Range Bounded Self-Adjoint Operator
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Consider a bounded, self-adjoint operator $T:Hrightarrow H$, $H$ is a Hilbert space, such that
beginalign*
langle Tx,xrangle geq beta ||x||^2
endalign*
where $beta>0$ is a constant. I know that the null space of this operator is simply $0$. However, I am having trouble showing that $mathcalR(T)$ is closed. How would I go about doing this?
functional-analysis hilbert-spaces
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
add a comment |Â
up vote
2
down vote
favorite
Consider a bounded, self-adjoint operator $T:Hrightarrow H$, $H$ is a Hilbert space, such that
beginalign*
langle Tx,xrangle geq beta ||x||^2
endalign*
where $beta>0$ is a constant. I know that the null space of this operator is simply $0$. However, I am having trouble showing that $mathcalR(T)$ is closed. How would I go about doing this?
functional-analysis hilbert-spaces
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider a bounded, self-adjoint operator $T:Hrightarrow H$, $H$ is a Hilbert space, such that
beginalign*
langle Tx,xrangle geq beta ||x||^2
endalign*
where $beta>0$ is a constant. I know that the null space of this operator is simply $0$. However, I am having trouble showing that $mathcalR(T)$ is closed. How would I go about doing this?
functional-analysis hilbert-spaces
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
Consider a bounded, self-adjoint operator $T:Hrightarrow H$, $H$ is a Hilbert space, such that
beginalign*
langle Tx,xrangle geq beta ||x||^2
endalign*
where $beta>0$ is a constant. I know that the null space of this operator is simply $0$. However, I am having trouble showing that $mathcalR(T)$ is closed. How would I go about doing this?
functional-analysis hilbert-spaces
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
asked Aug 15 at 4:29
Rebecca Hardenbrook
335310
335310
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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$beta |x|^2 leq |Tx| |x|$ so $beta |x| leq |Tx| $. If $Tx_n to y$ then $T(x_n-x_m) to 0$ so $beta |x_n-x_m| to 0$. Hence $x_n$ is Cauchy. if $x_n to x$ then $y =lim Tx_n =Tx$.
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$beta |x|^2 leq |Tx| |x|$ so $beta |x| leq |Tx| $. If $Tx_n to y$ then $T(x_n-x_m) to 0$ so $beta |x_n-x_m| to 0$. Hence $x_n$ is Cauchy. if $x_n to x$ then $y =lim Tx_n =Tx$.
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
add a comment |Â
up vote
1
down vote
$beta |x|^2 leq |Tx| |x|$ so $beta |x| leq |Tx| $. If $Tx_n to y$ then $T(x_n-x_m) to 0$ so $beta |x_n-x_m| to 0$. Hence $x_n$ is Cauchy. if $x_n to x$ then $y =lim Tx_n =Tx$.
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$beta |x|^2 leq |Tx| |x|$ so $beta |x| leq |Tx| $. If $Tx_n to y$ then $T(x_n-x_m) to 0$ so $beta |x_n-x_m| to 0$. Hence $x_n$ is Cauchy. if $x_n to x$ then $y =lim Tx_n =Tx$.
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
$beta |x|^2 leq |Tx| |x|$ so $beta |x| leq |Tx| $. If $Tx_n to y$ then $T(x_n-x_m) to 0$ so $beta |x_n-x_m| to 0$. Hence $x_n$ is Cauchy. if $x_n to x$ then $y =lim Tx_n =Tx$.
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 5:57
Kavi Rama Murthy
22.5k2933
22.5k2933
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
add a comment |Â
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong.
– Rebecca Hardenbrook
Aug 15 at 13:13
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
@RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence?
– Kavi Rama Murthy
Aug 15 at 23:21
add a comment |Â
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