conditions for existence of orthogonal function expansion
1
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 at 22:01
gilonik
113
|
show 3 more comments
1
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 at 22:01
gilonik
113
What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– gilonik
Nov 26 at 22:53
@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.
– DisintegratingByParts
Nov 27 at 0:28
|
show 3 more comments
1
1
1
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 at 22:01
gilonik
113
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 at 22:01
gilonik
113
asked Nov 26 at 22:01
gilonik
113
edited Nov 26 at 22:16
asked Nov 26 at 22:01
gilonik
113
asked Nov 26 at 22:01
gilonik
113
asked Nov 26 at 22:01
gilonik
113
113
What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– gilonik
Nov 26 at 22:53
@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.
– DisintegratingByParts
Nov 27 at 0:28
|
show 3 more comments
What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– gilonik
Nov 26 at 22:53
@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.
– DisintegratingByParts
Nov 27 at 0:28
What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– gilonik
Nov 26 at 22:53
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.
– gilonik
Nov 26 at 22:53
@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.
– DisintegratingByParts
Nov 27 at 0:28
@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.
– DisintegratingByParts
Nov 27 at 0:28
|
show 3 more comments
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What do you mean by Dirichlet conditions? What type of conditions are referring to?
– DisintegratingByParts
Nov 26 at 22:26
@DisintegratingByParts en.wikipedia.org/wiki/Dirichlet_conditions
– uniquesolution
Nov 26 at 22:29
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.
– DisintegratingByParts
Nov 26 at 22:35
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.
– gilonik
Nov 26 at 22:53
@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.
– DisintegratingByParts
Nov 27 at 0:28