Is there any sequence $a\_n$ of nonnegative numbers for which $\displaystyle\sum\_{n \geq 1}a\_n^2 <\infty$ and


$$\sum\_{n \geq 1}\left(\sum\_{k \geq 1}\frac{a\_{kn}}{k}\right)^2=\infty\quad?$$


See also <https://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010>



Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and  $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$?
Yes, such sequences exist.


In effect the problem concerns the linear operator, call it $T$, that maps any sequence $(a\_n)$ to the sequence whose $n$-th term is $\sum\_{k\geq1} a\_{kn}/k$. The problem asks whether there exists $a$ such that the $l^2$ norm $\|a\|\_2$ is finite but $\|Ta\|\_2 = \infty$. We show that such $a$ exist for each $l^q$ norm ($q \geq 1$). [I can't call it $l^p$ because I'll soon want to use $p$ for a generic prime number.]


The key is that $T$ can be regarded as an infinite tensor product (Kronecker product) of operators $T\_p$, where for each prime $p$ the operator $T\_p$ is convolution with $(1,p^{-1},p^{-2},p^{-3},\ldots)$ on $(a\_1,a\_p,a\_{p^2},a\_{p^3},\ldots)$. The operator $T\_p$ has norm $\sum\_{i=0}^\infty p^{-i} = p/(p-1)$, so $T = \otimes\_p T\_p$ should have norm $\prod\_p p/(p-1) = \infty$.


To get a construction from this, we first make for each $M>1$ a nonnegative sequence $\alpha = (\alpha\_n)$ such that $\|\alpha\|\_q < \infty$ and $T\alpha = M\alpha$. Since $\prod\_p p/(p-1)$ diverges, we can find $x$ large enough that $\prod\_{p\leq x} p/(p-1) > M$. Choose for each prime $p\leq x$ a positive number $z\_p$ smaller than 1 but close enough to 1 that $\prod\_{p\leq x} p/(p-z\_p) = M$. Now let $\alpha$ be the totally multiplicative function taking each $p\leq x$ to $z\_p$ and each $p>x$ to $0$. (That is, if $n=\prod\_{p\leq x} p^{e\_p}$ then $\alpha\_n = \prod\_{p \leq x} z\_p^{e\_p}$, while if $n$ has a prime factor bigger than $x$ than $\alpha\_n = 0$.) Then the $n$-th term of $T\alpha$ is
$$
\sum\_{k\geq 1} \frac{\alpha\_{kn}}{k} = \sum\_{k\geq 1} \frac{\alpha\_k \alpha\_n}{k}
= \alpha\_n \sum\_{k\geq 1} \frac{\alpha\_k}{k}
$$
because $\alpha$ is multiplicative, and $\sum\_{k\geq 1} \alpha\_k/k$ is the product over $p\leq x$ of geometric series
$$
\sum\_{e=0}^\infty \frac{\alpha\_{p^e}}{p^e} = \sum\_{e=0}^\infty (z\_p/p)^e = \frac{p}{p-z\_p},
$$
so
$$
T\alpha = \prod\_{p\leq x} \frac{p}{p-z\_p} \cdot \alpha = M \alpha;
$$
and $\|\alpha\|\_q < \infty$ because $\|\alpha\|\_q^q$ is the product of $\pi(x)$ convergent geometric series.


The conclusion is a standard argument (as already noted in the parallel *stackexchange* thread). For example, let $a^{(M)} = \alpha / \|\alpha\|$ where $\alpha$ is a sequence with $T\alpha = M\alpha$ as in the last paragraph. Then $\|a^{(M)}\| = 1$ and $Ta^{(M)} = M a^{(M)}$. Set
$$
a = \sum\_{m=1}^\infty \frac{a^{(4^m)}}{2^m} = \frac{a^{(4)}}{2} + \frac{a^{(16)}}{4} + \frac{a^{(64)}}{8} + \cdots .
$$
The $m$-th term has norm $1/2^m$, so the sum converges in $l^q$ to a vector of norm at most $\sum\_{m=1}^\infty 1/2^m = 1$. But since each term $a^{(4^m)} / 2^m$ is a nonnegative sequence, and $T$ has nonnegative coefficients, we have for each $m \geq 1$
$$
Ta > T\frac{a^{(4^m)}}{2^m} = 2^m a^{(4^m)};
$$
and $2^m a^{(4^m)}$ is a vector of norm $2^m$. Hence $\|Ta\|\_q = \infty$, **QED**.


