let $$a_n = c_n + \begin{cases} 1 & n \equiv 2 \mod 4 \\ -1 & n \equiv 3 \mod 4 \end{cases}$$ which is still valued in $$\{-1,0,1\}$$ and has the same averages as before except now $$\|a_{2n}\| = 1/2$$.

• It looks like you are missing some $i$-s in your $\exp$ terms. Commented Apr 24, 2020 at 23:56
• @user760870 once again this is a really cool example and I'm glad you put it, but it was for cases like this that I put the condition that $a_n$ is a sequence of integers. Commented Apr 25, 2020 at 1:02
• @MiloMoses Happy now? Commented Apr 25, 2020 at 3:09
• @user760870 yes! thank you, this is exactly the type of example I was looking for Commented Apr 25, 2020 at 5:30