Example of not continuous linear form? [closed]

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Q1) Let $E$ a Banach space of infinite dimension. Could someone gives me an example of linear form that is not continuous ? I can't find any.

Q2) Let $E$ and $F$ Banach spaces. If it exist (and if it doesn't exist, could you tell me why) give an example of a function $fin mathcal L(E,F)$ that is not continuous if

1) $dim(E)=infty $ and $dim(F)<infty $.

2) $dim(E)<infty $ and $dim(F)=infty $.

3) $dim(E)=infty $ and $dim(F)=infty $.

I know that if $dim(E)<infty $ and $dim(F)<infty $, then such function doesn't exist.

functional-analysis banach-spaces

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asked Jul 31 at 17:51

user330587

818310

closed as off-topic by José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele Aug 1 at 11:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Did you even bother to do a search here for “discontinuous linear form”?
  – José Carlos Santos
  Jul 31 at 17:58
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Q1) Let $E$ a Banach space of infinite dimension. Could someone gives me an example of linear form that is not continuous ? I can't find any.

Q2) Let $E$ and $F$ Banach spaces. If it exist (and if it doesn't exist, could you tell me why) give an example of a function $fin mathcal L(E,F)$ that is not continuous if

1) $dim(E)=infty $ and $dim(F)<infty $.

2) $dim(E)<infty $ and $dim(F)=infty $.

3) $dim(E)=infty $ and $dim(F)=infty $.

I know that if $dim(E)<infty $ and $dim(F)<infty $, then such function doesn't exist.

functional-analysis banach-spaces

share|cite|improve this question

asked Jul 31 at 17:51

user330587

818310

closed as off-topic by José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele Aug 1 at 11:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Did you even bother to do a search here for “discontinuous linear form”?
  – José Carlos Santos
  Jul 31 at 17:58
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Q1) Let $E$ a Banach space of infinite dimension. Could someone gives me an example of linear form that is not continuous ? I can't find any.

Q2) Let $E$ and $F$ Banach spaces. If it exist (and if it doesn't exist, could you tell me why) give an example of a function $fin mathcal L(E,F)$ that is not continuous if

1) $dim(E)=infty $ and $dim(F)<infty $.

2) $dim(E)<infty $ and $dim(F)=infty $.

3) $dim(E)=infty $ and $dim(F)=infty $.

I know that if $dim(E)<infty $ and $dim(F)<infty $, then such function doesn't exist.

functional-analysis banach-spaces

share|cite|improve this question

asked Jul 31 at 17:51

user330587

818310

Q1) Let $E$ a Banach space of infinite dimension. Could someone gives me an example of linear form that is not continuous ? I can't find any.

Q2) Let $E$ and $F$ Banach spaces. If it exist (and if it doesn't exist, could you tell me why) give an example of a function $fin mathcal L(E,F)$ that is not continuous if

1) $dim(E)=infty $ and $dim(F)<infty $.

2) $dim(E)<infty $ and $dim(F)=infty $.

3) $dim(E)=infty $ and $dim(F)=infty $.

I know that if $dim(E)<infty $ and $dim(F)<infty $, then such function doesn't exist.

functional-analysis banach-spaces

share|cite|improve this question

asked Jul 31 at 17:51

user330587

818310

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 17:51

user330587

818310

asked Jul 31 at 17:51

user330587

818310

asked Jul 31 at 17:51

user330587

818310

818310

closed as off-topic by José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele Aug 1 at 11:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele Aug 1 at 11:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Tyrone, Mathmo123, Taroccoesbrocco, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Did you even bother to do a search here for “discontinuous linear form”?
  – José Carlos Santos
  Jul 31 at 17:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Did you even bother to do a search here for “discontinuous linear form”?
  – José Carlos Santos
  Jul 31 at 17:58
  

  
  
  
  

1

1

Did you even bother to do a search here for “discontinuous linear form”?
– José Carlos Santos
Jul 31 at 17:58

Did you even bother to do a search here for “discontinuous linear form”?
– José Carlos Santos
Jul 31 at 17:58

add a comment | 

1 Answer
1

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up vote
2
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accepted

Q1) Take a basis $e_i_iin I$. By replacing $e_i$ by $e_i/|e_i|$ we can assume that $|e_i|=1$. Define $f(e_i)$ in any way you want and $f$ can be extended to a linear function. In particular, of you define $f(e_i)$ to take an unbounded set of values, then $f$ will be unbounded.

Q2) 1) If $F$ is the zero vector space, then all linear functions are zero, and therefore continuous. Otherwise, you can pick a vector $0neq vin F$ and define $g:Eto F$ sending $F(x)=f(x)v$, where $f$ is the linear function from (Q1).

Q2) 3) The same argument as the previous one.

Q2) 2) If $dim(E)<infty$ then for all linear $f:Eto F$ we have that $f(E)$ is finite dimensional, and therefore closed in $F$. It follows that $f$ is continuous if and only if $hatf:Eto f(F)$ defined by $hatf(x)=f(x)$ is continuous. But $f$ is a linear map between finite dimensional Banach spaces. It is therefore continuous.

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answered Jul 31 at 18:31

JessicaMcRae

1264

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Q1) Take a basis $e_i_iin I$. By replacing $e_i$ by $e_i/|e_i|$ we can assume that $|e_i|=1$. Define $f(e_i)$ in any way you want and $f$ can be extended to a linear function. In particular, of you define $f(e_i)$ to take an unbounded set of values, then $f$ will be unbounded.

Q2) 1) If $F$ is the zero vector space, then all linear functions are zero, and therefore continuous. Otherwise, you can pick a vector $0neq vin F$ and define $g:Eto F$ sending $F(x)=f(x)v$, where $f$ is the linear function from (Q1).

Q2) 3) The same argument as the previous one.

Q2) 2) If $dim(E)<infty$ then for all linear $f:Eto F$ we have that $f(E)$ is finite dimensional, and therefore closed in $F$. It follows that $f$ is continuous if and only if $hatf:Eto f(F)$ defined by $hatf(x)=f(x)$ is continuous. But $f$ is a linear map between finite dimensional Banach spaces. It is therefore continuous.

share|cite|improve this answer

answered Jul 31 at 18:31

JessicaMcRae

1264

add a comment | 

up vote
2
down vote



accepted

Q1) Take a basis $e_i_iin I$. By replacing $e_i$ by $e_i/|e_i|$ we can assume that $|e_i|=1$. Define $f(e_i)$ in any way you want and $f$ can be extended to a linear function. In particular, of you define $f(e_i)$ to take an unbounded set of values, then $f$ will be unbounded.

Q2) 1) If $F$ is the zero vector space, then all linear functions are zero, and therefore continuous. Otherwise, you can pick a vector $0neq vin F$ and define $g:Eto F$ sending $F(x)=f(x)v$, where $f$ is the linear function from (Q1).

Q2) 3) The same argument as the previous one.

Q2) 2) If $dim(E)<infty$ then for all linear $f:Eto F$ we have that $f(E)$ is finite dimensional, and therefore closed in $F$. It follows that $f$ is continuous if and only if $hatf:Eto f(F)$ defined by $hatf(x)=f(x)$ is continuous. But $f$ is a linear map between finite dimensional Banach spaces. It is therefore continuous.

share|cite|improve this answer

answered Jul 31 at 18:31

JessicaMcRae

1264

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Q1) Take a basis $e_i_iin I$. By replacing $e_i$ by $e_i/|e_i|$ we can assume that $|e_i|=1$. Define $f(e_i)$ in any way you want and $f$ can be extended to a linear function. In particular, of you define $f(e_i)$ to take an unbounded set of values, then $f$ will be unbounded.

Q2) 1) If $F$ is the zero vector space, then all linear functions are zero, and therefore continuous. Otherwise, you can pick a vector $0neq vin F$ and define $g:Eto F$ sending $F(x)=f(x)v$, where $f$ is the linear function from (Q1).

Q2) 3) The same argument as the previous one.

Q2) 2) If $dim(E)<infty$ then for all linear $f:Eto F$ we have that $f(E)$ is finite dimensional, and therefore closed in $F$. It follows that $f$ is continuous if and only if $hatf:Eto f(F)$ defined by $hatf(x)=f(x)$ is continuous. But $f$ is a linear map between finite dimensional Banach spaces. It is therefore continuous.

share|cite|improve this answer

answered Jul 31 at 18:31

JessicaMcRae

1264

Q1) Take a basis $e_i_iin I$. By replacing $e_i$ by $e_i/|e_i|$ we can assume that $|e_i|=1$. Define $f(e_i)$ in any way you want and $f$ can be extended to a linear function. In particular, of you define $f(e_i)$ to take an unbounded set of values, then $f$ will be unbounded.

Q2) 1) If $F$ is the zero vector space, then all linear functions are zero, and therefore continuous. Otherwise, you can pick a vector $0neq vin F$ and define $g:Eto F$ sending $F(x)=f(x)v$, where $f$ is the linear function from (Q1).

Q2) 3) The same argument as the previous one.

Q2) 2) If $dim(E)<infty$ then for all linear $f:Eto F$ we have that $f(E)$ is finite dimensional, and therefore closed in $F$. It follows that $f$ is continuous if and only if $hatf:Eto f(F)$ defined by $hatf(x)=f(x)$ is continuous. But $f$ is a linear map between finite dimensional Banach spaces. It is therefore continuous.

share|cite|improve this answer

answered Jul 31 at 18:31

JessicaMcRae

1264

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 18:31

JessicaMcRae

1264

answered Jul 31 at 18:31

JessicaMcRae

1264

answered Jul 31 at 18:31

JessicaMcRae

1264

1264

add a comment | 

add a comment |