Mixing
Question on the definition of almost periodic function
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
According to Bohr, the definition of the almost periodic function is:
A function $f:mathbbRrightarrow mathbbC$ is called almost periodic if it is continuous and if for every positive $epsilon$, there exists a positive number $l$ such that every closed interval of length $l$ contains an $epsilon$-almost period.
Intuitively, when $epsilon$ is getting smaller, the corresponding $l$ is getting larger. But is this always correct? If it is correct, may I ask how to prove it? Also, is there any relation between $epsilon$ and the corresponding $epsilon$-almost period?
ca.classical-analysis-and-odes harmonic-analysis
asked Aug 26 at 15:23
nanshan
185
add a comment |Â
up vote
3
down vote
favorite
According to Bohr, the definition of the almost periodic function is:
A function $f:mathbbRrightarrow mathbbC$ is called almost periodic if it is continuous and if for every positive $epsilon$, there exists a positive number $l$ such that every closed interval of length $l$ contains an $epsilon$-almost period.
Intuitively, when $epsilon$ is getting smaller, the corresponding $l$ is getting larger. But is this always correct? If it is correct, may I ask how to prove it? Also, is there any relation between $epsilon$ and the corresponding $epsilon$-almost period?
ca.classical-analysis-and-odes harmonic-analysis
asked Aug 26 at 15:23
nanshan
185
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
According to Bohr, the definition of the almost periodic function is:
A function $f:mathbbRrightarrow mathbbC$ is called almost periodic if it is continuous and if for every positive $epsilon$, there exists a positive number $l$ such that every closed interval of length $l$ contains an $epsilon$-almost period.
Intuitively, when $epsilon$ is getting smaller, the corresponding $l$ is getting larger. But is this always correct? If it is correct, may I ask how to prove it? Also, is there any relation between $epsilon$ and the corresponding $epsilon$-almost period?
ca.classical-analysis-and-odes harmonic-analysis
asked Aug 26 at 15:23
nanshan
185
According to Bohr, the definition of the almost periodic function is:
A function $f:mathbbRrightarrow mathbbC$ is called almost periodic if it is continuous and if for every positive $epsilon$, there exists a positive number $l$ such that every closed interval of length $l$ contains an $epsilon$-almost period.
Intuitively, when $epsilon$ is getting smaller, the corresponding $l$ is getting larger. But is this always correct? If it is correct, may I ask how to prove it? Also, is there any relation between $epsilon$ and the corresponding $epsilon$-almost period?
ca.classical-analysis-and-odes harmonic-analysis
asked Aug 26 at 15:23
nanshan
185
asked Aug 26 at 15:23
nanshan
185
asked Aug 26 at 15:23
nanshan
185
asked Aug 26 at 15:23
nanshan
185
185
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
It seems to me that if an interval is an $epsilon$-almost period then it is trivially an $epsilon'$-almost period for any $epsilon' >epsilon$. So as $epsilon$ gets smaller the set of intervals which contain $epsilon$-almost periods shrinks (i.e., fewer intervals have this property).
As to the second question, consider the function $sum_1^infty 2^-n e^ib_nx$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^-n$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $epsilon$.
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
It seems to me that if an interval is an $epsilon$-almost period then it is trivially an $epsilon'$-almost period for any $epsilon' >epsilon$. So as $epsilon$ gets smaller the set of intervals which contain $epsilon$-almost periods shrinks (i.e., fewer intervals have this property).
As to the second question, consider the function $sum_1^infty 2^-n e^ib_nx$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^-n$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $epsilon$.
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
add a comment |Â
up vote
6
down vote
accepted
It seems to me that if an interval is an $epsilon$-almost period then it is trivially an $epsilon'$-almost period for any $epsilon' >epsilon$. So as $epsilon$ gets smaller the set of intervals which contain $epsilon$-almost periods shrinks (i.e., fewer intervals have this property).
As to the second question, consider the function $sum_1^infty 2^-n e^ib_nx$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^-n$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $epsilon$.
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
It seems to me that if an interval is an $epsilon$-almost period then it is trivially an $epsilon'$-almost period for any $epsilon' >epsilon$. So as $epsilon$ gets smaller the set of intervals which contain $epsilon$-almost periods shrinks (i.e., fewer intervals have this property).
As to the second question, consider the function $sum_1^infty 2^-n e^ib_nx$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^-n$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $epsilon$.
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
It seems to me that if an interval is an $epsilon$-almost period then it is trivially an $epsilon'$-almost period for any $epsilon' >epsilon$. So as $epsilon$ gets smaller the set of intervals which contain $epsilon$-almost periods shrinks (i.e., fewer intervals have this property).
As to the second question, consider the function $sum_1^infty 2^-n e^ib_nx$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^-n$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $epsilon$.
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
answered Aug 26 at 15:51
Nik Weaver
17.9k142114
17.9k142114
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
add a comment |Â
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
(The series should be $sum 2^-ne^ix/b_n$.)
– Nik Weaver
Aug 30 at 15:56
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f309166%2fquestion-on-the-definition-of-almost-periodic-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
