Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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 –  no identity 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
1  
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

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.