completeness and the open mapping theorem
$begingroup$
Let $X,Y$ be normed vector spaces, I want to show that the open mapping theorem requires completeness of both spaces. So my question consists of two parts:
$textit{i)}$ Let $X$ be a Banach space and $Y$ a normed space and find a bounded surjective linear operator which is not open.
$textit{ii)}$ Let $X$ be a normed space and $Y$ a Banach space. Find a surjective linear operator which is not open.
For $textit{i)}$, I chose $X=(ell^1(mathbb{N}),||cdot||_1)$ and $Y=(ell^1(mathbb{N},||cdot||_infty)$ and $T:Xto Y$, $xmapsto x$. This map is clearly surjective and bounded, but how can I show this is not open? I wanted to check the image of the open unit ball $B$ in $X$, so see whether $T(B)$ is open but I got stuck.
For $textit{ii)}$, could anyone give me a hint which spaces I can use? I have not yet learend about Hamel basis, but I am allowed to use that there exists an unbounded linear functional on every infinite dimensional normed space.
functional-analysis complete-spaces open-map
asked Jan 7 at 9:46
JamesJames
873318
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be normed vector spaces, I want to show that the open mapping theorem requires completeness of both spaces. So my question consists of two parts:
$textit{i)}$ Let $X$ be a Banach space and $Y$ a normed space and find a bounded surjective linear operator which is not open.
$textit{ii)}$ Let $X$ be a normed space and $Y$ a Banach space. Find a surjective linear operator which is not open.
For $textit{i)}$, I chose $X=(ell^1(mathbb{N}),||cdot||_1)$ and $Y=(ell^1(mathbb{N},||cdot||_infty)$ and $T:Xto Y$, $xmapsto x$. This map is clearly surjective and bounded, but how can I show this is not open? I wanted to check the image of the open unit ball $B$ in $X$, so see whether $T(B)$ is open but I got stuck.
For $textit{ii)}$, could anyone give me a hint which spaces I can use? I have not yet learend about Hamel basis, but I am allowed to use that there exists an unbounded linear functional on every infinite dimensional normed space.
functional-analysis complete-spaces open-map
asked Jan 7 at 9:46
JamesJames
873318
$endgroup$
$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32
add a comment |
$begingroup$
Let $X,Y$ be normed vector spaces, I want to show that the open mapping theorem requires completeness of both spaces. So my question consists of two parts:
$textit{i)}$ Let $X$ be a Banach space and $Y$ a normed space and find a bounded surjective linear operator which is not open.
$textit{ii)}$ Let $X$ be a normed space and $Y$ a Banach space. Find a surjective linear operator which is not open.
For $textit{i)}$, I chose $X=(ell^1(mathbb{N}),||cdot||_1)$ and $Y=(ell^1(mathbb{N},||cdot||_infty)$ and $T:Xto Y$, $xmapsto x$. This map is clearly surjective and bounded, but how can I show this is not open? I wanted to check the image of the open unit ball $B$ in $X$, so see whether $T(B)$ is open but I got stuck.
For $textit{ii)}$, could anyone give me a hint which spaces I can use? I have not yet learend about Hamel basis, but I am allowed to use that there exists an unbounded linear functional on every infinite dimensional normed space.
functional-analysis complete-spaces open-map
asked Jan 7 at 9:46
JamesJames
873318
$endgroup$
Let $X,Y$ be normed vector spaces, I want to show that the open mapping theorem requires completeness of both spaces. So my question consists of two parts:
$textit{i)}$ Let $X$ be a Banach space and $Y$ a normed space and find a bounded surjective linear operator which is not open.
$textit{ii)}$ Let $X$ be a normed space and $Y$ a Banach space. Find a surjective linear operator which is not open.
For $textit{i)}$, I chose $X=(ell^1(mathbb{N}),||cdot||_1)$ and $Y=(ell^1(mathbb{N},||cdot||_infty)$ and $T:Xto Y$, $xmapsto x$. This map is clearly surjective and bounded, but how can I show this is not open? I wanted to check the image of the open unit ball $B$ in $X$, so see whether $T(B)$ is open but I got stuck.
For $textit{ii)}$, could anyone give me a hint which spaces I can use? I have not yet learend about Hamel basis, but I am allowed to use that there exists an unbounded linear functional on every infinite dimensional normed space.
functional-analysis complete-spaces open-map
functional-analysis complete-spaces open-map
asked Jan 7 at 9:46
JamesJames
873318
asked Jan 7 at 9:46
JamesJames
873318
asked Jan 7 at 9:46
JamesJames
873318
asked Jan 7 at 9:46
JamesJames
873318
asked Jan 7 at 9:46
JamesJames
873318
873318
$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32
add a comment |
$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32
$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,jin Bbb N$ let $x_{n,j}=1$ for $jleq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{jin Bbb N}$. Then $|x(n)|_{infty}=1$ and $|T^{-1}(x(n))|_1=|x(n)|_1=n.$
(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)
For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Yto Bbb R$ be linear and unbounded. Let $X={(y,g(y)): yin Y}$ and let $|(y,g(y))|_X=|y|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$sup_{0ne yin Y}frac {|f^{-1}(y)|_X}{|y|_Y}=sup_{0ne yin Y}frac {|y|_Y+|g(y)|}{|y|_Y}=infty$$ because $g$ is unbounded.
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
|
show 3 more comments
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$begingroup$
For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,jin Bbb N$ let $x_{n,j}=1$ for $jleq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{jin Bbb N}$. Then $|x(n)|_{infty}=1$ and $|T^{-1}(x(n))|_1=|x(n)|_1=n.$
(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)
For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Yto Bbb R$ be linear and unbounded. Let $X={(y,g(y)): yin Y}$ and let $|(y,g(y))|_X=|y|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$sup_{0ne yin Y}frac {|f^{-1}(y)|_X}{|y|_Y}=sup_{0ne yin Y}frac {|y|_Y+|g(y)|}{|y|_Y}=infty$$ because $g$ is unbounded.
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
|
show 3 more comments
$begingroup$
For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,jin Bbb N$ let $x_{n,j}=1$ for $jleq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{jin Bbb N}$. Then $|x(n)|_{infty}=1$ and $|T^{-1}(x(n))|_1=|x(n)|_1=n.$
(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)
For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Yto Bbb R$ be linear and unbounded. Let $X={(y,g(y)): yin Y}$ and let $|(y,g(y))|_X=|y|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$sup_{0ne yin Y}frac {|f^{-1}(y)|_X}{|y|_Y}=sup_{0ne yin Y}frac {|y|_Y+|g(y)|}{|y|_Y}=infty$$ because $g$ is unbounded.
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
|
show 3 more comments
$begingroup$
For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,jin Bbb N$ let $x_{n,j}=1$ for $jleq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{jin Bbb N}$. Then $|x(n)|_{infty}=1$ and $|T^{-1}(x(n))|_1=|x(n)|_1=n.$
(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)
For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Yto Bbb R$ be linear and unbounded. Let $X={(y,g(y)): yin Y}$ and let $|(y,g(y))|_X=|y|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$sup_{0ne yin Y}frac {|f^{-1}(y)|_X}{|y|_Y}=sup_{0ne yin Y}frac {|y|_Y+|g(y)|}{|y|_Y}=infty$$ because $g$ is unbounded.
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
For your example for (i), $T$ is continuous because $T$ is bounded and linear. A continuous open bijection is a homeomorphism. So if $T$ is open then $T^{-1}$ is continuous, and linear, and therefore bounded.... But $T^{-1}$ is NOT bounded. E.g. for $n,jin Bbb N$ let $x_{n,j}=1$ for $jleq n$ and $x_{n,j}=0$ for $j>n$, and let $x(n)=(x_{n,j})_{jin Bbb N}$. Then $|x(n)|_{infty}=1$ and $|T^{-1}(x(n))|_1=|x(n)|_1=n.$
(Recall that any linear map from any normed linear space to another normed linear space is continuous iff it is bounded.)
For (ii), let $Y$ be an infinite-dimensional real Banach space and let $g:Yto Bbb R$ be linear and unbounded. Let $X={(y,g(y)): yin Y}$ and let $|(y,g(y))|_X=|y|_Y+|g(y)|.$ Now let $f((y,g(y))=y.$ If $f$ were open then $f^{-1}$ would be continuous and hence $f^{-1}$ would be bounded . But $$sup_{0ne yin Y}frac {|f^{-1}(y)|_X}{|y|_Y}=sup_{0ne yin Y}frac {|y|_Y+|g(y)|}{|y|_Y}=infty$$ because $g$ is unbounded.
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
edited Jan 7 at 15:26
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 7 at 14:33
DanielWainfleetDanielWainfleet
35.1k31648
35.1k31648
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
|
show 3 more comments
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
BTW...I saw an example in Amer. Math. Monthly of a complex normed linear space $X$ with a continuous linear bijection $h:Xto X$ such that $h^{-1}$ is discontinuous.
$endgroup$
– DanielWainfleet
Jan 7 at 15:19
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
The construction in part (ii) partly came from musing on the Closed Graph Theorem, applied to $g$.
$endgroup$
– DanielWainfleet
Jan 7 at 15:37
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
How do we know that $X$ is not complete?
$endgroup$
– James
Jan 7 at 18:24
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
$begingroup$
Or is it due to the unboundedness of $g$?
$endgroup$
– James
Jan 7 at 18:42
1
1
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
$begingroup$
Yes. First, $f^{-1}$ is a function because $f$ is a bijection. Second, $f$ is a continuous bijection, so if $f$ were open then $f^{-1}$ would be continuous. Third, if $f^{-1}$ were continuous then $f^{-1}$ would be bounded
$endgroup$
– DanielWainfleet
Jan 9 at 4:02
|
show 3 more comments
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$begingroup$
Your question is confusing. The way it is stated you cannot choose $X$ and $Y$; they are given to you.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 11:54
$begingroup$
Oh i am sorry. What I mean is choose $X$ and $Y$ and give a map $T$ toch is subjective bounded but not open
$endgroup$
– James
Jan 7 at 11:56
$begingroup$
Thanks for the hint in your last sentence, so I could add part (ii) to my answer.
$endgroup$
– DanielWainfleet
Jan 7 at 15:32