Is $sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m...












4












$begingroup$


I would like to ask whether the expression



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}}$$



is finite?



Intuitively, this should be the case as away from the diagonal $n=m$ the exponential is rapidly decaying and on the diagonal, this expression is summable, but I cannot make it rigorous.



EDIT: The sum is over $n$ and $m$ both in $mathbb Z^2.$






real-analysis calculus sequences-and-series analysis







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$endgroup$












  • $begingroup$
    Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:01












  • $begingroup$
    sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:04










  • $begingroup$
    okay, I'll delete my ealier answer.
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:04
















4












$begingroup$


I would like to ask whether the expression



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}}$$



is finite?



Intuitively, this should be the case as away from the diagonal $n=m$ the exponential is rapidly decaying and on the diagonal, this expression is summable, but I cannot make it rigorous.



EDIT: The sum is over $n$ and $m$ both in $mathbb Z^2.$






real-analysis calculus sequences-and-series analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:01












  • $begingroup$
    sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:04










  • $begingroup$
    okay, I'll delete my ealier answer.
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:04














4












4








4


0



$begingroup$


I would like to ask whether the expression



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}}$$



is finite?



Intuitively, this should be the case as away from the diagonal $n=m$ the exponential is rapidly decaying and on the diagonal, this expression is summable, but I cannot make it rigorous.



EDIT: The sum is over $n$ and $m$ both in $mathbb Z^2.$






real-analysis calculus sequences-and-series analysis







share|cite|improve this question











$endgroup$




I would like to ask whether the expression



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}}$$



is finite?



Intuitively, this should be the case as away from the diagonal $n=m$ the exponential is rapidly decaying and on the diagonal, this expression is summable, but I cannot make it rigorous.



EDIT: The sum is over $n$ and $m$ both in $mathbb Z^2.$






real-analysis calculus sequences-and-series analysis




real-analysis calculus sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 19:12







Sascha

















asked Dec 31 '18 at 18:22









SaschaSascha

136318




136318












  • $begingroup$
    Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:01












  • $begingroup$
    sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:04










  • $begingroup$
    okay, I'll delete my ealier answer.
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:04


















  • $begingroup$
    Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:01












  • $begingroup$
    sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:04










  • $begingroup$
    okay, I'll delete my ealier answer.
    $endgroup$
    – Kenny Wong
    Dec 31 '18 at 19:04
















$begingroup$
Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
$endgroup$
– Kenny Wong
Dec 31 '18 at 19:01






$begingroup$
Can we please clarify: Do you mean that each of $n$ and $m$ is a vector in $mathbb Z^2$? Or do you mean $(n ,m) in mathbb Z^2$?
$endgroup$
– Kenny Wong
Dec 31 '18 at 19:01














$begingroup$
sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
$endgroup$
– Sascha
Dec 31 '18 at 19:04




$begingroup$
sorry, it is indeed that both $n$ amd $m$ are in $mathbb Z^2.$
$endgroup$
– Sascha
Dec 31 '18 at 19:04












$begingroup$
okay, I'll delete my ealier answer.
$endgroup$
– Kenny Wong
Dec 31 '18 at 19:04




$begingroup$
okay, I'll delete my ealier answer.
$endgroup$
– Kenny Wong
Dec 31 '18 at 19:04










1 Answer
1






active

oldest

votes


















3












$begingroup$

Note first that everything is positive.



Then



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} stackrel{m-n=k}{===}sum_{k in mathbb Z^2}sum_{m in mathbb Z^2} e^{-Vert k Vert} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} \stackrel{m-n=k}{===}sum_{k in mathbb Z^2}e^{-Vert k Vert} sum_{m in mathbb Z^2} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} $$



Now, for each $k in mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)



$$left(sum_{m in mathbb Z^2} frac{1}{1+Vert m Vert^{1+varepsilon}} frac{1}{1+Vert m+k Vert^{1+varepsilon}} right)^2 leq left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)left( sum_{m in mathbb Z^2} frac{1}{(1+Vert m+kVert^{1+varepsilon})^2} right)\=left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)^2 $$



Now, it is easy to argue that
$$C:= left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)< infty$$



Therefore,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for your help, it is a wonderful answer.
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:48











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Note first that everything is positive.



Then



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} stackrel{m-n=k}{===}sum_{k in mathbb Z^2}sum_{m in mathbb Z^2} e^{-Vert k Vert} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} \stackrel{m-n=k}{===}sum_{k in mathbb Z^2}e^{-Vert k Vert} sum_{m in mathbb Z^2} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} $$



Now, for each $k in mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)



$$left(sum_{m in mathbb Z^2} frac{1}{1+Vert m Vert^{1+varepsilon}} frac{1}{1+Vert m+k Vert^{1+varepsilon}} right)^2 leq left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)left( sum_{m in mathbb Z^2} frac{1}{(1+Vert m+kVert^{1+varepsilon})^2} right)\=left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)^2 $$



Now, it is easy to argue that
$$C:= left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)< infty$$



Therefore,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for your help, it is a wonderful answer.
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:48
















3












$begingroup$

Note first that everything is positive.



Then



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} stackrel{m-n=k}{===}sum_{k in mathbb Z^2}sum_{m in mathbb Z^2} e^{-Vert k Vert} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} \stackrel{m-n=k}{===}sum_{k in mathbb Z^2}e^{-Vert k Vert} sum_{m in mathbb Z^2} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} $$



Now, for each $k in mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)



$$left(sum_{m in mathbb Z^2} frac{1}{1+Vert m Vert^{1+varepsilon}} frac{1}{1+Vert m+k Vert^{1+varepsilon}} right)^2 leq left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)left( sum_{m in mathbb Z^2} frac{1}{(1+Vert m+kVert^{1+varepsilon})^2} right)\=left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)^2 $$



Now, it is easy to argue that
$$C:= left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)< infty$$



Therefore,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for your help, it is a wonderful answer.
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:48














3












3








3





$begingroup$

Note first that everything is positive.



Then



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} stackrel{m-n=k}{===}sum_{k in mathbb Z^2}sum_{m in mathbb Z^2} e^{-Vert k Vert} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} \stackrel{m-n=k}{===}sum_{k in mathbb Z^2}e^{-Vert k Vert} sum_{m in mathbb Z^2} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} $$



Now, for each $k in mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)



$$left(sum_{m in mathbb Z^2} frac{1}{1+Vert m Vert^{1+varepsilon}} frac{1}{1+Vert m+k Vert^{1+varepsilon}} right)^2 leq left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)left( sum_{m in mathbb Z^2} frac{1}{(1+Vert m+kVert^{1+varepsilon})^2} right)\=left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)^2 $$



Now, it is easy to argue that
$$C:= left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)< infty$$



Therefore,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$






share|cite|improve this answer









$endgroup$



Note first that everything is positive.



Then



$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} stackrel{m-n=k}{===}sum_{k in mathbb Z^2}sum_{m in mathbb Z^2} e^{-Vert k Vert} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} \stackrel{m-n=k}{===}sum_{k in mathbb Z^2}e^{-Vert k Vert} sum_{m in mathbb Z^2} frac{1}{1+Vert m+k Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} $$



Now, for each $k in mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)



$$left(sum_{m in mathbb Z^2} frac{1}{1+Vert m Vert^{1+varepsilon}} frac{1}{1+Vert m+k Vert^{1+varepsilon}} right)^2 leq left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)left( sum_{m in mathbb Z^2} frac{1}{(1+Vert m+kVert^{1+varepsilon})^2} right)\=left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)^2 $$



Now, it is easy to argue that
$$C:= left( sum_{m in mathbb Z^2} frac{1}{(1+Vert mVert^{1+varepsilon})^2} right)< infty$$



Therefore,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$







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share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 19:35









N. S.N. S.

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104k7114209












  • $begingroup$
    thank you very much for your help, it is a wonderful answer.
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:48


















  • $begingroup$
    thank you very much for your help, it is a wonderful answer.
    $endgroup$
    – Sascha
    Dec 31 '18 at 19:48
















$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48




$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48


















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