Under what condition on the space X, any Continuous operator will be Completely continuous.
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Categorise the spaces $X$ for which $B_{00}(X,X)=B(X,X)$, where
$B(X,X)$ is the set of bounded linear operators and $
B_{00}(X,X)$ the set of completely continuous operators, i.e. operators which take weak convergent sequences to strong convergence sequence.
functional-analysis operator-theory weak-convergence strong-convergence
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
add a comment |
up vote
1
down vote
favorite
Categorise the spaces $X$ for which $B_{00}(X,X)=B(X,X)$, where
$B(X,X)$ is the set of bounded linear operators and $
B_{00}(X,X)$ the set of completely continuous operators, i.e. operators which take weak convergent sequences to strong convergence sequence.
functional-analysis operator-theory weak-convergence strong-convergence
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Categorise the spaces $X$ for which $B_{00}(X,X)=B(X,X)$, where
$B(X,X)$ is the set of bounded linear operators and $
B_{00}(X,X)$ the set of completely continuous operators, i.e. operators which take weak convergent sequences to strong convergence sequence.
functional-analysis operator-theory weak-convergence strong-convergence
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
Categorise the spaces $X$ for which $B_{00}(X,X)=B(X,X)$, where
$B(X,X)$ is the set of bounded linear operators and $
B_{00}(X,X)$ the set of completely continuous operators, i.e. operators which take weak convergent sequences to strong convergence sequence.
functional-analysis operator-theory weak-convergence strong-convergence
functional-analysis operator-theory weak-convergence strong-convergence
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
edited Nov 14 at 13:59
Davide Giraudo
123k16149253
123k16149253
asked Nov 5 at 18:11
Ashis Pati
184
asked Nov 5 at 18:11
Ashis Pati
184
asked Nov 5 at 18:11
Ashis Pati
184
184
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
This is equivalent to $X$ having the Schur property (i.e., each weakly convergent sequence converges strongly). A prominent non-finite-dimensional example is $X=l_1$.
If $B(X)=B_{00}(X)$ then $operatorname{Id}in B_{00}(X)$ and $X$ has the Schur property.
Let $X$ have the Schur property then $B(X)=B_{00}(X)$ trivially.
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is equivalent to $X$ having the Schur property (i.e., each weakly convergent sequence converges strongly). A prominent non-finite-dimensional example is $X=l_1$.
If $B(X)=B_{00}(X)$ then $operatorname{Id}in B_{00}(X)$ and $X$ has the Schur property.
Let $X$ have the Schur property then $B(X)=B_{00}(X)$ trivially.
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
add a comment |
up vote
4
down vote
accepted
This is equivalent to $X$ having the Schur property (i.e., each weakly convergent sequence converges strongly). A prominent non-finite-dimensional example is $X=l_1$.
If $B(X)=B_{00}(X)$ then $operatorname{Id}in B_{00}(X)$ and $X$ has the Schur property.
Let $X$ have the Schur property then $B(X)=B_{00}(X)$ trivially.
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is equivalent to $X$ having the Schur property (i.e., each weakly convergent sequence converges strongly). A prominent non-finite-dimensional example is $X=l_1$.
If $B(X)=B_{00}(X)$ then $operatorname{Id}in B_{00}(X)$ and $X$ has the Schur property.
Let $X$ have the Schur property then $B(X)=B_{00}(X)$ trivially.
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
This is equivalent to $X$ having the Schur property (i.e., each weakly convergent sequence converges strongly). A prominent non-finite-dimensional example is $X=l_1$.
If $B(X)=B_{00}(X)$ then $operatorname{Id}in B_{00}(X)$ and $X$ has the Schur property.
Let $X$ have the Schur property then $B(X)=B_{00}(X)$ trivially.
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
edited Nov 14 at 14:00
Davide Giraudo
123k16149253
123k16149253
answered Nov 6 at 8:41
daw
23.8k1544
answered Nov 6 at 8:41
daw
23.8k1544
answered Nov 6 at 8:41
daw
23.8k1544
23.8k1544
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
add a comment |
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
Does there exist any other space having Schur Property?
– Ashis Pati
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
This is a good question, I have no idea.
– daw
2 days ago
add a comment |
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