Weak closure of subsets of the unitary sphere of a Banach space.
$begingroup$
Assume that $(X,|cdot|)$ is a Banach space with $|cdot|$ strictly convex. Define $S={xin X:|x|=1}$. Suppose that $varepsilon>0$ and $x_0in S$ and define
$$
B_varepsilon={xin X:|x-x_0|levarepsilon}cap S.
$$
Consider the following problem:
Can we find $yin X$ such that $|y|<1$ and any line which contains $y$ must intersect $B_varepsilon$?
Now consider the following possible solution:
Take a supporting hyperplane $H$ for the convex set $B={xin X:|x|le 1}$ at the point $x_0$. Once $|cdot|$ is strictly convex, one can find $zin Xsetminus{0}$ such the set
$$
H_z={h+z:hin H},
$$
has the property that $H_zcap B_varepsilonneq emptyset$ and $H_zcap B_varepsilon=H_zcap S$. Now let $Q$ be the region enclosed by $H_z$ and $B_varepsilon$. It follows that if $yin Q$, then any line which contains $x$ must intersect $B_varepsilon$.
Question 1: Is my proof correct?
Question 2: Do you know another proof?
Remark: If $X$ is an infinite dimensional Banach space, then every $yin Q$ belongs to the weak closure of $B_varepsilon$.
Remark 1: Cross-Posted: mathoverflow
functional-analysis banach-spaces weak-topology
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
$endgroup$
add a comment |
$begingroup$
Assume that $(X,|cdot|)$ is a Banach space with $|cdot|$ strictly convex. Define $S={xin X:|x|=1}$. Suppose that $varepsilon>0$ and $x_0in S$ and define
$$
B_varepsilon={xin X:|x-x_0|levarepsilon}cap S.
$$
Consider the following problem:
Can we find $yin X$ such that $|y|<1$ and any line which contains $y$ must intersect $B_varepsilon$?
Now consider the following possible solution:
Take a supporting hyperplane $H$ for the convex set $B={xin X:|x|le 1}$ at the point $x_0$. Once $|cdot|$ is strictly convex, one can find $zin Xsetminus{0}$ such the set
$$
H_z={h+z:hin H},
$$
has the property that $H_zcap B_varepsilonneq emptyset$ and $H_zcap B_varepsilon=H_zcap S$. Now let $Q$ be the region enclosed by $H_z$ and $B_varepsilon$. It follows that if $yin Q$, then any line which contains $x$ must intersect $B_varepsilon$.
Question 1: Is my proof correct?
Question 2: Do you know another proof?
Remark: If $X$ is an infinite dimensional Banach space, then every $yin Q$ belongs to the weak closure of $B_varepsilon$.
Remark 1: Cross-Posted: mathoverflow
functional-analysis banach-spaces weak-topology
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
$endgroup$
add a comment |
$begingroup$
Assume that $(X,|cdot|)$ is a Banach space with $|cdot|$ strictly convex. Define $S={xin X:|x|=1}$. Suppose that $varepsilon>0$ and $x_0in S$ and define
$$
B_varepsilon={xin X:|x-x_0|levarepsilon}cap S.
$$
Consider the following problem:
Can we find $yin X$ such that $|y|<1$ and any line which contains $y$ must intersect $B_varepsilon$?
Now consider the following possible solution:
Take a supporting hyperplane $H$ for the convex set $B={xin X:|x|le 1}$ at the point $x_0$. Once $|cdot|$ is strictly convex, one can find $zin Xsetminus{0}$ such the set
$$
H_z={h+z:hin H},
$$
has the property that $H_zcap B_varepsilonneq emptyset$ and $H_zcap B_varepsilon=H_zcap S$. Now let $Q$ be the region enclosed by $H_z$ and $B_varepsilon$. It follows that if $yin Q$, then any line which contains $x$ must intersect $B_varepsilon$.
Question 1: Is my proof correct?
Question 2: Do you know another proof?
Remark: If $X$ is an infinite dimensional Banach space, then every $yin Q$ belongs to the weak closure of $B_varepsilon$.
Remark 1: Cross-Posted: mathoverflow
functional-analysis banach-spaces weak-topology
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
$endgroup$
Assume that $(X,|cdot|)$ is a Banach space with $|cdot|$ strictly convex. Define $S={xin X:|x|=1}$. Suppose that $varepsilon>0$ and $x_0in S$ and define
$$
B_varepsilon={xin X:|x-x_0|levarepsilon}cap S.
$$
Consider the following problem:
Can we find $yin X$ such that $|y|<1$ and any line which contains $y$ must intersect $B_varepsilon$?
Now consider the following possible solution:
Take a supporting hyperplane $H$ for the convex set $B={xin X:|x|le 1}$ at the point $x_0$. Once $|cdot|$ is strictly convex, one can find $zin Xsetminus{0}$ such the set
$$
H_z={h+z:hin H},
$$
has the property that $H_zcap B_varepsilonneq emptyset$ and $H_zcap B_varepsilon=H_zcap S$. Now let $Q$ be the region enclosed by $H_z$ and $B_varepsilon$. It follows that if $yin Q$, then any line which contains $x$ must intersect $B_varepsilon$.
Question 1: Is my proof correct?
Question 2: Do you know another proof?
Remark: If $X$ is an infinite dimensional Banach space, then every $yin Q$ belongs to the weak closure of $B_varepsilon$.
Remark 1: Cross-Posted: mathoverflow
functional-analysis banach-spaces weak-topology
functional-analysis banach-spaces weak-topology
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
edited Dec 21 '18 at 10:04
Tomás
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
asked Dec 4 '18 at 15:37
TomásTomás
15.8k32177
15.8k32177
add a comment |
add a comment |
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