Why Hahn-Banach theorem is needed for the following theorem?
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
add a comment |
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
add a comment |
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
One theorem in Rudin's Real and complex analysis says the following:
If $X$ is a normed linear space and if $x_0 in X$, $x_0neq 0$, there is a bounded linear funcitonal $f$ on $X$, of norm 1, so that $f(x_0)=||x_0||$.
He uses the Hahn-Banach theorem to prove it, but why we cannot just say that $f(x)=||x||$ works?
functional-analysis
functional-analysis
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
asked Dec 3 '18 at 14:37
eigenvalueeigenvalue
11
11
add a comment |
add a comment |
2 Answers
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No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
No, we cannot, because that map is not a linear map.
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
answered Dec 3 '18 at 14:40
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
add a comment |
add a comment |
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
add a comment |
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
add a comment |
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
$f(x)=|x|$ is not linear!
There is no guarantee that $f(x+y) = |x+y|$ is equal to $f(x)+f(y) = |x|+|y|$.
In fact this is almost never happens.
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
answered Dec 3 '18 at 14:41
YankoYanko
6,469727
6,469727
add a comment |
add a comment |
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