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


*

*$\chi_q(n)$ only depends on $n \pmod q$.

*$\chi_q(0) = 1$.

*$\displaystyle{\sum_{n=0}^{p-1} \chi_q(n) = 0}$.


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:


*

*$\chi_q(n)$ only depends on $n \pmod q$.

*If $0 \le n \le p-2$, then $\chi_q(n) = (-1)^n$.

*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 $\|a_{2n}\| = 1/2$.
