The image of the weak topology with the canonical injection $J$
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Let $E$ be a Banach space. With the weak topology $sigma(E,E^*)$.
And let $J:Erightarrow E^{**} $ be the canonical injection.
Can we prove that the image of the weak topology with $J$ is exactly the trace of the topology $sigma(E^{**},E^*)$ on $J(E)$.
in other words $$J(sigma(E,E^*))=J(E)cap sigma(E^{**},E^*)$$
general-topology functional-analysis weak-topology
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
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add a comment |
$begingroup$
Let $E$ be a Banach space. With the weak topology $sigma(E,E^*)$.
And let $J:Erightarrow E^{**} $ be the canonical injection.
Can we prove that the image of the weak topology with $J$ is exactly the trace of the topology $sigma(E^{**},E^*)$ on $J(E)$.
in other words $$J(sigma(E,E^*))=J(E)cap sigma(E^{**},E^*)$$
general-topology functional-analysis weak-topology
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
$endgroup$
add a comment |
$begingroup$
Let $E$ be a Banach space. With the weak topology $sigma(E,E^*)$.
And let $J:Erightarrow E^{**} $ be the canonical injection.
Can we prove that the image of the weak topology with $J$ is exactly the trace of the topology $sigma(E^{**},E^*)$ on $J(E)$.
in other words $$J(sigma(E,E^*))=J(E)cap sigma(E^{**},E^*)$$
general-topology functional-analysis weak-topology
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
$endgroup$
Let $E$ be a Banach space. With the weak topology $sigma(E,E^*)$.
And let $J:Erightarrow E^{**} $ be the canonical injection.
Can we prove that the image of the weak topology with $J$ is exactly the trace of the topology $sigma(E^{**},E^*)$ on $J(E)$.
in other words $$J(sigma(E,E^*))=J(E)cap sigma(E^{**},E^*)$$
general-topology functional-analysis weak-topology
general-topology functional-analysis weak-topology
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
asked Jan 13 at 14:56
Anas BOUALIIAnas BOUALII
1417
1417
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, it is true that
$$
J:left(E,sigma(E,E^*)right)ni xmapsto J(x)in (J(E), sigma(E^{**},E^*)big|_{J(E)})
$$ is a homeomorphism where $J(x)in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $fin E^*$.
To see that $J$ is continuous, suppose a net ${x_i}_{iin I}$ in $E$ converges to $x$ weakly. Then, for all $fin E^*$, we have $limlimits_{iin I}J(x_i)(f) = limlimits_{iin I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $sigma(E^{**},E^*)big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.
Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $fin E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:Eto J(E)$ is surjective.
It remains to show that $J^{-1}:J(E)to E$ is continuous. Suppose ${J(x_i)}_{iin I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $fin E^*$, we have
$$
limlimits_{iin I}f(x_i) = limlimits_{iin I}J(x_i)(f) = J(x)(f)=f(x).
$$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:Eto J(E)$ is a homeomorphism.
answered Jan 14 at 20:09
SongSong
11.1k628
$endgroup$
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, it is true that
$$
J:left(E,sigma(E,E^*)right)ni xmapsto J(x)in (J(E), sigma(E^{**},E^*)big|_{J(E)})
$$ is a homeomorphism where $J(x)in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $fin E^*$.
To see that $J$ is continuous, suppose a net ${x_i}_{iin I}$ in $E$ converges to $x$ weakly. Then, for all $fin E^*$, we have $limlimits_{iin I}J(x_i)(f) = limlimits_{iin I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $sigma(E^{**},E^*)big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.
Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $fin E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:Eto J(E)$ is surjective.
It remains to show that $J^{-1}:J(E)to E$ is continuous. Suppose ${J(x_i)}_{iin I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $fin E^*$, we have
$$
limlimits_{iin I}f(x_i) = limlimits_{iin I}J(x_i)(f) = J(x)(f)=f(x).
$$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:Eto J(E)$ is a homeomorphism.
answered Jan 14 at 20:09
SongSong
11.1k628
$endgroup$
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
add a comment |
$begingroup$
Yes, it is true that
$$
J:left(E,sigma(E,E^*)right)ni xmapsto J(x)in (J(E), sigma(E^{**},E^*)big|_{J(E)})
$$ is a homeomorphism where $J(x)in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $fin E^*$.
To see that $J$ is continuous, suppose a net ${x_i}_{iin I}$ in $E$ converges to $x$ weakly. Then, for all $fin E^*$, we have $limlimits_{iin I}J(x_i)(f) = limlimits_{iin I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $sigma(E^{**},E^*)big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.
Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $fin E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:Eto J(E)$ is surjective.
It remains to show that $J^{-1}:J(E)to E$ is continuous. Suppose ${J(x_i)}_{iin I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $fin E^*$, we have
$$
limlimits_{iin I}f(x_i) = limlimits_{iin I}J(x_i)(f) = J(x)(f)=f(x).
$$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:Eto J(E)$ is a homeomorphism.
answered Jan 14 at 20:09
SongSong
11.1k628
$endgroup$
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
add a comment |
$begingroup$
Yes, it is true that
$$
J:left(E,sigma(E,E^*)right)ni xmapsto J(x)in (J(E), sigma(E^{**},E^*)big|_{J(E)})
$$ is a homeomorphism where $J(x)in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $fin E^*$.
To see that $J$ is continuous, suppose a net ${x_i}_{iin I}$ in $E$ converges to $x$ weakly. Then, for all $fin E^*$, we have $limlimits_{iin I}J(x_i)(f) = limlimits_{iin I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $sigma(E^{**},E^*)big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.
Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $fin E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:Eto J(E)$ is surjective.
It remains to show that $J^{-1}:J(E)to E$ is continuous. Suppose ${J(x_i)}_{iin I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $fin E^*$, we have
$$
limlimits_{iin I}f(x_i) = limlimits_{iin I}J(x_i)(f) = J(x)(f)=f(x).
$$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:Eto J(E)$ is a homeomorphism.
answered Jan 14 at 20:09
SongSong
11.1k628
$endgroup$
Yes, it is true that
$$
J:left(E,sigma(E,E^*)right)ni xmapsto J(x)in (J(E), sigma(E^{**},E^*)big|_{J(E)})
$$ is a homeomorphism where $J(x)in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $fin E^*$.
To see that $J$ is continuous, suppose a net ${x_i}_{iin I}$ in $E$ converges to $x$ weakly. Then, for all $fin E^*$, we have $limlimits_{iin I}J(x_i)(f) = limlimits_{iin I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $sigma(E^{**},E^*)big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.
Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $fin E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:Eto J(E)$ is surjective.
It remains to show that $J^{-1}:J(E)to E$ is continuous. Suppose ${J(x_i)}_{iin I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $fin E^*$, we have
$$
limlimits_{iin I}f(x_i) = limlimits_{iin I}J(x_i)(f) = J(x)(f)=f(x).
$$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:Eto J(E)$ is a homeomorphism.
answered Jan 14 at 20:09
SongSong
11.1k628
answered Jan 14 at 20:09
SongSong
11.1k628
answered Jan 14 at 20:09
SongSong
11.1k628
answered Jan 14 at 20:09
SongSong
11.1k628
11.1k628
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
add a comment |
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
$begingroup$
thank you that s totally helpful!
$endgroup$
– Anas BOUALII
Jan 15 at 11:45
add a comment |
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