# Open Mapping Theorem: counterexample

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?

• 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? – User43029 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