Geometric Hahn Banach implies Analytic Hahn Banach.
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I want to prove that the geometric Hahn Banach theorem implies the analytic one.
Edit:
To avoid confusion I will state the vesion of H.B theorems im familiar with:
Analytic H.B:
Let $X$ be a linear space(over $Bbb R$) and $Ysubset X$ a subspace. Let $p:X to Bbb R$ be a sub-additive function. Suppose $f:Yto Bbb R$ is a linear map s.t $f(y) le p(y)$ for all $yin Y$ then there exists an extention $g:Xto Bbb R$ of $f$ s.t. $g(x)le p(x) $ for all $xin X$.
Geometric H.B:
Let $X$ be a linear space. $K subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.
In my problem $X$ is NOT a normed space so there are no open sets.
Given $X$ a linear space and $Ysubset X$ a subspace $p:Xto Bbb R$ sub-additive, and $f:Yto Bbb R$ linear s.t $f(y)le p(y)$ for $yin Y$ we need to extend $f$ to $g:Xto Bbb R $ and $g(x)le p(x)$ fo r all $xin X$
So, we look at $X times Bbb R $and define $K = {(x,t) : t>p(x)}$.
The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).
Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)cap K = emptyset$.
So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.
Now my problem is to show that each $xin X$ has a unique $tin Bbb R$ s.t. $(x,t) in M$. (I need it in order to extend $f$).
I know that if we take $v_0notin M$ then each $v in X times Bbb R$ has a unique representation as $v = alpha v_0 + m$ for $m in M$ , this is because $M$ is a maximal subspace. not sure if that helps.
Thanks for helping!
functional-analysis
$endgroup$
|
show 1 more comment
$begingroup$
I want to prove that the geometric Hahn Banach theorem implies the analytic one.
Edit:
To avoid confusion I will state the vesion of H.B theorems im familiar with:
Analytic H.B:
Let $X$ be a linear space(over $Bbb R$) and $Ysubset X$ a subspace. Let $p:X to Bbb R$ be a sub-additive function. Suppose $f:Yto Bbb R$ is a linear map s.t $f(y) le p(y)$ for all $yin Y$ then there exists an extention $g:Xto Bbb R$ of $f$ s.t. $g(x)le p(x) $ for all $xin X$.
Geometric H.B:
Let $X$ be a linear space. $K subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.
In my problem $X$ is NOT a normed space so there are no open sets.
Given $X$ a linear space and $Ysubset X$ a subspace $p:Xto Bbb R$ sub-additive, and $f:Yto Bbb R$ linear s.t $f(y)le p(y)$ for $yin Y$ we need to extend $f$ to $g:Xto Bbb R $ and $g(x)le p(x)$ fo r all $xin X$
So, we look at $X times Bbb R $and define $K = {(x,t) : t>p(x)}$.
The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).
Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)cap K = emptyset$.
So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.
Now my problem is to show that each $xin X$ has a unique $tin Bbb R$ s.t. $(x,t) in M$. (I need it in order to extend $f$).
I know that if we take $v_0notin M$ then each $v in X times Bbb R$ has a unique representation as $v = alpha v_0 + m$ for $m in M$ , this is because $M$ is a maximal subspace. not sure if that helps.
Thanks for helping!
functional-analysis
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1
$begingroup$
How are you defining an interior point without reference to a topology?
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– user293794
Dec 25 '18 at 13:01
1
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
2
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55
|
show 1 more comment
$begingroup$
I want to prove that the geometric Hahn Banach theorem implies the analytic one.
Edit:
To avoid confusion I will state the vesion of H.B theorems im familiar with:
Analytic H.B:
Let $X$ be a linear space(over $Bbb R$) and $Ysubset X$ a subspace. Let $p:X to Bbb R$ be a sub-additive function. Suppose $f:Yto Bbb R$ is a linear map s.t $f(y) le p(y)$ for all $yin Y$ then there exists an extention $g:Xto Bbb R$ of $f$ s.t. $g(x)le p(x) $ for all $xin X$.
Geometric H.B:
Let $X$ be a linear space. $K subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.
In my problem $X$ is NOT a normed space so there are no open sets.
Given $X$ a linear space and $Ysubset X$ a subspace $p:Xto Bbb R$ sub-additive, and $f:Yto Bbb R$ linear s.t $f(y)le p(y)$ for $yin Y$ we need to extend $f$ to $g:Xto Bbb R $ and $g(x)le p(x)$ fo r all $xin X$
So, we look at $X times Bbb R $and define $K = {(x,t) : t>p(x)}$.
The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).
Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)cap K = emptyset$.
So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.
Now my problem is to show that each $xin X$ has a unique $tin Bbb R$ s.t. $(x,t) in M$. (I need it in order to extend $f$).
I know that if we take $v_0notin M$ then each $v in X times Bbb R$ has a unique representation as $v = alpha v_0 + m$ for $m in M$ , this is because $M$ is a maximal subspace. not sure if that helps.
Thanks for helping!
functional-analysis
$endgroup$
I want to prove that the geometric Hahn Banach theorem implies the analytic one.
Edit:
To avoid confusion I will state the vesion of H.B theorems im familiar with:
Analytic H.B:
Let $X$ be a linear space(over $Bbb R$) and $Ysubset X$ a subspace. Let $p:X to Bbb R$ be a sub-additive function. Suppose $f:Yto Bbb R$ is a linear map s.t $f(y) le p(y)$ for all $yin Y$ then there exists an extention $g:Xto Bbb R$ of $f$ s.t. $g(x)le p(x) $ for all $xin X$.
Geometric H.B:
Let $X$ be a linear space. $K subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.
In my problem $X$ is NOT a normed space so there are no open sets.
Given $X$ a linear space and $Ysubset X$ a subspace $p:Xto Bbb R$ sub-additive, and $f:Yto Bbb R$ linear s.t $f(y)le p(y)$ for $yin Y$ we need to extend $f$ to $g:Xto Bbb R $ and $g(x)le p(x)$ fo r all $xin X$
So, we look at $X times Bbb R $and define $K = {(x,t) : t>p(x)}$.
The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).
Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)cap K = emptyset$.
So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.
Now my problem is to show that each $xin X$ has a unique $tin Bbb R$ s.t. $(x,t) in M$. (I need it in order to extend $f$).
I know that if we take $v_0notin M$ then each $v in X times Bbb R$ has a unique representation as $v = alpha v_0 + m$ for $m in M$ , this is because $M$ is a maximal subspace. not sure if that helps.
Thanks for helping!
functional-analysis
functional-analysis
edited Jan 20 at 11:17
user123
asked Dec 24 '18 at 20:55
user123user123
1,360316
1,360316
1
$begingroup$
How are you defining an interior point without reference to a topology?
$endgroup$
– user293794
Dec 25 '18 at 13:01
1
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
2
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55
|
show 1 more comment
1
$begingroup$
How are you defining an interior point without reference to a topology?
$endgroup$
– user293794
Dec 25 '18 at 13:01
1
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
2
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55
1
1
$begingroup$
How are you defining an interior point without reference to a topology?
$endgroup$
– user293794
Dec 25 '18 at 13:01
$begingroup$
How are you defining an interior point without reference to a topology?
$endgroup$
– user293794
Dec 25 '18 at 13:01
1
1
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
2
2
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:={xin X:p(x)<1}$, $alpha:=sup_{yin Ycap K}{f(y)}$ and $D:={yin Y:f(y)=alpha}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $kin K$ and you can write explicit $epsilon$ for any $yin X$ to show the definition in the link you referred to holds).
If $fequiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $yin Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=alpha$. Note that $D=ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $Dcap K=emptyset$ (this is true thanks to the definition of $alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0notinhat D$). So $hat D-y_0$ is a maximal subspace which contains $ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $hat D-y_0$ and extend linearly. (Note that $Y=ker(f)oplus span{y_0}$ so this definition actually extends $f$).
$endgroup$
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
add a comment |
$begingroup$
To apply geometric Hahn-Banach, you need that the space in question is more than just a vector space - it must be a topological vector space, meaning a vector space with a topology such the vector space operations are continuous. So, if we hope to apply geometric Hahn-Banach to the space $Xtimes mathbb{R}$ we better figure out what its open sets are. We know what the topology on $mathbb{R}$ ought to be so what about the topology on $X$? When a vector space has a norm, we may give it a topology via that norm by defining open balls. Here, we are not lucky enough to have a norm but we do have the sub-additive functional $p$, which we should think of as "almost" a norm. This functional $p$ gives us a natural topology on $X$ that makes $K$ open in the product. One way to think about the way in which a norm $||cdot||:Yrightarrow mathbb{R}$ gives $Y$ a topology is to declare that the open sets of $Y$ are just $||cdot ||^{-1}(U)$ where $U$ is open in $mathbb{R}$. You should think about why this does in fact give a topology and why it is the same topology given by defining a basis of open balls. This is called the topology induced by the function $||cdot||$, and it works for any function. Thus, in the same way, we may define a topology on $X$ by declaring the open sets to be $p^{-1}(U)$, where $U$ is open in $mathbb{R}$. Now, we have a vector space with a topology, but we need to check that it is a topological vector space, so the extra requirement mentioned above that the vector space operations are compatible (i.e. continuous w/r/t) the topology. This should come from the fact that $p$ is not any old map, but a sub-additive linear functional. Once we have this we are free to apply geometric Hahn-Banach.
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$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
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– user123
Dec 25 '18 at 12:37
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both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
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– user123
Dec 25 '18 at 12:38
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
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– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
|
show 7 more comments
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2 Answers
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2 Answers
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$begingroup$
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:={xin X:p(x)<1}$, $alpha:=sup_{yin Ycap K}{f(y)}$ and $D:={yin Y:f(y)=alpha}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $kin K$ and you can write explicit $epsilon$ for any $yin X$ to show the definition in the link you referred to holds).
If $fequiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $yin Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=alpha$. Note that $D=ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $Dcap K=emptyset$ (this is true thanks to the definition of $alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0notinhat D$). So $hat D-y_0$ is a maximal subspace which contains $ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $hat D-y_0$ and extend linearly. (Note that $Y=ker(f)oplus span{y_0}$ so this definition actually extends $f$).
$endgroup$
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
add a comment |
$begingroup$
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:={xin X:p(x)<1}$, $alpha:=sup_{yin Ycap K}{f(y)}$ and $D:={yin Y:f(y)=alpha}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $kin K$ and you can write explicit $epsilon$ for any $yin X$ to show the definition in the link you referred to holds).
If $fequiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $yin Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=alpha$. Note that $D=ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $Dcap K=emptyset$ (this is true thanks to the definition of $alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0notinhat D$). So $hat D-y_0$ is a maximal subspace which contains $ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $hat D-y_0$ and extend linearly. (Note that $Y=ker(f)oplus span{y_0}$ so this definition actually extends $f$).
$endgroup$
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
add a comment |
$begingroup$
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:={xin X:p(x)<1}$, $alpha:=sup_{yin Ycap K}{f(y)}$ and $D:={yin Y:f(y)=alpha}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $kin K$ and you can write explicit $epsilon$ for any $yin X$ to show the definition in the link you referred to holds).
If $fequiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $yin Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=alpha$. Note that $D=ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $Dcap K=emptyset$ (this is true thanks to the definition of $alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0notinhat D$). So $hat D-y_0$ is a maximal subspace which contains $ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $hat D-y_0$ and extend linearly. (Note that $Y=ker(f)oplus span{y_0}$ so this definition actually extends $f$).
$endgroup$
Here's another approach, according to guidance in the book Functional Analysis by Prof.s Weiss, Lindenstrauss and Pazi (there's no English translation as far as I'm aware of, sorry) in page 217, Q.13:
Using the notation of the analytical H.B. you've written above, define $K:={xin X:p(x)<1}$, $alpha:=sup_{yin Ycap K}{f(y)}$ and $D:={yin Y:f(y)=alpha}$.
Note that $K$ is convex, and that every point of $K$ is an internal point (take $kin K$ and you can write explicit $epsilon$ for any $yin X$ to show the definition in the link you referred to holds).
If $fequiv 0$ it's easy to extend $f$ to be identically $0$ on all of $X$. Otherwise, there is a $yin Y$ for which $f(y)>0$, and as $Y$ is a vector space and $f$ is linear, we can scale this point to get $f(y_0)=alpha$. Note that $D=ker(f)+y_0$ so it is a plane (an affine subspace).
Lastly note that $Dcap K=emptyset$ (this is true thanks to the definition of $alpha$ as a supremum and $K$ as a preimage of an open interval).
Apply Geometric H.B. and achieve a hyperplane $hat D$ that contains $D$ and does not intersect with $K$ (specifically, $0notinhat D$). So $hat D-y_0$ is a maximal subspace which contains $ker(f)$ and does not contain $y_0$, so we may define the extension of $f$ as $0$ on $hat D-y_0$ and extend linearly. (Note that $Y=ker(f)oplus span{y_0}$ so this definition actually extends $f$).
edited Jan 21 at 15:30
answered Jan 20 at 9:20
et_let_l
1264
1264
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
add a comment |
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
In the defintion of $alpha$, did you mean to intersect $K$ instead of $M$?
$endgroup$
– S. R
Jan 21 at 12:47
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
$begingroup$
Yes! Thanks for spotting that. In the book, K is written as M and I started writing it the same way just to change all the M's to K's (or so I thought I did) after realising it would be more readable that way.
$endgroup$
– et_l
Jan 21 at 15:32
add a comment |
$begingroup$
To apply geometric Hahn-Banach, you need that the space in question is more than just a vector space - it must be a topological vector space, meaning a vector space with a topology such the vector space operations are continuous. So, if we hope to apply geometric Hahn-Banach to the space $Xtimes mathbb{R}$ we better figure out what its open sets are. We know what the topology on $mathbb{R}$ ought to be so what about the topology on $X$? When a vector space has a norm, we may give it a topology via that norm by defining open balls. Here, we are not lucky enough to have a norm but we do have the sub-additive functional $p$, which we should think of as "almost" a norm. This functional $p$ gives us a natural topology on $X$ that makes $K$ open in the product. One way to think about the way in which a norm $||cdot||:Yrightarrow mathbb{R}$ gives $Y$ a topology is to declare that the open sets of $Y$ are just $||cdot ||^{-1}(U)$ where $U$ is open in $mathbb{R}$. You should think about why this does in fact give a topology and why it is the same topology given by defining a basis of open balls. This is called the topology induced by the function $||cdot||$, and it works for any function. Thus, in the same way, we may define a topology on $X$ by declaring the open sets to be $p^{-1}(U)$, where $U$ is open in $mathbb{R}$. Now, we have a vector space with a topology, but we need to check that it is a topological vector space, so the extra requirement mentioned above that the vector space operations are compatible (i.e. continuous w/r/t) the topology. This should come from the fact that $p$ is not any old map, but a sub-additive linear functional. Once we have this we are free to apply geometric Hahn-Banach.
$endgroup$
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
|
show 7 more comments
$begingroup$
To apply geometric Hahn-Banach, you need that the space in question is more than just a vector space - it must be a topological vector space, meaning a vector space with a topology such the vector space operations are continuous. So, if we hope to apply geometric Hahn-Banach to the space $Xtimes mathbb{R}$ we better figure out what its open sets are. We know what the topology on $mathbb{R}$ ought to be so what about the topology on $X$? When a vector space has a norm, we may give it a topology via that norm by defining open balls. Here, we are not lucky enough to have a norm but we do have the sub-additive functional $p$, which we should think of as "almost" a norm. This functional $p$ gives us a natural topology on $X$ that makes $K$ open in the product. One way to think about the way in which a norm $||cdot||:Yrightarrow mathbb{R}$ gives $Y$ a topology is to declare that the open sets of $Y$ are just $||cdot ||^{-1}(U)$ where $U$ is open in $mathbb{R}$. You should think about why this does in fact give a topology and why it is the same topology given by defining a basis of open balls. This is called the topology induced by the function $||cdot||$, and it works for any function. Thus, in the same way, we may define a topology on $X$ by declaring the open sets to be $p^{-1}(U)$, where $U$ is open in $mathbb{R}$. Now, we have a vector space with a topology, but we need to check that it is a topological vector space, so the extra requirement mentioned above that the vector space operations are compatible (i.e. continuous w/r/t) the topology. This should come from the fact that $p$ is not any old map, but a sub-additive linear functional. Once we have this we are free to apply geometric Hahn-Banach.
$endgroup$
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
|
show 7 more comments
$begingroup$
To apply geometric Hahn-Banach, you need that the space in question is more than just a vector space - it must be a topological vector space, meaning a vector space with a topology such the vector space operations are continuous. So, if we hope to apply geometric Hahn-Banach to the space $Xtimes mathbb{R}$ we better figure out what its open sets are. We know what the topology on $mathbb{R}$ ought to be so what about the topology on $X$? When a vector space has a norm, we may give it a topology via that norm by defining open balls. Here, we are not lucky enough to have a norm but we do have the sub-additive functional $p$, which we should think of as "almost" a norm. This functional $p$ gives us a natural topology on $X$ that makes $K$ open in the product. One way to think about the way in which a norm $||cdot||:Yrightarrow mathbb{R}$ gives $Y$ a topology is to declare that the open sets of $Y$ are just $||cdot ||^{-1}(U)$ where $U$ is open in $mathbb{R}$. You should think about why this does in fact give a topology and why it is the same topology given by defining a basis of open balls. This is called the topology induced by the function $||cdot||$, and it works for any function. Thus, in the same way, we may define a topology on $X$ by declaring the open sets to be $p^{-1}(U)$, where $U$ is open in $mathbb{R}$. Now, we have a vector space with a topology, but we need to check that it is a topological vector space, so the extra requirement mentioned above that the vector space operations are compatible (i.e. continuous w/r/t) the topology. This should come from the fact that $p$ is not any old map, but a sub-additive linear functional. Once we have this we are free to apply geometric Hahn-Banach.
$endgroup$
To apply geometric Hahn-Banach, you need that the space in question is more than just a vector space - it must be a topological vector space, meaning a vector space with a topology such the vector space operations are continuous. So, if we hope to apply geometric Hahn-Banach to the space $Xtimes mathbb{R}$ we better figure out what its open sets are. We know what the topology on $mathbb{R}$ ought to be so what about the topology on $X$? When a vector space has a norm, we may give it a topology via that norm by defining open balls. Here, we are not lucky enough to have a norm but we do have the sub-additive functional $p$, which we should think of as "almost" a norm. This functional $p$ gives us a natural topology on $X$ that makes $K$ open in the product. One way to think about the way in which a norm $||cdot||:Yrightarrow mathbb{R}$ gives $Y$ a topology is to declare that the open sets of $Y$ are just $||cdot ||^{-1}(U)$ where $U$ is open in $mathbb{R}$. You should think about why this does in fact give a topology and why it is the same topology given by defining a basis of open balls. This is called the topology induced by the function $||cdot||$, and it works for any function. Thus, in the same way, we may define a topology on $X$ by declaring the open sets to be $p^{-1}(U)$, where $U$ is open in $mathbb{R}$. Now, we have a vector space with a topology, but we need to check that it is a topological vector space, so the extra requirement mentioned above that the vector space operations are compatible (i.e. continuous w/r/t) the topology. This should come from the fact that $p$ is not any old map, but a sub-additive linear functional. Once we have this we are free to apply geometric Hahn-Banach.
answered Dec 25 '18 at 12:36
user293794user293794
1,711613
1,711613
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
|
show 7 more comments
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
I learned a version of H.B theorem which do not use topological spaces..
$endgroup$
– user123
Dec 25 '18 at 12:37
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
$begingroup$
both versions (geometric and analytic one ) assumes only that $X$ is a linear space.
$endgroup$
– user123
Dec 25 '18 at 12:38
1
1
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
That is not true. Please look up geometric Hahn-Banach again, for instance on wikipedia or the math overflow post you linked. It requires a topology on the space - after all the statement of the theorem refers to open sets. Analytic Hahn-Banach does not require that the space begins with a topology, but to prove it from geometric we endow the space with a topology, which is the only way we can apply geometric.
$endgroup$
– user293794
Dec 25 '18 at 12:40
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
what's not true? this is how i learned it, and what im trying to prove..
$endgroup$
– user123
Dec 25 '18 at 12:41
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
$begingroup$
first statement is : $X$ is a linear space and $p$ sub-additive and $Y$ a subspace with $f$ linear on $Y$ dominated by $p$ then there exists a linear extension to f that is still dominated by p ..
$endgroup$
– user123
Dec 25 '18 at 12:42
|
show 7 more comments
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1
$begingroup$
How are you defining an interior point without reference to a topology?
$endgroup$
– user293794
Dec 25 '18 at 13:01
1
$begingroup$
planetmath.org/internalpoint
$endgroup$
– user123
Dec 25 '18 at 13:02
2
$begingroup$
@Liad You already posted this question, got an answer, then deleted that post, only to ask it yet again, here. Do not delete a post after receiving an answer to it, and above all, do not delete a post, only to ask the same equation. You are free to edit and improve the original post, but not to repost the same question you yourself deleted for whatever reason.
$endgroup$
– amWhy
Dec 28 '18 at 23:05
$begingroup$
@amWhy you are right, i got an answer based on the wrong fact that $X$ is a normed space, so i thought re-posting the question empathizing the fact that $X$ is just a linear space.
$endgroup$
– user123
Dec 29 '18 at 9:33
$begingroup$
I think some of the confusion between the versions arises from a mistake in the phrasing of the Geometric H.B. for a general vector space: it should read "each point in K is an INTERNAL point" rather than "interior". Then the condition makes sense without any topology on X. Later on in the second step of the proof ("The fact that K...") you refer to the fact that we need to prove all of K's points are internal points (as the phrasing that I'm familiar with goes), and not interior points (as the current phrasing of your question reads), which makes more sense.
$endgroup$
– et_l
Jan 19 at 17:55