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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)?

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Hint: It is suffices to find a continuous and linear bijection from a normed space to a Banach space which isn't a homeomorphism. –  commenter Nov 20 '12 at 2:35
@commenter : Of course, I think I know that by know. I'm asking for an example. I couldn't find one. I didn't see many examples besides $\ell^p(K)$ and $L^p(K)$ with different norms. –  Patrick Da Silva Nov 20 '12 at 2:42
If you drop assumption that $F$ is complete, then result is true –  Norbert Nov 20 '12 at 8:17
@Norbert : Definitely not. Think about it. If you say it's true without a complete $F$, can you at least show it with a complete $F$? I'm actually asking for a proof. –  Patrick Da Silva Nov 20 '12 at 8:58
Of course you are right. For a linear homeomorphism cannot exist between a space $A$ which is Banach and a space $B$ which is not, because linear and continuous operators between normed spaces are automatically Lipschitz. In your example, the space $\text{Im}(T)$ is Banach while $E/\ker T$ needs not be. I'll think at a concrete example. –  Giuseppe Negro Nov 20 '12 at 11:40
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1 Answer 1

up vote 6 down vote accepted

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 $\{e_n\}_{n \in \mathbb{N}}$.

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Yes! That was precisely the kind of example I wanted to pop out : the identity map between a normed space and itself but with two different norms in the domain and image. I'm glad you chose this precise example! Thank you =D +1 & check –  Patrick Da Silva Nov 20 '12 at 12:27
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