Open unit ball in integral norm is open in supremum norm
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I have to show the following
I know I have to find an $epsilon$ such that $B_infty(f,epsilon)$ is in $B_1(0,1) forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question.
I think I am missing something.
Kindly help thanks and regards
real-analysis functional-analysis normed-spaces
asked Nov 17 at 13:01
Devendra Singh Rana
744216
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up vote
1
down vote
favorite
I have to show the following
I know I have to find an $epsilon$ such that $B_infty(f,epsilon)$ is in $B_1(0,1) forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question.
I think I am missing something.
Kindly help thanks and regards
real-analysis functional-analysis normed-spaces
asked Nov 17 at 13:01
Devendra Singh Rana
744216
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to show the following
I know I have to find an $epsilon$ such that $B_infty(f,epsilon)$ is in $B_1(0,1) forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question.
I think I am missing something.
Kindly help thanks and regards
real-analysis functional-analysis normed-spaces
asked Nov 17 at 13:01
Devendra Singh Rana
744216
I have to show the following
I know I have to find an $epsilon$ such that $B_infty(f,epsilon)$ is in $B_1(0,1) forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question.
I think I am missing something.
Kindly help thanks and regards
real-analysis functional-analysis normed-spaces
real-analysis functional-analysis normed-spaces
asked Nov 17 at 13:01
Devendra Singh Rana
744216
asked Nov 17 at 13:01
Devendra Singh Rana
744216
edited Nov 17 at 13:06
asked Nov 17 at 13:01
Devendra Singh Rana
744216
asked Nov 17 at 13:01
Devendra Singh Rana
744216
asked Nov 17 at 13:01
Devendra Singh Rana
744216
744216
add a comment |
add a comment |
1 Answer
1
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1
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accepted
Let $f in B_1(0,1)$. Set $delta=min{ ||f||_1 , 1 - ||f||_1 }$.
Now conside $B_0(f,delta/2) = { gin C[0,1]: ||f-g||_infty < delta /2 }$.
Now $gin B_0(f, delta/2)$, then $||f-g||_1<||f-g||_infty<delta/2$
$||g||_1 le ||f||_1+||f-g||_1 < ||f||_1 + delta/2 < 1$
Hence, $gin B_1(0,1) Rightarrow B_0(f,delta/2) subset B_1(0,1)$
Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_infty)$.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
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
accepted
Let $f in B_1(0,1)$. Set $delta=min{ ||f||_1 , 1 - ||f||_1 }$.
Now conside $B_0(f,delta/2) = { gin C[0,1]: ||f-g||_infty < delta /2 }$.
Now $gin B_0(f, delta/2)$, then $||f-g||_1<||f-g||_infty<delta/2$
$||g||_1 le ||f||_1+||f-g||_1 < ||f||_1 + delta/2 < 1$
Hence, $gin B_1(0,1) Rightarrow B_0(f,delta/2) subset B_1(0,1)$
Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_infty)$.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
add a comment |
up vote
1
down vote
accepted
Let $f in B_1(0,1)$. Set $delta=min{ ||f||_1 , 1 - ||f||_1 }$.
Now conside $B_0(f,delta/2) = { gin C[0,1]: ||f-g||_infty < delta /2 }$.
Now $gin B_0(f, delta/2)$, then $||f-g||_1<||f-g||_infty<delta/2$
$||g||_1 le ||f||_1+||f-g||_1 < ||f||_1 + delta/2 < 1$
Hence, $gin B_1(0,1) Rightarrow B_0(f,delta/2) subset B_1(0,1)$
Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_infty)$.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f in B_1(0,1)$. Set $delta=min{ ||f||_1 , 1 - ||f||_1 }$.
Now conside $B_0(f,delta/2) = { gin C[0,1]: ||f-g||_infty < delta /2 }$.
Now $gin B_0(f, delta/2)$, then $||f-g||_1<||f-g||_infty<delta/2$
$||g||_1 le ||f||_1+||f-g||_1 < ||f||_1 + delta/2 < 1$
Hence, $gin B_1(0,1) Rightarrow B_0(f,delta/2) subset B_1(0,1)$
Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_infty)$.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $f in B_1(0,1)$. Set $delta=min{ ||f||_1 , 1 - ||f||_1 }$.
Now conside $B_0(f,delta/2) = { gin C[0,1]: ||f-g||_infty < delta /2 }$.
Now $gin B_0(f, delta/2)$, then $||f-g||_1<||f-g||_infty<delta/2$
$||g||_1 le ||f||_1+||f-g||_1 < ||f||_1 + delta/2 < 1$
Hence, $gin B_1(0,1) Rightarrow B_0(f,delta/2) subset B_1(0,1)$
Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_infty)$.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 17 at 13:40
Offlaw
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 17 at 13:40
Offlaw
2138
answered Nov 17 at 13:40
Offlaw
2138
2138
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
add a comment |
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Can you explain the Third last line of inequalities @Offlaw
– Devendra Singh Rana
Nov 17 at 13:45
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
Just triangle inequality.
– Offlaw
Nov 17 at 14:01
add a comment |
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