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


2 Answers 2


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.

  • $\begingroup$ In line 3, did you mean "the sequence whose $n$-th term is..."? $\endgroup$
    – Alon Amit
    Jun 11, 2011 at 6:50
  • $\begingroup$ @Alon Amit: yes, thanks. I'll make the edit. $\endgroup$ Jun 11, 2011 at 13:23
  • 4
    $\begingroup$ +1, very nice. I really like the way you look at it as a tensor product of operators. $\endgroup$ Jun 11, 2011 at 18:30
  • 4
    $\begingroup$ @Eric - thanks! The tensor-product description turns out not to be explicitly used in the proof (which indeed generalizes to some operators, such as the multiplicative convolution with $1/(k \log k)$, that do not factor as tensor products), but it did suggest where to find counterexamples. $\endgroup$ Jun 11, 2011 at 20:08
  • 1
    $\begingroup$ I don't think we need Tensor product of operators. Call the expression $f(x_{n})$. It suffices to find $\{x_{n}\}$'s with $|x_{n}|=1$ and $f(x_n)$ unbounded. For a large $N$, let $x_{N}= N^{-\frac{N}{2}}$ if $N=\prod_{i=1}^{N} p_{i}^{a_{i}}$ where $a_{i}<N$ (where $p_i$ is the i-th prime) and $0$ otherwise. $\endgroup$
    – C.S.
    Nov 7, 2017 at 11:09

In the recent issue of Matematikai Lapok, the report on the Schweitzer contest attributes the problem to Zoltan Buczolich and Julien Bremont.


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