On possible closures of the derivative operator
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Let $frac{partial^2}{partial x^2}: L^2([0,1]) to L^2([0,1])$ be the one-dimensional Laplacian, considered as an unbounded, densley defined operator with domain $mathcal D(frac{partial^2}{partial x^2}) = C^infty([0,1])$ the smooth funtions without boundary conditions.
Question 1: Can one explicitly describe the domain $mathcal D((frac{partial^2}{partial x^2})^*)$ of the adjoint operator ?
By means of partial integration, one easily checks that
a) $B := { f in C^2([0,1]): f(0) = f'(0) = f(1) = f'(1) = 0 } subseteq mathcal D((frac{partial^2}{partial x^2})^*)$, so that $(frac{partial^2}{partial x^2})^*$ is densely defined (in particular, $frac{partial^2}{partial x^2}$ is closable).
b) $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$.
c) The restriction $frac{partial^2}{partial x^2}|_{B}$ is symmetric and densley defined, hence closable.
By c), we may consider the minimal closure $overline{frac{partial^2}{partial x^2}|_{B}}$. Now since $(frac{partial^2}{partial x^2})^*$ is a closed extension of $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$ by b), we must have $overline{frac{partial^2}{partial x^2}|_{B}} subset (frac{partial^2}{partial x^2})^*$.
Question 2: Do we have the equality of closed operators $(frac{partial^2}{partial x^2})^* = overline{frac{partial^2}{partial x^2}|_{B}}$. If not, what is their difference ?
functional-analysis ordinary-differential-equations pde
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
$endgroup$
add a comment |
$begingroup$
Let $frac{partial^2}{partial x^2}: L^2([0,1]) to L^2([0,1])$ be the one-dimensional Laplacian, considered as an unbounded, densley defined operator with domain $mathcal D(frac{partial^2}{partial x^2}) = C^infty([0,1])$ the smooth funtions without boundary conditions.
Question 1: Can one explicitly describe the domain $mathcal D((frac{partial^2}{partial x^2})^*)$ of the adjoint operator ?
By means of partial integration, one easily checks that
a) $B := { f in C^2([0,1]): f(0) = f'(0) = f(1) = f'(1) = 0 } subseteq mathcal D((frac{partial^2}{partial x^2})^*)$, so that $(frac{partial^2}{partial x^2})^*$ is densely defined (in particular, $frac{partial^2}{partial x^2}$ is closable).
b) $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$.
c) The restriction $frac{partial^2}{partial x^2}|_{B}$ is symmetric and densley defined, hence closable.
By c), we may consider the minimal closure $overline{frac{partial^2}{partial x^2}|_{B}}$. Now since $(frac{partial^2}{partial x^2})^*$ is a closed extension of $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$ by b), we must have $overline{frac{partial^2}{partial x^2}|_{B}} subset (frac{partial^2}{partial x^2})^*$.
Question 2: Do we have the equality of closed operators $(frac{partial^2}{partial x^2})^* = overline{frac{partial^2}{partial x^2}|_{B}}$. If not, what is their difference ?
functional-analysis ordinary-differential-equations pde
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
$endgroup$
$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20
add a comment |
$begingroup$
Let $frac{partial^2}{partial x^2}: L^2([0,1]) to L^2([0,1])$ be the one-dimensional Laplacian, considered as an unbounded, densley defined operator with domain $mathcal D(frac{partial^2}{partial x^2}) = C^infty([0,1])$ the smooth funtions without boundary conditions.
Question 1: Can one explicitly describe the domain $mathcal D((frac{partial^2}{partial x^2})^*)$ of the adjoint operator ?
By means of partial integration, one easily checks that
a) $B := { f in C^2([0,1]): f(0) = f'(0) = f(1) = f'(1) = 0 } subseteq mathcal D((frac{partial^2}{partial x^2})^*)$, so that $(frac{partial^2}{partial x^2})^*$ is densely defined (in particular, $frac{partial^2}{partial x^2}$ is closable).
b) $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$.
c) The restriction $frac{partial^2}{partial x^2}|_{B}$ is symmetric and densley defined, hence closable.
By c), we may consider the minimal closure $overline{frac{partial^2}{partial x^2}|_{B}}$. Now since $(frac{partial^2}{partial x^2})^*$ is a closed extension of $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$ by b), we must have $overline{frac{partial^2}{partial x^2}|_{B}} subset (frac{partial^2}{partial x^2})^*$.
Question 2: Do we have the equality of closed operators $(frac{partial^2}{partial x^2})^* = overline{frac{partial^2}{partial x^2}|_{B}}$. If not, what is their difference ?
functional-analysis ordinary-differential-equations pde
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
$endgroup$
Let $frac{partial^2}{partial x^2}: L^2([0,1]) to L^2([0,1])$ be the one-dimensional Laplacian, considered as an unbounded, densley defined operator with domain $mathcal D(frac{partial^2}{partial x^2}) = C^infty([0,1])$ the smooth funtions without boundary conditions.
Question 1: Can one explicitly describe the domain $mathcal D((frac{partial^2}{partial x^2})^*)$ of the adjoint operator ?
By means of partial integration, one easily checks that
a) $B := { f in C^2([0,1]): f(0) = f'(0) = f(1) = f'(1) = 0 } subseteq mathcal D((frac{partial^2}{partial x^2})^*)$, so that $(frac{partial^2}{partial x^2})^*$ is densely defined (in particular, $frac{partial^2}{partial x^2}$ is closable).
b) $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$.
c) The restriction $frac{partial^2}{partial x^2}|_{B}$ is symmetric and densley defined, hence closable.
By c), we may consider the minimal closure $overline{frac{partial^2}{partial x^2}|_{B}}$. Now since $(frac{partial^2}{partial x^2})^*$ is a closed extension of $(frac{partial^2}{partial x^2})^*|_{B} = frac{partial^2}{partial x^2}|_{B}$ by b), we must have $overline{frac{partial^2}{partial x^2}|_{B}} subset (frac{partial^2}{partial x^2})^*$.
Question 2: Do we have the equality of closed operators $(frac{partial^2}{partial x^2})^* = overline{frac{partial^2}{partial x^2}|_{B}}$. If not, what is their difference ?
functional-analysis ordinary-differential-equations pde
functional-analysis ordinary-differential-equations pde
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
edited Dec 12 '18 at 16:20
Berni Waterman
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
asked Dec 12 '18 at 15:44
Berni WatermanBerni Waterman
998714
998714
$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20
add a comment |
$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20
$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20
add a comment |
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$begingroup$
Q2 : Yes, the two operators are the same.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 18:08
$begingroup$
(b) is not true.
$endgroup$
– DisintegratingByParts
Dec 12 '18 at 23:20