Mixing
Second derivative operator closed on $AC^2[0,1]$
$begingroup$
Consider the Hilbert space $mathcal H=L^2(0,1)$ and let $D:=AC^2[0,1]$ be the space of functions $fin C^1[0,1]$ such that $f'$ is absolutely continuous and the weak derivative $f''$ is in $ L^{2}(0,1)$. Define the operator $$T:Dto mathcal H; , T= -frac{d^2}{dx^2}.$$
I want to prove that $T$ is closed, i.e. the graph of $T, Gamma(T)$ is closed. To that end, let $(varphi, psi) in overline{Gamma(T)}$ and $(varphi_n)_{nin mathbb N}$ such that $varphi_n stackrel{nto infty}{longrightarrow} varphi$ and $Tvarphi_n stackrel{nto infty}{longrightarrow} psi$ in $L^2$. I need to prove $varphi in D$ and $Tvarphi = psi$. Here I am stuck. I tried a lot playing with integration by parts and other stuff. I am also not sure what I can say about the first derivative of $varphi$. Any help appreciated!
functional-analysis operator-theory mathematical-physics
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
$endgroup$
add a comment |
$begingroup$
Consider the Hilbert space $mathcal H=L^2(0,1)$ and let $D:=AC^2[0,1]$ be the space of functions $fin C^1[0,1]$ such that $f'$ is absolutely continuous and the weak derivative $f''$ is in $ L^{2}(0,1)$. Define the operator $$T:Dto mathcal H; , T= -frac{d^2}{dx^2}.$$
I want to prove that $T$ is closed, i.e. the graph of $T, Gamma(T)$ is closed. To that end, let $(varphi, psi) in overline{Gamma(T)}$ and $(varphi_n)_{nin mathbb N}$ such that $varphi_n stackrel{nto infty}{longrightarrow} varphi$ and $Tvarphi_n stackrel{nto infty}{longrightarrow} psi$ in $L^2$. I need to prove $varphi in D$ and $Tvarphi = psi$. Here I am stuck. I tried a lot playing with integration by parts and other stuff. I am also not sure what I can say about the first derivative of $varphi$. Any help appreciated!
functional-analysis operator-theory mathematical-physics
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
$endgroup$
1
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14
add a comment |
$begingroup$
Consider the Hilbert space $mathcal H=L^2(0,1)$ and let $D:=AC^2[0,1]$ be the space of functions $fin C^1[0,1]$ such that $f'$ is absolutely continuous and the weak derivative $f''$ is in $ L^{2}(0,1)$. Define the operator $$T:Dto mathcal H; , T= -frac{d^2}{dx^2}.$$
I want to prove that $T$ is closed, i.e. the graph of $T, Gamma(T)$ is closed. To that end, let $(varphi, psi) in overline{Gamma(T)}$ and $(varphi_n)_{nin mathbb N}$ such that $varphi_n stackrel{nto infty}{longrightarrow} varphi$ and $Tvarphi_n stackrel{nto infty}{longrightarrow} psi$ in $L^2$. I need to prove $varphi in D$ and $Tvarphi = psi$. Here I am stuck. I tried a lot playing with integration by parts and other stuff. I am also not sure what I can say about the first derivative of $varphi$. Any help appreciated!
functional-analysis operator-theory mathematical-physics
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
$endgroup$
Consider the Hilbert space $mathcal H=L^2(0,1)$ and let $D:=AC^2[0,1]$ be the space of functions $fin C^1[0,1]$ such that $f'$ is absolutely continuous and the weak derivative $f''$ is in $ L^{2}(0,1)$. Define the operator $$T:Dto mathcal H; , T= -frac{d^2}{dx^2}.$$
I want to prove that $T$ is closed, i.e. the graph of $T, Gamma(T)$ is closed. To that end, let $(varphi, psi) in overline{Gamma(T)}$ and $(varphi_n)_{nin mathbb N}$ such that $varphi_n stackrel{nto infty}{longrightarrow} varphi$ and $Tvarphi_n stackrel{nto infty}{longrightarrow} psi$ in $L^2$. I need to prove $varphi in D$ and $Tvarphi = psi$. Here I am stuck. I tried a lot playing with integration by parts and other stuff. I am also not sure what I can say about the first derivative of $varphi$. Any help appreciated!
functional-analysis operator-theory mathematical-physics
functional-analysis operator-theory mathematical-physics
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
edited Jan 30 at 9:06
YuiTo Cheng
2,1863937
2,1863937
asked Jan 29 at 16:44
Staki42Staki42
1,154618
asked Jan 29 at 16:44
Staki42Staki42
1,154618
asked Jan 29 at 16:44
Staki42Staki42
1,154618
1,154618
1
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14
add a comment |
1
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14
1
1
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)int_{0}^{x}f(t)tdt+xint_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
begin{align}
(Sf)' &= (1-x)f(x)x-int_0^xf(t)tdt-xf(x)(1-x)+int_x^1f(t)(1-t)dt \
&= -int_0^x f(t)tdt+int_x^1f(t)(1-t)dt.
end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) ;;; a.e..
$$
Therefore $Sfinmathcal{D}(T)$ for all $fin L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $fin mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
begin{align}
langle T_e f,grangle -langle f,T_e grangle &= -int_0^1 (f''g-fg'')dx \
& = -(f'g-fg')|_{0}^{1}.
end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $finmathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) inmathcal{D}(T)
$$
and
begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \
+&E_0(f)T_e x+E_1(f)T_e(1-x)
end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
|
show 2 more comments
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092406%2fsecond-derivative-operator-closed-on-ac20-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)int_{0}^{x}f(t)tdt+xint_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
begin{align}
(Sf)' &= (1-x)f(x)x-int_0^xf(t)tdt-xf(x)(1-x)+int_x^1f(t)(1-t)dt \
&= -int_0^x f(t)tdt+int_x^1f(t)(1-t)dt.
end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) ;;; a.e..
$$
Therefore $Sfinmathcal{D}(T)$ for all $fin L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $fin mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
begin{align}
langle T_e f,grangle -langle f,T_e grangle &= -int_0^1 (f''g-fg'')dx \
& = -(f'g-fg')|_{0}^{1}.
end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $finmathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) inmathcal{D}(T)
$$
and
begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \
+&E_0(f)T_e x+E_1(f)T_e(1-x)
end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
|
show 2 more comments
$begingroup$
The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)int_{0}^{x}f(t)tdt+xint_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
begin{align}
(Sf)' &= (1-x)f(x)x-int_0^xf(t)tdt-xf(x)(1-x)+int_x^1f(t)(1-t)dt \
&= -int_0^x f(t)tdt+int_x^1f(t)(1-t)dt.
end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) ;;; a.e..
$$
Therefore $Sfinmathcal{D}(T)$ for all $fin L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $fin mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
begin{align}
langle T_e f,grangle -langle f,T_e grangle &= -int_0^1 (f''g-fg'')dx \
& = -(f'g-fg')|_{0}^{1}.
end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $finmathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) inmathcal{D}(T)
$$
and
begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \
+&E_0(f)T_e x+E_1(f)T_e(1-x)
end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
|
show 2 more comments
$begingroup$
The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)int_{0}^{x}f(t)tdt+xint_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
begin{align}
(Sf)' &= (1-x)f(x)x-int_0^xf(t)tdt-xf(x)(1-x)+int_x^1f(t)(1-t)dt \
&= -int_0^x f(t)tdt+int_x^1f(t)(1-t)dt.
end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) ;;; a.e..
$$
Therefore $Sfinmathcal{D}(T)$ for all $fin L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $fin mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
begin{align}
langle T_e f,grangle -langle f,T_e grangle &= -int_0^1 (f''g-fg'')dx \
& = -(f'g-fg')|_{0}^{1}.
end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $finmathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) inmathcal{D}(T)
$$
and
begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \
+&E_0(f)T_e x+E_1(f)T_e(1-x)
end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by
$$
Sf=(1-x)int_{0}^{x}f(t)tdt+xint_{x}^{1}f(t)(1-t)dt.
$$
To see this, note that, for all $f in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.:
begin{align}
(Sf)' &= (1-x)f(x)x-int_0^xf(t)tdt-xf(x)(1-x)+int_x^1f(t)(1-t)dt \
&= -int_0^x f(t)tdt+int_x^1f(t)(1-t)dt.
end{align}
Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'in L^2$ and
$$
(Sf)'' = -f(x)x-f(x)(1-x)=-f(x) ;;; a.e..
$$
Therefore $Sfinmathcal{D}(T)$ for all $fin L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $fin mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.
Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider
begin{align}
langle T_e f,grangle -langle f,T_e grangle &= -int_0^1 (f''g-fg'')dx \
& = -(f'g-fg')|_{0}^{1}.
end{align}
By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $finmathcal{D}(T_e)$, one has
$$
f-E_0(f)x-E_1(f)(1-x) inmathcal{D}(T)
$$
and
begin{align}
T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\
+&E_0(f)T_e x+E_1(f)T_e(1-x) \
= S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \
+&E_0(f)T_e x+E_1(f)T_e(1-x)
end{align}
Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
edited Jan 31 at 1:59
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
answered Jan 30 at 0:56
DisintegratingByPartsDisintegratingByParts
60.3k42681
60.3k42681
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
|
show 2 more comments
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
Wow, you are great! One question: How does it follow that my original operator on the smaller domain $D$ without boundary conditions is closed too?
$endgroup$
– Staki42
Jan 30 at 8:10
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
The operator domain with endpoint restrictions is of co-codimension 2 in the one without endpoint restrictions. The larger graph is also closed because of this.
$endgroup$
– DisintegratingByParts
Jan 30 at 11:14
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
I see! Could you maybe provide a reference for this fact? It sounds very interesting and I'd like to see a proof. Thanks for helping out! :)
$endgroup$
– Staki42
Jan 30 at 12:18
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
$begingroup$
@Staki42 : If $finmathcal{D}(T)$, then $g=STf$ satisfies $Tg=Tf$ which gives $T(g-f)=0$, which implies that there are constants $A,B$ such that $g-f=Ax+B(1-x)$. So $f=g+Ax+B(1-x)$ where $ginmathcal{D}(S)$ and $A,B$ are constants. That means $T$ is a 2-dimensional extension of $S$.
$endgroup$
– DisintegratingByParts
Jan 30 at 16:07
1
1
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
$begingroup$
Very helpful. Appreciated!
$endgroup$
– Staki42
Jan 30 at 23:44
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092406%2fsecond-derivative-operator-closed-on-ac20-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
How are you defining convergence in $AC^2$?
$endgroup$
– Umberto P.
Jan 29 at 23:24
$begingroup$
I consider it as a subspace of $L^2(0,1)$, with the $lVert cdot rVert_2 $-norm.
$endgroup$
– Staki42
Jan 30 at 0:14