conditions for existence of orthogonal function expansion
1
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Do the Dirichlet conditions for Fourier series apply to orthogonal functions generally (with perhaps an appropriate weight function) over an interval [a,b]? What about Laguerre and Hermite, which are defined over an infinite interval?
functional-analysis
asked Nov 26 '18 at 22:01
gilonikgilonik
113
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show 3 more comments
1
$begingroup$
Do the Dirichlet conditions for Fourier series apply to orthogonal functions generally (with perhaps an appropriate weight function) over an interval [a,b]? What about Laguerre and Hermite, which are defined over an infinite interval?
functional-analysis
asked Nov 26 '18 at 22:01
gilonikgilonik
113
$endgroup$
$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28
|
show 3 more comments
1
1
1
$begingroup$
Do the Dirichlet conditions for Fourier series apply to orthogonal functions generally (with perhaps an appropriate weight function) over an interval [a,b]? What about Laguerre and Hermite, which are defined over an infinite interval?
functional-analysis
asked Nov 26 '18 at 22:01
gilonikgilonik
113
$endgroup$
Do the Dirichlet conditions for Fourier series apply to orthogonal functions generally (with perhaps an appropriate weight function) over an interval [a,b]? What about Laguerre and Hermite, which are defined over an infinite interval?
functional-analysis
functional-analysis
asked Nov 26 '18 at 22:01
gilonikgilonik
113
asked Nov 26 '18 at 22:01
gilonikgilonik
113
edited Nov 26 '18 at 22:16
gilonik
asked Nov 26 '18 at 22:01
gilonikgilonik
113
asked Nov 26 '18 at 22:01
gilonikgilonik
113
asked Nov 26 '18 at 22:01
gilonikgilonik
113
113
$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28
|
show 3 more comments
$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28
$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28
|
show 3 more comments
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$begingroup$
What do you mean by Dirichlet conditions? What type of conditions are referring to?
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:26
$begingroup$
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
$endgroup$
– uniquesolution
Nov 26 '18 at 22:29
$begingroup$
Then, yes, Dirichlet conditions will give convergence of the orthogonal expansions. On the infinite interval, you need some condition of integrability or square integrability. Classical proofs of this are not easy for the Hermite case, but the result holds.
$endgroup$
– DisintegratingByParts
Nov 26 '18 at 22:35
$begingroup$
A condition of integrability is required even for Fourier. On an infinite interval, then, I would expect absolute integrability of a function multiplied by the square-root of the weight function to be required.
$endgroup$
– gilonik
Nov 26 '18 at 22:53
$begingroup$
@gilonik : The Dirichlet condition require local integrability in order to make sense. A global condition such as $L^1$ is also needed. The global condition just needs to be good enough to argue the Riemann-Lebesgue type of lemma to force vanishing of the tail integrals in the limit.
$endgroup$
– DisintegratingByParts
Nov 27 '18 at 0:28