Mixing
Converse of the Spectral Theorem
$begingroup$
It is known that the spectrum of a compact operator $T in B(X)$, where $X$ is an infinite-dimensional Banach space, is given by
$sigma(T)=sigma_p(T) cup {0}$
and $0$ is the only accumulation point of $sigma(T)$.
Answers to similar questions suggest that the converse is true if $X$ is a Hilbert space and $T$ is self-adjoint.
Does the converse also hold in this more general setting or does it fail for Banach spaces?
functional-analysis banach-spaces spectral-theory
asked Jan 16 at 6:25
Max93Max93
31529
$endgroup$
add a comment |
$begingroup$
It is known that the spectrum of a compact operator $T in B(X)$, where $X$ is an infinite-dimensional Banach space, is given by
$sigma(T)=sigma_p(T) cup {0}$
and $0$ is the only accumulation point of $sigma(T)$.
Answers to similar questions suggest that the converse is true if $X$ is a Hilbert space and $T$ is self-adjoint.
Does the converse also hold in this more general setting or does it fail for Banach spaces?
functional-analysis banach-spaces spectral-theory
asked Jan 16 at 6:25
Max93Max93
31529
$endgroup$
$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07
add a comment |
$begingroup$
It is known that the spectrum of a compact operator $T in B(X)$, where $X$ is an infinite-dimensional Banach space, is given by
$sigma(T)=sigma_p(T) cup {0}$
and $0$ is the only accumulation point of $sigma(T)$.
Answers to similar questions suggest that the converse is true if $X$ is a Hilbert space and $T$ is self-adjoint.
Does the converse also hold in this more general setting or does it fail for Banach spaces?
functional-analysis banach-spaces spectral-theory
asked Jan 16 at 6:25
Max93Max93
31529
$endgroup$
It is known that the spectrum of a compact operator $T in B(X)$, where $X$ is an infinite-dimensional Banach space, is given by
$sigma(T)=sigma_p(T) cup {0}$
and $0$ is the only accumulation point of $sigma(T)$.
Answers to similar questions suggest that the converse is true if $X$ is a Hilbert space and $T$ is self-adjoint.
Does the converse also hold in this more general setting or does it fail for Banach spaces?
functional-analysis banach-spaces spectral-theory
functional-analysis banach-spaces spectral-theory
asked Jan 16 at 6:25
Max93Max93
31529
asked Jan 16 at 6:25
Max93Max93
31529
asked Jan 16 at 6:25
Max93Max93
31529
asked Jan 16 at 6:25
Max93Max93
31529
asked Jan 16 at 6:25
Max93Max93
31529
31529
$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07
add a comment |
$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07
$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07
$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07
add a comment |
1 Answer
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votes
$begingroup$
The converse does not hold, even in Hilbert spaces! Let us take $X = ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a in ell^infty$, i.e.,
$$T_a , x = (a_1 , x_1, a_2 , x_2, ldots ).$$
Properties of this operator:
$T_a$ is always self-adjoint.- The point spectrum is ${a_1, a_2, ldots}$.
$T_a$ is compact iff $a_n to 0$.
To give a precise example, we consider
$$a_n = begin{cases} 0 & text{if $n = 1$}\1 & text{if $n>1$ is odd}\ 2/n & text{if $n$ is even}.end{cases}$$
Then, $T_a$ is not compact, but $sigma(T_a) = sigma_p(T_a) = {1/n mid n in mathbb N} cup {0}$.
I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $lambda in sigma_p setminus{0}$ are finite-dimensional.
answered Jan 16 at 7:35
gerwgerw
19.6k11334
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add a comment |
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1 Answer
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$begingroup$
The converse does not hold, even in Hilbert spaces! Let us take $X = ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a in ell^infty$, i.e.,
$$T_a , x = (a_1 , x_1, a_2 , x_2, ldots ).$$
Properties of this operator:
$T_a$ is always self-adjoint.- The point spectrum is ${a_1, a_2, ldots}$.
$T_a$ is compact iff $a_n to 0$.
To give a precise example, we consider
$$a_n = begin{cases} 0 & text{if $n = 1$}\1 & text{if $n>1$ is odd}\ 2/n & text{if $n$ is even}.end{cases}$$
Then, $T_a$ is not compact, but $sigma(T_a) = sigma_p(T_a) = {1/n mid n in mathbb N} cup {0}$.
I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $lambda in sigma_p setminus{0}$ are finite-dimensional.
answered Jan 16 at 7:35
gerwgerw
19.6k11334
$endgroup$
add a comment |
$begingroup$
The converse does not hold, even in Hilbert spaces! Let us take $X = ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a in ell^infty$, i.e.,
$$T_a , x = (a_1 , x_1, a_2 , x_2, ldots ).$$
Properties of this operator:
$T_a$ is always self-adjoint.- The point spectrum is ${a_1, a_2, ldots}$.
$T_a$ is compact iff $a_n to 0$.
To give a precise example, we consider
$$a_n = begin{cases} 0 & text{if $n = 1$}\1 & text{if $n>1$ is odd}\ 2/n & text{if $n$ is even}.end{cases}$$
Then, $T_a$ is not compact, but $sigma(T_a) = sigma_p(T_a) = {1/n mid n in mathbb N} cup {0}$.
I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $lambda in sigma_p setminus{0}$ are finite-dimensional.
answered Jan 16 at 7:35
gerwgerw
19.6k11334
$endgroup$
add a comment |
$begingroup$
The converse does not hold, even in Hilbert spaces! Let us take $X = ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a in ell^infty$, i.e.,
$$T_a , x = (a_1 , x_1, a_2 , x_2, ldots ).$$
Properties of this operator:
$T_a$ is always self-adjoint.- The point spectrum is ${a_1, a_2, ldots}$.
$T_a$ is compact iff $a_n to 0$.
To give a precise example, we consider
$$a_n = begin{cases} 0 & text{if $n = 1$}\1 & text{if $n>1$ is odd}\ 2/n & text{if $n$ is even}.end{cases}$$
Then, $T_a$ is not compact, but $sigma(T_a) = sigma_p(T_a) = {1/n mid n in mathbb N} cup {0}$.
I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $lambda in sigma_p setminus{0}$ are finite-dimensional.
answered Jan 16 at 7:35
gerwgerw
19.6k11334
$endgroup$
The converse does not hold, even in Hilbert spaces! Let us take $X = ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a in ell^infty$, i.e.,
$$T_a , x = (a_1 , x_1, a_2 , x_2, ldots ).$$
Properties of this operator:
$T_a$ is always self-adjoint.- The point spectrum is ${a_1, a_2, ldots}$.
$T_a$ is compact iff $a_n to 0$.
To give a precise example, we consider
$$a_n = begin{cases} 0 & text{if $n = 1$}\1 & text{if $n>1$ is odd}\ 2/n & text{if $n$ is even}.end{cases}$$
Then, $T_a$ is not compact, but $sigma(T_a) = sigma_p(T_a) = {1/n mid n in mathbb N} cup {0}$.
I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $lambda in sigma_p setminus{0}$ are finite-dimensional.
answered Jan 16 at 7:35
gerwgerw
19.6k11334
answered Jan 16 at 7:35
gerwgerw
19.6k11334
answered Jan 16 at 7:35
gerwgerw
19.6k11334
answered Jan 16 at 7:35
gerwgerw
19.6k11334
19.6k11334
add a comment |
add a comment |
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$begingroup$
Actually, more is true. For $lambdainsigma(T)setminus{0}$, the null space $ker (T-lambda)$ should be of finite dimensional if $T$ is compact.
$endgroup$
– Song
Jan 16 at 9:07