If $|f(x) | leq |x|$, then the sup norm $| f | = 1$
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Recall $| f | = sup {|f| : x in M, | x | leq 1 }$, then if $|f(x) | leq |x|$, then the sup norm $| f | = 1$?
This was used in a proof in Rudin's Real and Complex analysis in his Hahn Banach theorem proof.
Doesn't this only show $| f | leq 1$? How do I get the other inequality?
real-analysis
edited Nov 17 at 11:53
Paul Frost
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
|
show 4 more comments
up vote
0
down vote
favorite
Recall $| f | = sup {|f| : x in M, | x | leq 1 }$, then if $|f(x) | leq |x|$, then the sup norm $| f | = 1$?
This was used in a proof in Rudin's Real and Complex analysis in his Hahn Banach theorem proof.
Doesn't this only show $| f | leq 1$? How do I get the other inequality?
real-analysis
edited Nov 17 at 11:53
Paul Frost
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
Must be a typo.
– Randall
Nov 17 at 1:59
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
I can, hold on.
– Hawk
Nov 17 at 2:09
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Recall $| f | = sup {|f| : x in M, | x | leq 1 }$, then if $|f(x) | leq |x|$, then the sup norm $| f | = 1$?
This was used in a proof in Rudin's Real and Complex analysis in his Hahn Banach theorem proof.
Doesn't this only show $| f | leq 1$? How do I get the other inequality?
real-analysis
edited Nov 17 at 11:53
Paul Frost
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
Recall $| f | = sup {|f| : x in M, | x | leq 1 }$, then if $|f(x) | leq |x|$, then the sup norm $| f | = 1$?
This was used in a proof in Rudin's Real and Complex analysis in his Hahn Banach theorem proof.
Doesn't this only show $| f | leq 1$? How do I get the other inequality?
real-analysis
real-analysis
edited Nov 17 at 11:53
Paul Frost
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
edited Nov 17 at 11:53
Paul Frost
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
edited Nov 17 at 11:53
Paul Frost
7,9541527
edited Nov 17 at 11:53
Paul Frost
7,9541527
edited Nov 17 at 11:53
Paul Frost
7,9541527
7,9541527
asked Nov 17 at 1:46
Hawk
5,4381138102
asked Nov 17 at 1:46
Hawk
5,4381138102
asked Nov 17 at 1:46
Hawk
5,4381138102
5,4381138102
Must be a typo.
– Randall
Nov 17 at 1:59
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
I can, hold on.
– Hawk
Nov 17 at 2:09
|
show 4 more comments
Must be a typo.
– Randall
Nov 17 at 1:59
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
I can, hold on.
– Hawk
Nov 17 at 2:09
Must be a typo.
– Randall
Nov 17 at 1:59
Must be a typo.
– Randall
Nov 17 at 1:59
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
I can, hold on.
– Hawk
Nov 17 at 2:09
I can, hold on.
– Hawk
Nov 17 at 2:09
|
show 4 more comments
2 Answers
2
active
oldest
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up vote
1
down vote
As stated in the text, if $|f| = 0$, then $fequiv 0$ the desired extension is $Fequiv 0$.
Otherwise, replace $f$ with $tilde f=dfrac{f}{|f|}$ and proceed with the proof. At the end, you get an extension $tilde F$ for $tilde f$. Then the desired extension for $f$ is just $F(x) := |f|cdot tilde F(x)$, which works by linearity.
answered Nov 17 at 2:48
Dosidis
41129
add a comment |
up vote
1
down vote
Since $M_1supset M,$ if $|f|=1geq |f_1|$ then $$1geq |f_1|= sup {frac {|f(y)|}{|y|}:0ne yin M_1}geq$$ $$geq sup {frac {|f(y)|}{|y|}:0ne yin M}=$$ $$=|f|=1.$$
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As stated in the text, if $|f| = 0$, then $fequiv 0$ the desired extension is $Fequiv 0$.
Otherwise, replace $f$ with $tilde f=dfrac{f}{|f|}$ and proceed with the proof. At the end, you get an extension $tilde F$ for $tilde f$. Then the desired extension for $f$ is just $F(x) := |f|cdot tilde F(x)$, which works by linearity.
answered Nov 17 at 2:48
Dosidis
41129
add a comment |
up vote
1
down vote
As stated in the text, if $|f| = 0$, then $fequiv 0$ the desired extension is $Fequiv 0$.
Otherwise, replace $f$ with $tilde f=dfrac{f}{|f|}$ and proceed with the proof. At the end, you get an extension $tilde F$ for $tilde f$. Then the desired extension for $f$ is just $F(x) := |f|cdot tilde F(x)$, which works by linearity.
answered Nov 17 at 2:48
Dosidis
41129
add a comment |
up vote
1
down vote
up vote
1
down vote
As stated in the text, if $|f| = 0$, then $fequiv 0$ the desired extension is $Fequiv 0$.
Otherwise, replace $f$ with $tilde f=dfrac{f}{|f|}$ and proceed with the proof. At the end, you get an extension $tilde F$ for $tilde f$. Then the desired extension for $f$ is just $F(x) := |f|cdot tilde F(x)$, which works by linearity.
answered Nov 17 at 2:48
Dosidis
41129
As stated in the text, if $|f| = 0$, then $fequiv 0$ the desired extension is $Fequiv 0$.
Otherwise, replace $f$ with $tilde f=dfrac{f}{|f|}$ and proceed with the proof. At the end, you get an extension $tilde F$ for $tilde f$. Then the desired extension for $f$ is just $F(x) := |f|cdot tilde F(x)$, which works by linearity.
answered Nov 17 at 2:48
Dosidis
41129
answered Nov 17 at 2:48
Dosidis
41129
answered Nov 17 at 2:48
Dosidis
41129
answered Nov 17 at 2:48
Dosidis
41129
41129
add a comment |
add a comment |
up vote
1
down vote
Since $M_1supset M,$ if $|f|=1geq |f_1|$ then $$1geq |f_1|= sup {frac {|f(y)|}{|y|}:0ne yin M_1}geq$$ $$geq sup {frac {|f(y)|}{|y|}:0ne yin M}=$$ $$=|f|=1.$$
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
add a comment |
up vote
1
down vote
Since $M_1supset M,$ if $|f|=1geq |f_1|$ then $$1geq |f_1|= sup {frac {|f(y)|}{|y|}:0ne yin M_1}geq$$ $$geq sup {frac {|f(y)|}{|y|}:0ne yin M}=$$ $$=|f|=1.$$
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $M_1supset M,$ if $|f|=1geq |f_1|$ then $$1geq |f_1|= sup {frac {|f(y)|}{|y|}:0ne yin M_1}geq$$ $$geq sup {frac {|f(y)|}{|y|}:0ne yin M}=$$ $$=|f|=1.$$
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
Since $M_1supset M,$ if $|f|=1geq |f_1|$ then $$1geq |f_1|= sup {frac {|f(y)|}{|y|}:0ne yin M_1}geq$$ $$geq sup {frac {|f(y)|}{|y|}:0ne yin M}=$$ $$=|f|=1.$$
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
answered Nov 17 at 11:46
DanielWainfleet
33.7k31647
33.7k31647
add a comment |
add a comment |
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Must be a typo.
– Randall
Nov 17 at 1:59
I don't think this is true for unspecified $f$
– Apocalypse
Nov 17 at 2:08
@Apocalypse it is a linear functional.
– Hawk
Nov 17 at 2:08
Can you provide more context for how this statement is used in the proof? Perhaps the fact that $lVert frVertge 1$ is implicitly true based on some other property of $f$. (For example, if $f$ is defined so that it is an extension to a larger domain of a function with norm $1$, then the norm of $f$ would have to be at least $1$.)
– Carmeister
Nov 17 at 2:08
I can, hold on.
– Hawk
Nov 17 at 2:09