Let $a_{n,m} \geq 0$ for all $n,m$. Suppose there exists an $M$, such that for any $T$, $\frac{1}{T}\sum_{n,m=1}^Ta_{n,m}\leq M$. Can we find another constant $C$, such that $a_{n,m}\leq C\ \forall n,m$?
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See https://math.stackexchange.com/a/1105424/42969 for a one-dimensional counterexample, which can easily be extended to a two-dimensional counterexample. – Martin R Jul 06 '22 at 08:27
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1Consider the sequence M,0,2M,0,0,3M,0,0,0,4M,0,0,0,0,5M,.... – Jaap Scherphuis Jul 06 '22 at 08:28
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Ah sorry, did not notice that $T$ is not fixed. – Mr. Gandalf Sauron Jul 06 '22 at 08:29
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@JaapScherphuis What if we require $a_{n,m} >0$ for all $n,m$ – Daaaaa Jul 06 '22 at 08:34
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You can make everything non-zero by adding for example $2^{-(n+m)}$ to each $a_{n,m}$ which has a finite sum. – Jaap Scherphuis Jul 06 '22 at 08:40
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@MartinR Hi, sorry I see one-dimensional case works, but I cannot easily figure out the two-dimensional counterexample from this since it is divided by T not $T^2$ – Daaaaa Jul 06 '22 at 08:53
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@JaapScherphuis the original sequence does not have bound sum divided by T. – Daaaaa Jul 06 '22 at 08:54
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Just put the one-dimensional sequence from my first comment along the diagonal $a_{n,n}$, and let all non-diagonal entries be $0$. – Jaap Scherphuis Jul 06 '22 at 08:58
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@JaapScherphuis Ah I see Thanks. – Daaaaa Jul 06 '22 at 09:02