Fdtxjr

Can the druid cantrip Thorn Whip really defeat a water weird this easily?

Do I need to leave some extra space available on the disk which my database log files reside, for log backup operations to successfully occur?

How could a female member of a species produce eggs unto death?

What is the dot in “1.2.4."

Word for a person who has no opinion about whether god exists

Single word request: Harming the benefactor

What to do when during a meeting client people start to fight (even physically) with each others?

Is all copper pipe pretty much the same?

Time dilation for a moving electronic clock

Touchscreen-controlled dentist office snowman collector game

When were linguistics departments first established

Replacing Windows 7 security updates with anti-virus?

Make a transparent 448*448 image

Best approach to update all entries in a list that is paginated?

How can I discourage/prevent PCs from using door choke-points?

Prove that the total distance is minimised (when travelling across the longest path)

validation vs test vs training accuracy, which one to compare for claiming overfit?

Time travel short story where dinosaur doesn't taste like chicken

Should we release the security issues we found in our product as CVE or we can just update those on weekly release notes?

What does it mean when multiple 々 marks follow a 、?

Does splitting a potentially monolithic application into several smaller ones help prevent bugs?

Counter-example to the existence of left Bousfield localization of combinatorial model category

Playing ONE triplet (not three)

Best mythical creature to use as livestock?

Convergence in the topology of $L^2_textloc$ implies convergence in $B^2$?

Necessary conditions for “convergence in mean” of a sequence of integrable functions.Convergence in $L^1_loc$ implies convergence almost everywhereIs $L^1_loc(mathbbR)$ complete with the norm $|f|=sup_xin mathbbRint_x^x+1|f(y)|dy$Convergence of Functions when viewed as Distributions and other Convergence ConditionsIf $fin L^1_loc$, does it mean that $f$ is continuous on $mathbbR$?show that $ f=lim_j to infty f_j(x) text a.e. everywhere $Besicovitch almost periodic functions with seminorm zeroLet $u(x_1, x_2)=minx_2$. For which $p$ it holds that $uin W^1, p_textloc(mathbbR^2)$?Is the space of bounded continuous functions on the reals with the topology of uniform convergence on compact sets fully barrelled?Mean value of an almost periodic function

1

$begingroup$

Let $f_n$ be a sequence of functions in $L^2_textloc(mathbbR)$ which converge to a function $fin L^2_textloc(mathbbR)$ in the topology of $L^2_textloc(mathbbR)$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $KsubsetmathbbR$.

A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_Ttoinftyfrac12Tint_-T^T|f(x)|^2dyright)^1/2$. Clearly, all Besicovitch almost periodic functions are locally square integrable.

Assuming that $f_n,f$ are Besicovitch almost periodic, can we also conclude that $f_nto f$ in the seminorm of almost periodic functions, i.e, $|f_n-f|_2to 0$?

functional-analysis lebesgue-integral almost-periodic-functions

share|cite|improve this question

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

$endgroup$

add a comment | 

1

$begingroup$

Let $f_n$ be a sequence of functions in $L^2_textloc(mathbbR)$ which converge to a function $fin L^2_textloc(mathbbR)$ in the topology of $L^2_textloc(mathbbR)$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $KsubsetmathbbR$.

A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_Ttoinftyfrac12Tint_-T^T|f(x)|^2dyright)^1/2$. Clearly, all Besicovitch almost periodic functions are locally square integrable.

Assuming that $f_n,f$ are Besicovitch almost periodic, can we also conclude that $f_nto f$ in the seminorm of almost periodic functions, i.e, $|f_n-f|_2to 0$?

functional-analysis lebesgue-integral almost-periodic-functions

share|cite|improve this question

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

$endgroup$

add a comment | 

$begingroup$

Let $f_n$ be a sequence of functions in $L^2_textloc(mathbbR)$ which converge to a function $fin L^2_textloc(mathbbR)$ in the topology of $L^2_textloc(mathbbR)$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $KsubsetmathbbR$.

A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_Ttoinftyfrac12Tint_-T^T|f(x)|^2dyright)^1/2$. Clearly, all Besicovitch almost periodic functions are locally square integrable.

Assuming that $f_n,f$ are Besicovitch almost periodic, can we also conclude that $f_nto f$ in the seminorm of almost periodic functions, i.e, $|f_n-f|_2to 0$?

functional-analysis lebesgue-integral almost-periodic-functions

share|cite|improve this question

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

$endgroup$

share|cite|improve this question

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

share|cite|improve this question

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

edited Jan 16 at 12:59

Bernard

122k741116

edited Jan 16 at 12:59

Bernard

122k741116

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

asked Jan 16 at 12:30

Tanuj DipshikhaTanuj Dipshikha

199210

1 Answer
1

active

oldest

votes

2

$begingroup$

No, convergence in $L_loc^2(mathbbR)$ does not imply convergence in the seminorm. Consider
$$ f_n(x) = sinleft( frac2pi xn right) $$
Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_loc^2(mathbbR)$ to the zero function (which of course is Besicovitch almost periodic itself). However, we do not have convergence in the seminorm as (now we use that $f_n$ is $n$-periodic)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$

share|cite|improve this answer

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sin(x fracn+1n)$ is another example. To repair it we can try replacing $|f_n-f|_2$ by things like $| (f_n-f) ast phi_m|_2$ where $phi_m(x) = m e^-pi m^2 x^2$
  $endgroup$
  – reuns
  Mar 10 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I would guess that fixing a common frequency module which has finite rank would allow us to conclude as well (however, I am not completely sure).
  $endgroup$
  – Severin Schraven
  Mar 10 at 20:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant replacing $ frac12m int_-m^m |f_n(x)-f(x)|^2dx$ by $int_-infty^infty (f_n-f)astphi_m(x) e^-pi x^2/m^2 dx$. The convolution by $phi_m$ is to repair your example, the multiplication by $e^-pi x^2/m^2$ is to repair mine.
  $endgroup$
  – reuns
  Mar 10 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I meant that if we assumed in addition that all the function in the sequence are quasiperiodic with the same frequency module, then the result might be true.
  $endgroup$
  – Severin Schraven
  Mar 10 at 21:08
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075672%2fconvergence-in-the-topology-of-l2-textloc-implies-convergence-in-b2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

No, convergence in $L_loc^2(mathbbR)$ does not imply convergence in the seminorm. Consider
$$ f_n(x) = sinleft( frac2pi xn right) $$
Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_loc^2(mathbbR)$ to the zero function (which of course is Besicovitch almost periodic itself). However, we do not have convergence in the seminorm as (now we use that $f_n$ is $n$-periodic)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$

share|cite|improve this answer

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sin(x fracn+1n)$ is another example. To repair it we can try replacing $|f_n-f|_2$ by things like $| (f_n-f) ast phi_m|_2$ where $phi_m(x) = m e^-pi m^2 x^2$
  $endgroup$
  – reuns
  Mar 10 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I would guess that fixing a common frequency module which has finite rank would allow us to conclude as well (however, I am not completely sure).
  $endgroup$
  – Severin Schraven
  Mar 10 at 20:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant replacing $ frac12m int_-m^m |f_n(x)-f(x)|^2dx$ by $int_-infty^infty (f_n-f)astphi_m(x) e^-pi x^2/m^2 dx$. The convolution by $phi_m$ is to repair your example, the multiplication by $e^-pi x^2/m^2$ is to repair mine.
  $endgroup$
  – reuns
  Mar 10 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I meant that if we assumed in addition that all the function in the sequence are quasiperiodic with the same frequency module, then the result might be true.
  $endgroup$
  – Severin Schraven
  Mar 10 at 21:08
  

  
  
  
  

add a comment | 

2

$begingroup$

No, convergence in $L_loc^2(mathbbR)$ does not imply convergence in the seminorm. Consider
$$ f_n(x) = sinleft( frac2pi xn right) $$
Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_loc^2(mathbbR)$ to the zero function (which of course is Besicovitch almost periodic itself). However, we do not have convergence in the seminorm as (now we use that $f_n$ is $n$-periodic)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$

share|cite|improve this answer

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sin(x fracn+1n)$ is another example. To repair it we can try replacing $|f_n-f|_2$ by things like $| (f_n-f) ast phi_m|_2$ where $phi_m(x) = m e^-pi m^2 x^2$
  $endgroup$
  – reuns
  Mar 10 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I would guess that fixing a common frequency module which has finite rank would allow us to conclude as well (however, I am not completely sure).
  $endgroup$
  – Severin Schraven
  Mar 10 at 20:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant replacing $ frac12m int_-m^m |f_n(x)-f(x)|^2dx$ by $int_-infty^infty (f_n-f)astphi_m(x) e^-pi x^2/m^2 dx$. The convolution by $phi_m$ is to repair your example, the multiplication by $e^-pi x^2/m^2$ is to repair mine.
  $endgroup$
  – reuns
  Mar 10 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I meant that if we assumed in addition that all the function in the sequence are quasiperiodic with the same frequency module, then the result might be true.
  $endgroup$
  – Severin Schraven
  Mar 10 at 21:08
  

  
  
  
  

add a comment | 

$begingroup$

No, convergence in $L_loc^2(mathbbR)$ does not imply convergence in the seminorm. Consider
$$ f_n(x) = sinleft( frac2pi xn right) $$
Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_loc^2(mathbbR)$ to the zero function (which of course is Besicovitch almost periodic itself). However, we do not have convergence in the seminorm as (now we use that $f_n$ is $n$-periodic)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$

share|cite|improve this answer

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

$endgroup$

share|cite|improve this answer

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935

answered Mar 10 at 19:55

Severin SchravenSeverin Schraven

6,4251935