Mixing
Prove that there is no self-adjoint extension using deficiency indices
$begingroup$
Consider an operator $P =-ifrac{d}{dx} : dom(P) to L^2(mathbb{R}^+)$ where
$$ dom(P) = { f in mathcal{D}(mathbb{R}^+) : f(0)=0}$$
where $mathcal{D}(mathbb{R}^+)$ - smooth compactly supported functions (test-functions).
I need to prove that this operator has no self-adjoint extension by calculating deficiency indices.
But I'm stuck calculating those indices by their definition:
$$ n_+ (P) = dim(im(P+i)^perp) \
n_-(P)=dim(im(P-i)^perp)$$
So I have:
$$im(P+i)^perp = { phi in L^2(mathbb{R}^+) | <phi, -ifrac{d}{dx}f+if>=0, forall f in dom(P)}$$
That is, a set of such $phi in L^2(mathbb{R}^+)$ such that:
$$ int^infty_0 bar{phi}(x)(-ifrac{df}{dx}+if)dx=0$$
How do I proceed from here? I would try to do integration by parts, but general $phi$ does not have to be differentiable (only square integrable).
I need to show somehow that one of the deficiency indices is not equal to zero.
linear-algebra operator-theory spectral-theory adjoint-operators
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
$endgroup$
add a comment |
$begingroup$
Consider an operator $P =-ifrac{d}{dx} : dom(P) to L^2(mathbb{R}^+)$ where
$$ dom(P) = { f in mathcal{D}(mathbb{R}^+) : f(0)=0}$$
where $mathcal{D}(mathbb{R}^+)$ - smooth compactly supported functions (test-functions).
I need to prove that this operator has no self-adjoint extension by calculating deficiency indices.
But I'm stuck calculating those indices by their definition:
$$ n_+ (P) = dim(im(P+i)^perp) \
n_-(P)=dim(im(P-i)^perp)$$
So I have:
$$im(P+i)^perp = { phi in L^2(mathbb{R}^+) | <phi, -ifrac{d}{dx}f+if>=0, forall f in dom(P)}$$
That is, a set of such $phi in L^2(mathbb{R}^+)$ such that:
$$ int^infty_0 bar{phi}(x)(-ifrac{df}{dx}+if)dx=0$$
How do I proceed from here? I would try to do integration by parts, but general $phi$ does not have to be differentiable (only square integrable).
I need to show somehow that one of the deficiency indices is not equal to zero.
linear-algebra operator-theory spectral-theory adjoint-operators
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
$endgroup$
add a comment |
$begingroup$
Consider an operator $P =-ifrac{d}{dx} : dom(P) to L^2(mathbb{R}^+)$ where
$$ dom(P) = { f in mathcal{D}(mathbb{R}^+) : f(0)=0}$$
where $mathcal{D}(mathbb{R}^+)$ - smooth compactly supported functions (test-functions).
I need to prove that this operator has no self-adjoint extension by calculating deficiency indices.
But I'm stuck calculating those indices by their definition:
$$ n_+ (P) = dim(im(P+i)^perp) \
n_-(P)=dim(im(P-i)^perp)$$
So I have:
$$im(P+i)^perp = { phi in L^2(mathbb{R}^+) | <phi, -ifrac{d}{dx}f+if>=0, forall f in dom(P)}$$
That is, a set of such $phi in L^2(mathbb{R}^+)$ such that:
$$ int^infty_0 bar{phi}(x)(-ifrac{df}{dx}+if)dx=0$$
How do I proceed from here? I would try to do integration by parts, but general $phi$ does not have to be differentiable (only square integrable).
I need to show somehow that one of the deficiency indices is not equal to zero.
linear-algebra operator-theory spectral-theory adjoint-operators
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
$endgroup$
Consider an operator $P =-ifrac{d}{dx} : dom(P) to L^2(mathbb{R}^+)$ where
$$ dom(P) = { f in mathcal{D}(mathbb{R}^+) : f(0)=0}$$
where $mathcal{D}(mathbb{R}^+)$ - smooth compactly supported functions (test-functions).
I need to prove that this operator has no self-adjoint extension by calculating deficiency indices.
But I'm stuck calculating those indices by their definition:
$$ n_+ (P) = dim(im(P+i)^perp) \
n_-(P)=dim(im(P-i)^perp)$$
So I have:
$$im(P+i)^perp = { phi in L^2(mathbb{R}^+) | <phi, -ifrac{d}{dx}f+if>=0, forall f in dom(P)}$$
That is, a set of such $phi in L^2(mathbb{R}^+)$ such that:
$$ int^infty_0 bar{phi}(x)(-ifrac{df}{dx}+if)dx=0$$
How do I proceed from here? I would try to do integration by parts, but general $phi$ does not have to be differentiable (only square integrable).
I need to show somehow that one of the deficiency indices is not equal to zero.
linear-algebra operator-theory spectral-theory adjoint-operators
linear-algebra operator-theory spectral-theory adjoint-operators
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
edited Jan 14 at 1:05
Sergey Dylda
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
asked Jan 14 at 0:49
Sergey DyldaSergey Dylda
1416
1416
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The closure $overline{P}$ of $P$ in $L^2[0,infty)$ has a domain $mathcal{D}(overline{P})$ consisting of every $fin L^2[0,infty)$ that is equal a.e. to an absolutely continuous function $tilde{f}in L^2[0,infty)$ such that $tilde{f}'in L^2[0,infty)$ and $tilde{f}(0)=0$. The ajdoint $P^*$ has the same action and domain except that $tilde{f}(0)$ is unconstrained.
Because of the homogeneous endpoint condition $overline{P}$ has no non-trivial eigenfunctions. However $P^* e^{-x}=-ie^{-x}$ does hold, and $e^{-x}in L^2[0,infty)$. $P^* f = if$ has no non-trivial solutions $finmathcal{D}(P^*)$ because $e^{x}notin L^2[0,infty)$.
Summarizing,
$$
mathcal{R}(P-iI)^{perp}= mathcal{N}(P^*+iI)=[{ e^{-x}}] \
mathcal{R}(P+iI)^{perp}= mathcal{N}(P^*-iI)=[{0}].
$$
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
The closure $overline{P}$ of $P$ in $L^2[0,infty)$ has a domain $mathcal{D}(overline{P})$ consisting of every $fin L^2[0,infty)$ that is equal a.e. to an absolutely continuous function $tilde{f}in L^2[0,infty)$ such that $tilde{f}'in L^2[0,infty)$ and $tilde{f}(0)=0$. The ajdoint $P^*$ has the same action and domain except that $tilde{f}(0)$ is unconstrained.
Because of the homogeneous endpoint condition $overline{P}$ has no non-trivial eigenfunctions. However $P^* e^{-x}=-ie^{-x}$ does hold, and $e^{-x}in L^2[0,infty)$. $P^* f = if$ has no non-trivial solutions $finmathcal{D}(P^*)$ because $e^{x}notin L^2[0,infty)$.
Summarizing,
$$
mathcal{R}(P-iI)^{perp}= mathcal{N}(P^*+iI)=[{ e^{-x}}] \
mathcal{R}(P+iI)^{perp}= mathcal{N}(P^*-iI)=[{0}].
$$
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
add a comment |
$begingroup$
The closure $overline{P}$ of $P$ in $L^2[0,infty)$ has a domain $mathcal{D}(overline{P})$ consisting of every $fin L^2[0,infty)$ that is equal a.e. to an absolutely continuous function $tilde{f}in L^2[0,infty)$ such that $tilde{f}'in L^2[0,infty)$ and $tilde{f}(0)=0$. The ajdoint $P^*$ has the same action and domain except that $tilde{f}(0)$ is unconstrained.
Because of the homogeneous endpoint condition $overline{P}$ has no non-trivial eigenfunctions. However $P^* e^{-x}=-ie^{-x}$ does hold, and $e^{-x}in L^2[0,infty)$. $P^* f = if$ has no non-trivial solutions $finmathcal{D}(P^*)$ because $e^{x}notin L^2[0,infty)$.
Summarizing,
$$
mathcal{R}(P-iI)^{perp}= mathcal{N}(P^*+iI)=[{ e^{-x}}] \
mathcal{R}(P+iI)^{perp}= mathcal{N}(P^*-iI)=[{0}].
$$
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
add a comment |
$begingroup$
The closure $overline{P}$ of $P$ in $L^2[0,infty)$ has a domain $mathcal{D}(overline{P})$ consisting of every $fin L^2[0,infty)$ that is equal a.e. to an absolutely continuous function $tilde{f}in L^2[0,infty)$ such that $tilde{f}'in L^2[0,infty)$ and $tilde{f}(0)=0$. The ajdoint $P^*$ has the same action and domain except that $tilde{f}(0)$ is unconstrained.
Because of the homogeneous endpoint condition $overline{P}$ has no non-trivial eigenfunctions. However $P^* e^{-x}=-ie^{-x}$ does hold, and $e^{-x}in L^2[0,infty)$. $P^* f = if$ has no non-trivial solutions $finmathcal{D}(P^*)$ because $e^{x}notin L^2[0,infty)$.
Summarizing,
$$
mathcal{R}(P-iI)^{perp}= mathcal{N}(P^*+iI)=[{ e^{-x}}] \
mathcal{R}(P+iI)^{perp}= mathcal{N}(P^*-iI)=[{0}].
$$
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
The closure $overline{P}$ of $P$ in $L^2[0,infty)$ has a domain $mathcal{D}(overline{P})$ consisting of every $fin L^2[0,infty)$ that is equal a.e. to an absolutely continuous function $tilde{f}in L^2[0,infty)$ such that $tilde{f}'in L^2[0,infty)$ and $tilde{f}(0)=0$. The ajdoint $P^*$ has the same action and domain except that $tilde{f}(0)$ is unconstrained.
Because of the homogeneous endpoint condition $overline{P}$ has no non-trivial eigenfunctions. However $P^* e^{-x}=-ie^{-x}$ does hold, and $e^{-x}in L^2[0,infty)$. $P^* f = if$ has no non-trivial solutions $finmathcal{D}(P^*)$ because $e^{x}notin L^2[0,infty)$.
Summarizing,
$$
mathcal{R}(P-iI)^{perp}= mathcal{N}(P^*+iI)=[{ e^{-x}}] \
mathcal{R}(P+iI)^{perp}= mathcal{N}(P^*-iI)=[{0}].
$$
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 14 at 7:07
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
add a comment |
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
Thanks! How do I prove that $P^*$ has the same action and domain?
$endgroup$
– Sergey Dylda
Jan 14 at 10:36
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
$begingroup$
@SergeyDylda Start with the adjoint relation, meaning $ginmathcal{D}(P^*)$ iff $int_0^{infty}(-if'(t))overline{g(t)}dt = int_0^{infty}f(t)overline{L^*g}dt$ holds for all $finmathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=int_r^s overline{L^*g} dt$ a.e. for $ginmathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation.
$endgroup$
– DisintegratingByParts
Jan 14 at 17:24
add a comment |
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