Open Mapping Theorem fails when space is not complete
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I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely
For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?
functional-analysis complete-spaces open-map
asked Jan 8 at 17:34
JamesJames
1,002320
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add a comment |
$begingroup$
I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely
For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?
functional-analysis complete-spaces open-map
asked Jan 8 at 17:34
JamesJames
1,002320
$endgroup$
add a comment |
$begingroup$
I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely
For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?
functional-analysis complete-spaces open-map
asked Jan 8 at 17:34
JamesJames
1,002320
$endgroup$
I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $X$ is Banach, $Y$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely
For $X$ a normed space, $Y$ Banach, find a surjective bounded linear operator $T:Xto Y$ which is not open. I wanted to let $Y$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $Y$. What can I take for $X$, which map can I use and how can I show this map is not open?
functional-analysis complete-spaces open-map
functional-analysis complete-spaces open-map
asked Jan 8 at 17:34
JamesJames
1,002320
asked Jan 8 at 17:34
JamesJames
1,002320
asked Jan 8 at 17:34
JamesJames
1,002320
asked Jan 8 at 17:34
JamesJames
1,002320
asked Jan 8 at 17:34
JamesJames
1,002320
1,002320
add a comment |
add a comment |
1 Answer
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I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
|T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
|T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
answered Jan 8 at 18:10
JamesJames
1,002320
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
|T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
|T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
answered Jan 8 at 18:10
JamesJames
1,002320
$endgroup$
add a comment |
$begingroup$
I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
|T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
|T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
answered Jan 8 at 18:10
JamesJames
1,002320
$endgroup$
add a comment |
$begingroup$
I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
|T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
|T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
answered Jan 8 at 18:10
JamesJames
1,002320
$endgroup$
I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||cdot||_Y)$ be an infinite dimensional Banach space and let $g:Ytomathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||cdot||_g||)$ where $||y||_g:=||y||_Y+|g(y)|$ for $yin Y$. One can easily check that $||cdot||_g$ defines indeed a norm. Let $T:Xto Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
|T|_{op}=sup{||Tx||_Y,big|,xin Y,||x||_gleq 1}=sup{||x||_Y,big|,xin Y,||x||_gleq 1}leqsup{||x||_Y+|g(x)|,big|,xin Y ||x||_gleq 1}=sup{||x||_g,big|,xin Y ,||x||_gleq 1}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
|T^{-1}|_{op}=sup{||T^{-1}(x)||_Y,big|,||y||_Yleq 1}=sup{||y||_Y+|g(y)|,big|,||y||_Yleq 1}=1+sup{|g(y)|,big|,||y||_Y}=1+|g|_{op}=infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
answered Jan 8 at 18:10
JamesJames
1,002320
answered Jan 8 at 18:10
JamesJames
1,002320
answered Jan 8 at 18:10
JamesJames
1,002320
answered Jan 8 at 18:10
JamesJames
1,002320
1,002320
add a comment |
add a comment |
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