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Open Mapping Theorem

Xedi

Posted: Submitted by Xedi on 1 July 2008 - 11:50pm.

Joined: 2007-10-03
Posts: 397

I'm reading a proof of the Open Mapping Theorem, and it says at the very beginning :

$\displaystyle \text{Let } A : E \rightarrow F \text{ be a linear, continuous, surjective mapping of Banach spaces.}$

To prove that A is an open mapping (ie that open sets are mapped to open sets), it suffices to show that :

$\displaystyle A\left(\overline{B_1\left(0\right)}\right) \text{contains an open ball centered at 0 in F.}$

(where $B_1(0)$ is the unit open ball (in $E$ ) and $\overline{X}$ is the closure of $X$ )

I don't see that at all, where does that come from ?

Newtonswig

Posted: 2 July 2008 - 1:24am

Joined: 2006-10-01
Posts: 432

That's just the magic of linearity- by adding vectors and multiplying by scalars, we can get any closed ball we want from the contents of the original one. The process of translation/dilation by a fixed (non-zero) scalar will not affect the topology.

On the other hand, I fear your assumptions may be a little off- try the natural inclusion $\mathbb{R}^2 \rightarrow \mathbb{R}^3$ ... I think you may mean surjective..

Xedi

Posted: 2 July 2008 - 1:25am

Joined: 2007-10-03
Posts: 397

Ah, I figured it out.

Suppose that r is small enough so that $B_r(0) \subset A\left(\overline{B_1(0)}\right)$

Then $B_{tr}(0) \subset A\left(\overline{B_t(0)}\right)$ by linearity.

For any open set $U$ and any $x \in U$ , take t such that $B_t(x) \subset U$ .

We then have that $f\left(B_{tr}(x)\right) \subset B_t(x)$ , and this proves that f is open.

Not totally sure why we're using the closed unit ball at the beginning though, maybe just for convenience in the rest of the proof, I suppose any neighborhood of the unit ball would've done ?

Edit : Ah, I see you answered Tom, yes indeed surjective is the correct assumption to make. And that example with the natural inclusion made me realise why the theorem is true, thanks.

Colin

Posted: 2 July 2008 - 7:55am

Joined: 2007-10-17
Posts: 109

Oh it's so nice to know that you're still there for us Tom, for the time being at least. Will you have much time left for us when you're off doing guage theory etc..?