How to show that the set of functions E={f(0)=f(1)=0} is closed in C[0, 1] with respect to the $infty$ norm?...
To do it I need to show that all limit points of E are inside C[0, 1]. How can it be showed?
One option is to show that addition of E is an open set.
The second is to build some sequence which has a limit in E, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that E contains all its limit points:
1) Using the theorem: $f_0(x)$ - a limit point of the set E $subset$ C[0, 1] $Rightarrow$ $Leftarrow$ $exists${$f_n$} $subset$ E: $f_n to f_0, n to infty, f_n neq f_0$ $forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
asked Dec 26 at 3:31
shpindler
42
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put on hold as off-topic by Saad, Eevee Trainer, Chinnapparaj R, John Douma, Paul Frost Dec 26 at 23:33
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To do it I need to show that all limit points of E are inside C[0, 1]. How can it be showed?
One option is to show that addition of E is an open set.
The second is to build some sequence which has a limit in E, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that E contains all its limit points:
1) Using the theorem: $f_0(x)$ - a limit point of the set E $subset$ C[0, 1] $Rightarrow$ $Leftarrow$ $exists${$f_n$} $subset$ E: $f_n to f_0, n to infty, f_n neq f_0$ $forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
asked Dec 26 at 3:31
shpindler
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Saad, Eevee Trainer, Chinnapparaj R, John Douma, Paul Frost Dec 26 at 23:33
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Chinnapparaj R, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
To do it I need to show that all limit points of E are inside C[0, 1]. How can it be showed?
One option is to show that addition of E is an open set.
The second is to build some sequence which has a limit in E, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that E contains all its limit points:
1) Using the theorem: $f_0(x)$ - a limit point of the set E $subset$ C[0, 1] $Rightarrow$ $Leftarrow$ $exists${$f_n$} $subset$ E: $f_n to f_0, n to infty, f_n neq f_0$ $forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
asked Dec 26 at 3:31
shpindler
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
To do it I need to show that all limit points of E are inside C[0, 1]. How can it be showed?
One option is to show that addition of E is an open set.
The second is to build some sequence which has a limit in E, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that E contains all its limit points:
1) Using the theorem: $f_0(x)$ - a limit point of the set E $subset$ C[0, 1] $Rightarrow$ $Leftarrow$ $exists${$f_n$} $subset$ E: $f_n to f_0, n to infty, f_n neq f_0$ $forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
functional-analysis
asked Dec 26 at 3:31
shpindler
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 at 3:31
shpindler
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
asked Dec 26 at 3:31
shpindler
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 at 3:31
shpindler
42
asked Dec 26 at 3:31
shpindler
42
42
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shpindler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Saad, Eevee Trainer, Chinnapparaj R, John Douma, Paul Frost Dec 26 at 23:33
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Chinnapparaj R, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Saad, Eevee Trainer, Chinnapparaj R, John Douma, Paul Frost Dec 26 at 23:33
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Chinnapparaj R, Paul Frost
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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oldest
votes
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 at 12:26
Hermione
18119
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 at 12:26
Hermione
18119
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
add a comment |
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 at 12:26
Hermione
18119
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
add a comment |
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 at 12:26
Hermione
18119
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 at 12:26
Hermione
18119
edited Dec 26 at 12:33
answered Dec 26 at 12:26
Hermione
18119
answered Dec 26 at 12:26
Hermione
18119
answered Dec 26 at 12:26
Hermione
18119
18119
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
add a comment |
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
I would add for completeness that $to$ means converges in supremum norm.
– Jonas Lenz
Dec 26 at 12:29
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
Your are right, now it is more clear. Thanks!
– Hermione
Dec 26 at 12:34
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
– shpindler
Dec 26 at 15:25
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
Yes @shpindler, you are right
– Hermione
Dec 26 at 17:05
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
@Hermione, thanks!
– shpindler
Dec 26 at 17:24
add a comment |
