Mixing
Spherical Harmonics expansion for $fin L^{2}(S^{n-1})$.
$begingroup$
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
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
$endgroup$
add a comment |
$begingroup$
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
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
$endgroup$
add a comment |
$begingroup$
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
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
$endgroup$
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
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
asked Nov 19 '17 at 12:24
A.Zoran A.Zoran
314
314
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
$endgroup$
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2527434%2fspherical-harmonics-expansion-for-f-in-l2sn-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
$endgroup$
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.
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
answered Jan 19 at 20:13
paul garrettpaul garrett
32k362118
32k362118
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2527434%2fspherical-harmonics-expansion-for-f-in-l2sn-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
