Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
See this post on MO. – David Mitra Jan 31 '13 at 21:22
@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! ;-) – André Caldas Jan 31 '13 at 21:32
@AndréCaldas is there any hint why l1 with the sup norm is not complete? – user 123456 Feb 17 '15 at 20:30
@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 $c_0 \subset \ell^\infty$. – André Caldas Feb 22 '15 at 20:03
up vote 1 down vote accepted

Solution is given on MO.

Thanks to David Mitra who pointed out this in a comment.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.