Riesz representation and inverse operator.
$begingroup$
In class, my professor went through the following construction: Let $Omega$ be a bounded domain and define $X$ to be $H_0^1(Omega)$ or $H^1(Omega)$. We also define $A: X to X^prime$ (where $X^prime$ denotes the dual space of $X$) by the duality pairing
$$
langle Af, grangle = int_Omega nabla f cdot nabla g
$$
We will also view $L^2(Omega)$ as a subspace of $X^prime$. Now, there exists $t_0$ such that for all $t>t_0$,
$$
(cdot, cdot) : Xtimes X to mathbb{R}, quad
(f, g) = langle Af, g rangle + tlangle f, grangle_{L^2(Omega)}
$$
defines an inner product on $X$. For each such $t$, it follows from the Riesz representation theorem that $A+tmathrm{I} : Xto X^prime$ is invertible.
I am very confused as to how the Reisz representation theorem applies here and how one can deduce this. Any input is appreciated.
functional-analysis pde sobolev-spaces
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
$endgroup$
add a comment |
$begingroup$
In class, my professor went through the following construction: Let $Omega$ be a bounded domain and define $X$ to be $H_0^1(Omega)$ or $H^1(Omega)$. We also define $A: X to X^prime$ (where $X^prime$ denotes the dual space of $X$) by the duality pairing
$$
langle Af, grangle = int_Omega nabla f cdot nabla g
$$
We will also view $L^2(Omega)$ as a subspace of $X^prime$. Now, there exists $t_0$ such that for all $t>t_0$,
$$
(cdot, cdot) : Xtimes X to mathbb{R}, quad
(f, g) = langle Af, g rangle + tlangle f, grangle_{L^2(Omega)}
$$
defines an inner product on $X$. For each such $t$, it follows from the Riesz representation theorem that $A+tmathrm{I} : Xto X^prime$ is invertible.
I am very confused as to how the Reisz representation theorem applies here and how one can deduce this. Any input is appreciated.
functional-analysis pde sobolev-spaces
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
$endgroup$
$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01
add a comment |
$begingroup$
In class, my professor went through the following construction: Let $Omega$ be a bounded domain and define $X$ to be $H_0^1(Omega)$ or $H^1(Omega)$. We also define $A: X to X^prime$ (where $X^prime$ denotes the dual space of $X$) by the duality pairing
$$
langle Af, grangle = int_Omega nabla f cdot nabla g
$$
We will also view $L^2(Omega)$ as a subspace of $X^prime$. Now, there exists $t_0$ such that for all $t>t_0$,
$$
(cdot, cdot) : Xtimes X to mathbb{R}, quad
(f, g) = langle Af, g rangle + tlangle f, grangle_{L^2(Omega)}
$$
defines an inner product on $X$. For each such $t$, it follows from the Riesz representation theorem that $A+tmathrm{I} : Xto X^prime$ is invertible.
I am very confused as to how the Reisz representation theorem applies here and how one can deduce this. Any input is appreciated.
functional-analysis pde sobolev-spaces
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
$endgroup$
In class, my professor went through the following construction: Let $Omega$ be a bounded domain and define $X$ to be $H_0^1(Omega)$ or $H^1(Omega)$. We also define $A: X to X^prime$ (where $X^prime$ denotes the dual space of $X$) by the duality pairing
$$
langle Af, grangle = int_Omega nabla f cdot nabla g
$$
We will also view $L^2(Omega)$ as a subspace of $X^prime$. Now, there exists $t_0$ such that for all $t>t_0$,
$$
(cdot, cdot) : Xtimes X to mathbb{R}, quad
(f, g) = langle Af, g rangle + tlangle f, grangle_{L^2(Omega)}
$$
defines an inner product on $X$. For each such $t$, it follows from the Riesz representation theorem that $A+tmathrm{I} : Xto X^prime$ is invertible.
I am very confused as to how the Reisz representation theorem applies here and how one can deduce this. Any input is appreciated.
functional-analysis pde sobolev-spaces
functional-analysis pde sobolev-spaces
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
edited Nov 26 '18 at 18:44
Quoka
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
asked Nov 26 '18 at 17:52
QuokaQuoka
1,240212
1,240212
$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01
add a comment |
$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01
$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.
The inner product on $H_0^1(Omega)$ is given by $(f,g) = displaystyle int_Omega nabla f cdot nabla g , dx$.
The operator $A+I : H_0^1(Omega) to H_0^1(Omega)'$ is defined by
$$langle (A + I)f,g rangle = int_Omega nabla f cdot nabla g + fg , dx.$$
The Poincare inequality gives you
$$int_Omega f^2 , dx le C int_Omega |nabla f|^2 , dx$$ for all $f in H_0^1(Omega)$. Consequently for fixed $f$ the functional $Lg = langle (A+I)f,g rangle$ defines a bounded linear functional on $H_0^1(Omega)$. The Riesz representation theorem provides you with a unique function $h in H_0^1(Omega)$ satisfying
$$Lg = (h,g)$$ for all $g in H_0^1(Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(Omega)$ to itself, with
$$(A+I)f = h.$$
I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $langle (A+I)f,g rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $displaystyle int_Omega |nabla f|^2 + f^2 , dx = 0$, giving you $f=0$.
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
$endgroup$
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
add a comment |
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$begingroup$
For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.
The inner product on $H_0^1(Omega)$ is given by $(f,g) = displaystyle int_Omega nabla f cdot nabla g , dx$.
The operator $A+I : H_0^1(Omega) to H_0^1(Omega)'$ is defined by
$$langle (A + I)f,g rangle = int_Omega nabla f cdot nabla g + fg , dx.$$
The Poincare inequality gives you
$$int_Omega f^2 , dx le C int_Omega |nabla f|^2 , dx$$ for all $f in H_0^1(Omega)$. Consequently for fixed $f$ the functional $Lg = langle (A+I)f,g rangle$ defines a bounded linear functional on $H_0^1(Omega)$. The Riesz representation theorem provides you with a unique function $h in H_0^1(Omega)$ satisfying
$$Lg = (h,g)$$ for all $g in H_0^1(Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(Omega)$ to itself, with
$$(A+I)f = h.$$
I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $langle (A+I)f,g rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $displaystyle int_Omega |nabla f|^2 + f^2 , dx = 0$, giving you $f=0$.
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
$endgroup$
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
add a comment |
$begingroup$
For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.
The inner product on $H_0^1(Omega)$ is given by $(f,g) = displaystyle int_Omega nabla f cdot nabla g , dx$.
The operator $A+I : H_0^1(Omega) to H_0^1(Omega)'$ is defined by
$$langle (A + I)f,g rangle = int_Omega nabla f cdot nabla g + fg , dx.$$
The Poincare inequality gives you
$$int_Omega f^2 , dx le C int_Omega |nabla f|^2 , dx$$ for all $f in H_0^1(Omega)$. Consequently for fixed $f$ the functional $Lg = langle (A+I)f,g rangle$ defines a bounded linear functional on $H_0^1(Omega)$. The Riesz representation theorem provides you with a unique function $h in H_0^1(Omega)$ satisfying
$$Lg = (h,g)$$ for all $g in H_0^1(Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(Omega)$ to itself, with
$$(A+I)f = h.$$
I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $langle (A+I)f,g rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $displaystyle int_Omega |nabla f|^2 + f^2 , dx = 0$, giving you $f=0$.
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
$endgroup$
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
add a comment |
$begingroup$
For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.
The inner product on $H_0^1(Omega)$ is given by $(f,g) = displaystyle int_Omega nabla f cdot nabla g , dx$.
The operator $A+I : H_0^1(Omega) to H_0^1(Omega)'$ is defined by
$$langle (A + I)f,g rangle = int_Omega nabla f cdot nabla g + fg , dx.$$
The Poincare inequality gives you
$$int_Omega f^2 , dx le C int_Omega |nabla f|^2 , dx$$ for all $f in H_0^1(Omega)$. Consequently for fixed $f$ the functional $Lg = langle (A+I)f,g rangle$ defines a bounded linear functional on $H_0^1(Omega)$. The Riesz representation theorem provides you with a unique function $h in H_0^1(Omega)$ satisfying
$$Lg = (h,g)$$ for all $g in H_0^1(Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(Omega)$ to itself, with
$$(A+I)f = h.$$
I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $langle (A+I)f,g rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $displaystyle int_Omega |nabla f|^2 + f^2 , dx = 0$, giving you $f=0$.
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
$endgroup$
For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.
The inner product on $H_0^1(Omega)$ is given by $(f,g) = displaystyle int_Omega nabla f cdot nabla g , dx$.
The operator $A+I : H_0^1(Omega) to H_0^1(Omega)'$ is defined by
$$langle (A + I)f,g rangle = int_Omega nabla f cdot nabla g + fg , dx.$$
The Poincare inequality gives you
$$int_Omega f^2 , dx le C int_Omega |nabla f|^2 , dx$$ for all $f in H_0^1(Omega)$. Consequently for fixed $f$ the functional $Lg = langle (A+I)f,g rangle$ defines a bounded linear functional on $H_0^1(Omega)$. The Riesz representation theorem provides you with a unique function $h in H_0^1(Omega)$ satisfying
$$Lg = (h,g)$$ for all $g in H_0^1(Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(Omega)$ to itself, with
$$(A+I)f = h.$$
I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $langle (A+I)f,g rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $displaystyle int_Omega |nabla f|^2 + f^2 , dx = 0$, giving you $f=0$.
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
answered Nov 27 '18 at 0:48
Umberto P.Umberto P.
38.9k13064
38.9k13064
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
add a comment |
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
$begingroup$
Thanks! I mainly got lost with the notation and it helps to see the steps written out :-)
$endgroup$
– Quoka
Nov 27 '18 at 0:51
add a comment |
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$begingroup$
Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia
$endgroup$
– Javi
Nov 26 '18 at 18:00
$begingroup$
I think you mean $(cdot,cdot)$ defines an inner product on $X$, not $V$.
$endgroup$
– user10354138
Nov 26 '18 at 18:01