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Proximal kernel if and only if there exists $x$ of norm 1 such that $Lambda x = |Lambda|$, $X$ a Banach space, $Lambdain X^*$
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I am working through the proof of the following proposition V.4.4.7 in Conway's functional analysis.
$colorblueLambda$
Now the beginning of the proof goes as follows:
$implies$ Let $M := textkerLambda$. Define $f: X/M to mathbbF$ via $f(x+M) = Lambda x$, so clearly $f$ is a linear functional with $|f| = |Lambda|$. Since $textdim(X/M)=1$, there is $x in X$ with $|x+M| = 1$ and $f(x+M) = |f|$. Etc.
The rest of the forward direction from there is very straightforward, but I do not understand this last sentence. It is basically saying that $sup_ f(x)$ is attained, but it is not obvious to me why a continuous linear map from technically "$mathbbR$ to $mathbbR$" would give us this attainment.
Similarly, in the backwards direction:
$impliedby$ Let $x_0$ be the element such that the hypothesis holds. If $x in X$ with $|x+M| = alpha$, then $|alpha^-1 x+ M| = 1$. But it is also true that $|x_0 + M| = 1$ (I'm assuming the reason for this is the answer to my previous question). Then since $textdimX/M = 1$, there is a $beta in mathbbF$ with $|beta|=1$ such that $alpha^-1 x+M = beta(x_0 + M)$. Etc.
I once again am not sure why this last sentence is true. Maybe I am missing something obvious, but I'm not sure. Any help or hints to point me in the right direction would be much appreciated.
functional-analysis banach-spaces dual-spaces
asked Jul 22 at 14:55
mathishard.butweloveit
1069
add a comment |Â
up vote
1
down vote
favorite
I am working through the proof of the following proposition V.4.4.7 in Conway's functional analysis.
$colorblueLambda$
Now the beginning of the proof goes as follows:
$implies$ Let $M := textkerLambda$. Define $f: X/M to mathbbF$ via $f(x+M) = Lambda x$, so clearly $f$ is a linear functional with $|f| = |Lambda|$. Since $textdim(X/M)=1$, there is $x in X$ with $|x+M| = 1$ and $f(x+M) = |f|$. Etc.
The rest of the forward direction from there is very straightforward, but I do not understand this last sentence. It is basically saying that $sup_ f(x)$ is attained, but it is not obvious to me why a continuous linear map from technically "$mathbbR$ to $mathbbR$" would give us this attainment.
Similarly, in the backwards direction:
$impliedby$ Let $x_0$ be the element such that the hypothesis holds. If $x in X$ with $|x+M| = alpha$, then $|alpha^-1 x+ M| = 1$. But it is also true that $|x_0 + M| = 1$ (I'm assuming the reason for this is the answer to my previous question). Then since $textdimX/M = 1$, there is a $beta in mathbbF$ with $|beta|=1$ such that $alpha^-1 x+M = beta(x_0 + M)$. Etc.
I once again am not sure why this last sentence is true. Maybe I am missing something obvious, but I'm not sure. Any help or hints to point me in the right direction would be much appreciated.
functional-analysis banach-spaces dual-spaces
asked Jul 22 at 14:55
mathishard.butweloveit
1069
What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am working through the proof of the following proposition V.4.4.7 in Conway's functional analysis.
$colorblueLambda$
Now the beginning of the proof goes as follows:
$implies$ Let $M := textkerLambda$. Define $f: X/M to mathbbF$ via $f(x+M) = Lambda x$, so clearly $f$ is a linear functional with $|f| = |Lambda|$. Since $textdim(X/M)=1$, there is $x in X$ with $|x+M| = 1$ and $f(x+M) = |f|$. Etc.
The rest of the forward direction from there is very straightforward, but I do not understand this last sentence. It is basically saying that $sup_ f(x)$ is attained, but it is not obvious to me why a continuous linear map from technically "$mathbbR$ to $mathbbR$" would give us this attainment.
Similarly, in the backwards direction:
$impliedby$ Let $x_0$ be the element such that the hypothesis holds. If $x in X$ with $|x+M| = alpha$, then $|alpha^-1 x+ M| = 1$. But it is also true that $|x_0 + M| = 1$ (I'm assuming the reason for this is the answer to my previous question). Then since $textdimX/M = 1$, there is a $beta in mathbbF$ with $|beta|=1$ such that $alpha^-1 x+M = beta(x_0 + M)$. Etc.
I once again am not sure why this last sentence is true. Maybe I am missing something obvious, but I'm not sure. Any help or hints to point me in the right direction would be much appreciated.
functional-analysis banach-spaces dual-spaces
asked Jul 22 at 14:55
mathishard.butweloveit
1069
I am working through the proof of the following proposition V.4.4.7 in Conway's functional analysis.
$colorblueLambda$
Now the beginning of the proof goes as follows:
$implies$ Let $M := textkerLambda$. Define $f: X/M to mathbbF$ via $f(x+M) = Lambda x$, so clearly $f$ is a linear functional with $|f| = |Lambda|$. Since $textdim(X/M)=1$, there is $x in X$ with $|x+M| = 1$ and $f(x+M) = |f|$. Etc.
The rest of the forward direction from there is very straightforward, but I do not understand this last sentence. It is basically saying that $sup_ f(x)$ is attained, but it is not obvious to me why a continuous linear map from technically "$mathbbR$ to $mathbbR$" would give us this attainment.
Similarly, in the backwards direction:
$impliedby$ Let $x_0$ be the element such that the hypothesis holds. If $x in X$ with $|x+M| = alpha$, then $|alpha^-1 x+ M| = 1$. But it is also true that $|x_0 + M| = 1$ (I'm assuming the reason for this is the answer to my previous question). Then since $textdimX/M = 1$, there is a $beta in mathbbF$ with $|beta|=1$ such that $alpha^-1 x+M = beta(x_0 + M)$. Etc.
I once again am not sure why this last sentence is true. Maybe I am missing something obvious, but I'm not sure. Any help or hints to point me in the right direction would be much appreciated.
functional-analysis banach-spaces dual-spaces
asked Jul 22 at 14:55
mathishard.butweloveit
1069
edited Jul 22 at 15:07
asked Jul 22 at 14:55
mathishard.butweloveit
1069
asked Jul 22 at 14:55
mathishard.butweloveit
1069
asked Jul 22 at 14:55
mathishard.butweloveit
1069
1069
What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02
add a comment |Â
What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02
What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
For the first point, since $dim(X/M)=1<infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $x_n+M$ in this sphere such that $f(x_n+M)to|f|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=|f|$.
For the second point, since $dim(X/M)=1$, we have $X/M$ is isomorphic to $mathbb F$. Let $g:X/Mtomathbb F$ be such an isomorphism, and put $beta=fracf(alpha^-1x+M)f(x_0+M)$. Then since $f((alpha^-1x+M)-beta(x_0+M))=0$, we must have $(alpha^-1x+M)-beta(x_0+M)=0$.
answered Jul 22 at 15:17
Aweygan
11.9k21437
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For the first point, since $dim(X/M)=1<infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $x_n+M$ in this sphere such that $f(x_n+M)to|f|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=|f|$.
For the second point, since $dim(X/M)=1$, we have $X/M$ is isomorphic to $mathbb F$. Let $g:X/Mtomathbb F$ be such an isomorphism, and put $beta=fracf(alpha^-1x+M)f(x_0+M)$. Then since $f((alpha^-1x+M)-beta(x_0+M))=0$, we must have $(alpha^-1x+M)-beta(x_0+M)=0$.
answered Jul 22 at 15:17
Aweygan
11.9k21437
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
add a comment |Â
up vote
1
down vote
accepted
For the first point, since $dim(X/M)=1<infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $x_n+M$ in this sphere such that $f(x_n+M)to|f|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=|f|$.
For the second point, since $dim(X/M)=1$, we have $X/M$ is isomorphic to $mathbb F$. Let $g:X/Mtomathbb F$ be such an isomorphism, and put $beta=fracf(alpha^-1x+M)f(x_0+M)$. Then since $f((alpha^-1x+M)-beta(x_0+M))=0$, we must have $(alpha^-1x+M)-beta(x_0+M)=0$.
answered Jul 22 at 15:17
Aweygan
11.9k21437
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For the first point, since $dim(X/M)=1<infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $x_n+M$ in this sphere such that $f(x_n+M)to|f|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=|f|$.
For the second point, since $dim(X/M)=1$, we have $X/M$ is isomorphic to $mathbb F$. Let $g:X/Mtomathbb F$ be such an isomorphism, and put $beta=fracf(alpha^-1x+M)f(x_0+M)$. Then since $f((alpha^-1x+M)-beta(x_0+M))=0$, we must have $(alpha^-1x+M)-beta(x_0+M)=0$.
answered Jul 22 at 15:17
Aweygan
11.9k21437
For the first point, since $dim(X/M)=1<infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $x_n+M$ in this sphere such that $f(x_n+M)to|f|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=|f|$.
For the second point, since $dim(X/M)=1$, we have $X/M$ is isomorphic to $mathbb F$. Let $g:X/Mtomathbb F$ be such an isomorphism, and put $beta=fracf(alpha^-1x+M)f(x_0+M)$. Then since $f((alpha^-1x+M)-beta(x_0+M))=0$, we must have $(alpha^-1x+M)-beta(x_0+M)=0$.
answered Jul 22 at 15:17
Aweygan
11.9k21437
edited Jul 22 at 15:35
answered Jul 22 at 15:17
Aweygan
11.9k21437
answered Jul 22 at 15:17
Aweygan
11.9k21437
answered Jul 22 at 15:17
Aweygan
11.9k21437
11.9k21437
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
add a comment |Â
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
1
1
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
Whoops, sorry for the accidental downvote. Thank you very much for the answer; I will make note there is a slight typo in the definition of $beta$ for future visitors to this post. $beta = fracf(alpha^-1x+ M)f(x_0+M)$
– mathishard.butweloveit
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
ahh yes, thank for spotting that typo! I will edit my answer.
– Aweygan
Jul 22 at 15:34
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
And no worries, I didn't even notice the downvote.
– Aweygan
Jul 22 at 15:35
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
I actually have one more question, as I didn't notice this before. It doesn't seem clear that $|beta|= 1$, since there isn't a direct relationship between $f(alpha^-1x + M)$ and $|f| = f(x_0+M)$.
– mathishard.butweloveit
Jul 22 at 15:47
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
Never mind, got it
– mathishard.butweloveit
Jul 22 at 15:52
add a comment |Â
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What is the definition of a proximal subspace?
– Aweygan
Jul 22 at 15:01
A linear subspace $M subset X$ is proximal if for every $xin X$, there exists a $yin M$ such that $|x-y| = inf_zin M |x-z|$. In other words, there exists a closest point.
– mathishard.butweloveit
Jul 22 at 15:02