Weak closure of unit sphere is unit ball - a question about the hypotheses.
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A homework problem I recall from functional analysis was to prove that the weak closure of the unit sphere, $S$, in an infinite-dimensional real normed vector space is the unit ball, $B$.
Looking back at what I turned in, I argued as follows:
Note that $S$ would be weakly dense in $B$ if, for any nonempty (relatively) weakly open subset $Usubset B$, one has $Scap Uneqemptyset$. Let $U$ be such a subset and let $x_{0}in Usubset B$. Fixing $epsilon>0$ and $x^{*}in X^{*}$, one has by continuity, that the inverse image
$$V_{*}^{epsilon}:=(x^{*})^{-1}[(langle x^{*},x_{0}rangle-epsilon,langle x^{*},x_{0}rangle+epsilon)]$$
is weakly open, and hence, $Ucap V_{*}^{epsilon}$ is (relatively) weakly open in $B$, and contains $x_{0}$. As long as $x^{*}$ does not vanish identically, it's kernel has codimension $1$, so since $text{dim}(X)=infty$, one must have that $text{ker}(x^{*})$ is nontrivial. Then, finding a nonzero $xiintext{ker}(x^{*})$, one has
$$x_{0}+txiin S$$
for some $tinmathbb{R}$. Finally, this yields
$$|langle x^{*},x_{0}rangle-langle x^{*},x_{0}+txirangle|=|t|cdot|langle x^{*},xirangle|=0<epsilon$$
which means $x_{0}+txiin V_{*}^{epsilon}$.
Now, I have two questions:
- If we knew that $V_{*}^{epsilon}subset U$, we'd be done. Why can we assume this? (It seems in some of the proofs I've seen elsewhere, this is assumed WLOG)
- Why do we need $text{dim}(X)=infty$? We are using the fact that
$$X/text{ker}(x^{*})congmathbb{R}$$
so if the kernel were trivial, wouldn't this still be a contradiction as long as $text{dim}(X)geq 2$?
functional-analysis
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A homework problem I recall from functional analysis was to prove that the weak closure of the unit sphere, $S$, in an infinite-dimensional real normed vector space is the unit ball, $B$.
Looking back at what I turned in, I argued as follows:
Note that $S$ would be weakly dense in $B$ if, for any nonempty (relatively) weakly open subset $Usubset B$, one has $Scap Uneqemptyset$. Let $U$ be such a subset and let $x_{0}in Usubset B$. Fixing $epsilon>0$ and $x^{*}in X^{*}$, one has by continuity, that the inverse image
$$V_{*}^{epsilon}:=(x^{*})^{-1}[(langle x^{*},x_{0}rangle-epsilon,langle x^{*},x_{0}rangle+epsilon)]$$
is weakly open, and hence, $Ucap V_{*}^{epsilon}$ is (relatively) weakly open in $B$, and contains $x_{0}$. As long as $x^{*}$ does not vanish identically, it's kernel has codimension $1$, so since $text{dim}(X)=infty$, one must have that $text{ker}(x^{*})$ is nontrivial. Then, finding a nonzero $xiintext{ker}(x^{*})$, one has
$$x_{0}+txiin S$$
for some $tinmathbb{R}$. Finally, this yields
$$|langle x^{*},x_{0}rangle-langle x^{*},x_{0}+txirangle|=|t|cdot|langle x^{*},xirangle|=0<epsilon$$
which means $x_{0}+txiin V_{*}^{epsilon}$.
Now, I have two questions:
- If we knew that $V_{*}^{epsilon}subset U$, we'd be done. Why can we assume this? (It seems in some of the proofs I've seen elsewhere, this is assumed WLOG)
- Why do we need $text{dim}(X)=infty$? We are using the fact that
$$X/text{ker}(x^{*})congmathbb{R}$$
so if the kernel were trivial, wouldn't this still be a contradiction as long as $text{dim}(X)geq 2$?
functional-analysis
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A homework problem I recall from functional analysis was to prove that the weak closure of the unit sphere, $S$, in an infinite-dimensional real normed vector space is the unit ball, $B$.
Looking back at what I turned in, I argued as follows:
Note that $S$ would be weakly dense in $B$ if, for any nonempty (relatively) weakly open subset $Usubset B$, one has $Scap Uneqemptyset$. Let $U$ be such a subset and let $x_{0}in Usubset B$. Fixing $epsilon>0$ and $x^{*}in X^{*}$, one has by continuity, that the inverse image
$$V_{*}^{epsilon}:=(x^{*})^{-1}[(langle x^{*},x_{0}rangle-epsilon,langle x^{*},x_{0}rangle+epsilon)]$$
is weakly open, and hence, $Ucap V_{*}^{epsilon}$ is (relatively) weakly open in $B$, and contains $x_{0}$. As long as $x^{*}$ does not vanish identically, it's kernel has codimension $1$, so since $text{dim}(X)=infty$, one must have that $text{ker}(x^{*})$ is nontrivial. Then, finding a nonzero $xiintext{ker}(x^{*})$, one has
$$x_{0}+txiin S$$
for some $tinmathbb{R}$. Finally, this yields
$$|langle x^{*},x_{0}rangle-langle x^{*},x_{0}+txirangle|=|t|cdot|langle x^{*},xirangle|=0<epsilon$$
which means $x_{0}+txiin V_{*}^{epsilon}$.
Now, I have two questions:
- If we knew that $V_{*}^{epsilon}subset U$, we'd be done. Why can we assume this? (It seems in some of the proofs I've seen elsewhere, this is assumed WLOG)
- Why do we need $text{dim}(X)=infty$? We are using the fact that
$$X/text{ker}(x^{*})congmathbb{R}$$
so if the kernel were trivial, wouldn't this still be a contradiction as long as $text{dim}(X)geq 2$?
functional-analysis
A homework problem I recall from functional analysis was to prove that the weak closure of the unit sphere, $S$, in an infinite-dimensional real normed vector space is the unit ball, $B$.
Looking back at what I turned in, I argued as follows:
Note that $S$ would be weakly dense in $B$ if, for any nonempty (relatively) weakly open subset $Usubset B$, one has $Scap Uneqemptyset$. Let $U$ be such a subset and let $x_{0}in Usubset B$. Fixing $epsilon>0$ and $x^{*}in X^{*}$, one has by continuity, that the inverse image
$$V_{*}^{epsilon}:=(x^{*})^{-1}[(langle x^{*},x_{0}rangle-epsilon,langle x^{*},x_{0}rangle+epsilon)]$$
is weakly open, and hence, $Ucap V_{*}^{epsilon}$ is (relatively) weakly open in $B$, and contains $x_{0}$. As long as $x^{*}$ does not vanish identically, it's kernel has codimension $1$, so since $text{dim}(X)=infty$, one must have that $text{ker}(x^{*})$ is nontrivial. Then, finding a nonzero $xiintext{ker}(x^{*})$, one has
$$x_{0}+txiin S$$
for some $tinmathbb{R}$. Finally, this yields
$$|langle x^{*},x_{0}rangle-langle x^{*},x_{0}+txirangle|=|t|cdot|langle x^{*},xirangle|=0<epsilon$$
which means $x_{0}+txiin V_{*}^{epsilon}$.
Now, I have two questions:
- If we knew that $V_{*}^{epsilon}subset U$, we'd be done. Why can we assume this? (It seems in some of the proofs I've seen elsewhere, this is assumed WLOG)
- Why do we need $text{dim}(X)=infty$? We are using the fact that
$$X/text{ker}(x^{*})congmathbb{R}$$
so if the kernel were trivial, wouldn't this still be a contradiction as long as $text{dim}(X)geq 2$?
functional-analysis
functional-analysis
asked Nov 17 at 18:39
JWP_HTX
385213
385213
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Nevermind - I have answered my own questions.
$V_{w}=Big{{} bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] text{ }Big{|}text{ } x_{1}^{*},ldots, x_{n}^{*}in X^{*}Big{}}$ is a base for the weak topology on $X$. Thus, we may find some $Vin V_{w}$ so that $x_{0}in Vsubset U$, and in particular, this means that for some $epsilon>0$, we have that
$$V_{epsilon}=Big{{}xin X text{ } Big{|}text{ } |langle x_{j}^{*},x_{0}-xrangle|<epsilon text{ for all } j=1,ldots,nBig{}}subset U$$The fact that $text{dim}(X)=infty$ is then required to find a nonzero $xiinbigcap_{j=1}^{n}text{ker}(x^{*}_{j})$, and then we may proceed as above.
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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up vote
0
down vote
Nevermind - I have answered my own questions.
$V_{w}=Big{{} bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] text{ }Big{|}text{ } x_{1}^{*},ldots, x_{n}^{*}in X^{*}Big{}}$ is a base for the weak topology on $X$. Thus, we may find some $Vin V_{w}$ so that $x_{0}in Vsubset U$, and in particular, this means that for some $epsilon>0$, we have that
$$V_{epsilon}=Big{{}xin X text{ } Big{|}text{ } |langle x_{j}^{*},x_{0}-xrangle|<epsilon text{ for all } j=1,ldots,nBig{}}subset U$$The fact that $text{dim}(X)=infty$ is then required to find a nonzero $xiinbigcap_{j=1}^{n}text{ker}(x^{*}_{j})$, and then we may proceed as above.
add a comment |
up vote
0
down vote
Nevermind - I have answered my own questions.
$V_{w}=Big{{} bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] text{ }Big{|}text{ } x_{1}^{*},ldots, x_{n}^{*}in X^{*}Big{}}$ is a base for the weak topology on $X$. Thus, we may find some $Vin V_{w}$ so that $x_{0}in Vsubset U$, and in particular, this means that for some $epsilon>0$, we have that
$$V_{epsilon}=Big{{}xin X text{ } Big{|}text{ } |langle x_{j}^{*},x_{0}-xrangle|<epsilon text{ for all } j=1,ldots,nBig{}}subset U$$The fact that $text{dim}(X)=infty$ is then required to find a nonzero $xiinbigcap_{j=1}^{n}text{ker}(x^{*}_{j})$, and then we may proceed as above.
add a comment |
up vote
0
down vote
up vote
0
down vote
Nevermind - I have answered my own questions.
$V_{w}=Big{{} bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] text{ }Big{|}text{ } x_{1}^{*},ldots, x_{n}^{*}in X^{*}Big{}}$ is a base for the weak topology on $X$. Thus, we may find some $Vin V_{w}$ so that $x_{0}in Vsubset U$, and in particular, this means that for some $epsilon>0$, we have that
$$V_{epsilon}=Big{{}xin X text{ } Big{|}text{ } |langle x_{j}^{*},x_{0}-xrangle|<epsilon text{ for all } j=1,ldots,nBig{}}subset U$$The fact that $text{dim}(X)=infty$ is then required to find a nonzero $xiinbigcap_{j=1}^{n}text{ker}(x^{*}_{j})$, and then we may proceed as above.
Nevermind - I have answered my own questions.
$V_{w}=Big{{} bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] text{ }Big{|}text{ } x_{1}^{*},ldots, x_{n}^{*}in X^{*}Big{}}$ is a base for the weak topology on $X$. Thus, we may find some $Vin V_{w}$ so that $x_{0}in Vsubset U$, and in particular, this means that for some $epsilon>0$, we have that
$$V_{epsilon}=Big{{}xin X text{ } Big{|}text{ } |langle x_{j}^{*},x_{0}-xrangle|<epsilon text{ for all } j=1,ldots,nBig{}}subset U$$The fact that $text{dim}(X)=infty$ is then required to find a nonzero $xiinbigcap_{j=1}^{n}text{ker}(x^{*}_{j})$, and then we may proceed as above.
edited Nov 18 at 19:32
answered Nov 17 at 23:25
JWP_HTX
385213
385213
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