A closed convex set in a complex Banach space.
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Let $X$ be a Banach space over the field of complex numbers. Let $K$ be a subset in $X$ which contains all linear combinations $Z_1x+Z_2y$ for all $Z_1,Z_2in mathbb C$ with $Re(Z_1)geq0$ & $Re(Z_2)geq0$ and for all $x,y in K$.
I want to prove or disprove that $K$ is closed convex in $X$. I have proved $K$ is convex, but I am stuck to prove or disprove that $K$ is closed.
functional-analysis banach-spaces normed-spaces
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Let $X$ be a Banach space over the field of complex numbers. Let $K$ be a subset in $X$ which contains all linear combinations $Z_1x+Z_2y$ for all $Z_1,Z_2in mathbb C$ with $Re(Z_1)geq0$ & $Re(Z_2)geq0$ and for all $x,y in K$.
I want to prove or disprove that $K$ is closed convex in $X$. I have proved $K$ is convex, but I am stuck to prove or disprove that $K$ is closed.
functional-analysis banach-spaces normed-spaces
Think about an open angle or cone..
– Berci
Nov 21 at 11:23
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
Let $X$ be a Banach space over the field of complex numbers. Let $K$ be a subset in $X$ which contains all linear combinations $Z_1x+Z_2y$ for all $Z_1,Z_2in mathbb C$ with $Re(Z_1)geq0$ & $Re(Z_2)geq0$ and for all $x,y in K$.
I want to prove or disprove that $K$ is closed convex in $X$. I have proved $K$ is convex, but I am stuck to prove or disprove that $K$ is closed.
functional-analysis banach-spaces normed-spaces
Let $X$ be a Banach space over the field of complex numbers. Let $K$ be a subset in $X$ which contains all linear combinations $Z_1x+Z_2y$ for all $Z_1,Z_2in mathbb C$ with $Re(Z_1)geq0$ & $Re(Z_2)geq0$ and for all $x,y in K$.
I want to prove or disprove that $K$ is closed convex in $X$. I have proved $K$ is convex, but I am stuck to prove or disprove that $K$ is closed.
functional-analysis banach-spaces normed-spaces
functional-analysis banach-spaces normed-spaces
asked Nov 21 at 11:03
Infinity
602314
602314
Think about an open angle or cone..
– Berci
Nov 21 at 11:23
add a comment |
Think about an open angle or cone..
– Berci
Nov 21 at 11:23
Think about an open angle or cone..
– Berci
Nov 21 at 11:23
Think about an open angle or cone..
– Berci
Nov 21 at 11:23
add a comment |
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Not every linear subspace of a Banach space is closed so the answer is NO. [ Eg. $X=l^{1}$, $K$ is the space of all sequences with only finite number of non-zero entries].
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Not every linear subspace of a Banach space is closed so the answer is NO. [ Eg. $X=l^{1}$, $K$ is the space of all sequences with only finite number of non-zero entries].
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
add a comment |
up vote
2
down vote
Not every linear subspace of a Banach space is closed so the answer is NO. [ Eg. $X=l^{1}$, $K$ is the space of all sequences with only finite number of non-zero entries].
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
add a comment |
up vote
2
down vote
up vote
2
down vote
Not every linear subspace of a Banach space is closed so the answer is NO. [ Eg. $X=l^{1}$, $K$ is the space of all sequences with only finite number of non-zero entries].
Not every linear subspace of a Banach space is closed so the answer is NO. [ Eg. $X=l^{1}$, $K$ is the space of all sequences with only finite number of non-zero entries].
answered Nov 21 at 12:05
Kavi Rama Murthy
47.7k31854
47.7k31854
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
add a comment |
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
But K in my question is not a subspace.
– Infinity
Nov 21 at 13:56
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
@Infinity Any linear subspace satisfies your conditions on $K$.
– Kavi Rama Murthy
Nov 21 at 23:09
add a comment |
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Think about an open angle or cone..
– Berci
Nov 21 at 11:23