Mixing
Large radius limit for a differential operator on a circle
$begingroup$
Let $(A_r)_{rgeq0}$ be a one-parameter family of linear operators, with $A_r$ being the (weak) first derivative operator on $L^2(S_r)$, $S_r$ being the one-dimensional circle, with a multiplicative factor $-i$ to ensure symmetry. Each $A_r$ is known to be self-adjoint on its Hilbert space.
Intuitively, one may expect $A_r$ to "converge" to the first-derivative operator on the real line (again with a factor $-i$), say $A_infty$, which is self-adjoint as well. However, all operators $A_r$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.
Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $A_rto A_infty?$
operator-theory hilbert-spaces differential-operators
asked Jan 30 at 11:27
DavideDavide
106
$endgroup$
add a comment |
$begingroup$
Let $(A_r)_{rgeq0}$ be a one-parameter family of linear operators, with $A_r$ being the (weak) first derivative operator on $L^2(S_r)$, $S_r$ being the one-dimensional circle, with a multiplicative factor $-i$ to ensure symmetry. Each $A_r$ is known to be self-adjoint on its Hilbert space.
Intuitively, one may expect $A_r$ to "converge" to the first-derivative operator on the real line (again with a factor $-i$), say $A_infty$, which is self-adjoint as well. However, all operators $A_r$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.
Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $A_rto A_infty?$
operator-theory hilbert-spaces differential-operators
asked Jan 30 at 11:27
DavideDavide
106
$endgroup$
add a comment |
$begingroup$
Let $(A_r)_{rgeq0}$ be a one-parameter family of linear operators, with $A_r$ being the (weak) first derivative operator on $L^2(S_r)$, $S_r$ being the one-dimensional circle, with a multiplicative factor $-i$ to ensure symmetry. Each $A_r$ is known to be self-adjoint on its Hilbert space.
Intuitively, one may expect $A_r$ to "converge" to the first-derivative operator on the real line (again with a factor $-i$), say $A_infty$, which is self-adjoint as well. However, all operators $A_r$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.
Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $A_rto A_infty?$
operator-theory hilbert-spaces differential-operators
asked Jan 30 at 11:27
DavideDavide
106
$endgroup$
Let $(A_r)_{rgeq0}$ be a one-parameter family of linear operators, with $A_r$ being the (weak) first derivative operator on $L^2(S_r)$, $S_r$ being the one-dimensional circle, with a multiplicative factor $-i$ to ensure symmetry. Each $A_r$ is known to be self-adjoint on its Hilbert space.
Intuitively, one may expect $A_r$ to "converge" to the first-derivative operator on the real line (again with a factor $-i$), say $A_infty$, which is self-adjoint as well. However, all operators $A_r$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.
Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $A_rto A_infty?$
operator-theory hilbert-spaces differential-operators
operator-theory hilbert-spaces differential-operators
asked Jan 30 at 11:27
DavideDavide
106
asked Jan 30 at 11:27
DavideDavide
106
edited Jan 30 at 11:41
Davide
asked Jan 30 at 11:27
DavideDavide
106
asked Jan 30 at 11:27
DavideDavide
106
asked Jan 30 at 11:27
DavideDavide
106
106
add a comment |
add a comment |
1 Answer
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The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as
$$
rmathbb S^1={(rcos theta, r sin theta), :, thetain (-pi, pi)}, $$
so the operator you refer to is
$$
ipartial_theta f := ipartial_theta [f(rcos theta, r sin theta)], $$
where $fin C^infty(mathbb R^2)$. The chain rule shows that
$$ipartial_theta= ir (-sin theta, cos theta)cdot nabla, $$ so, in some sense, it diverges as $rto infty$.
Thus, I expect that the right operator to consider is the normalized one, $frac{ipartial_theta}{r}.$
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
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1 Answer
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1 Answer
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$begingroup$
The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as
$$
rmathbb S^1={(rcos theta, r sin theta), :, thetain (-pi, pi)}, $$
so the operator you refer to is
$$
ipartial_theta f := ipartial_theta [f(rcos theta, r sin theta)], $$
where $fin C^infty(mathbb R^2)$. The chain rule shows that
$$ipartial_theta= ir (-sin theta, cos theta)cdot nabla, $$ so, in some sense, it diverges as $rto infty$.
Thus, I expect that the right operator to consider is the normalized one, $frac{ipartial_theta}{r}.$
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
$endgroup$
add a comment |
$begingroup$
The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as
$$
rmathbb S^1={(rcos theta, r sin theta), :, thetain (-pi, pi)}, $$
so the operator you refer to is
$$
ipartial_theta f := ipartial_theta [f(rcos theta, r sin theta)], $$
where $fin C^infty(mathbb R^2)$. The chain rule shows that
$$ipartial_theta= ir (-sin theta, cos theta)cdot nabla, $$ so, in some sense, it diverges as $rto infty$.
Thus, I expect that the right operator to consider is the normalized one, $frac{ipartial_theta}{r}.$
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
$endgroup$
add a comment |
$begingroup$
The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as
$$
rmathbb S^1={(rcos theta, r sin theta), :, thetain (-pi, pi)}, $$
so the operator you refer to is
$$
ipartial_theta f := ipartial_theta [f(rcos theta, r sin theta)], $$
where $fin C^infty(mathbb R^2)$. The chain rule shows that
$$ipartial_theta= ir (-sin theta, cos theta)cdot nabla, $$ so, in some sense, it diverges as $rto infty$.
Thus, I expect that the right operator to consider is the normalized one, $frac{ipartial_theta}{r}.$
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
$endgroup$
The solution is to write down explicit expressions in coordinates. The circle of radius $r$ is parametrized as
$$
rmathbb S^1={(rcos theta, r sin theta), :, thetain (-pi, pi)}, $$
so the operator you refer to is
$$
ipartial_theta f := ipartial_theta [f(rcos theta, r sin theta)], $$
where $fin C^infty(mathbb R^2)$. The chain rule shows that
$$ipartial_theta= ir (-sin theta, cos theta)cdot nabla, $$ so, in some sense, it diverges as $rto infty$.
Thus, I expect that the right operator to consider is the normalized one, $frac{ipartial_theta}{r}.$
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
edited Jan 30 at 12:25
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
answered Jan 30 at 11:50
Giuseppe NegroGiuseppe Negro
17.6k332127
17.6k332127
add a comment |
add a comment |
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