TITLE: Functional analysis and linear homeomorphism
QUESTION [7 upvotes]: Here's a question in one of our exercises list : 
Let $E$ be a normed vector space, $F$ be a Banach space and $T : E \to F$ a continuous linear application. Define 
$$
E/\mathrm{Ker} (T) = \{ [x] \}
$$
where $[x]$ is the equivalence class of $x \in E$, i.e. $[x] = \{ x + y \, | \, y \in \mathrm{Ker} (T )\}$. This space is a normed vector space with the following norm : 
$$
\Vert[x]\Vert = \inf \{ \Vert y \Vert \, | \, y \in [x] \}. 
$$
Define $[T] : E / \mathrm{Ker}(T) \to \mathrm{Im}(T)$ by $[T]([x]) = T(x)$. Suppose that $\mathrm{Im}(T)$ is closed. Show that $[T]$ is an homeomorphism (i.e. its inverse is linear and continuous.
Now here's the deal ; this question turns out to be hard (and most probably false) because my teacher did a typo and hence forgot to mention that $E$ was supposed to be a Banach space for this question to work out. So I worked for a few days on it and only managed to show the following : 

$[T]$ is an homeomorphism if and only if $E/\mathrm{Ker}(T)$ is a Banach space. 
If $E/\mathrm{Ker}(T)$ is not Banach (i.e. not complete), there exists a Cauchy sequence $\{ [y_n] \}$ such that $\Vert [y_n] \Vert \to A > 0$ but $[T]([y_n]) \to 0$. 

Now here is my question. 
Can anyone find a counter example to show that this exercise is false (which is most probably the case), or prove it otherwise (which I have no hope in doing)?

REPLY [7 votes]: Take a discontinuous linear functional $\varphi \colon E \to \mathbb{R}$ on the infinite-dimensional Banach space $(E,\lVert \cdot \rVert_{E})$. Define a new norm $\lVert x \rVert_{\rm new} = \lVert x\rVert_E + \lvert \varphi(x)\rvert$ on $E$. Then the identity map $T \colon (E,\lVert \cdot \rVert_{\rm new}) \to (E, \lVert \cdot \rVert)$ provides a counterexample.
Observe that $(E,\lVert \cdot \rVert_{\rm new})$ can't be complete by the open mapping theorem: otherwise $T^{-1}$ would be continuous and hence $\varphi \circ T^{-1} = \varphi$ would be continuous on $(E,\lVert \cdot \rVert_{E})$.
To construct a discontinuous linear functional, choose a linearly independent sequence of unit vectors $(e_n)_{n \in \mathbb{N}}$. Set $\varphi(e_n) = n e_n$ and let $\varphi = 0$ on a complement of the linear span of $\{e_n\}_{n \in \mathbb{N}}$.