There exists a point with a bigger distance from a point than a compact
3
$begingroup$
Let $X$ be a Banach space and fix a convex compact set $Ksubseteq X$ with $0notin K$.
Is it true that there exists $x in X$ such that
$
|x|>max_{k in K} |x-k|?
$
The answer is positive in the case $X=mathbf{R}^k$ (and, more generally, in Hilbert spaces): draw a "suitable" line passing through the origin which intersects $K$ and choose a point $x$ which is sufficiently far away from $0$, on the side of $K$ (see details below).
What about the general case in Banach spaces? (A previous version of the question was in the context of locally convex topological vector spaces with a compatible translation-invariant metric; cf. comments of user8268.)
general-topology banach-spaces compactness
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
|
show 5 more comments
3
$begingroup$
Let $X$ be a Banach space and fix a convex compact set $Ksubseteq X$ with $0notin K$.
Is it true that there exists $x in X$ such that
$
|x|>max_{k in K} |x-k|?
$
The answer is positive in the case $X=mathbf{R}^k$ (and, more generally, in Hilbert spaces): draw a "suitable" line passing through the origin which intersects $K$ and choose a point $x$ which is sufficiently far away from $0$, on the side of $K$ (see details below).
What about the general case in Banach spaces? (A previous version of the question was in the context of locally convex topological vector spaces with a compatible translation-invariant metric; cf. comments of user8268.)
general-topology banach-spaces compactness
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
1
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
2
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29
|
show 5 more comments
3
3
3
$begingroup$
Let $X$ be a Banach space and fix a convex compact set $Ksubseteq X$ with $0notin K$.
Is it true that there exists $x in X$ such that
$
|x|>max_{k in K} |x-k|?
$
The answer is positive in the case $X=mathbf{R}^k$ (and, more generally, in Hilbert spaces): draw a "suitable" line passing through the origin which intersects $K$ and choose a point $x$ which is sufficiently far away from $0$, on the side of $K$ (see details below).
What about the general case in Banach spaces? (A previous version of the question was in the context of locally convex topological vector spaces with a compatible translation-invariant metric; cf. comments of user8268.)
general-topology banach-spaces compactness
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
Let $X$ be a Banach space and fix a convex compact set $Ksubseteq X$ with $0notin K$.
Is it true that there exists $x in X$ such that
$
|x|>max_{k in K} |x-k|?
$
The answer is positive in the case $X=mathbf{R}^k$ (and, more generally, in Hilbert spaces): draw a "suitable" line passing through the origin which intersects $K$ and choose a point $x$ which is sufficiently far away from $0$, on the side of $K$ (see details below).
What about the general case in Banach spaces? (A previous version of the question was in the context of locally convex topological vector spaces with a compatible translation-invariant metric; cf. comments of user8268.)
general-topology banach-spaces compactness
general-topology banach-spaces compactness
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
edited Dec 24 '18 at 14:32
Paolo Leonetti
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
asked Dec 23 '18 at 23:09
Paolo LeonettiPaolo Leonetti
11.5k21550
11.5k21550
$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
1
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
2
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29
|
show 5 more comments
$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
1
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
2
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29
$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
1
1
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
2
2
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29
|
show 5 more comments
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$begingroup$
is $d$ supposed to be translation-invariant?
$endgroup$
– user8268
Dec 23 '18 at 23:18
$begingroup$
There exists a compatible metric $d$ which is translation-invariant provided that $X$ is first countable: if needed, assume it
$endgroup$
– Paolo Leonetti
Dec 23 '18 at 23:20
$begingroup$
it's just that without that assumption it is not true, even for $X=Bbb R$ (and a suitable $d$)
$endgroup$
– user8268
Dec 23 '18 at 23:22
1
$begingroup$
@PaulFrost Indeed it was just a sketch. Let $H$ be a $n-1$-dimensional hyperplane separating $0$ and $K$ (this is clearly geometrically; formally, use the Hahn-Banach theorem). Draw the line $L$ passing through $0$ and perpendicular to $H$. The projection of $K$ on $H$ is bounded, hence contained in a $n-1$ dimensional open ball $B$ with center $L cap H$. Draw a $n$-dimensional open ball passing through $B$ with center $x$ on $L$ sufficiently far from $0$, and you're done (since it contains $K$ and not $0$).
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 13:48
2
$begingroup$
Well, it is essential just we want to replicate the above proof as it is: in a Hilbert space, let $H$ be the separating (closed) hyperplane, then there exists the unique point $yin H$ which is nearest to $0$ (here we need completeness). $K$ is compact hence bounded, hence the projection of $K$ on $H$ is contained in a ball $B$ with center $y$. Now it is easy to show that there exists a ball with center $alpha y$ passing through $B$, containing $K$ and not containing $0$. What about the case of Banach spaces (without the notion of orthogonality)?
$endgroup$
– Paolo Leonetti
Dec 24 '18 at 14:29