Mixing
A subspace $F$ of $E$ is dense in $E$ $iff$ $F^o = {0}$
$begingroup$
Problem: Let $E$ is a normed space and $A subset E$. Define $$A^o := {y in
A': |y(x)| leq 1, forall x in A}.$$ in which, $A'$ is the dual
space of $A$ with $A' = L(A,mathbb{K})$ under norm $$||y|| :=
displaystylesup_{||x|| le 1} |y(x)|.$$
With this definition, show that
$A^o$ is convex, balanced, closed in $A'$.- If $A$ is convex, balanced then $overline{A} = (A^o)^o.$
- A subspace $F$ of $E$ is dense in $E$ $iff$ $F^o = {0}.$
My attempt:
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$.
$Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$.
So, $A^o$ is convex, balanced.
$forall {y_n} in A^o: y_n longrightarrow y_0 in A'Rightarrow y_0 in A^o$.
$forall x in A$ since $y_n longrightarrow y_0 Rightarrow y_n(x) longrightarrow y_0(x) Rightarrow |y_n(x)| longrightarrow |y_0(x)|.$
Since $y_n in A^o$ then $|y_n (x)| leq 1, forall n Rightarrow |y_0 (x)| le 1, forall x in A Rightarrow y_0 in A^o$.
So, $A^o$ is closed.
- We'll show that $overline{A} subset A^{oo}$ and $A^{oo} subset overline{A}$.
Choose $x in A Rightarrow |y(x)| le 1, forall y in A^o Rightarrow x in (A^o)^o$.
$$Rightarrow sup_{y in A^o} |x(y)| = sup_{y in A^o} |y(x)| le 1.$$
$Rightarrow A subset A^{oo} Rightarrow overline{A} subset A^{oo}$ since $overline{A}$ is the smallest closed set including $A$.
$forall xi in E setminus overline{A}$, exists a $epsilon > 0 : B(xi,2 epsilon) cap overline{A} = emptyset$.
$Rightarrow exists y^* in E' : |y(x)|<1, forall x in A.$
$Rightarrow y^* in A^o.$
Is that right? How to show that the finally cases? Thank all!
functional-analysis proof-verification
asked Jan 15 at 15:42
MinhMinh
1959
$endgroup$
add a comment |
$begingroup$
Problem: Let $E$ is a normed space and $A subset E$. Define $$A^o := {y in
A': |y(x)| leq 1, forall x in A}.$$ in which, $A'$ is the dual
space of $A$ with $A' = L(A,mathbb{K})$ under norm $$||y|| :=
displaystylesup_{||x|| le 1} |y(x)|.$$
With this definition, show that
$A^o$ is convex, balanced, closed in $A'$.- If $A$ is convex, balanced then $overline{A} = (A^o)^o.$
- A subspace $F$ of $E$ is dense in $E$ $iff$ $F^o = {0}.$
My attempt:
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$.
$Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$.
So, $A^o$ is convex, balanced.
$forall {y_n} in A^o: y_n longrightarrow y_0 in A'Rightarrow y_0 in A^o$.
$forall x in A$ since $y_n longrightarrow y_0 Rightarrow y_n(x) longrightarrow y_0(x) Rightarrow |y_n(x)| longrightarrow |y_0(x)|.$
Since $y_n in A^o$ then $|y_n (x)| leq 1, forall n Rightarrow |y_0 (x)| le 1, forall x in A Rightarrow y_0 in A^o$.
So, $A^o$ is closed.
- We'll show that $overline{A} subset A^{oo}$ and $A^{oo} subset overline{A}$.
Choose $x in A Rightarrow |y(x)| le 1, forall y in A^o Rightarrow x in (A^o)^o$.
$$Rightarrow sup_{y in A^o} |x(y)| = sup_{y in A^o} |y(x)| le 1.$$
$Rightarrow A subset A^{oo} Rightarrow overline{A} subset A^{oo}$ since $overline{A}$ is the smallest closed set including $A$.
$forall xi in E setminus overline{A}$, exists a $epsilon > 0 : B(xi,2 epsilon) cap overline{A} = emptyset$.
$Rightarrow exists y^* in E' : |y(x)|<1, forall x in A.$
$Rightarrow y^* in A^o.$
Is that right? How to show that the finally cases? Thank all!
functional-analysis proof-verification
asked Jan 15 at 15:42
MinhMinh
1959
$endgroup$
add a comment |
$begingroup$
Problem: Let $E$ is a normed space and $A subset E$. Define $$A^o := {y in
A': |y(x)| leq 1, forall x in A}.$$ in which, $A'$ is the dual
space of $A$ with $A' = L(A,mathbb{K})$ under norm $$||y|| :=
displaystylesup_{||x|| le 1} |y(x)|.$$
With this definition, show that
$A^o$ is convex, balanced, closed in $A'$.- If $A$ is convex, balanced then $overline{A} = (A^o)^o.$
- A subspace $F$ of $E$ is dense in $E$ $iff$ $F^o = {0}.$
My attempt:
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$.
$Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$.
So, $A^o$ is convex, balanced.
$forall {y_n} in A^o: y_n longrightarrow y_0 in A'Rightarrow y_0 in A^o$.
$forall x in A$ since $y_n longrightarrow y_0 Rightarrow y_n(x) longrightarrow y_0(x) Rightarrow |y_n(x)| longrightarrow |y_0(x)|.$
Since $y_n in A^o$ then $|y_n (x)| leq 1, forall n Rightarrow |y_0 (x)| le 1, forall x in A Rightarrow y_0 in A^o$.
So, $A^o$ is closed.
- We'll show that $overline{A} subset A^{oo}$ and $A^{oo} subset overline{A}$.
Choose $x in A Rightarrow |y(x)| le 1, forall y in A^o Rightarrow x in (A^o)^o$.
$$Rightarrow sup_{y in A^o} |x(y)| = sup_{y in A^o} |y(x)| le 1.$$
$Rightarrow A subset A^{oo} Rightarrow overline{A} subset A^{oo}$ since $overline{A}$ is the smallest closed set including $A$.
$forall xi in E setminus overline{A}$, exists a $epsilon > 0 : B(xi,2 epsilon) cap overline{A} = emptyset$.
$Rightarrow exists y^* in E' : |y(x)|<1, forall x in A.$
$Rightarrow y^* in A^o.$
Is that right? How to show that the finally cases? Thank all!
functional-analysis proof-verification
asked Jan 15 at 15:42
MinhMinh
1959
$endgroup$
Problem: Let $E$ is a normed space and $A subset E$. Define $$A^o := {y in
A': |y(x)| leq 1, forall x in A}.$$ in which, $A'$ is the dual
space of $A$ with $A' = L(A,mathbb{K})$ under norm $$||y|| :=
displaystylesup_{||x|| le 1} |y(x)|.$$
With this definition, show that
$A^o$ is convex, balanced, closed in $A'$.- If $A$ is convex, balanced then $overline{A} = (A^o)^o.$
- A subspace $F$ of $E$ is dense in $E$ $iff$ $F^o = {0}.$
My attempt:
$forall y_1,y_2 in A^o; forall alpha,beta in mathbb{K}, |alpha| + |beta| leq 1 Rightarrow alpha y_1 + beta y_2 in A$ i.e $forall x in A, |(alpha y_1 + beta y_2)(x)| = |alpha y_1 (x) + beta y_2 (x)| le |alpha||y_1 (x)| + |beta||y_2 (x)| le |alpha|.1 + |beta|.1 le 1$.
$Rightarrow sup_{x in A} (alpha y_1 + beta y_2) le 1 Rightarrow alpha y_1 + beta y_2 in A^o$.
So, $A^o$ is convex, balanced.
$forall {y_n} in A^o: y_n longrightarrow y_0 in A'Rightarrow y_0 in A^o$.
$forall x in A$ since $y_n longrightarrow y_0 Rightarrow y_n(x) longrightarrow y_0(x) Rightarrow |y_n(x)| longrightarrow |y_0(x)|.$
Since $y_n in A^o$ then $|y_n (x)| leq 1, forall n Rightarrow |y_0 (x)| le 1, forall x in A Rightarrow y_0 in A^o$.
So, $A^o$ is closed.
- We'll show that $overline{A} subset A^{oo}$ and $A^{oo} subset overline{A}$.
Choose $x in A Rightarrow |y(x)| le 1, forall y in A^o Rightarrow x in (A^o)^o$.
$$Rightarrow sup_{y in A^o} |x(y)| = sup_{y in A^o} |y(x)| le 1.$$
$Rightarrow A subset A^{oo} Rightarrow overline{A} subset A^{oo}$ since $overline{A}$ is the smallest closed set including $A$.
$forall xi in E setminus overline{A}$, exists a $epsilon > 0 : B(xi,2 epsilon) cap overline{A} = emptyset$.
$Rightarrow exists y^* in E' : |y(x)|<1, forall x in A.$
$Rightarrow y^* in A^o.$
Is that right? How to show that the finally cases? Thank all!
functional-analysis proof-verification
functional-analysis proof-verification
asked Jan 15 at 15:42
MinhMinh
1959
asked Jan 15 at 15:42
MinhMinh
1959
edited Jan 16 at 5:46
Minh
asked Jan 15 at 15:42
MinhMinh
1959
asked Jan 15 at 15:42
MinhMinh
1959
asked Jan 15 at 15:42
MinhMinh
1959
1959
add a comment |
add a comment |
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