Open Mapping Theorem: counterexample
$begingroup$
The Open Mapping Theorem says that a linear continuous
surjection between Banach spaces is an open mapping.
I am writing some lecture notes on the Open Mapping Theorem.
I guess it would be nice to have some counterexamples.
After all, how can you appreciate it's meaning without a
nice counterexample showing how the conclusion could fail
and why the conclusion is not obvious at all.
Let $ell^1 subset mathbb{R}^infty$ be the set of sequences
$(a_1, a_2, dotsc)$, such that $sum |a_j| < infty$.
If we consider the $ell^1$ norm $|cdot|_1$ and
the supremum norm $|cdot|_s$, then,
$(ell^1, |cdot|_1)$ is complete,
while $(ell^1, |cdot|_s)$ is not complete.
In this case, the identity
$$
begin{array}{rrcl}
mathrm{id}:& (ell^1, |cdot|_1)& to &(ell^1, |cdot|_s)
\
& x & mapsto & x
end{array}
$$
is a continuous bijection but it is not open.
I want a counterexample in the opposite direction.
That is, I want a linear continuous bijection
$T: E to F$ between normed spaces $E$ and $F$
such that $F$ is Banach but $T$ is not open.
This is equivalent to finding a vector space
$E$ with non-equivalent norms $|cdot|_c$ and $|cdot|_n$,
such that $E$ is complete when considered the norm $|cdot|_c$,
and such that
$$
|cdot|_c
leq
|cdot|_n.
$$
The Open Mapping Theorem implies that $|cdot|_n$ is
not complete.
So, is anyone aware of such a counterexample?
functional-analysis examples-counterexamples normed-spaces
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
$endgroup$
add a comment |
$begingroup$
The Open Mapping Theorem says that a linear continuous
surjection between Banach spaces is an open mapping.
I am writing some lecture notes on the Open Mapping Theorem.
I guess it would be nice to have some counterexamples.
After all, how can you appreciate it's meaning without a
nice counterexample showing how the conclusion could fail
and why the conclusion is not obvious at all.
Let $ell^1 subset mathbb{R}^infty$ be the set of sequences
$(a_1, a_2, dotsc)$, such that $sum |a_j| < infty$.
If we consider the $ell^1$ norm $|cdot|_1$ and
the supremum norm $|cdot|_s$, then,
$(ell^1, |cdot|_1)$ is complete,
while $(ell^1, |cdot|_s)$ is not complete.
In this case, the identity
$$
begin{array}{rrcl}
mathrm{id}:& (ell^1, |cdot|_1)& to &(ell^1, |cdot|_s)
\
& x & mapsto & x
end{array}
$$
is a continuous bijection but it is not open.
I want a counterexample in the opposite direction.
That is, I want a linear continuous bijection
$T: E to F$ between normed spaces $E$ and $F$
such that $F$ is Banach but $T$ is not open.
This is equivalent to finding a vector space
$E$ with non-equivalent norms $|cdot|_c$ and $|cdot|_n$,
such that $E$ is complete when considered the norm $|cdot|_c$,
and such that
$$
|cdot|_c
leq
|cdot|_n.
$$
The Open Mapping Theorem implies that $|cdot|_n$ is
not complete.
So, is anyone aware of such a counterexample?
functional-analysis examples-counterexamples normed-spaces
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
$endgroup$
4
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03
add a comment |
$begingroup$
The Open Mapping Theorem says that a linear continuous
surjection between Banach spaces is an open mapping.
I am writing some lecture notes on the Open Mapping Theorem.
I guess it would be nice to have some counterexamples.
After all, how can you appreciate it's meaning without a
nice counterexample showing how the conclusion could fail
and why the conclusion is not obvious at all.
Let $ell^1 subset mathbb{R}^infty$ be the set of sequences
$(a_1, a_2, dotsc)$, such that $sum |a_j| < infty$.
If we consider the $ell^1$ norm $|cdot|_1$ and
the supremum norm $|cdot|_s$, then,
$(ell^1, |cdot|_1)$ is complete,
while $(ell^1, |cdot|_s)$ is not complete.
In this case, the identity
$$
begin{array}{rrcl}
mathrm{id}:& (ell^1, |cdot|_1)& to &(ell^1, |cdot|_s)
\
& x & mapsto & x
end{array}
$$
is a continuous bijection but it is not open.
I want a counterexample in the opposite direction.
That is, I want a linear continuous bijection
$T: E to F$ between normed spaces $E$ and $F$
such that $F$ is Banach but $T$ is not open.
This is equivalent to finding a vector space
$E$ with non-equivalent norms $|cdot|_c$ and $|cdot|_n$,
such that $E$ is complete when considered the norm $|cdot|_c$,
and such that
$$
|cdot|_c
leq
|cdot|_n.
$$
The Open Mapping Theorem implies that $|cdot|_n$ is
not complete.
So, is anyone aware of such a counterexample?
functional-analysis examples-counterexamples normed-spaces
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
$endgroup$
The Open Mapping Theorem says that a linear continuous
surjection between Banach spaces is an open mapping.
I am writing some lecture notes on the Open Mapping Theorem.
I guess it would be nice to have some counterexamples.
After all, how can you appreciate it's meaning without a
nice counterexample showing how the conclusion could fail
and why the conclusion is not obvious at all.
Let $ell^1 subset mathbb{R}^infty$ be the set of sequences
$(a_1, a_2, dotsc)$, such that $sum |a_j| < infty$.
If we consider the $ell^1$ norm $|cdot|_1$ and
the supremum norm $|cdot|_s$, then,
$(ell^1, |cdot|_1)$ is complete,
while $(ell^1, |cdot|_s)$ is not complete.
In this case, the identity
$$
begin{array}{rrcl}
mathrm{id}:& (ell^1, |cdot|_1)& to &(ell^1, |cdot|_s)
\
& x & mapsto & x
end{array}
$$
is a continuous bijection but it is not open.
I want a counterexample in the opposite direction.
That is, I want a linear continuous bijection
$T: E to F$ between normed spaces $E$ and $F$
such that $F$ is Banach but $T$ is not open.
This is equivalent to finding a vector space
$E$ with non-equivalent norms $|cdot|_c$ and $|cdot|_n$,
such that $E$ is complete when considered the norm $|cdot|_c$,
and such that
$$
|cdot|_c
leq
|cdot|_n.
$$
The Open Mapping Theorem implies that $|cdot|_n$ is
not complete.
So, is anyone aware of such a counterexample?
functional-analysis examples-counterexamples normed-spaces
functional-analysis examples-counterexamples normed-spaces
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
asked Jan 31 '13 at 21:12
André CaldasAndré Caldas
3,4171227
3,4171227
4
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03
add a comment |
4
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03
4
4
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03
add a comment |
1 Answer
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$begingroup$
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.
$endgroup$
add a comment |
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$begingroup$
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.
$endgroup$
add a comment |
$begingroup$
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.
$endgroup$
add a comment |
$begingroup$
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.
$endgroup$
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.
edited Apr 13 '17 at 12:58
community wiki
2 revs
Norbert
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4
$begingroup$
See this post on MO.
$endgroup$
– David Mitra
Jan 31 '13 at 21:22
$begingroup$
@DavidMitra: Thank you! I didn't think it would be on MO... Not only it was answered there... in the MO the question is much better written as well! ;-)
$endgroup$
– André Caldas
Jan 31 '13 at 21:32
$begingroup$
@AndréCaldas is there any hint why l1 with the sup norm is not complete?
$endgroup$
– Charles
Feb 17 '15 at 20:30
$begingroup$
@Charles: take any sequence $a_n in mathbb{R}$ with $sum a_n = infty$ and $a_n rightarrow 0$. The sequence $A_n = (a_1, dotsc a_n, 0, 0, 0, dotsc)$ belongs to $ell^1$. The sequence $A_n$ is Cauchy with the supremum norm since $a_n rightarrow 0$. It actually converges to $(a_1, a_2, dotsc) in ell^infty setminus ell^1$. In fact, the closure of $ell^1$ is $c_0 subset ell^infty$.
$endgroup$
– André Caldas
Feb 22 '15 at 20:03