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Constant “periodization†of a function
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Let $w$ be a rapidly decaying function on $mathbbR$ such that
$$ sum_n in mathbbZ w(x+n) = 0$$
for all $x in mathbbR$. Does that imply that $w$ is identically zero? What if we assume that $w$ is continuous?
real-analysis sequences-and-series
asked 4 hours ago
Matthias Ludewig
4,07511543
add a comment |Â
up vote
1
down vote
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Let $w$ be a rapidly decaying function on $mathbbR$ such that
$$ sum_n in mathbbZ w(x+n) = 0$$
for all $x in mathbbR$. Does that imply that $w$ is identically zero? What if we assume that $w$ is continuous?
real-analysis sequences-and-series
asked 4 hours ago
Matthias Ludewig
4,07511543
The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
1
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $w$ be a rapidly decaying function on $mathbbR$ such that
$$ sum_n in mathbbZ w(x+n) = 0$$
for all $x in mathbbR$. Does that imply that $w$ is identically zero? What if we assume that $w$ is continuous?
real-analysis sequences-and-series
asked 4 hours ago
Matthias Ludewig
4,07511543
Let $w$ be a rapidly decaying function on $mathbbR$ such that
$$ sum_n in mathbbZ w(x+n) = 0$$
for all $x in mathbbR$. Does that imply that $w$ is identically zero? What if we assume that $w$ is continuous?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked 4 hours ago
Matthias Ludewig
4,07511543
asked 4 hours ago
Matthias Ludewig
4,07511543
asked 4 hours ago
Matthias Ludewig
4,07511543
asked 4 hours ago
Matthias Ludewig
4,07511543
asked 4 hours ago
Matthias Ludewig
4,07511543
4,07511543
The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
1
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago
add a comment |Â
The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
1
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago
The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
1
1
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
No, the Haar wavelet is a counter example (i.e. $w(x) = chi_[0,1[(x) - chi_[1,2[(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
answered 4 hours ago
Dirk
6,81943264
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No, the Haar wavelet is a counter example (i.e. $w(x) = chi_[0,1[(x) - chi_[1,2[(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
answered 4 hours ago
Dirk
6,81943264
add a comment |Â
up vote
3
down vote
accepted
No, the Haar wavelet is a counter example (i.e. $w(x) = chi_[0,1[(x) - chi_[1,2[(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
answered 4 hours ago
Dirk
6,81943264
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No, the Haar wavelet is a counter example (i.e. $w(x) = chi_[0,1[(x) - chi_[1,2[(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
answered 4 hours ago
Dirk
6,81943264
No, the Haar wavelet is a counter example (i.e. $w(x) = chi_[0,1[(x) - chi_[1,2[(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
answered 4 hours ago
Dirk
6,81943264
answered 4 hours ago
Dirk
6,81943264
answered 4 hours ago
Dirk
6,81943264
answered 4 hours ago
Dirk
6,81943264
6,81943264
add a comment |Â
add a comment |Â
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The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $lim_mtoinftysum_-m^m$?
– Gerry Myerson
4 hours ago
Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value.
– Matthias Ludewig
2 hours ago
1
Indeed if you assume say that $w$ is also smooth this periodization equals $sum hat w(n) e(n x)$, so the condition is equivalent to $hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them).
– Rodrigo
48 mins ago