Mixing
Every nonempty closed subset $Msubset mathbb C$ is the spectrum of a closed operator
$begingroup$
Several questions on this cite concern themselves with the operator $T:ell^2(mathbb N)toell^2(mathbb N)$ where $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$. Here the sequence ${r_n:ninmathbb N}subset Msubsetmathbb C$ is dense in the closed and bounded set $M$. These questions concern themselves with the continuous nature of the operator $T$ as well as the fact that $sigma(T)=M$. Some of these questions include the following:
Show that $A:ell^2(mathbb N)toell^2(mathbb N)$ where $A(e_n)=lambda_ne_n$ is bounded.
Prove $forall$ compact $M: M subset Cquad exists A:l_2rightarrow l_2, sigma(A)=M$
Operator whose spectrum is given compact set
One will note that rather than having $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$, we could also have our operator defined by $T(e_n)_{n=1}^infty=(r_ne_n)_{n=1}^infty$, where ${e_n}_{n=1}^infty$ is the usual basis for $ell^2(mathbb N)$.
I am interested in this example for the case when $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is potentially unbounded but at least closed. The following is an excerpt from Unbounded Self-adjoint Operators on Hilbert Space by Konrad Schmüdgen. The example contained within the excerpt details that an arbitrary nonempty closed subset $Msubsetmathbb C$ is the spectrum for some closed operator.
In trying to tackle the closed analogue to the hitherto often discussed bounded case, I am finding it hard to grasp and show the following points:
- What difference is there - if any - in requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ over stipulating that $Msubsetmathbb C$ be compact? As noted in the opening, we often encounter the requirement that $Msubsetmathbb C$ be compact when $T$ is bounded. Interestingly, requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ just ensures that $T$ is bounded - so are both stipulations equivalent?
- How does one use the fact that $(r_nx_n)inell^2(mathbb N)$ in the definition of $mathcal D(T)$ to show that $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is closed? I feel that this is obvious since $T$ acts on all of $ell^2(mathbb N)$, but how do I demonstrate that the limit, $x=(x_m)_{m=1}^inftyinell^2(mathbb N)$, of any convergent sequence $(x_n)_{n=1}^inftysubsetmathcal D(T)$ satisfies that '$(r_mx_m)inell^2(mathbb N)$'?
- What is the difference between defining $Tx_n=r_nx_n$ and defining $Te_n=r_ne_n$, as has been propositioned in similar questions? In particular, I feel that this point is best considered when taken with the next question ...
- How, exactly, do we see that ${r_n:ninmathbb N}subset M$ is contained within $sigma(T)$? In particular, I don't see how $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ has anything to do with the eigenvalue problem; in particular, if we were interested in showing that this was included in the spectrum why don't we look at $Tx=lambda x=(lambda x_1, lambda x_2, lambda x_3, dots)$? It just seems to me that regarding $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ seems to say that this whole sequence $(r_n)_{n=1}^infty$ is an eigenvalue, rather than that each of its components is an eigenvalue. Is this where it makes more sense to look at the operator $Te_n=(r_ne_n)?$
sequences-and-series functional-analysis operator-theory hilbert-spaces unbounded-operators
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
add a comment |
$begingroup$
Several questions on this cite concern themselves with the operator $T:ell^2(mathbb N)toell^2(mathbb N)$ where $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$. Here the sequence ${r_n:ninmathbb N}subset Msubsetmathbb C$ is dense in the closed and bounded set $M$. These questions concern themselves with the continuous nature of the operator $T$ as well as the fact that $sigma(T)=M$. Some of these questions include the following:
Show that $A:ell^2(mathbb N)toell^2(mathbb N)$ where $A(e_n)=lambda_ne_n$ is bounded.
Prove $forall$ compact $M: M subset Cquad exists A:l_2rightarrow l_2, sigma(A)=M$
Operator whose spectrum is given compact set
One will note that rather than having $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$, we could also have our operator defined by $T(e_n)_{n=1}^infty=(r_ne_n)_{n=1}^infty$, where ${e_n}_{n=1}^infty$ is the usual basis for $ell^2(mathbb N)$.
I am interested in this example for the case when $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is potentially unbounded but at least closed. The following is an excerpt from Unbounded Self-adjoint Operators on Hilbert Space by Konrad Schmüdgen. The example contained within the excerpt details that an arbitrary nonempty closed subset $Msubsetmathbb C$ is the spectrum for some closed operator.
In trying to tackle the closed analogue to the hitherto often discussed bounded case, I am finding it hard to grasp and show the following points:
- What difference is there - if any - in requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ over stipulating that $Msubsetmathbb C$ be compact? As noted in the opening, we often encounter the requirement that $Msubsetmathbb C$ be compact when $T$ is bounded. Interestingly, requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ just ensures that $T$ is bounded - so are both stipulations equivalent?
- How does one use the fact that $(r_nx_n)inell^2(mathbb N)$ in the definition of $mathcal D(T)$ to show that $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is closed? I feel that this is obvious since $T$ acts on all of $ell^2(mathbb N)$, but how do I demonstrate that the limit, $x=(x_m)_{m=1}^inftyinell^2(mathbb N)$, of any convergent sequence $(x_n)_{n=1}^inftysubsetmathcal D(T)$ satisfies that '$(r_mx_m)inell^2(mathbb N)$'?
- What is the difference between defining $Tx_n=r_nx_n$ and defining $Te_n=r_ne_n$, as has been propositioned in similar questions? In particular, I feel that this point is best considered when taken with the next question ...
- How, exactly, do we see that ${r_n:ninmathbb N}subset M$ is contained within $sigma(T)$? In particular, I don't see how $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ has anything to do with the eigenvalue problem; in particular, if we were interested in showing that this was included in the spectrum why don't we look at $Tx=lambda x=(lambda x_1, lambda x_2, lambda x_3, dots)$? It just seems to me that regarding $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ seems to say that this whole sequence $(r_n)_{n=1}^infty$ is an eigenvalue, rather than that each of its components is an eigenvalue. Is this where it makes more sense to look at the operator $Te_n=(r_ne_n)?$
sequences-and-series functional-analysis operator-theory hilbert-spaces unbounded-operators
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
add a comment |
$begingroup$
Several questions on this cite concern themselves with the operator $T:ell^2(mathbb N)toell^2(mathbb N)$ where $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$. Here the sequence ${r_n:ninmathbb N}subset Msubsetmathbb C$ is dense in the closed and bounded set $M$. These questions concern themselves with the continuous nature of the operator $T$ as well as the fact that $sigma(T)=M$. Some of these questions include the following:
Show that $A:ell^2(mathbb N)toell^2(mathbb N)$ where $A(e_n)=lambda_ne_n$ is bounded.
Prove $forall$ compact $M: M subset Cquad exists A:l_2rightarrow l_2, sigma(A)=M$
Operator whose spectrum is given compact set
One will note that rather than having $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$, we could also have our operator defined by $T(e_n)_{n=1}^infty=(r_ne_n)_{n=1}^infty$, where ${e_n}_{n=1}^infty$ is the usual basis for $ell^2(mathbb N)$.
I am interested in this example for the case when $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is potentially unbounded but at least closed. The following is an excerpt from Unbounded Self-adjoint Operators on Hilbert Space by Konrad Schmüdgen. The example contained within the excerpt details that an arbitrary nonempty closed subset $Msubsetmathbb C$ is the spectrum for some closed operator.
In trying to tackle the closed analogue to the hitherto often discussed bounded case, I am finding it hard to grasp and show the following points:
- What difference is there - if any - in requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ over stipulating that $Msubsetmathbb C$ be compact? As noted in the opening, we often encounter the requirement that $Msubsetmathbb C$ be compact when $T$ is bounded. Interestingly, requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ just ensures that $T$ is bounded - so are both stipulations equivalent?
- How does one use the fact that $(r_nx_n)inell^2(mathbb N)$ in the definition of $mathcal D(T)$ to show that $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is closed? I feel that this is obvious since $T$ acts on all of $ell^2(mathbb N)$, but how do I demonstrate that the limit, $x=(x_m)_{m=1}^inftyinell^2(mathbb N)$, of any convergent sequence $(x_n)_{n=1}^inftysubsetmathcal D(T)$ satisfies that '$(r_mx_m)inell^2(mathbb N)$'?
- What is the difference between defining $Tx_n=r_nx_n$ and defining $Te_n=r_ne_n$, as has been propositioned in similar questions? In particular, I feel that this point is best considered when taken with the next question ...
- How, exactly, do we see that ${r_n:ninmathbb N}subset M$ is contained within $sigma(T)$? In particular, I don't see how $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ has anything to do with the eigenvalue problem; in particular, if we were interested in showing that this was included in the spectrum why don't we look at $Tx=lambda x=(lambda x_1, lambda x_2, lambda x_3, dots)$? It just seems to me that regarding $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ seems to say that this whole sequence $(r_n)_{n=1}^infty$ is an eigenvalue, rather than that each of its components is an eigenvalue. Is this where it makes more sense to look at the operator $Te_n=(r_ne_n)?$
sequences-and-series functional-analysis operator-theory hilbert-spaces unbounded-operators
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
Several questions on this cite concern themselves with the operator $T:ell^2(mathbb N)toell^2(mathbb N)$ where $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$. Here the sequence ${r_n:ninmathbb N}subset Msubsetmathbb C$ is dense in the closed and bounded set $M$. These questions concern themselves with the continuous nature of the operator $T$ as well as the fact that $sigma(T)=M$. Some of these questions include the following:
Show that $A:ell^2(mathbb N)toell^2(mathbb N)$ where $A(e_n)=lambda_ne_n$ is bounded.
Prove $forall$ compact $M: M subset Cquad exists A:l_2rightarrow l_2, sigma(A)=M$
Operator whose spectrum is given compact set
One will note that rather than having $T(x_n)_{n=1}^infty=(r_nx_n)_{n=1}^infty$, we could also have our operator defined by $T(e_n)_{n=1}^infty=(r_ne_n)_{n=1}^infty$, where ${e_n}_{n=1}^infty$ is the usual basis for $ell^2(mathbb N)$.
I am interested in this example for the case when $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is potentially unbounded but at least closed. The following is an excerpt from Unbounded Self-adjoint Operators on Hilbert Space by Konrad Schmüdgen. The example contained within the excerpt details that an arbitrary nonempty closed subset $Msubsetmathbb C$ is the spectrum for some closed operator.
In trying to tackle the closed analogue to the hitherto often discussed bounded case, I am finding it hard to grasp and show the following points:
- What difference is there - if any - in requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ over stipulating that $Msubsetmathbb C$ be compact? As noted in the opening, we often encounter the requirement that $Msubsetmathbb C$ be compact when $T$ is bounded. Interestingly, requiring that $mathcal D(T)={(x_n)inell^2(mathbb N):(r_nx_n)inell^2(mathbb N)}$ just ensures that $T$ is bounded - so are both stipulations equivalent?
- How does one use the fact that $(r_nx_n)inell^2(mathbb N)$ in the definition of $mathcal D(T)$ to show that $T:mathcal D(T)subseteq ell^2(mathbb N)toell^2(mathbb N)$ is closed? I feel that this is obvious since $T$ acts on all of $ell^2(mathbb N)$, but how do I demonstrate that the limit, $x=(x_m)_{m=1}^inftyinell^2(mathbb N)$, of any convergent sequence $(x_n)_{n=1}^inftysubsetmathcal D(T)$ satisfies that '$(r_mx_m)inell^2(mathbb N)$'?
- What is the difference between defining $Tx_n=r_nx_n$ and defining $Te_n=r_ne_n$, as has been propositioned in similar questions? In particular, I feel that this point is best considered when taken with the next question ...
- How, exactly, do we see that ${r_n:ninmathbb N}subset M$ is contained within $sigma(T)$? In particular, I don't see how $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ has anything to do with the eigenvalue problem; in particular, if we were interested in showing that this was included in the spectrum why don't we look at $Tx=lambda x=(lambda x_1, lambda x_2, lambda x_3, dots)$? It just seems to me that regarding $Tx_n=(r_nx_n)=(r_1x_1, r_2x_2, r_3x_3,dots)$ seems to say that this whole sequence $(r_n)_{n=1}^infty$ is an eigenvalue, rather than that each of its components is an eigenvalue. Is this where it makes more sense to look at the operator $Te_n=(r_ne_n)?$
sequences-and-series functional-analysis operator-theory hilbert-spaces unbounded-operators
sequences-and-series functional-analysis operator-theory hilbert-spaces unbounded-operators
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Jan 25 at 17:40
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
1,055717
add a comment |
add a comment |
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