Mixing
Show that $A:ell^2(mathbb N)toell^2(mathbb N)$ where $A(e_n)=lambda_ne_n$ is bounded.
$begingroup$
Let $Csubsetmathbb C$ be closed. As $mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset ${lambda_n:ninmathbb N}subset C$ dense in $C$.
In the first comment to the question here the claim is made that the operator $A:ell^2(mathbb N)toell^2(mathbb N)$, the action of which is $A(e_n)=lambda_ne_n$, is bounded. Here, ${e_n:ninmathbb N}subsetell^2(mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.
Here is my working so far. Recall that in any Hilbert space, $mathcal H$, with an orthonormal basis, any $xinmathcal H$ can be uniquely expressed in the form, $$x=sum_{n=1}^inftylangle x,e_nrangle e_n.$$ Now, consider the quantity,
$$begin{align*}
|Ax|&=left|Aleft(sum_{n=1}^inftylangle x,e_nrangle e_nright)right|\
&=left|sum_{n=1}^inftylangle x,e_nrangle Ae_nright| &&text{(assuming A linear)}\
&=left|sum_{n=1}^inftylangle x,e_nrangle lambda_ne_nright|\
&lesum_{n=1}^infty|langle x,e_nrangle lambda_ne_n| &&text{(triangle inequality)}\
&=sum_{n=1}^infty|langle x,e_nrangle| |lambda_n||e_n|\
&le|x|sum_{n=1}^infty |lambda_n||e_n|^2 &&text{(Cauchy-Schwarz).}\
end{align*}$$
What I thought here to do was to bound $|lambda_n|$ by $M:=max_{ninmathbb N}{lambda_n}$ so as to obtain,
$$|x|sum_{n=1}^infty |lambda_n||e_n|^2le M|x|sum_{n=1}^infty |e_n|^2.$$
But as the norm of each $e_ninell^2(mathbb N)$ is unity the series we are left with diverges.
As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?
sequences-and-series functional-analysis operator-theory hilbert-spaces norm
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
$endgroup$
add a comment |
$begingroup$
Let $Csubsetmathbb C$ be closed. As $mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset ${lambda_n:ninmathbb N}subset C$ dense in $C$.
In the first comment to the question here the claim is made that the operator $A:ell^2(mathbb N)toell^2(mathbb N)$, the action of which is $A(e_n)=lambda_ne_n$, is bounded. Here, ${e_n:ninmathbb N}subsetell^2(mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.
Here is my working so far. Recall that in any Hilbert space, $mathcal H$, with an orthonormal basis, any $xinmathcal H$ can be uniquely expressed in the form, $$x=sum_{n=1}^inftylangle x,e_nrangle e_n.$$ Now, consider the quantity,
$$begin{align*}
|Ax|&=left|Aleft(sum_{n=1}^inftylangle x,e_nrangle e_nright)right|\
&=left|sum_{n=1}^inftylangle x,e_nrangle Ae_nright| &&text{(assuming A linear)}\
&=left|sum_{n=1}^inftylangle x,e_nrangle lambda_ne_nright|\
&lesum_{n=1}^infty|langle x,e_nrangle lambda_ne_n| &&text{(triangle inequality)}\
&=sum_{n=1}^infty|langle x,e_nrangle| |lambda_n||e_n|\
&le|x|sum_{n=1}^infty |lambda_n||e_n|^2 &&text{(Cauchy-Schwarz).}\
end{align*}$$
What I thought here to do was to bound $|lambda_n|$ by $M:=max_{ninmathbb N}{lambda_n}$ so as to obtain,
$$|x|sum_{n=1}^infty |lambda_n||e_n|^2le M|x|sum_{n=1}^infty |e_n|^2.$$
But as the norm of each $e_ninell^2(mathbb N)$ is unity the series we are left with diverges.
As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?
sequences-and-series functional-analysis operator-theory hilbert-spaces norm
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
$endgroup$
$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12
add a comment |
$begingroup$
Let $Csubsetmathbb C$ be closed. As $mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset ${lambda_n:ninmathbb N}subset C$ dense in $C$.
In the first comment to the question here the claim is made that the operator $A:ell^2(mathbb N)toell^2(mathbb N)$, the action of which is $A(e_n)=lambda_ne_n$, is bounded. Here, ${e_n:ninmathbb N}subsetell^2(mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.
Here is my working so far. Recall that in any Hilbert space, $mathcal H$, with an orthonormal basis, any $xinmathcal H$ can be uniquely expressed in the form, $$x=sum_{n=1}^inftylangle x,e_nrangle e_n.$$ Now, consider the quantity,
$$begin{align*}
|Ax|&=left|Aleft(sum_{n=1}^inftylangle x,e_nrangle e_nright)right|\
&=left|sum_{n=1}^inftylangle x,e_nrangle Ae_nright| &&text{(assuming A linear)}\
&=left|sum_{n=1}^inftylangle x,e_nrangle lambda_ne_nright|\
&lesum_{n=1}^infty|langle x,e_nrangle lambda_ne_n| &&text{(triangle inequality)}\
&=sum_{n=1}^infty|langle x,e_nrangle| |lambda_n||e_n|\
&le|x|sum_{n=1}^infty |lambda_n||e_n|^2 &&text{(Cauchy-Schwarz).}\
end{align*}$$
What I thought here to do was to bound $|lambda_n|$ by $M:=max_{ninmathbb N}{lambda_n}$ so as to obtain,
$$|x|sum_{n=1}^infty |lambda_n||e_n|^2le M|x|sum_{n=1}^infty |e_n|^2.$$
But as the norm of each $e_ninell^2(mathbb N)$ is unity the series we are left with diverges.
As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?
sequences-and-series functional-analysis operator-theory hilbert-spaces norm
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
$endgroup$
Let $Csubsetmathbb C$ be closed. As $mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset ${lambda_n:ninmathbb N}subset C$ dense in $C$.
In the first comment to the question here the claim is made that the operator $A:ell^2(mathbb N)toell^2(mathbb N)$, the action of which is $A(e_n)=lambda_ne_n$, is bounded. Here, ${e_n:ninmathbb N}subsetell^2(mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.
Here is my working so far. Recall that in any Hilbert space, $mathcal H$, with an orthonormal basis, any $xinmathcal H$ can be uniquely expressed in the form, $$x=sum_{n=1}^inftylangle x,e_nrangle e_n.$$ Now, consider the quantity,
$$begin{align*}
|Ax|&=left|Aleft(sum_{n=1}^inftylangle x,e_nrangle e_nright)right|\
&=left|sum_{n=1}^inftylangle x,e_nrangle Ae_nright| &&text{(assuming A linear)}\
&=left|sum_{n=1}^inftylangle x,e_nrangle lambda_ne_nright|\
&lesum_{n=1}^infty|langle x,e_nrangle lambda_ne_n| &&text{(triangle inequality)}\
&=sum_{n=1}^infty|langle x,e_nrangle| |lambda_n||e_n|\
&le|x|sum_{n=1}^infty |lambda_n||e_n|^2 &&text{(Cauchy-Schwarz).}\
end{align*}$$
What I thought here to do was to bound $|lambda_n|$ by $M:=max_{ninmathbb N}{lambda_n}$ so as to obtain,
$$|x|sum_{n=1}^infty |lambda_n||e_n|^2le M|x|sum_{n=1}^infty |e_n|^2.$$
But as the norm of each $e_ninell^2(mathbb N)$ is unity the series we are left with diverges.
As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?
sequences-and-series functional-analysis operator-theory hilbert-spaces norm
sequences-and-series functional-analysis operator-theory hilbert-spaces norm
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
asked Jan 22 at 16:54
Jeremy Jeffrey JamesJeremy Jeffrey James
1,013717
1,013717
$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12
add a comment |
$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12
$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence ${lambda_n}$ is not bounded, your $A$ will not be bounded.
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=sup_n|lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have
$$
left|sum_{n=1}^Mlambdalangle x,e_nrangle,e_nright|^2=sum_{n=1}^M|lambda_n|^2,|langle x,e_nrangle|^2leq c^2,sum_{n=1}^M ,|langle x,e_nrangle|^2leq c^2,|x|^2.
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
|
show 1 more comment
$begingroup$
You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, ldots) in ell^2$,
$A x = (lambda_1 x_1, lambda_2 x_2, ldots)$ with
$$ |A x |^2 = sum_{n=1}^infty |lambda_n|^2 |x_n|^2 le M^2 sum_{n=1}^infty |x_n|^2 = M^2 |x|^2 $$
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
$endgroup$
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence ${lambda_n}$ is not bounded, your $A$ will not be bounded.
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=sup_n|lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have
$$
left|sum_{n=1}^Mlambdalangle x,e_nrangle,e_nright|^2=sum_{n=1}^M|lambda_n|^2,|langle x,e_nrangle|^2leq c^2,sum_{n=1}^M ,|langle x,e_nrangle|^2leq c^2,|x|^2.
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
|
show 1 more comment
$begingroup$
It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence ${lambda_n}$ is not bounded, your $A$ will not be bounded.
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=sup_n|lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have
$$
left|sum_{n=1}^Mlambdalangle x,e_nrangle,e_nright|^2=sum_{n=1}^M|lambda_n|^2,|langle x,e_nrangle|^2leq c^2,sum_{n=1}^M ,|langle x,e_nrangle|^2leq c^2,|x|^2.
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
|
show 1 more comment
$begingroup$
It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence ${lambda_n}$ is not bounded, your $A$ will not be bounded.
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=sup_n|lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have
$$
left|sum_{n=1}^Mlambdalangle x,e_nrangle,e_nright|^2=sum_{n=1}^M|lambda_n|^2,|langle x,e_nrangle|^2leq c^2,sum_{n=1}^M ,|langle x,e_nrangle|^2leq c^2,|x|^2.
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence ${lambda_n}$ is not bounded, your $A$ will not be bounded.
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=sup_n|lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have
$$
left|sum_{n=1}^Mlambdalangle x,e_nrangle,e_nright|^2=sum_{n=1}^M|lambda_n|^2,|langle x,e_nrangle|^2leq c^2,sum_{n=1}^M ,|langle x,e_nrangle|^2leq c^2,|x|^2.
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
edited Jan 28 at 11:46
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 22 at 16:58
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
|
show 1 more comment
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular.
$endgroup$
– Klaus
Jan 22 at 17:02
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
$begingroup$
Taking $A$ inside the sum is just down to $A$ being linear, no?
$endgroup$
– Jeremy Jeffrey James
Jan 22 at 17:03
1
1
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@JeremyJeffreyJames Not if the sum is infinite, i.e. a series.
$endgroup$
– lisyarus
Jan 22 at 17:04
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
$begingroup$
@Klaus: good point. The estimate is the same, though.
$endgroup$
– Martin Argerami
Jan 22 at 17:59
1
1
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
$begingroup$
They are the span of an orthonormal basis.
$endgroup$
– Martin Argerami
Jan 28 at 11:49
|
show 1 more comment
$begingroup$
You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, ldots) in ell^2$,
$A x = (lambda_1 x_1, lambda_2 x_2, ldots)$ with
$$ |A x |^2 = sum_{n=1}^infty |lambda_n|^2 |x_n|^2 le M^2 sum_{n=1}^infty |x_n|^2 = M^2 |x|^2 $$
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
$endgroup$
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
add a comment |
$begingroup$
You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, ldots) in ell^2$,
$A x = (lambda_1 x_1, lambda_2 x_2, ldots)$ with
$$ |A x |^2 = sum_{n=1}^infty |lambda_n|^2 |x_n|^2 le M^2 sum_{n=1}^infty |x_n|^2 = M^2 |x|^2 $$
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
$endgroup$
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
add a comment |
$begingroup$
You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, ldots) in ell^2$,
$A x = (lambda_1 x_1, lambda_2 x_2, ldots)$ with
$$ |A x |^2 = sum_{n=1}^infty |lambda_n|^2 |x_n|^2 le M^2 sum_{n=1}^infty |x_n|^2 = M^2 |x|^2 $$
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
$endgroup$
You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, ldots) in ell^2$,
$A x = (lambda_1 x_1, lambda_2 x_2, ldots)$ with
$$ |A x |^2 = sum_{n=1}^infty |lambda_n|^2 |x_n|^2 le M^2 sum_{n=1}^infty |x_n|^2 = M^2 |x|^2 $$
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
answered Jan 22 at 17:01
Robert IsraelRobert Israel
327k23215469
327k23215469
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
add a comment |
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, ldots) in ell^2$ then $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (lambda_1 x_1, lambda_2 x_2, ldots)$, then we can deduce that the whole sequence $(lambda_n)_{n=1}^infty$ is in $sigma(A)$?
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:35
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
$begingroup$
Because this does not mirror the usual eigenvalue problem of finding $lambdainmathbb C$ for which $Ax=lambda x=(lambda x_1, lambda x_2, lambda x_3 ...)$
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 16:36
1
1
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
$lambda_n in sigma(A)$ because $A e_n = lambda_n e_n$.
$endgroup$
– Robert Israel
Jan 25 at 17:57
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
$begingroup$
I see, I wasn't thinking correctly.
$endgroup$
– Jeremy Jeffrey James
Jan 25 at 18:59
add a comment |
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$begingroup$
Take $C=mathbb N$, which is closed. $mathbb N$ is a countable dense subset of $C=mathbb N$, while the operator $Ae_n = n cdot e_n$ is clearly unbounded.
$endgroup$
– lisyarus
Jan 22 at 17:01
$begingroup$
I get the idea, but $overline{mathbb N}neqmathbb C$?
$endgroup$
– Jeremy Jeffrey James
Jan 23 at 11:12