Mixing
Show that an operator is symmetric but not selfadjoint.
$begingroup$
I am stuck with the following exercise:
Show that the operator $A= -d^2x$ with $D(A) ={f in L_2[0,1]:f,f' in C[0,1] ,with, f'' in L_2,, f(0)=f(1)=0
, f'(0)=f'(1)=0 }$
Is symmetric but not self-adjoint.
I am aware of all the definitions, but I just can't figure it out.
Thanks a lot for your help.
functional-analysis operator-theory self-adjoint-operators
edited Feb 3 at 10:40
pitariver
469213
asked Feb 3 at 10:24
YuheYuhe
1089
$endgroup$
add a comment |
$begingroup$
I am stuck with the following exercise:
Show that the operator $A= -d^2x$ with $D(A) ={f in L_2[0,1]:f,f' in C[0,1] ,with, f'' in L_2,, f(0)=f(1)=0
, f'(0)=f'(1)=0 }$
Is symmetric but not self-adjoint.
I am aware of all the definitions, but I just can't figure it out.
Thanks a lot for your help.
functional-analysis operator-theory self-adjoint-operators
edited Feb 3 at 10:40
pitariver
469213
asked Feb 3 at 10:24
YuheYuhe
1089
$endgroup$
$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45
add a comment |
$begingroup$
I am stuck with the following exercise:
Show that the operator $A= -d^2x$ with $D(A) ={f in L_2[0,1]:f,f' in C[0,1] ,with, f'' in L_2,, f(0)=f(1)=0
, f'(0)=f'(1)=0 }$
Is symmetric but not self-adjoint.
I am aware of all the definitions, but I just can't figure it out.
Thanks a lot for your help.
functional-analysis operator-theory self-adjoint-operators
edited Feb 3 at 10:40
pitariver
469213
asked Feb 3 at 10:24
YuheYuhe
1089
$endgroup$
I am stuck with the following exercise:
Show that the operator $A= -d^2x$ with $D(A) ={f in L_2[0,1]:f,f' in C[0,1] ,with, f'' in L_2,, f(0)=f(1)=0
, f'(0)=f'(1)=0 }$
Is symmetric but not self-adjoint.
I am aware of all the definitions, but I just can't figure it out.
Thanks a lot for your help.
functional-analysis operator-theory self-adjoint-operators
functional-analysis operator-theory self-adjoint-operators
edited Feb 3 at 10:40
pitariver
469213
asked Feb 3 at 10:24
YuheYuhe
1089
edited Feb 3 at 10:40
pitariver
469213
asked Feb 3 at 10:24
YuheYuhe
1089
edited Feb 3 at 10:40
pitariver
469213
edited Feb 3 at 10:40
pitariver
469213
edited Feb 3 at 10:40
pitariver
469213
469213
asked Feb 3 at 10:24
YuheYuhe
1089
asked Feb 3 at 10:24
YuheYuhe
1089
asked Feb 3 at 10:24
YuheYuhe
1089
1089
$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45
add a comment |
$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45
$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45
$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $gin L^2[0,1]$ be such that $gin C^1[0,1]$ and $g'$ is absolutely continuous with $g''in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that
$$
langle Af,grangle = langle f,-g''rangle,;; finmathcal{D}(A), ginmathcal{D}'.
$$
In particular, the above holds for all $ginmathcal{D}(A)$ because $mathcal{D}(A)subset mathcal{D}'$. Hence $A$ is symmetric.
Because this holds for all $finmathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $ginmathcal{D}(A^*)$ and $A^*g=-g''$. Therefore
$$
mathcal{D}(A) subsetneqmathcal{D}'subseteqmathcal{D}(A^*),
$$
which shows that $A$ is symmetric, but not self-adjoint.
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $gin L^2[0,1]$ be such that $gin C^1[0,1]$ and $g'$ is absolutely continuous with $g''in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that
$$
langle Af,grangle = langle f,-g''rangle,;; finmathcal{D}(A), ginmathcal{D}'.
$$
In particular, the above holds for all $ginmathcal{D}(A)$ because $mathcal{D}(A)subset mathcal{D}'$. Hence $A$ is symmetric.
Because this holds for all $finmathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $ginmathcal{D}(A^*)$ and $A^*g=-g''$. Therefore
$$
mathcal{D}(A) subsetneqmathcal{D}'subseteqmathcal{D}(A^*),
$$
which shows that $A$ is symmetric, but not self-adjoint.
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
add a comment |
$begingroup$
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $gin L^2[0,1]$ be such that $gin C^1[0,1]$ and $g'$ is absolutely continuous with $g''in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that
$$
langle Af,grangle = langle f,-g''rangle,;; finmathcal{D}(A), ginmathcal{D}'.
$$
In particular, the above holds for all $ginmathcal{D}(A)$ because $mathcal{D}(A)subset mathcal{D}'$. Hence $A$ is symmetric.
Because this holds for all $finmathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $ginmathcal{D}(A^*)$ and $A^*g=-g''$. Therefore
$$
mathcal{D}(A) subsetneqmathcal{D}'subseteqmathcal{D}(A^*),
$$
which shows that $A$ is symmetric, but not self-adjoint.
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
add a comment |
$begingroup$
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $gin L^2[0,1]$ be such that $gin C^1[0,1]$ and $g'$ is absolutely continuous with $g''in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that
$$
langle Af,grangle = langle f,-g''rangle,;; finmathcal{D}(A), ginmathcal{D}'.
$$
In particular, the above holds for all $ginmathcal{D}(A)$ because $mathcal{D}(A)subset mathcal{D}'$. Hence $A$ is symmetric.
Because this holds for all $finmathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $ginmathcal{D}(A^*)$ and $A^*g=-g''$. Therefore
$$
mathcal{D}(A) subsetneqmathcal{D}'subseteqmathcal{D}(A^*),
$$
which shows that $A$ is symmetric, but not self-adjoint.
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $gin L^2[0,1]$ be such that $gin C^1[0,1]$ and $g'$ is absolutely continuous with $g''in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that
$$
langle Af,grangle = langle f,-g''rangle,;; finmathcal{D}(A), ginmathcal{D}'.
$$
In particular, the above holds for all $ginmathcal{D}(A)$ because $mathcal{D}(A)subset mathcal{D}'$. Hence $A$ is symmetric.
Because this holds for all $finmathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $ginmathcal{D}(A^*)$ and $A^*g=-g''$. Therefore
$$
mathcal{D}(A) subsetneqmathcal{D}'subseteqmathcal{D}(A^*),
$$
which shows that $A$ is symmetric, but not self-adjoint.
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
edited Feb 3 at 22:27
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Feb 3 at 21:44
DisintegratingByPartsDisintegratingByParts
60.4k42681
60.4k42681
add a comment |
add a comment |
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$begingroup$
Show that $langle Af,grangle=langle f,-g''rangle$ for any $g$ such that $g,g'in C[0,1]$ and $g'$ absolutely continuous with $g''in L^2$. So such $g$ is in $mathcal{D}(A^*)$ and $A^*g=-g''$. Conclude that the domain of $A^*$ is strictly larger than that of $A$.
$endgroup$
– DisintegratingByParts
Feb 3 at 13:45