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...
$begingroup$
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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add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$
$endgroup$
$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48
add a comment |
Your Answer
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1 Answer
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$begingroup$
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,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$
$endgroup$
$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48
add a comment |
$begingroup$
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,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$
$endgroup$
$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48
add a comment |
$begingroup$
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,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$
$endgroup$
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,
$$sum_{n,m in mathbb Z^2} e^{-Vert n-m Vert} frac{1}{1+Vert n Vert^{1+varepsilon}} frac{1}{1+Vert m Vert^{1+varepsilon}} leq sum_{k in mathbb Z^2}e^{-Vert k Vert} cdot C <infty $$
answered Dec 31 '18 at 19:35
N. S.N. S.
104k7114209
104k7114209
$begingroup$
thank you very much for your help, it is a wonderful answer.
$endgroup$
– Sascha
Dec 31 '18 at 19:48
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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