Mixing
$X^*$ with $weak^*$-topology is first category in itself when $X$ is infinite dimensional Fréchet space
I am reading Rudin's Functional Analysis and stuck at the problem in exercise 11 on page 87 : Let $X$ be infinite dimensional Fréchet space , then $X^*$ with it's $weak^*$-topology is of first category in itself. Here is my attempt.
Consider the a basic open set at origin corresponding to ${x_1,...,x_n}subseteq X$ of $weak^*$-topology on $X^*$ given by ${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}$. Now let $yin X$ and consider a particular $weak^*$ nbd ${x^*in X^*: |x^*(y)|<1}$ which contains a basic nbd i.e. for some $x_1,x_2,...,x_nin X$ and $epsilon>0$ we have $${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}subseteq {x^*in X^*: |x^*(y)|<1}.$$ Now this implies that if $z^*in X^*$ and $|z^*(x_i)|=0, forall x_i$ with $i=1,2,...,n$ then $tz^* in {x^*in X^*: |x^*(y)|<1}$ for each $t$ in scalar field $Bbb K$ and this implies $|tz^*(y)|<1, forall tin Bbb Kimplies z^*=0$. Now for $xin X$ defining $f_x:X^*rightarrow Bbb K$ by $f_x(Lambda)=Lambda(x), forall Lambda in X^*$ we have $$ker(f_y)supseteq bigcap_{i=1}^n ker(f_{x_i})$$ so that $f_y$ can be written as linear combination of $f_{x_i},i=1,2,...,n$ and this implies $y$ can be written as linear combination of $x_1,x_2,...,x_n$.
Any suggestion or solution will help me . Thanks in advance.
general-topology functional-analysis operator-theory baire-category weak-topology
asked Nov 20 '18 at 20:24
Mathlover
1326
add a comment |
I am reading Rudin's Functional Analysis and stuck at the problem in exercise 11 on page 87 : Let $X$ be infinite dimensional Fréchet space , then $X^*$ with it's $weak^*$-topology is of first category in itself. Here is my attempt.
Consider the a basic open set at origin corresponding to ${x_1,...,x_n}subseteq X$ of $weak^*$-topology on $X^*$ given by ${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}$. Now let $yin X$ and consider a particular $weak^*$ nbd ${x^*in X^*: |x^*(y)|<1}$ which contains a basic nbd i.e. for some $x_1,x_2,...,x_nin X$ and $epsilon>0$ we have $${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}subseteq {x^*in X^*: |x^*(y)|<1}.$$ Now this implies that if $z^*in X^*$ and $|z^*(x_i)|=0, forall x_i$ with $i=1,2,...,n$ then $tz^* in {x^*in X^*: |x^*(y)|<1}$ for each $t$ in scalar field $Bbb K$ and this implies $|tz^*(y)|<1, forall tin Bbb Kimplies z^*=0$. Now for $xin X$ defining $f_x:X^*rightarrow Bbb K$ by $f_x(Lambda)=Lambda(x), forall Lambda in X^*$ we have $$ker(f_y)supseteq bigcap_{i=1}^n ker(f_{x_i})$$ so that $f_y$ can be written as linear combination of $f_{x_i},i=1,2,...,n$ and this implies $y$ can be written as linear combination of $x_1,x_2,...,x_n$.
Any suggestion or solution will help me . Thanks in advance.
general-topology functional-analysis operator-theory baire-category weak-topology
asked Nov 20 '18 at 20:24
Mathlover
1326
add a comment |
I am reading Rudin's Functional Analysis and stuck at the problem in exercise 11 on page 87 : Let $X$ be infinite dimensional Fréchet space , then $X^*$ with it's $weak^*$-topology is of first category in itself. Here is my attempt.
Consider the a basic open set at origin corresponding to ${x_1,...,x_n}subseteq X$ of $weak^*$-topology on $X^*$ given by ${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}$. Now let $yin X$ and consider a particular $weak^*$ nbd ${x^*in X^*: |x^*(y)|<1}$ which contains a basic nbd i.e. for some $x_1,x_2,...,x_nin X$ and $epsilon>0$ we have $${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}subseteq {x^*in X^*: |x^*(y)|<1}.$$ Now this implies that if $z^*in X^*$ and $|z^*(x_i)|=0, forall x_i$ with $i=1,2,...,n$ then $tz^* in {x^*in X^*: |x^*(y)|<1}$ for each $t$ in scalar field $Bbb K$ and this implies $|tz^*(y)|<1, forall tin Bbb Kimplies z^*=0$. Now for $xin X$ defining $f_x:X^*rightarrow Bbb K$ by $f_x(Lambda)=Lambda(x), forall Lambda in X^*$ we have $$ker(f_y)supseteq bigcap_{i=1}^n ker(f_{x_i})$$ so that $f_y$ can be written as linear combination of $f_{x_i},i=1,2,...,n$ and this implies $y$ can be written as linear combination of $x_1,x_2,...,x_n$.
Any suggestion or solution will help me . Thanks in advance.
general-topology functional-analysis operator-theory baire-category weak-topology
asked Nov 20 '18 at 20:24
Mathlover
1326
I am reading Rudin's Functional Analysis and stuck at the problem in exercise 11 on page 87 : Let $X$ be infinite dimensional Fréchet space , then $X^*$ with it's $weak^*$-topology is of first category in itself. Here is my attempt.
Consider the a basic open set at origin corresponding to ${x_1,...,x_n}subseteq X$ of $weak^*$-topology on $X^*$ given by ${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}$. Now let $yin X$ and consider a particular $weak^*$ nbd ${x^*in X^*: |x^*(y)|<1}$ which contains a basic nbd i.e. for some $x_1,x_2,...,x_nin X$ and $epsilon>0$ we have $${x^*in X^* : |x^*(x_i)|<epsilon, forall i=1,2,..,n}subseteq {x^*in X^*: |x^*(y)|<1}.$$ Now this implies that if $z^*in X^*$ and $|z^*(x_i)|=0, forall x_i$ with $i=1,2,...,n$ then $tz^* in {x^*in X^*: |x^*(y)|<1}$ for each $t$ in scalar field $Bbb K$ and this implies $|tz^*(y)|<1, forall tin Bbb Kimplies z^*=0$. Now for $xin X$ defining $f_x:X^*rightarrow Bbb K$ by $f_x(Lambda)=Lambda(x), forall Lambda in X^*$ we have $$ker(f_y)supseteq bigcap_{i=1}^n ker(f_{x_i})$$ so that $f_y$ can be written as linear combination of $f_{x_i},i=1,2,...,n$ and this implies $y$ can be written as linear combination of $x_1,x_2,...,x_n$.
Any suggestion or solution will help me . Thanks in advance.
general-topology functional-analysis operator-theory baire-category weak-topology
general-topology functional-analysis operator-theory baire-category weak-topology
asked Nov 20 '18 at 20:24
Mathlover
1326
asked Nov 20 '18 at 20:24
Mathlover
1326
asked Nov 20 '18 at 20:24
Mathlover
1326
asked Nov 20 '18 at 20:24
Mathlover
1326
asked Nov 20 '18 at 20:24
Mathlover
1326
1326
add a comment |
add a comment |
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Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^circ$ are weak$^*$-closed with $X^*=bigcup_{ninmathbb N} U_n^circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)
answered Nov 21 '18 at 8:40
Jochen
6,675922
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1 Answer
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Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^circ$ are weak$^*$-closed with $X^*=bigcup_{ninmathbb N} U_n^circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)
answered Nov 21 '18 at 8:40
Jochen
6,675922
add a comment |
Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^circ$ are weak$^*$-closed with $X^*=bigcup_{ninmathbb N} U_n^circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)
answered Nov 21 '18 at 8:40
Jochen
6,675922
add a comment |
Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^circ$ are weak$^*$-closed with $X^*=bigcup_{ninmathbb N} U_n^circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)
answered Nov 21 '18 at 8:40
Jochen
6,675922
Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^circ$ are weak$^*$-closed with $X^*=bigcup_{ninmathbb N} U_n^circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)
answered Nov 21 '18 at 8:40
Jochen
6,675922
answered Nov 21 '18 at 8:40
Jochen
6,675922
answered Nov 21 '18 at 8:40
Jochen
6,675922
answered Nov 21 '18 at 8:40
Jochen
6,675922
6,675922
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