TITLE: Prove or disprove that $\sum_{n=1}^{\infty}\frac{\lambda(n)\mathbb{E}_{n\in\mathbb{N}}[a_n]}{n}=0$ for any choice of $(a_n)$
QUESTION [21 upvotes]: I've been obsessed with this one problem for many months now, and today is the sad day that I admit to myself I won't be able to solve it and I need your help. The problem is simple. We let
$$\mathbb{E}_{n\in\mathbb{N}}[f(n)]:=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(n)$$
denote the expected value of a function $f(n)$ (if it exists). This means that for some fixed choice of sequence $(a_n)_{n=1}^{\infty}$ the quantities $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ will give the "average value of every $k$" elements. For example, $\mathbb{E}_{n\in\mathbb{N}}[a_{3n}]$ will be the average of $a_{3},a_{6},a_{9}$ etc... . If all of these quantities exist, we will call a sequence $a_n$ to be "sequentially summable". The problem is as follows:

Show that for any bounded sequentially summable $a_n$
\begin{equation}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}=0\tag{1}\end{equation}
where $\lambda(n)$ denotes the Liouville function

Swapping which side the term $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ is on, this gives an absolutely lovely formula for $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ from $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ ($k\geq 2$) which can be appreciated by even people who have done no mathematics. Similarly, I would expect that the conjecture holds for $\lambda(n)$ replaced by its "twin" $\mu(n)$, the Mobius function. Throughout the rest of this question, I will give all of the partial results and a general outline of how they are obtained.
This first partial results answers the question "why on earth would these values be 0???":

Partial Result $\#1$: For any bounded sequentially summable sequence $a_n$, it holds that
\begin{equation}\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^s}=\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\mu(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^{s}}=0\tag{2}\end{equation}
and thus if the sum in (1) converges it must converge to $0$

This result is obtained by inverting the summation order and exploiting the fact that
$$\sum_{d|n}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)>0$$
which means that the triangle inequality does not "lose" any sign cancelation. The following partial result is much stronger:

Partial Result $\#2$: There exists an absolute constant $c_0$ such that for any sequentially summable sequence $a_n$ and $N>0$
\begin{equation}\left|\sum_{k=1}^N\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}\right|<c_0m\tag{3}\end{equation}
where
$$m^2=\limsup_{N\to\infty}\frac{1}{N}\sum_{n=1}^Na^2_{n}$$

An outline of the proof of this result is given in this Math.SE question, but in essence the result comes from the fact that the boundedness of
\begin{equation}\left|\sum_{k=1}^N\frac{f(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}\right|\end{equation}
for some function $f(k)$ is essentially equivalent to tight enough bounds on the partial sums
$$\sum_{n=1}^{N}\frac{f(mk)}{k}$$
for all values of $m<N$, which $\mu(n)$ definitely has. The boundedness in Partial Result $\#2$ can be translated to bounds when $\mu(n)$ is replaced by $\lambda(n)$ by exploiting the identity $\lambda(n)=\sum_{d^2|n}\mu\left(\frac{n}{d^2}\right)$ and the relative uniformity of the bound w.r.t out choice of $a_n$. Here is the next partial result

Partial Result $\#3$: If, for any bounded sequentially summable sequence $a_n$, we have that
\begin{equation}
\lim_{N\to\infty}\frac{1}{N}\sum_{k=1}^{N}\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]=0
\end{equation}
then our conjecute (1) holds.

This is a very important partial result since it is often much easier to show that the average of coefficients is zero than to show that they converge when summed with a factor of $\frac{1}{k}$. This result is done in a "smoothing" manner where Partial Result $\#3$ can be considered as control over short intervals and Partial Result $\#2$ can be considered as control over large intervals which we can bound as error.
At this point I will note that the condition that $a_n$ be bounded is quite important. For example, if $a_n=\Lambda(n)$ is the Von-Magnolt function then by the PNT $\mathbb{E}_{n\in\mathbb{N}}[a_n]=1$ but since $kn$ will have "few" prime powers for $k>1$ we have that $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]=0$ and thus our sum converges, but not to $0$.
I add as well that
$$\sum_{k=1}^{N}\frac{\mu(k)}{k}\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}=\frac{a_1}{N}$$
due to simple inversion of summation order, and so since
$$\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}\approx \mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$$
we get further intuition for the result.

REPLY [10 votes]: Here is a more classical (more streamlined) version of Terry's nice argument. It avoids any reference to generalized limits (hence the axiom of choice).
We shall only use that $a_n$ is sequentially summable and bounded in square mean:
$$\sum_{n\leq N}|a_n|^2\ll N.$$
By assumption, the limits
$$f(k):=\lim_{N\to\infty}\frac{1}{N}\sum_{n\leq N}a_{kn}$$
exist, and we want to show that
$$\sum_{k=1}^\infty\frac{\lambda(k)f(k)}{k}=0.\tag{$\ast$}$$
Lemma. The multiplicative convolution $g:=\mu\ast f$ satisfies
$$\sum_{q=1}^\infty\frac{|g(q)|^2}{\varphi(q)}<\infty.$$
Proof. For the exponential sums
$$S_N(x):=\sum_{n\leq N} a_n e(nx),$$
we observe that
$$\sum_{q\mid k}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q)=\sum_{c\bmod k}S_N(c/k)=k\sum_{\substack{n\leq N\\k\mid n}}a_n=k\sum_{m\leq N/k}a_{km}.$$
Dividing by $N$ and letting $N\to\infty$, the right hand side tends to $f(k)$. Therefore,
$$f(k)=\lim_{N\to\infty}\frac{1}{N}\sum_{q\mid k}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q).$$
It follows that
$$g(q)=\lim_{N\to\infty}\frac{1}{N}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q).$$
In particular, the limit on the right hand side exists. Using the Cauchy-Schwarz inequality and the large sieve inequality, we infer
$$\sum_{q\leq Q}\frac{|g(q)|^2}{\varphi(q)}\leq\limsup_{N\to\infty}\frac{1}{N^2}\sum_{q\leq Q}\sideset{}{^*}\sum_{b\bmod q}|S_N(b/q)|^2\leq\limsup_{N\to\infty}\frac{1}{N}\sum_{n\leq N}|a_n|^2\ll 1.$$
Letting $Q$ tend to infinity, the inequality in the lemma follows. $\Box$
For $g$ as in the Lemma, we have $f=1\ast g$, hence also
$$\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}=\sum_{k\leq K}\frac{\lambda(k)}{k}\sum_{q\mid k}g(q)=\sum_{q\leq K}g(q)\sum_{\substack{k\leq K\\q\mid k}}\frac{\lambda(k)}{k}
=\sum_{q\leq K}\frac{g(q)\lambda(q)}{q}\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}.$$
In particular,
$$\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|\leq
\sum_{q\leq K}\frac{|g(q)|}{q}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|.$$
Let us fix a large $L>0$. We cut the $q$-sum into two parts $q\leq K/L$ and $q>K/L$, and we apply Cauchy-Schwarz to both parts:
$$\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|^2\ll
\left(\sum_{q\leq K/L}\frac{|g(q)|^2}{\varphi(q)}\right)\left(\sum_{q\leq K/L}\frac{\varphi(q)}{q^2}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|^2\right)
+\left(\sum_{K/L<q\leq K}\frac{|g(q)|^2}{\varphi(q)}\right)\left(\sum_{K/L<q\leq K}\frac{\varphi(q)}{q^2}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|^2\right).$$
Now we let $K\to\infty$, and apply the Lemma. On the right hand side, the first $q$-sum is $O(1)$, the third $q$-sum is $o_L(1)$, and the fourth $q$-sum is $O_L(1)$. As a result,
$$\limsup_{K\to\infty}\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|^2
\ll\limsup_{K\to\infty}\sum_{q\leq K/L}\frac{\varphi(q)}{q^2}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|^2.$$
By the Prime Number Theorem, the $\ell$-sum is $\ll 1/\log(K/q)$, where of course $K/q\geq L$. By applying a dyadic decomposition for $q$, we conclude that
$$\limsup_{K\to\infty}\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|^2
\ll\frac{1}{\log L}.$$
Letting $L\to\infty$, we see that the left hand side is zero. We have proved $(\ast)$.