Mixing
$f(x_0)=0 forall f in X^*$ ($X^*$ is topological dual) then $x_0=0$.
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
$endgroup$
add a comment |
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
$endgroup$
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
$begingroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
$endgroup$
Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.
I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.
The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?
functional-analysis
functional-analysis
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
edited Jan 12 at 17:58
David C. Ullrich
60.7k43994
60.7k43994
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
asked Jan 12 at 17:48
roi_saumonroi_saumon
56438
56438
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
3
3
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
answered Jan 12 at 19:00
NikiforosNikiforos
363
$endgroup$
add a comment |
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$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
answered Jan 12 at 19:00
NikiforosNikiforos
363
$endgroup$
add a comment |
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
answered Jan 12 at 19:00
NikiforosNikiforos
363
$endgroup$
add a comment |
$begingroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
answered Jan 12 at 19:00
NikiforosNikiforos
363
$endgroup$
If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.
answered Jan 12 at 19:00
NikiforosNikiforos
363
edited Jan 12 at 19:07
answered Jan 12 at 19:00
NikiforosNikiforos
363
answered Jan 12 at 19:00
NikiforosNikiforos
363
answered Jan 12 at 19:00
NikiforosNikiforos
363
363
add a comment |
add a comment |
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$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52
1
$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56
$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14