Mixing
Unbounded self-adjoint operator with compact resolvent: Expansion
Given an unbounded self-adjoint operator $Tcolon H supset mathrm{dom}(T) to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form an orthonormal basis of $H,$ the eigenvalues $lambda_n in mathbb R$ with $|lambda_n | to infty$ and $Te_n = lambda_n e_n$ for $n in mathbb N.$ My question: Does a representation
$$Tx = sum_{n=1}^infty lambda_n (x,e_n)_H e_n$$
for $x in mathrm{dom}(T) subset H$ hold as for compact operators?
Thanks!
EDIT: I think that the answer is quite simple:
Let $x in mathrm{dom}(T) subset H$ be fixed. It holds $Tx in H$ and hence $Tx$ can be represented by
$$Tx = sum_{n=1}^infty (Tx,e_n)_H e_n = sum_{n=1}^infty (x,Te_n)_H e_n = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$
Is this correct?
spectral-theory compact-operators unbounded-operators
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
add a comment |
Given an unbounded self-adjoint operator $Tcolon H supset mathrm{dom}(T) to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form an orthonormal basis of $H,$ the eigenvalues $lambda_n in mathbb R$ with $|lambda_n | to infty$ and $Te_n = lambda_n e_n$ for $n in mathbb N.$ My question: Does a representation
$$Tx = sum_{n=1}^infty lambda_n (x,e_n)_H e_n$$
for $x in mathrm{dom}(T) subset H$ hold as for compact operators?
Thanks!
EDIT: I think that the answer is quite simple:
Let $x in mathrm{dom}(T) subset H$ be fixed. It holds $Tx in H$ and hence $Tx$ can be represented by
$$Tx = sum_{n=1}^infty (Tx,e_n)_H e_n = sum_{n=1}^infty (x,Te_n)_H e_n = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$
Is this correct?
spectral-theory compact-operators unbounded-operators
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30
add a comment |
Given an unbounded self-adjoint operator $Tcolon H supset mathrm{dom}(T) to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form an orthonormal basis of $H,$ the eigenvalues $lambda_n in mathbb R$ with $|lambda_n | to infty$ and $Te_n = lambda_n e_n$ for $n in mathbb N.$ My question: Does a representation
$$Tx = sum_{n=1}^infty lambda_n (x,e_n)_H e_n$$
for $x in mathrm{dom}(T) subset H$ hold as for compact operators?
Thanks!
EDIT: I think that the answer is quite simple:
Let $x in mathrm{dom}(T) subset H$ be fixed. It holds $Tx in H$ and hence $Tx$ can be represented by
$$Tx = sum_{n=1}^infty (Tx,e_n)_H e_n = sum_{n=1}^infty (x,Te_n)_H e_n = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$
Is this correct?
spectral-theory compact-operators unbounded-operators
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
Given an unbounded self-adjoint operator $Tcolon H supset mathrm{dom}(T) to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form an orthonormal basis of $H,$ the eigenvalues $lambda_n in mathbb R$ with $|lambda_n | to infty$ and $Te_n = lambda_n e_n$ for $n in mathbb N.$ My question: Does a representation
$$Tx = sum_{n=1}^infty lambda_n (x,e_n)_H e_n$$
for $x in mathrm{dom}(T) subset H$ hold as for compact operators?
Thanks!
EDIT: I think that the answer is quite simple:
Let $x in mathrm{dom}(T) subset H$ be fixed. It holds $Tx in H$ and hence $Tx$ can be represented by
$$Tx = sum_{n=1}^infty (Tx,e_n)_H e_n = sum_{n=1}^infty (x,Te_n)_H e_n = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$
Is this correct?
spectral-theory compact-operators unbounded-operators
spectral-theory compact-operators unbounded-operators
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
edited Nov 22 '18 at 12:11
eierkopf
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
asked Nov 21 '18 at 22:48
eierkopfeierkopf
113
113
It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30
add a comment |
It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30
It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30
add a comment |
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It should be exactly the content of the spectral theorem in this particular case. Also, $xinmathrm{dom}(T)$ should be equivalent to the sum $sum |lambda_n (x, e_n)|^2$ being finite. I say "should" because I am not really checking it, I am just relying on my memory and intuition.
– Giuseppe Negro
Nov 22 '18 at 12:21
I think so too, because the Spectral Theorem states with the unique spectral measure $E_T$ that $$ T = int_{mathbb R} lambda mathrm d E_T(lambda).$$
– eierkopf
Nov 22 '18 at 21:09
For $x in mathrm{dom} (T)$ this gives $$ Tx = int_{mathbb R} lambda , mathrm d E_T(lambda)x = sum_{n=1}^infty lambda_n (x,e_n)_H e_n.$$ Right?
– eierkopf
Nov 22 '18 at 21:15
That's what I thought.
– Giuseppe Negro
Nov 23 '18 at 8:30