Show that C [a,b] is a Banach space
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Let $C [a,b]$ be a set of all real-valued functions $x(t),y(t),...$ which are functions of an independent real variable $t$ and are defined and continuous on a closed interval $[a,b]$. Show that $C [a,b]$ is a Banach space with norm given by
$||x|| = max_{t∈[a,b]}|x(t)|$.
banach-spaces
edited Nov 27 at 13:07
mathnoob
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asked Nov 27 at 12:59
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up vote
-2
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Let $C [a,b]$ be a set of all real-valued functions $x(t),y(t),...$ which are functions of an independent real variable $t$ and are defined and continuous on a closed interval $[a,b]$. Show that $C [a,b]$ is a Banach space with norm given by
$||x|| = max_{t∈[a,b]}|x(t)|$.
banach-spaces
edited Nov 27 at 13:07
mathnoob
1,673322
asked Nov 27 at 12:59
Thanks for answering
44
1
Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $C [a,b]$ be a set of all real-valued functions $x(t),y(t),...$ which are functions of an independent real variable $t$ and are defined and continuous on a closed interval $[a,b]$. Show that $C [a,b]$ is a Banach space with norm given by
$||x|| = max_{t∈[a,b]}|x(t)|$.
banach-spaces
edited Nov 27 at 13:07
mathnoob
1,673322
asked Nov 27 at 12:59
Thanks for answering
44
Let $C [a,b]$ be a set of all real-valued functions $x(t),y(t),...$ which are functions of an independent real variable $t$ and are defined and continuous on a closed interval $[a,b]$. Show that $C [a,b]$ is a Banach space with norm given by
$||x|| = max_{t∈[a,b]}|x(t)|$.
banach-spaces
banach-spaces
edited Nov 27 at 13:07
mathnoob
1,673322
asked Nov 27 at 12:59
Thanks for answering
44
edited Nov 27 at 13:07
mathnoob
1,673322
asked Nov 27 at 12:59
Thanks for answering
44
edited Nov 27 at 13:07
mathnoob
1,673322
edited Nov 27 at 13:07
mathnoob
1,673322
edited Nov 27 at 13:07
mathnoob
1,673322
1,673322
asked Nov 27 at 12:59
Thanks for answering
44
asked Nov 27 at 12:59
Thanks for answering
44
asked Nov 27 at 12:59
Thanks for answering
44
44
1
Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21
add a comment |
1
Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21
1
1
Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21
add a comment |
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Can you at least verify that it is a vector space, and that the norm given actually qualifies as one? This should be straightforward verification of the definitions. If you have any confusion, please mention it above.
– астон вілла олоф мэллбэрг
Nov 27 at 13:00
Another fact you have probably already seen: the uniform limit of a sequence of continuous functions is continuous.
– GEdgar
Nov 27 at 13:21