Convergence in a normed vector space - Linear operator [closed]
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Having $X$ a normed vector space. If $f$ is a linear operator from $X$ to $ℝ$ and is not continuous in $0$ (element of $X$) , how can we show that there exists a sequence $x_n$ that converges to $0$ for which we have $f(x_n) = 1$ (for all $n$ element of $ℕ$).
Any help would be greatly appreciated, thank you.
normed-spaces
edited Nov 25 at 22:19
Mason
1,8411527
asked Nov 25 at 22:07
mimi
125
closed as off-topic by John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh Nov 26 at 8:22
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." – John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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down vote
favorite
Having $X$ a normed vector space. If $f$ is a linear operator from $X$ to $ℝ$ and is not continuous in $0$ (element of $X$) , how can we show that there exists a sequence $x_n$ that converges to $0$ for which we have $f(x_n) = 1$ (for all $n$ element of $ℕ$).
Any help would be greatly appreciated, thank you.
normed-spaces
edited Nov 25 at 22:19
Mason
1,8411527
asked Nov 25 at 22:07
mimi
125
closed as off-topic by John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh Nov 26 at 8:22
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." – John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Having $X$ a normed vector space. If $f$ is a linear operator from $X$ to $ℝ$ and is not continuous in $0$ (element of $X$) , how can we show that there exists a sequence $x_n$ that converges to $0$ for which we have $f(x_n) = 1$ (for all $n$ element of $ℕ$).
Any help would be greatly appreciated, thank you.
normed-spaces
edited Nov 25 at 22:19
Mason
1,8411527
asked Nov 25 at 22:07
mimi
125
Having $X$ a normed vector space. If $f$ is a linear operator from $X$ to $ℝ$ and is not continuous in $0$ (element of $X$) , how can we show that there exists a sequence $x_n$ that converges to $0$ for which we have $f(x_n) = 1$ (for all $n$ element of $ℕ$).
Any help would be greatly appreciated, thank you.
normed-spaces
normed-spaces
edited Nov 25 at 22:19
Mason
1,8411527
asked Nov 25 at 22:07
mimi
125
edited Nov 25 at 22:19
Mason
1,8411527
asked Nov 25 at 22:07
mimi
125
edited Nov 25 at 22:19
Mason
1,8411527
edited Nov 25 at 22:19
Mason
1,8411527
edited Nov 25 at 22:19
Mason
1,8411527
1,8411527
asked Nov 25 at 22:07
mimi
125
asked Nov 25 at 22:07
mimi
125
asked Nov 25 at 22:07
mimi
125
125
closed as off-topic by John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh Nov 26 at 8:22
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." – John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh Nov 26 at 8:22
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." – John B, Cesareo, Chinnapparaj R, user10354138, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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By definition of a limit, construct a sequence $x_n$ such that $x_nrightarrow 0$ but $|f(x_n)|geqdelta>0$. Here we use $f(0)=0$. Then rescale by letting
$$u_n=dfrac{x_n}{|f(x_n)|}$$
Clearly $|f(u_n)|=1$ by linearity of $f$ but $u_nrightarrow 0$ since $x_n$ does and $|f(x_n)|geqdelta$.
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By definition of a limit, construct a sequence $x_n$ such that $x_nrightarrow 0$ but $|f(x_n)|geqdelta>0$. Here we use $f(0)=0$. Then rescale by letting
$$u_n=dfrac{x_n}{|f(x_n)|}$$
Clearly $|f(u_n)|=1$ by linearity of $f$ but $u_nrightarrow 0$ since $x_n$ does and $|f(x_n)|geqdelta$.
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
add a comment |
up vote
1
down vote
By definition of a limit, construct a sequence $x_n$ such that $x_nrightarrow 0$ but $|f(x_n)|geqdelta>0$. Here we use $f(0)=0$. Then rescale by letting
$$u_n=dfrac{x_n}{|f(x_n)|}$$
Clearly $|f(u_n)|=1$ by linearity of $f$ but $u_nrightarrow 0$ since $x_n$ does and $|f(x_n)|geqdelta$.
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
add a comment |
up vote
1
down vote
up vote
1
down vote
By definition of a limit, construct a sequence $x_n$ such that $x_nrightarrow 0$ but $|f(x_n)|geqdelta>0$. Here we use $f(0)=0$. Then rescale by letting
$$u_n=dfrac{x_n}{|f(x_n)|}$$
Clearly $|f(u_n)|=1$ by linearity of $f$ but $u_nrightarrow 0$ since $x_n$ does and $|f(x_n)|geqdelta$.
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
By definition of a limit, construct a sequence $x_n$ such that $x_nrightarrow 0$ but $|f(x_n)|geqdelta>0$. Here we use $f(0)=0$. Then rescale by letting
$$u_n=dfrac{x_n}{|f(x_n)|}$$
Clearly $|f(u_n)|=1$ by linearity of $f$ but $u_nrightarrow 0$ since $x_n$ does and $|f(x_n)|geqdelta$.
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
answered Nov 25 at 22:22
Olivier Moschetta
2,7661411
2,7661411
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
add a comment |
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Where did you use f(0) = 0 ?
– mimi
Nov 28 at 18:47
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
Since $f$ is not continuous at $0$ then there exists $delta>0$ such that for every $ninmathbb{N}$ there exists $x=x_n$ with $|x_n|leq 1/n$ and $|f(x_n)-f(0)|geqdelta$. Since $f(0)=0$ this gives $|f(x_n)|geqdelta$.
– Olivier Moschetta
Nov 28 at 18:51
add a comment |
