Mixing
Approximating a Lipschitz function by spherical harmonics
$begingroup$
Let $f:mathbb{S}^nto mathbb{R}$ be a Lipschitz function (i.e. so
$$
Vert fVert_{L}=sup_{xin mathbb{S}^n} |f(x)|+sup_{xneq yin mathbb{S}^n} frac{|f(x)-f(y)|}{d_{mathbb{S}^n}(x,y)}<infty.
$$
Can one find a sequence $S_N(f)$ consisting of the linear combination of $N$ spherical harmonics so that
$$
Vert f-S_N(f)Vert_0=sup_{xin mathbb{S}^n} |f(x)-S_N(f)(x)|to 0
$$
and
so for $N$ sufficiently large
$$
Vert S_N(f)Vert_Lleq 2 Vert fVert_L.
$$
In other words, can $f$ be uniformly approximated by a finite linear combination of sphereical harmonics whose Lipschitz norm is uniformly bounded (by twice the Lipschitz norm of $f$ but this is not so important). I believe this can be shown when $n=1$ by using the Fejer Kernel.
harmonic-analysis
asked Jan 18 at 20:33
RBega2RBega2
1333
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{S}^nto mathbb{R}$ be a Lipschitz function (i.e. so
$$
Vert fVert_{L}=sup_{xin mathbb{S}^n} |f(x)|+sup_{xneq yin mathbb{S}^n} frac{|f(x)-f(y)|}{d_{mathbb{S}^n}(x,y)}<infty.
$$
Can one find a sequence $S_N(f)$ consisting of the linear combination of $N$ spherical harmonics so that
$$
Vert f-S_N(f)Vert_0=sup_{xin mathbb{S}^n} |f(x)-S_N(f)(x)|to 0
$$
and
so for $N$ sufficiently large
$$
Vert S_N(f)Vert_Lleq 2 Vert fVert_L.
$$
In other words, can $f$ be uniformly approximated by a finite linear combination of sphereical harmonics whose Lipschitz norm is uniformly bounded (by twice the Lipschitz norm of $f$ but this is not so important). I believe this can be shown when $n=1$ by using the Fejer Kernel.
harmonic-analysis
asked Jan 18 at 20:33
RBega2RBega2
1333
$endgroup$
$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50
add a comment |
$begingroup$
Let $f:mathbb{S}^nto mathbb{R}$ be a Lipschitz function (i.e. so
$$
Vert fVert_{L}=sup_{xin mathbb{S}^n} |f(x)|+sup_{xneq yin mathbb{S}^n} frac{|f(x)-f(y)|}{d_{mathbb{S}^n}(x,y)}<infty.
$$
Can one find a sequence $S_N(f)$ consisting of the linear combination of $N$ spherical harmonics so that
$$
Vert f-S_N(f)Vert_0=sup_{xin mathbb{S}^n} |f(x)-S_N(f)(x)|to 0
$$
and
so for $N$ sufficiently large
$$
Vert S_N(f)Vert_Lleq 2 Vert fVert_L.
$$
In other words, can $f$ be uniformly approximated by a finite linear combination of sphereical harmonics whose Lipschitz norm is uniformly bounded (by twice the Lipschitz norm of $f$ but this is not so important). I believe this can be shown when $n=1$ by using the Fejer Kernel.
harmonic-analysis
asked Jan 18 at 20:33
RBega2RBega2
1333
$endgroup$
Let $f:mathbb{S}^nto mathbb{R}$ be a Lipschitz function (i.e. so
$$
Vert fVert_{L}=sup_{xin mathbb{S}^n} |f(x)|+sup_{xneq yin mathbb{S}^n} frac{|f(x)-f(y)|}{d_{mathbb{S}^n}(x,y)}<infty.
$$
Can one find a sequence $S_N(f)$ consisting of the linear combination of $N$ spherical harmonics so that
$$
Vert f-S_N(f)Vert_0=sup_{xin mathbb{S}^n} |f(x)-S_N(f)(x)|to 0
$$
and
so for $N$ sufficiently large
$$
Vert S_N(f)Vert_Lleq 2 Vert fVert_L.
$$
In other words, can $f$ be uniformly approximated by a finite linear combination of sphereical harmonics whose Lipschitz norm is uniformly bounded (by twice the Lipschitz norm of $f$ but this is not so important). I believe this can be shown when $n=1$ by using the Fejer Kernel.
harmonic-analysis
harmonic-analysis
asked Jan 18 at 20:33
RBega2RBega2
1333
asked Jan 18 at 20:33
RBega2RBega2
1333
asked Jan 18 at 20:33
RBega2RBega2
1333
asked Jan 18 at 20:33
RBega2RBega2
1333
asked Jan 18 at 20:33
RBega2RBega2
1333
1333
$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50
add a comment |
$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50
$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50
add a comment |
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$begingroup$
The spherical harmonics are the eigenfunctions of an Hermitian operator, right? If your Lipschitz function is in the same space, or even sometimes not, the eigenfunctions of an Hermitian operator are orthogonal and complete - that might be all you need.
$endgroup$
– Adrian Keister
Jan 18 at 20:47
$begingroup$
Oh, wait: you need a finite combination. That's harder. Never mind.
$endgroup$
– Adrian Keister
Jan 18 at 20:48
$begingroup$
I don't care so much about the finite combination, but more about the uniform Lipschitz bound (which does not readily follow from using the $L^2$ theory as far as I can tell).
$endgroup$
– RBega2
Jan 18 at 20:50