Convergence in the topology of $L^2_text{loc}$ implies convergence in $B^2$?












1












$begingroup$


Let $f_n$ be a sequence of functions in $L^2_text{loc}(mathbb{R})$ which converge to a function $fin L^2_text{loc}(mathbb{R})$ in the topology of $L^2_text{loc}(mathbb{R})$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $Ksubsetmathbb{R}$.



A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_{Ttoinfty}frac{1}{2T}int_{-T}^T|f(x)|^2dyright)^{1/2}$. 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







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$endgroup$

















    1












    $begingroup$


    Let $f_n$ be a sequence of functions in $L^2_text{loc}(mathbb{R})$ which converge to a function $fin L^2_text{loc}(mathbb{R})$ in the topology of $L^2_text{loc}(mathbb{R})$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $Ksubsetmathbb{R}$.



    A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_{Ttoinfty}frac{1}{2T}int_{-T}^T|f(x)|^2dyright)^{1/2}$. 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











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f_n$ be a sequence of functions in $L^2_text{loc}(mathbb{R})$ which converge to a function $fin L^2_text{loc}(mathbb{R})$ in the topology of $L^2_text{loc}(mathbb{R})$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $Ksubsetmathbb{R}$.



      A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_{Ttoinfty}frac{1}{2T}int_{-T}^T|f(x)|^2dyright)^{1/2}$. 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











      $endgroup$




      Let $f_n$ be a sequence of functions in $L^2_text{loc}(mathbb{R})$ which converge to a function $fin L^2_text{loc}(mathbb{R})$ in the topology of $L^2_text{loc}(mathbb{R})$, i.e., $f_nto f$ in $L^2(K)$ for all compact subsets $Ksubsetmathbb{R}$.



      A function is said to be Besicovitch almost periodic if it is the limit of trigonometrical polynomials in the seminorm $|f|_2=left(lim_{Ttoinfty}frac{1}{2T}int_{-T}^T|f(x)|^2dyright)^{1/2}$. 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






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 16 at 12:59









      Bernard

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      124k741117










      asked Jan 16 at 12:30









      Tanuj DipshikhaTanuj Dipshikha

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          1 Answer
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          $begingroup$

          No, convergence in $L_{loc}^2(mathbb{R})$ does not imply convergence in the seminorm. Consider
          $$ f_n(x) = sinleft( frac{2pi x}{n} right) $$
          Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(mathbb{R})$ 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)
          $$ vert f_n - 0 vert_2^2 = frac{1}{2n} int_{-n}^{n} vert f_n(x) vert^2 dx = frac{1}{2} int_{-1}^{1} vert sin(2pi y) vert^2 dy =1. $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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












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          $begingroup$

          No, convergence in $L_{loc}^2(mathbb{R})$ does not imply convergence in the seminorm. Consider
          $$ f_n(x) = sinleft( frac{2pi x}{n} right) $$
          Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(mathbb{R})$ 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)
          $$ vert f_n - 0 vert_2^2 = frac{1}{2n} int_{-n}^{n} vert f_n(x) vert^2 dx = frac{1}{2} int_{-1}^{1} vert sin(2pi y) vert^2 dy =1. $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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
















          2












          $begingroup$

          No, convergence in $L_{loc}^2(mathbb{R})$ does not imply convergence in the seminorm. Consider
          $$ f_n(x) = sinleft( frac{2pi x}{n} right) $$
          Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(mathbb{R})$ 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)
          $$ vert f_n - 0 vert_2^2 = frac{1}{2n} int_{-n}^{n} vert f_n(x) vert^2 dx = frac{1}{2} int_{-1}^{1} vert sin(2pi y) vert^2 dy =1. $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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














          2












          2








          2





          $begingroup$

          No, convergence in $L_{loc}^2(mathbb{R})$ does not imply convergence in the seminorm. Consider
          $$ f_n(x) = sinleft( frac{2pi x}{n} right) $$
          Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(mathbb{R})$ 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)
          $$ vert f_n - 0 vert_2^2 = frac{1}{2n} int_{-n}^{n} vert f_n(x) vert^2 dx = frac{1}{2} int_{-1}^{1} vert sin(2pi y) vert^2 dy =1. $$






          share|cite|improve this answer









          $endgroup$



          No, convergence in $L_{loc}^2(mathbb{R})$ does not imply convergence in the seminorm. Consider
          $$ f_n(x) = sinleft( frac{2pi x}{n} right) $$
          Clearly $f_n$ is Besicovitch almost periodic. Furthermore, it converges locally uniformly to the zero function and hence it converges in $L_{loc}^2(mathbb{R})$ 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)
          $$ vert f_n - 0 vert_2^2 = frac{1}{2n} int_{-n}^{n} vert f_n(x) vert^2 dx = frac{1}{2} int_{-1}^{1} vert sin(2pi y) vert^2 dy =1. $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 10 at 19:55









          Severin SchravenSeverin Schraven

          6,7702936




          6,7702936












          • $begingroup$
            $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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


















          • $begingroup$
            $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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
















          $begingroup$
          $sin(x frac{n+1}{n})$ 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






          $begingroup$
          $sin(x frac{n+1}{n})$ 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 $ frac{1}{2m} 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 $ frac{1}{2m} 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


















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