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
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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
$begingroup$
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
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
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
edited Jan 16 at 12:59
Bernard
122k741116
asked Jan 16 at 12:30
Tanuj DipshikhaTanuj Dipshikha
199210
$endgroup$
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
functional-analysis lebesgue-integral almost-periodic-functions
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
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
edited Jan 16 at 12:59
Bernard
122k741116
122k741116
asked Jan 16 at 12:30
Tanuj DipshikhaTanuj Dipshikha
199210
asked Jan 16 at 12:30
Tanuj DipshikhaTanuj Dipshikha
199210
asked Jan 16 at 12:30
Tanuj DipshikhaTanuj Dipshikha
199210
199210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$
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 |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$
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)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$
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)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$
answered Mar 10 at 19:55
Severin SchravenSeverin Schraven
6,4251935
$endgroup$
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)
$$ vert f_n - 0 vert_2^2 = frac12n int_-n^n vert f_n(x) vert^2 dx = frac12 int_-1^1 vert sin(2pi y) vert^2 dy =1. $$
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
answered Mar 10 at 19:55
Severin SchravenSeverin Schraven
6,4251935
6,4251935
$begingroup$
$sin(x fracn+1n)$ 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 |
$begingroup$
$sin(x fracn+1n)$ 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
$begingroup$
$sin(x fracn+1n)$ 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
$begingroup$
$sin(x fracn+1n)$ 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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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 |
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