Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:55:34Z http://mathoverflow.net/feeds/question/67436 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67436/is-there-any-sequence-a-n-of-nonnegative-numbers-for-which-sum-n-geq-1a-n Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$? a_MSE_user 2011-06-10T12:31:12Z 2012-01-13T07:07:57Z <p>Is there any sequence $a_n$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 &lt;\infty$ and</p> <p>$$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$</p> <p>See also <a href="http://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010" rel="nofollow">http://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010</a></p> http://mathoverflow.net/questions/67436/is-there-any-sequence-a-n-of-nonnegative-numbers-for-which-sum-n-geq-1a-n/67487#67487 Answer by Noam D. Elkies for Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$? Noam D. Elkies 2011-06-11T04:41:25Z 2011-06-11T20:06:25Z <p>Yes, such sequences exist.</p> <p>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.]</p> <p>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 $\prod_p p/(p-1) = \infty$.</p> <p>To get a construction from this, we first make for each $M>1$ a nonnegative sequence $\alpha = (\alpha_n)$ such that $\|\alpha\|_q &lt; \infty$ 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 $\|\alpha\|_q &lt; \infty$ because $\|\alpha\|_q^q$ is the product of $\pi(x)$ convergent geometric series.</p> <p>The conclusion is a standard argument (as already noted in the parallel <em>stackexchange</em> 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 $\|Ta\|_q = \infty$, <strong>QED</strong>.</p> http://mathoverflow.net/questions/67436/is-there-any-sequence-a-n-of-nonnegative-numbers-for-which-sum-n-geq-1a-n/85559#85559 Answer by Péter Komjáth for Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$? Péter Komjáth 2012-01-13T07:07:57Z 2012-01-13T07:07:57Z <p>In the recent issue of Matematikai Lapok, the report on the Schweitzer contest attributes the problem to Zoltan Buczolich and Julien Bremont. </p>