Mixing
Fredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$.
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Let $X$ and $Y$ be Hilbert spaces with respective inner products $langle , rangle_X,Y$. Let $A:X rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $(,)$ on $Y$. Take $y in Y$, and define the map
$l_y : x' in X mapsto (A x', y)$.
Assume this last map is bounded, then by Riesz's theorem, there exists a unique $x in X$ such that $l_y(x') = langle x' ,x rangle_X$. We define the formal adjoint of $A$, denoted as $A^*$ by $A^*(y) = x$. Note that this coincides with the adjoint of $A$ when $(,) = langle , rangle_Y$.
Is it true that if $AA^*$ is a Fredholm operator then $A$ is Fredholm as well?
A particular case I am interested take $d: L^2_k(mathbbR) rightarrow L^2_k-1(mathbbR)$ (where $L^2_k$ denotes the Sobolev space with $k$ weak derivatives in $L^2$) and take $(,)$ as the $L^2(mathbbR)$ inner product.
real-analysis functional-analysis
asked Jul 27 at 7:40
inquisitor
699621
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Let $X$ and $Y$ be Hilbert spaces with respective inner products $langle , rangle_X,Y$. Let $A:X rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $(,)$ on $Y$. Take $y in Y$, and define the map
$l_y : x' in X mapsto (A x', y)$.
Assume this last map is bounded, then by Riesz's theorem, there exists a unique $x in X$ such that $l_y(x') = langle x' ,x rangle_X$. We define the formal adjoint of $A$, denoted as $A^*$ by $A^*(y) = x$. Note that this coincides with the adjoint of $A$ when $(,) = langle , rangle_Y$.
Is it true that if $AA^*$ is a Fredholm operator then $A$ is Fredholm as well?
A particular case I am interested take $d: L^2_k(mathbbR) rightarrow L^2_k-1(mathbbR)$ (where $L^2_k$ denotes the Sobolev space with $k$ weak derivatives in $L^2$) and take $(,)$ as the $L^2(mathbbR)$ inner product.
real-analysis functional-analysis
asked Jul 27 at 7:40
inquisitor
699621
This question has an open bounty worth +50
reputation from inquisitor ending ending at 2018-08-13 16:24:34Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $X$ and $Y$ be Hilbert spaces with respective inner products $langle , rangle_X,Y$. Let $A:X rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $(,)$ on $Y$. Take $y in Y$, and define the map
$l_y : x' in X mapsto (A x', y)$.
Assume this last map is bounded, then by Riesz's theorem, there exists a unique $x in X$ such that $l_y(x') = langle x' ,x rangle_X$. We define the formal adjoint of $A$, denoted as $A^*$ by $A^*(y) = x$. Note that this coincides with the adjoint of $A$ when $(,) = langle , rangle_Y$.
Is it true that if $AA^*$ is a Fredholm operator then $A$ is Fredholm as well?
A particular case I am interested take $d: L^2_k(mathbbR) rightarrow L^2_k-1(mathbbR)$ (where $L^2_k$ denotes the Sobolev space with $k$ weak derivatives in $L^2$) and take $(,)$ as the $L^2(mathbbR)$ inner product.
real-analysis functional-analysis
asked Jul 27 at 7:40
inquisitor
699621
Let $X$ and $Y$ be Hilbert spaces with respective inner products $langle , rangle_X,Y$. Let $A:X rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $(,)$ on $Y$. Take $y in Y$, and define the map
$l_y : x' in X mapsto (A x', y)$.
Assume this last map is bounded, then by Riesz's theorem, there exists a unique $x in X$ such that $l_y(x') = langle x' ,x rangle_X$. We define the formal adjoint of $A$, denoted as $A^*$ by $A^*(y) = x$. Note that this coincides with the adjoint of $A$ when $(,) = langle , rangle_Y$.
Is it true that if $AA^*$ is a Fredholm operator then $A$ is Fredholm as well?
A particular case I am interested take $d: L^2_k(mathbbR) rightarrow L^2_k-1(mathbbR)$ (where $L^2_k$ denotes the Sobolev space with $k$ weak derivatives in $L^2$) and take $(,)$ as the $L^2(mathbbR)$ inner product.
real-analysis functional-analysis
asked Jul 27 at 7:40
inquisitor
699621
edited Jul 27 at 10:02
asked Jul 27 at 7:40
inquisitor
699621
asked Jul 27 at 7:40
inquisitor
699621
asked Jul 27 at 7:40
inquisitor
699621
699621
This question has an open bounty worth +50
reputation from inquisitor ending ending at 2018-08-13 16:24:34Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from inquisitor ending ending at 2018-08-13 16:24:34Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16
 |Â
show 2 more comments
What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16
What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16
 |Â
show 2 more comments
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What do you mean by formal adjoint of $A$ in this context?
– Jan Bohr
Jul 27 at 8:19
I mean in the sense of the second line, the one that satisfies: $int_X (Bf,g) = int_X (f,B^*g)$ for all $f, g$, where $(cdot, cdot)$ is a non-degenerate bilinear (or complex in one variable in case it maps to $mathbbC$) map. I guess I am assuming that it exists. But solving it for the specific case of the $A$ given above would be already helpful. thanks
– inquisitor
Jul 27 at 8:28
- Why do you make a difference between adjoint and formal adjoint? - Don't you mean $(f,g)=bar f g$? Or do you want another inner product? - What is $L^2_1$?
– Jan Bohr
Jul 27 at 8:37
yes, I meant that one, I should have specified that (I have edited it now). $L^2_1$ is the Sobolev space with a first weak derivative in $L^2$
– inquisitor
Jul 27 at 8:46
You still have to say what you mean by formal adjoint and whether $A^*$ is different from $A'$.
– Jan Bohr
Jul 27 at 9:16