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Compact operator on invariant subspace (not necessarily closed) is compact
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I am looking at a problem in Conway's functional analysis text. It is problem II.5.7, which states: "If $T$ is compact and $mathscrM$ is an invariant subspace for $T$, show that $T|_mathscrM$ is compact." I take this to mean that for every bounded sequence $x_m_minmathbbN$ in $mathscrM$, we need to show that there is a subsequence $x_m_k_kinmathbbN$ such that $T|_mathscrMx_m_k = Tx_m_k$ converges to some element $y$, and $y$ must be in $mathscrM$. If $mathscrM$ is closed, this is pretty much trivial. However, I am confused as to how to handle it when $mathscrM$ is not closed.
Now there are seemingly various answers to this problem on StackExchange, but they all seem to ignore this fact that $y$ must be in $mathscrM$. My question is: am I misunderstanding the problem, or is it stated incorrectly? Thanks.
functional-analysis compact-operators invariant-subspace
asked Jul 15 at 23:10
mathishard.butweloveit
1069
add a comment |Â
up vote
2
down vote
favorite
I am looking at a problem in Conway's functional analysis text. It is problem II.5.7, which states: "If $T$ is compact and $mathscrM$ is an invariant subspace for $T$, show that $T|_mathscrM$ is compact." I take this to mean that for every bounded sequence $x_m_minmathbbN$ in $mathscrM$, we need to show that there is a subsequence $x_m_k_kinmathbbN$ such that $T|_mathscrMx_m_k = Tx_m_k$ converges to some element $y$, and $y$ must be in $mathscrM$. If $mathscrM$ is closed, this is pretty much trivial. However, I am confused as to how to handle it when $mathscrM$ is not closed.
Now there are seemingly various answers to this problem on StackExchange, but they all seem to ignore this fact that $y$ must be in $mathscrM$. My question is: am I misunderstanding the problem, or is it stated incorrectly? Thanks.
functional-analysis compact-operators invariant-subspace
asked Jul 15 at 23:10
mathishard.butweloveit
1069
1
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
1
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am looking at a problem in Conway's functional analysis text. It is problem II.5.7, which states: "If $T$ is compact and $mathscrM$ is an invariant subspace for $T$, show that $T|_mathscrM$ is compact." I take this to mean that for every bounded sequence $x_m_minmathbbN$ in $mathscrM$, we need to show that there is a subsequence $x_m_k_kinmathbbN$ such that $T|_mathscrMx_m_k = Tx_m_k$ converges to some element $y$, and $y$ must be in $mathscrM$. If $mathscrM$ is closed, this is pretty much trivial. However, I am confused as to how to handle it when $mathscrM$ is not closed.
Now there are seemingly various answers to this problem on StackExchange, but they all seem to ignore this fact that $y$ must be in $mathscrM$. My question is: am I misunderstanding the problem, or is it stated incorrectly? Thanks.
functional-analysis compact-operators invariant-subspace
asked Jul 15 at 23:10
mathishard.butweloveit
1069
I am looking at a problem in Conway's functional analysis text. It is problem II.5.7, which states: "If $T$ is compact and $mathscrM$ is an invariant subspace for $T$, show that $T|_mathscrM$ is compact." I take this to mean that for every bounded sequence $x_m_minmathbbN$ in $mathscrM$, we need to show that there is a subsequence $x_m_k_kinmathbbN$ such that $T|_mathscrMx_m_k = Tx_m_k$ converges to some element $y$, and $y$ must be in $mathscrM$. If $mathscrM$ is closed, this is pretty much trivial. However, I am confused as to how to handle it when $mathscrM$ is not closed.
Now there are seemingly various answers to this problem on StackExchange, but they all seem to ignore this fact that $y$ must be in $mathscrM$. My question is: am I misunderstanding the problem, or is it stated incorrectly? Thanks.
functional-analysis compact-operators invariant-subspace
asked Jul 15 at 23:10
mathishard.butweloveit
1069
edited Jul 15 at 23:24
asked Jul 15 at 23:10
mathishard.butweloveit
1069
asked Jul 15 at 23:10
mathishard.butweloveit
1069
asked Jul 15 at 23:10
mathishard.butweloveit
1069
1069
1
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
1
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31
add a comment |Â
1
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
1
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31
1
1
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
1
1
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : ell^2 to ell^2$ given by $T(x_n)_n = left(fracx_nnright)_n$. Then $T$ is compact and $c_00$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = left(frac12, frac14, ldots, frac12^n, 0, 0, ldotsright) in c_00$. Then $(y_n)_n$ converges to the vector $left(frac12^nright)_n in ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $left(frac1n2^nright)_n in ell^2 setminus c_00$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_00$.
answered Jul 15 at 23:38
mechanodroid
22.3k52041
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
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
accepted
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : ell^2 to ell^2$ given by $T(x_n)_n = left(fracx_nnright)_n$. Then $T$ is compact and $c_00$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = left(frac12, frac14, ldots, frac12^n, 0, 0, ldotsright) in c_00$. Then $(y_n)_n$ converges to the vector $left(frac12^nright)_n in ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $left(frac1n2^nright)_n in ell^2 setminus c_00$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_00$.
answered Jul 15 at 23:38
mechanodroid
22.3k52041
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
add a comment |Â
up vote
1
down vote
accepted
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : ell^2 to ell^2$ given by $T(x_n)_n = left(fracx_nnright)_n$. Then $T$ is compact and $c_00$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = left(frac12, frac14, ldots, frac12^n, 0, 0, ldotsright) in c_00$. Then $(y_n)_n$ converges to the vector $left(frac12^nright)_n in ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $left(frac1n2^nright)_n in ell^2 setminus c_00$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_00$.
answered Jul 15 at 23:38
mechanodroid
22.3k52041
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : ell^2 to ell^2$ given by $T(x_n)_n = left(fracx_nnright)_n$. Then $T$ is compact and $c_00$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = left(frac12, frac14, ldots, frac12^n, 0, 0, ldotsright) in c_00$. Then $(y_n)_n$ converges to the vector $left(frac12^nright)_n in ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $left(frac1n2^nright)_n in ell^2 setminus c_00$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_00$.
answered Jul 15 at 23:38
mechanodroid
22.3k52041
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : ell^2 to ell^2$ given by $T(x_n)_n = left(fracx_nnright)_n$. Then $T$ is compact and $c_00$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = left(frac12, frac14, ldots, frac12^n, 0, 0, ldotsright) in c_00$. Then $(y_n)_n$ converges to the vector $left(frac12^nright)_n in ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $left(frac1n2^nright)_n in ell^2 setminus c_00$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_00$.
answered Jul 15 at 23:38
mechanodroid
22.3k52041
edited Jul 16 at 8:46
answered Jul 15 at 23:38
mechanodroid
22.3k52041
answered Jul 15 at 23:38
mechanodroid
22.3k52041
answered Jul 15 at 23:38
mechanodroid
22.3k52041
22.3k52041
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
add a comment |Â
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
I am little confused by the notation $(x_n)_n$ could you elaborate? I am thinking these $n$'s are independent of each other?
– mathishard.butweloveit
Jul 15 at 23:49
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
@user707959 I changed the notation to $(x_n)_n$ and $(y_n)_n$. In $T(x_n)_n = left(fracx_nnright)_n$, $(x_n)_n$ represents an element of $ell^2$, i.e. it is a sequence of scalars. On the other hand, $(y_n)_n$ is a sequence in $ell^2$, i.e. every $y_n$ is an element of $ell^2$, given as $y_n(k) = begincases frac12^n, &k le n \ 0, & k > n endcases$ where $y_n = (y_n(k))_k$ so $y(k)$ is the $k$-th element of the sequence $y_n$.
– mechanodroid
Jul 16 at 8:46
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
Ok, this all makes sense, thank you! This was the intuition I had in a sequence sense, so thank you for making it concrete.
– mathishard.butweloveit
Jul 16 at 14:40
add a comment |Â
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1
Your counterexample is really anything but. An invariant subspace of a bounded operator $T$ is a closed linear subspace $mathscr M$ such that $Tmathscr Msubsetmathscr M$, and $(0,1)$ is not a linear subspace of $mathbb R$.
– Aweygan
Jul 15 at 23:18
Yes, you are correct.
– mathishard.butweloveit
Jul 15 at 23:20
1
I reiterate: An invariant subspace is a closed linear subspace such that...
– Aweygan
Jul 15 at 23:28
Ah so it was most certainly a misunderstanding of the definition, thank you!
– mathishard.butweloveit
Jul 15 at 23:31
You're welcome, glad to help!
– Aweygan
Jul 15 at 23:31