Spherical Harmonics expansion for $fin L^{2}(S^{n-1})$.












2












$begingroup$


Let $fin L^{2}(S^{n-1})$ where $S^{n-1}$ is the unit sphere of $Bbb{R}^n$.



As it is known $f$ has the following spherical Harmonics expansion (the convergence is $L^{2}(S^{n-1})$ in :



$f(w)=sum_{j} Y_j(w)$ where $Y_jin H_j$ with $H_j$ is the space of spherical harmonics of degree $j$.



Now let $fin C(S^{n-1})$ (space of continuous functions on S^{n-1} ) then $fin L^{2}(S^{n-1})$.



Hence $f(w)=sum_{j} Y_j(w)$.



My question can we say that the series in question converge uniformly and absolutely on $S^{n-1}$.






fourier-analysis harmonic-analysis spherical-harmonics







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $fin L^{2}(S^{n-1})$ where $S^{n-1}$ is the unit sphere of $Bbb{R}^n$.



    As it is known $f$ has the following spherical Harmonics expansion (the convergence is $L^{2}(S^{n-1})$ in :



    $f(w)=sum_{j} Y_j(w)$ where $Y_jin H_j$ with $H_j$ is the space of spherical harmonics of degree $j$.



    Now let $fin C(S^{n-1})$ (space of continuous functions on S^{n-1} ) then $fin L^{2}(S^{n-1})$.



    Hence $f(w)=sum_{j} Y_j(w)$.



    My question can we say that the series in question converge uniformly and absolutely on $S^{n-1}$.






    fourier-analysis harmonic-analysis spherical-harmonics







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $fin L^{2}(S^{n-1})$ where $S^{n-1}$ is the unit sphere of $Bbb{R}^n$.



      As it is known $f$ has the following spherical Harmonics expansion (the convergence is $L^{2}(S^{n-1})$ in :



      $f(w)=sum_{j} Y_j(w)$ where $Y_jin H_j$ with $H_j$ is the space of spherical harmonics of degree $j$.



      Now let $fin C(S^{n-1})$ (space of continuous functions on S^{n-1} ) then $fin L^{2}(S^{n-1})$.



      Hence $f(w)=sum_{j} Y_j(w)$.



      My question can we say that the series in question converge uniformly and absolutely on $S^{n-1}$.






      fourier-analysis harmonic-analysis spherical-harmonics







      share|cite|improve this question









      $endgroup$




      Let $fin L^{2}(S^{n-1})$ where $S^{n-1}$ is the unit sphere of $Bbb{R}^n$.



      As it is known $f$ has the following spherical Harmonics expansion (the convergence is $L^{2}(S^{n-1})$ in :



      $f(w)=sum_{j} Y_j(w)$ where $Y_jin H_j$ with $H_j$ is the space of spherical harmonics of degree $j$.



      Now let $fin C(S^{n-1})$ (space of continuous functions on S^{n-1} ) then $fin L^{2}(S^{n-1})$.



      Hence $f(w)=sum_{j} Y_j(w)$.



      My question can we say that the series in question converge uniformly and absolutely on $S^{n-1}$.






      fourier-analysis harmonic-analysis spherical-harmonics




      fourier-analysis harmonic-analysis spherical-harmonics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 19 '17 at 12:24









      A.Zoran A.Zoran

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          No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.



          On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $Delta^k f$ where (e.g.) $Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.






          share|cite|improve this answer









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            $begingroup$

            No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.



            On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $Delta^k f$ where (e.g.) $Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.



              On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $Delta^k f$ where (e.g.) $Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.



                On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $Delta^k f$ where (e.g.) $Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.






                share|cite|improve this answer









                $endgroup$



                No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.



                On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $Delta^k f$ where (e.g.) $Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 20:13









                paul garrettpaul garrett

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                32k362118






























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