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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.2012-11-20
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    @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.2012-11-20
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    If you drop assumption that $F$ is complete, then result is true2012-11-20
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    @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.2012-11-20
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    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.2012-11-20
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    @Giuseppe : Thanks for the heads up, but the faith part was really pretty much taken care of ; I asked the question mostly because I wanted an example to stop thinking about this problem.2012-11-20
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    @PatrickDaSilva Yes I was mistaken. In fact $[T]$ is a homeomorphism on its image if and only if $T$ is an open mapping on its image.2012-11-21
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    @Norbert : Don't try to just vaguely use your memory of your course here ; the open mapping theorem works when both the domain and range are Banach spaces. Therefore continuity in one direction is just equivalent to being an open mapping in the other direction, i.e. if $T : E \to F$ is bijective, it is continuous iff $T^{-1} : F \to E$ is an open mapping, and $T^{-1}$ is continuous iff $T$ is an open mapping.2012-11-21
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    @PatrickDaSilva I didn't mentioned use of open mapping theorem, I just reuired $T$ to be open map on its image. The proof of the fact I stated in my last comment can be made without open mapping theorem.2012-11-21
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    @Norbert : Sorry, I misread your comment. You're right. Thanks for explaining yourself.2012-11-22

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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 & check2012-11-20