Closedness of $R(lambda -T)$ when $T$ closed symmetric operator
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Let $T$ be a closed symmetric operator defined on a Hilbert space.
I want to proof that the Rang $R(lambda - T)$ is closed if and only if $Im(lambda)neq 0$.
I used the sequences technic to proof that if $Im(lambda)neq 0$ then the equality hold, but I couldn't see why it is not true if $Im(lambda)= 0$ !!!
I proceed like the following
if $T$ is closed operator so $(lambda - T)$ too, then $R(lambda - T)$ is close.
Could someone give me a counter-example or any clarification, please?
functional-analysis operator-theory operator-algebras spectral-theory
asked Dec 27 '17 at 10:21
The_lost
1,052318
 |Â
show 5 more comments
up vote
1
down vote
favorite
Let $T$ be a closed symmetric operator defined on a Hilbert space.
I want to proof that the Rang $R(lambda - T)$ is closed if and only if $Im(lambda)neq 0$.
I used the sequences technic to proof that if $Im(lambda)neq 0$ then the equality hold, but I couldn't see why it is not true if $Im(lambda)= 0$ !!!
I proceed like the following
if $T$ is closed operator so $(lambda - T)$ too, then $R(lambda - T)$ is close.
Could someone give me a counter-example or any clarification, please?
functional-analysis operator-theory operator-algebras spectral-theory
asked Dec 27 '17 at 10:21
The_lost
1,052318
It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $T$ be a closed symmetric operator defined on a Hilbert space.
I want to proof that the Rang $R(lambda - T)$ is closed if and only if $Im(lambda)neq 0$.
I used the sequences technic to proof that if $Im(lambda)neq 0$ then the equality hold, but I couldn't see why it is not true if $Im(lambda)= 0$ !!!
I proceed like the following
if $T$ is closed operator so $(lambda - T)$ too, then $R(lambda - T)$ is close.
Could someone give me a counter-example or any clarification, please?
functional-analysis operator-theory operator-algebras spectral-theory
asked Dec 27 '17 at 10:21
The_lost
1,052318
Let $T$ be a closed symmetric operator defined on a Hilbert space.
I want to proof that the Rang $R(lambda - T)$ is closed if and only if $Im(lambda)neq 0$.
I used the sequences technic to proof that if $Im(lambda)neq 0$ then the equality hold, but I couldn't see why it is not true if $Im(lambda)= 0$ !!!
I proceed like the following
if $T$ is closed operator so $(lambda - T)$ too, then $R(lambda - T)$ is close.
Could someone give me a counter-example or any clarification, please?
functional-analysis operator-theory operator-algebras spectral-theory
asked Dec 27 '17 at 10:21
The_lost
1,052318
edited Aug 14 at 10:59
asked Dec 27 '17 at 10:21
The_lost
1,052318
asked Dec 27 '17 at 10:21
The_lost
1,052318
asked Dec 27 '17 at 10:21
The_lost
1,052318
1,052318
It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33
 |Â
show 5 more comments
It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33
It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33
 |Â
show 5 more comments
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It's not necessarily true for a selfadjoint operator $T$, and such a $T$ is symmetric.
– DisintegratingByParts
Dec 27 '17 at 15:00
Thank you sir, yes we can show that that rang is closed even if $Im(lambda)=0$ for exemple when $T$ is self-adjoint and $lambda = 0 $ :)
– The_lost
Dec 27 '17 at 15:21
Let $lambda_n$ be a sequence of real numbers. Define $Tf=sum_n=1^inftylambda_n langle f,e_nrangle e_n$, where $ e_n _n=1^infty$ is an orthonormal basis of $H$. Then $(T-lambda I)$ does not have a closed range for any real $lambda$ that is a cluster point of $ lambda_n $.
– DisintegratingByParts
Dec 27 '17 at 16:54
If $T=T^*$ and $T-lambda I$ has trivial null space for some real $lambda$, then $mathcalR(T-lambda I)^perp=mathcalN(T-lambda)$ shows that $T-lambda I$ has dense range. If that range is also closed, then $T-lambda I$ will be invertible, which means that $lambdainrho(T)$.
– DisintegratingByParts
Dec 27 '17 at 16:58
Thank you, i think i am not in that advanced level as your explication!, can you please tell me where is the problem if $lambda in rho(T)$ ?
– The_lost
Dec 27 '17 at 17:33