Let $t>0$ and $\delta\in\big(0,\frac12\big)$ be fixed. For any $k\in\mathbb{N}$ let $I_k,J_k\in\mathbb{N}$ be finite subsets of natural numbers with cardinalities denoted as $|I_k|,|J_k|$, respectively. Now define numbers

$\hspace{70pt}C_k:=\max\Big\{\max\limits_{i\in I_k}i^{2t}\sum\limits_{j\in J_k}j^{2t},\,\,\max\limits_{j\in J_k}j^{2t}\sum\limits_{i\in I_k}i^{2t}\Big\}$


$\hspace{70pt}D_k:=\max\Big\{|I_k|\max\limits_{j\in J_k}j^{4t+2\delta},\,\,|J_k|\max\limits_{i\in I_k}i^{4t+2\delta}\Big\}$.

Aim: choose the subsets $|I_k,|J_k|$ so that


  • $\begingroup$ A substantial simplification: we can suppose that $I_k = \{a_k, a_k+1, \ldots, b_k\}, J_k:= \{c_k, c_k+1, \ldots, d_k\}$ are made of consecutive numbers. Indeed, note that if you keep the maximum of the two sets fixed and you "shift up" all the numbers, $D_k$ remains invaried and $C_k$ grows. So the problem is now a four variable problem. Using integral approximation for sums one can get a almost precise formula for everything. In particular $C_k$ has the form $b_k^{2t} \frac{(d_k+1)^{2t+1}-c_k^{2t+1}}{2t+1}$ or the other one, which is analogous. $\endgroup$ – Andrea Marino Jun 18 at 9:21
  • $\begingroup$ Another hint: such choice is not always possible (but not always impossible either). Anyway, voting to close here and move to MSE. $\endgroup$ – fedja Jun 18 at 12:49
  • $\begingroup$ Couldn't find a counterexample myself for some $\delta,t$. Have you? $\endgroup$ – Andrea Marino Jun 18 at 14:01
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    $\begingroup$ @AndreaMarino Yes. Just use your own idea and replace all sums by length times the maximum. Then you'll have two inequalities for 4 integers to satisfy plus the trivial bounds like $\max_{j\in J}j\ge |J|$. If I haven't made a stupid mistake, the answer is that the construction is possible if and only if $2t\frac{2\delta+\delta^2}{1+\delta}<1$. $\endgroup$ – fedja Jun 18 at 14:18
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    $\begingroup$ @Krzysztof OK, done. But in all honesty, this is a rather trivial exercise in elementary analysis, so if you know such scary words as "cohomology" and have trouble with this question, then you should ask yourself if your education is lopsided a bit too much (some lopsidedness is certainly OK, nobody escapes it. The question is where to draw the line :-) $\endgroup$ – fedja Jun 19 at 16:44

As @Krzysztof noticed, there was a mistake in the original post, which was that I was looking at $2t+2\delta$ and perceived it (for some unknown reason) as $2t(1+\delta)$. So here is the correctedversion:

Let $I,J$ be the interval lengths and $i,j$ be the top numbers in the intervals. Then the sum of integers in $I$ to the power $s>0$ is $i^sI$ up to a constant factor. What you want is $$ i^{2t}j^{2t}J\gg \max(Ij^{4t+2\delta},Ji^{4t+2\delta})\, $$ i.e., $i^{2t}J\ge Ij^{2t+2\delta}$ and $j^{2t}\gg i^{2t+2\delta}$\,. Clearly, we should take $I=1$ to make the task easier. The second inequality gives $i^{2t}\ll j^{2t\frac t{t+\delta}}$, so, plugging it into the first inequality, we see that we want $$ J\gg j^{2t[\frac{t+\delta}t-\frac t{t+\delta}]}\,. $$ This is compatible with the trivial restriction $j\ge J$ for large $j$ if and only if the power on the RHS is less than $1$.

  • $\begingroup$ Thank you for your answer. There is however one thing that still bothers me. How do you extract $i^{2t}\ll j^{2t(1+\delta)^{-1}}$ from $j^{2t}\gg i^{2t+2\delta}$? Besides, if you take $I$ to be a singleton consisting of $i=k^{1-\frac{2\delta}{24+2\delta}}$ (I mean the integer part of it) and $J$ consisting of $k$ numbers of the form $k^{1+\frac{1-2\delta}{4t+2\delta}}-j,\,j=1,\ldots,k$ then you get $R_k\approx k^{2t\frac{1-4\delta}{4t+2\delta}}$ which tends in $k$ to infinity for any positive $t$ as long as $\delta<\frac14$. $\endgroup$ – Krzysztof Jun 21 at 10:07
  • $\begingroup$ @Krzysztof Indeed, You are completely right: I made a stupid mistake in elementary algebra. It should be $i^{2t}\ll j^{2t\frac t{t+\delta}}$, etc.. I'll correct later today. Have to go now :-). $\endgroup$ – fedja Jun 21 at 14:30

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