weak topology has less open set than strong topology (in Banach spaces). Why?

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Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $mathcal T_W$ the weak topology on $E$.

1) I call "Dual topological" the element of $E^*=mathcal L(E,mathbb R)$ that are continuous (wrt the strong topology that I denote $mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'subsetneq E^*$, i.e. there are linear functional that are not continuous.

2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written that all open of the weak topology are also open in the strong topology.

3) Then it's written that the weak topology is strictly thicker than the strong topology in the sense that there are less open.

Question : For me, 1) is not coherent with 2) and 3). If $E'subset E^*$ and that $mathcal T_W$ makes all element of $E^*$ continuous, there are more continuous function with refer to the weak topology, and thus, it should has more open set no ? If we consider $E$ with a topology $T$ and $mathcal F(E,mathbb R)$ the set of function $f:Eto mathbb R$, a priori, thiner is $T$ and more element of $mathcal F(E,mathbb R)$ are going to be continuous, and thus, more open there are no ? So with this argument, weak topology should has more open than strong topology. Could someone give me more explications ?

general-topology functional-analysis weak-topology

share|cite|improve this question

edited Jul 31 at 16:22

asked Jul 31 at 14:18

user330587

818310

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
  – Lee Mosher
  Jul 31 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $mathcal T_W$ the weak topology on $E$.

1) I call "Dual topological" the element of $E^*=mathcal L(E,mathbb R)$ that are continuous (wrt the strong topology that I denote $mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'subsetneq E^*$, i.e. there are linear functional that are not continuous.

2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written that all open of the weak topology are also open in the strong topology.

3) Then it's written that the weak topology is strictly thicker than the strong topology in the sense that there are less open.

Question : For me, 1) is not coherent with 2) and 3). If $E'subset E^*$ and that $mathcal T_W$ makes all element of $E^*$ continuous, there are more continuous function with refer to the weak topology, and thus, it should has more open set no ? If we consider $E$ with a topology $T$ and $mathcal F(E,mathbb R)$ the set of function $f:Eto mathbb R$, a priori, thiner is $T$ and more element of $mathcal F(E,mathbb R)$ are going to be continuous, and thus, more open there are no ? So with this argument, weak topology should has more open than strong topology. Could someone give me more explications ?

general-topology functional-analysis weak-topology

share|cite|improve this question

edited Jul 31 at 16:22

asked Jul 31 at 14:18

user330587

818310

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
  – Lee Mosher
  Jul 31 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $mathcal T_W$ the weak topology on $E$.

1) I call "Dual topological" the element of $E^*=mathcal L(E,mathbb R)$ that are continuous (wrt the strong topology that I denote $mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'subsetneq E^*$, i.e. there are linear functional that are not continuous.

2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written that all open of the weak topology are also open in the strong topology.

3) Then it's written that the weak topology is strictly thicker than the strong topology in the sense that there are less open.

Question : For me, 1) is not coherent with 2) and 3). If $E'subset E^*$ and that $mathcal T_W$ makes all element of $E^*$ continuous, there are more continuous function with refer to the weak topology, and thus, it should has more open set no ? If we consider $E$ with a topology $T$ and $mathcal F(E,mathbb R)$ the set of function $f:Eto mathbb R$, a priori, thiner is $T$ and more element of $mathcal F(E,mathbb R)$ are going to be continuous, and thus, more open there are no ? So with this argument, weak topology should has more open than strong topology. Could someone give me more explications ?

general-topology functional-analysis weak-topology

share|cite|improve this question

edited Jul 31 at 16:22

asked Jul 31 at 14:18

user330587

818310

Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $mathcal T_W$ the weak topology on $E$.

1) I call "Dual topological" the element of $E^*=mathcal L(E,mathbb R)$ that are continuous (wrt the strong topology that I denote $mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'subsetneq E^*$, i.e. there are linear functional that are not continuous.

2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written that all open of the weak topology are also open in the strong topology.

3) Then it's written that the weak topology is strictly thicker than the strong topology in the sense that there are less open.

Question : For me, 1) is not coherent with 2) and 3). If $E'subset E^*$ and that $mathcal T_W$ makes all element of $E^*$ continuous, there are more continuous function with refer to the weak topology, and thus, it should has more open set no ? If we consider $E$ with a topology $T$ and $mathcal F(E,mathbb R)$ the set of function $f:Eto mathbb R$, a priori, thiner is $T$ and more element of $mathcal F(E,mathbb R)$ are going to be continuous, and thus, more open there are no ? So with this argument, weak topology should has more open than strong topology. Could someone give me more explications ?

general-topology functional-analysis weak-topology

share|cite|improve this question

edited Jul 31 at 16:22

asked Jul 31 at 14:18

user330587

818310

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 16:22

edited Jul 31 at 16:22

edited Jul 31 at 16:22

asked Jul 31 at 14:18

user330587

818310

asked Jul 31 at 14:18

user330587

818310

asked Jul 31 at 14:18

user330587

818310

818310

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
  – Lee Mosher
  Jul 31 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
  – Lee Mosher
  Jul 31 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

1

1

The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
– Lee Mosher
Jul 31 at 14:25

The phrase "all open of the weak topology are open in the strong topology are also open in the weak topology" makes little sense, and looks like it has been miscopied.
– Lee Mosher
Jul 31 at 14:25

@LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
– user330587
Jul 31 at 16:19

@LeeMosher : Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
– user330587
Jul 31 at 16:19

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.




• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh ! I see. Thanks a lot for your answer, that helps me a lot.
  – user330587
  Jul 31 at 17:46
  

  
  
  
  

add a comment | 

up vote
1
down vote

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.




• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh ! I see. Thanks a lot for your answer, that helps me a lot.
  – user330587
  Jul 31 at 17:46
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.




• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh ! I see. Thanks a lot for your answer, that helps me a lot.
  – user330587
  Jul 31 at 17:46
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.




• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.




• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 17:26

Surb

36.2k84274

answered Jul 31 at 17:26

Surb

36.2k84274

answered Jul 31 at 17:26

Surb

36.2k84274

36.2k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh ! I see. Thanks a lot for your answer, that helps me a lot.
  – user330587
  Jul 31 at 17:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh ! I see. Thanks a lot for your answer, that helps me a lot.
  – user330587
  Jul 31 at 17:46
  

  
  
  
  

Oh ! I see. Thanks a lot for your answer, that helps me a lot.
– user330587
Jul 31 at 17:46

Oh ! I see. Thanks a lot for your answer, that helps me a lot.
– user330587
Jul 31 at 17:46

add a comment | 

up vote
1
down vote

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 15:11

Pedro

9,70322962

answered Jul 31 at 15:11

Pedro

9,70322962

answered Jul 31 at 15:11

Pedro

9,70322962

9,70322962

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
  – user330587
  Jul 31 at 16:19
  

  
  
  
  

Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
– user330587
Jul 31 at 16:19

Yes exactly, thanks for the reference. I made indeed a mistakes that I corrected it. Ok, so indeed, this sentence says that there are more open in the strong topology than in the weak one. But What about the 3rd argument ? i.e. if there are more continuous function in the weak topology, it should has more open, no ?
– user330587
Jul 31 at 16:19

add a comment | 

 

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