$H$ Hilbert space, $T$ symmetric bounded linear, when is $H=R(T) oplus N(T)$?
$begingroup$
I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?
functional-analysis operator-theory hilbert-spaces orthogonality direct-sum
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
$endgroup$
add a comment |
$begingroup$
I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?
functional-analysis operator-theory hilbert-spaces orthogonality direct-sum
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
$endgroup$
add a comment |
$begingroup$
I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?
functional-analysis operator-theory hilbert-spaces orthogonality direct-sum
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
$endgroup$
I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{perp}$. Now I was asking myself whether $H=R(T) oplus N(T)$.
On wiki I saw that if $X$ is closed in $H$ then $H=X oplus X^{perp}$.
So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?
functional-analysis operator-theory hilbert-spaces orthogonality direct-sum
functional-analysis operator-theory hilbert-spaces orthogonality direct-sum
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
edited Jan 8 at 15:18
Davide Giraudo
125k16150261
125k16150261
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
asked Jan 8 at 14:26
roi_saumonroi_saumon
45628
45628
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.
In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
$$
|Tx|geq c|x|
$$
for all $xin N(T)^perp$.
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
$endgroup$
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.
In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
$$
|Tx|geq c|x|
$$
for all $xin N(T)^perp$.
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
$endgroup$
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
|
show 4 more comments
$begingroup$
The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.
In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
$$
|Tx|geq c|x|
$$
for all $xin N(T)^perp$.
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
$endgroup$
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
|
show 4 more comments
$begingroup$
The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.
In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
$$
|Tx|geq c|x|
$$
for all $xin N(T)^perp$.
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
$endgroup$
The kernel of a bounded linear operator is always closed (it's the preimage of the closed set ${0}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^perp$, which you wrongfully claim to hold in general.
In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that
$$
|Tx|geq c|x|
$$
for all $xin N(T)^perp$.
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
answered Jan 8 at 14:35
MaoWaoMaoWao
2,543617
2,543617
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
|
show 4 more comments
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
Just curious, do you have an example of a bounded linear map $T : H to H$ on an incomplete inner product space $H$ such that $R(T)$ is closed but $H ne R(T) oplus N(T)$?
$endgroup$
– mechanodroid
Jan 8 at 14:40
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@MaoWao Oh, right, if T is bounded and linear it is continuous so the preimage of the closed set ${0}$ is also closed. But also, I claimed $R(T)=N(T)^{perp}$ if T is symmetric linear and bounded. Isn't this true?
$endgroup$
– roi_saumon
Jan 8 at 14:48
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@roi_saumon As I said, the equality $R(T)=N(T)^perp$ holds for a symmetric bounded linear operator if and only if $R(T)$ is closed. Remember that the orthogonal complement of a set is always closed.
$endgroup$
– MaoWao
Jan 8 at 14:50
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
$begingroup$
@mechanodroid Not off the top of my head, no. I will think about it.
$endgroup$
– MaoWao
Jan 8 at 14:51
1
1
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
$begingroup$
@mechanodroid: the function $z(t)=t^{3/4}$ is in $overline {R(T)}$ btu not in $R(T)$: as $Tf_n to z$ for $f_n(t) = max(n,t^{-1/4})$.
$endgroup$
– daw
Jan 9 at 14:36
|
show 4 more comments
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