Distributional derivative only generalization?
$begingroup$
Suppose $H_1$ and $H_2$ are Hilbert spaces which lie between test-functions and distributions, i.e.
$$
mathscr{D}(mathbb{R}) subset H_i subsetmathscr{D}'(mathbb{R}),
$$
such that the inclusions are continuous. The derivative $d/dx$ can be viewed as continuous linear maps $Dcolonmathscr{D}(mathbb{R}) rightarrow mathscr{D}(mathbb{R})$ and $tilde Dcolonmathscr{D}'(mathbb{R}) rightarrow mathscr{D}'(mathbb{R})$ respectively and with this notation $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$.
Question: If $Tcolon H_1rightarrow H_2$ is bounded and $Trestriction_{mathscr{D}(mathbb{R})} = D$, does it follow that $T=tilde Drestriction_{H_1}$?
An example for this situation would be the Sobolev spaces $H_1=W^{2,k}(mathbb{R})$ and $H_2=W^{2,k-1}(mathbb{R})$ and there the assertion is true.
I've tried to look at it abstracty, so to forget about the derivative and ask $D$, $tilde D$ to be arbitrary continuous linear maps with $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$. Then the only (?) thing we can use is the density of $mathscr{D}(mathbb{R}) subset mathscr{D}'(mathbb{R})$. But this does not seem to be enough.
Surely if also $mathscr{D}(mathbb{R}) subset H_i $ was dense the situation would be clear. While this is the case for the Sobolev spaces above, I don't want to assume that because it fails if $mathbb{R}$ is replaced by $(0,1)$.
Another sufficient condition would be to ask $C^infty(mathbb{R})cap H_i subset H_i$ to be dense. This is true for Sobolev spaces on $(0,1)$ but it is not true for Sobolev spaces on badly behaved domains in $mathbb{R}^n$.
real-analysis functional-analysis distribution-theory
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
$endgroup$
add a comment |
$begingroup$
Suppose $H_1$ and $H_2$ are Hilbert spaces which lie between test-functions and distributions, i.e.
$$
mathscr{D}(mathbb{R}) subset H_i subsetmathscr{D}'(mathbb{R}),
$$
such that the inclusions are continuous. The derivative $d/dx$ can be viewed as continuous linear maps $Dcolonmathscr{D}(mathbb{R}) rightarrow mathscr{D}(mathbb{R})$ and $tilde Dcolonmathscr{D}'(mathbb{R}) rightarrow mathscr{D}'(mathbb{R})$ respectively and with this notation $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$.
Question: If $Tcolon H_1rightarrow H_2$ is bounded and $Trestriction_{mathscr{D}(mathbb{R})} = D$, does it follow that $T=tilde Drestriction_{H_1}$?
An example for this situation would be the Sobolev spaces $H_1=W^{2,k}(mathbb{R})$ and $H_2=W^{2,k-1}(mathbb{R})$ and there the assertion is true.
I've tried to look at it abstracty, so to forget about the derivative and ask $D$, $tilde D$ to be arbitrary continuous linear maps with $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$. Then the only (?) thing we can use is the density of $mathscr{D}(mathbb{R}) subset mathscr{D}'(mathbb{R})$. But this does not seem to be enough.
Surely if also $mathscr{D}(mathbb{R}) subset H_i $ was dense the situation would be clear. While this is the case for the Sobolev spaces above, I don't want to assume that because it fails if $mathbb{R}$ is replaced by $(0,1)$.
Another sufficient condition would be to ask $C^infty(mathbb{R})cap H_i subset H_i$ to be dense. This is true for Sobolev spaces on $(0,1)$ but it is not true for Sobolev spaces on badly behaved domains in $mathbb{R}^n$.
real-analysis functional-analysis distribution-theory
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
$endgroup$
$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10
add a comment |
$begingroup$
Suppose $H_1$ and $H_2$ are Hilbert spaces which lie between test-functions and distributions, i.e.
$$
mathscr{D}(mathbb{R}) subset H_i subsetmathscr{D}'(mathbb{R}),
$$
such that the inclusions are continuous. The derivative $d/dx$ can be viewed as continuous linear maps $Dcolonmathscr{D}(mathbb{R}) rightarrow mathscr{D}(mathbb{R})$ and $tilde Dcolonmathscr{D}'(mathbb{R}) rightarrow mathscr{D}'(mathbb{R})$ respectively and with this notation $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$.
Question: If $Tcolon H_1rightarrow H_2$ is bounded and $Trestriction_{mathscr{D}(mathbb{R})} = D$, does it follow that $T=tilde Drestriction_{H_1}$?
An example for this situation would be the Sobolev spaces $H_1=W^{2,k}(mathbb{R})$ and $H_2=W^{2,k-1}(mathbb{R})$ and there the assertion is true.
I've tried to look at it abstracty, so to forget about the derivative and ask $D$, $tilde D$ to be arbitrary continuous linear maps with $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$. Then the only (?) thing we can use is the density of $mathscr{D}(mathbb{R}) subset mathscr{D}'(mathbb{R})$. But this does not seem to be enough.
Surely if also $mathscr{D}(mathbb{R}) subset H_i $ was dense the situation would be clear. While this is the case for the Sobolev spaces above, I don't want to assume that because it fails if $mathbb{R}$ is replaced by $(0,1)$.
Another sufficient condition would be to ask $C^infty(mathbb{R})cap H_i subset H_i$ to be dense. This is true for Sobolev spaces on $(0,1)$ but it is not true for Sobolev spaces on badly behaved domains in $mathbb{R}^n$.
real-analysis functional-analysis distribution-theory
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
$endgroup$
Suppose $H_1$ and $H_2$ are Hilbert spaces which lie between test-functions and distributions, i.e.
$$
mathscr{D}(mathbb{R}) subset H_i subsetmathscr{D}'(mathbb{R}),
$$
such that the inclusions are continuous. The derivative $d/dx$ can be viewed as continuous linear maps $Dcolonmathscr{D}(mathbb{R}) rightarrow mathscr{D}(mathbb{R})$ and $tilde Dcolonmathscr{D}'(mathbb{R}) rightarrow mathscr{D}'(mathbb{R})$ respectively and with this notation $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$.
Question: If $Tcolon H_1rightarrow H_2$ is bounded and $Trestriction_{mathscr{D}(mathbb{R})} = D$, does it follow that $T=tilde Drestriction_{H_1}$?
An example for this situation would be the Sobolev spaces $H_1=W^{2,k}(mathbb{R})$ and $H_2=W^{2,k-1}(mathbb{R})$ and there the assertion is true.
I've tried to look at it abstracty, so to forget about the derivative and ask $D$, $tilde D$ to be arbitrary continuous linear maps with $tilde Drestriction_{mathscr{D}(mathbb{R})}=D$. Then the only (?) thing we can use is the density of $mathscr{D}(mathbb{R}) subset mathscr{D}'(mathbb{R})$. But this does not seem to be enough.
Surely if also $mathscr{D}(mathbb{R}) subset H_i $ was dense the situation would be clear. While this is the case for the Sobolev spaces above, I don't want to assume that because it fails if $mathbb{R}$ is replaced by $(0,1)$.
Another sufficient condition would be to ask $C^infty(mathbb{R})cap H_i subset H_i$ to be dense. This is true for Sobolev spaces on $(0,1)$ but it is not true for Sobolev spaces on badly behaved domains in $mathbb{R}^n$.
real-analysis functional-analysis distribution-theory
real-analysis functional-analysis distribution-theory
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
edited Dec 23 '18 at 17:55
Daniele Tampieri
2,62221022
2,62221022
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
asked Jun 15 '18 at 9:38
Jan BohrJan Bohr
3,3071521
3,3071521
$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10
add a comment |
$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10
$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10
add a comment |
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$begingroup$
I am trying to follow a way I have in mind to prove your statement, but I would like to know if I am understanding correctly your question: if, by trying a reductio ad absurdum, I assume that $Tneqtilde Drestriction_{H_1}$, does this mean that there should exists an isomorphism $tilde{tilde{D}}:H_2to H_2$ such that $$T=tilde{tilde{D}}restriction_{H_1}?$$
$endgroup$
– Daniele Tampieri
Feb 17 at 22:16
$begingroup$
I don't see where this $tilde{tilde D}$ should come from. You also seem to assume that $H_1subset H_2$, I didn't do that initially, but if it helps I'm happy to see your idea.
$endgroup$
– Jan Bohr
Feb 18 at 11:00
$begingroup$
I have asked you more info since I would like to avoid this kind of situation (even if it was not bad at all for me). What I'd like to do is to see if your question can be solved by using a folclore result of distribution theory (recently proved by Langenbruch and Voigt) that it doesn't exists a Banach space $B$ invariant under differentiation such that $mathscr{D}(Omega)subset Bsubset mathscr{D}^prime(Omega)$.
$endgroup$
– Daniele Tampieri
Feb 18 at 13:10