Example of a linear onto map which is not open

Question

We know that every linear open map between normed spaces is onto. This fact actually motivates the Open Mapping theorem which gives extra assumptions for converse to hold true. But I am unable to construct counterexample for the first fact, i.e. I am looking for a linear map between normed spaces which is onto but not open. Any hint for such map. Thanks.

Answer 1

Consider the identity $$I : (\ell^1, \|\cdot\|_1) \to (\ell^1, \|\cdot\|_\infty)$$ which is a continuous bijection since $\|\cdot\|_\infty \le \|\cdot\|_1$. However, it is not open.

Indeed, if $I$ were open, it would imply that $I^{-1}$ is bounded i.e. that $\|\cdot\|_1$ is bounded by $\|\cdot\|_\infty$, which is false.

Answer 2

Consider $\mathfrak{c}_{00} = \{\text{sequences } f: \mathbb{N} \to \mathbb{R} \text{ with } f(n) \neq 0 \text{ for finitely many } $n$ \}$, normed by supremum. A Hamel basis for $\mathfrak{c}_{00}$ is given by the singleton sequences $f_n : m \to \delta_{mn}$ for $n \in \mathbb{N}$.

Then consider the map $g: \mathfrak{c}_{00} \to \mathfrak{c}_{00}$ given by $g(f)(n) = \frac{1}{n} f(n)$. Then $g$ is linear, $||g|| = 1$ (and hence $g$ is continuous), and $g$ is onto as for any sequence $f: \mathbb{N} \to \mathbb{R}$, the sequence $\tilde{f} : n \to nf(n)$ is still an element of $\mathfrak{c}_{00}$ and satisfies $g(\tilde{f}) = f$. However, $g$ is not open, because for any $f$ with $||f|| < 1$, $$g(f)(n) = \frac{1}{n} f(n) < 1/n.$$ So $g(B(0,1))$ contains no neighborhood of $0$.

It's worth mentioning here that we can see exactly why this specific example fails in the completion of $\mathfrak{c}_{00}$. The completion of $\mathfrak{c}_0$ has an explicit description given by $\mathfrak{c}_0 = \{\text{sequences } f: \mathbb{N} \to \mathbb{R} \text{ so } \lim_{n \to \infty}f(n) = 0\}.$ If we define $g$ as above on $\mathfrak{c}_{0}$, $g$ is not onto because $(1, \frac{1}{2}, \frac{1}{3}, ...) \in \mathfrak{c}_0$, but if it were to happen that $g(f)(n) = \frac{1}{n}$ for all $n$, then $f(n)$ would need to be $1$ for all $n$. This is impossible for $f \in \mathfrak{c}_0$.

Answer 3

Let $V$ be the set of all $f:\Bbb N\to \Bbb R$ such that $\{n\in \Bbb N:f(n)\ne 0\}$ is finite.

For $f,g\in V$ let $\|f-g\|=\max_{n\in \Bbb N}|f(n)-g(n)|. $

And (as usual) for $f,g\in V$ and $n\in \Bbb N$ and $r\in \Bbb R$ let $(f+g)(n)=f(n)+g(n)$ and $(rf)(n)=r\cdot f(n).$

Let $M:V\to V$ where $(M(f))(n)=f(n)/n^2$ for $f\in V$ and $n\in \Bbb N.$

$M$ is linear ( & continuous ). $M$ is also a bijection, so if $M$ was an open map then $M^{-1}$ would be continuous. But $M^{-1}:V\to V$ is linear but unbounded so $M^{-1}$ cannot be continuous.