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?
Best Answer
Solution is given on MO.
Thanks to David Mitra who pointed out this in a comment.