18
$\begingroup$

Let $\Omega$ be an open, bounded from $\mathbb{R}^n$ and $f: \overline{\Omega} \rightarrow \overline{\Omega}$ a contiuous function such that $f(x)=x, \forall x \in \partial \Omega$.

Prove that $f(\overline{\Omega})=\overline{\Omega}$.

how to solve it ? any idea please ?

$\endgroup$
9
  • 1
    $\begingroup$ Did you mean that $\Omega\subset\mathbb{R}$? $\endgroup$
    – Marra
    Feb 23, 2013 at 14:44
  • 1
    $\begingroup$ $\Omega \subset \mathbb{R}^n$ sorry $\endgroup$
    – Vrouvrou
    Feb 23, 2013 at 14:57
  • $\begingroup$ If we divide $\overline{\Omega}$ in it's connected components, I think we just have to show that if $U$ is an component, then $f(int(U))= U$ where $int$ is interior. Is this true? $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:18
  • $\begingroup$ @Tomás Yes, but you probably want to look at the components of $\Omega$. Otherwise $int(U)$ doesn't need to be connected. I think one can show fairly easily that it suffices to prove the result for a connected set. $\endgroup$
    – JSchlather
    Feb 23, 2013 at 15:50
  • $\begingroup$ For the sufficiency it is enough to prove that $f(int (U))\subset U$. $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:59

3 Answers 3

13
+100
$\begingroup$

The problem can be solved using degree theory.

Suppose otherwise, that $f(\overline{\Omega})\subset\overline{\Omega}$ is strict. Since $f(\partial\Omega)=\partial{\Omega}\subset f(\overline{\Omega})$, then there is some $p\in\Omega$ such that $p\notin f(\Omega)$.

Since $p\notin\partial\Omega=f(\partial\Omega)=\mathbb{I}(\partial\Omega)$, then $\text{deg}(f,\Omega,p)=\text{deg}(\mathbb{I},\Omega,p)=1$ by the Poincare-Bohl theorem.

But the basic properties of degree give that $\text{deg}(f,\Omega,p)\neq0\Longrightarrow\exists x\in\Omega$ such that $f(x)=p$, so we get a contradiction.

$\endgroup$
4
  • $\begingroup$ What happens if $p\in f(\bar\Omega)\setminus\bar\Omega$ instead? $\endgroup$
    – Asaf Karagila
    Mar 2, 2013 at 2:41
  • $\begingroup$ $f(\overline{\Omega})\backslash\overline{\Omega}=\emptyset$ since $f:\overline{\Omega}\longrightarrow\overline{\Omega}$. $\endgroup$
    – mdg
    Mar 2, 2013 at 2:48
  • $\begingroup$ D'oh, thanks! :-) $\endgroup$
    – Asaf Karagila
    Mar 2, 2013 at 2:51
  • 1
    $\begingroup$ The various techniques which constitute the area of nonlinear (functional) analysis are all equally impressive and useful, and I believe deserve more attention and study than they get...! $\endgroup$
    – mdg
    Mar 2, 2013 at 20:09
1
$\begingroup$

Note that $\Omega$ can be written as a countable union of disjoint open intervals, i.e. $\Omega=\cup I_n$, where $I_n=(a_n,b_n)$. Then $f(a_n)=a_n$ and $f(b_n)=b_n$ and because $f$ is continous, $f$ satisfies the intermediate value property, hence, for each $u\in [a_n,b_n]$ with $f(a_n)\leq u\leq f(b_n)$, you can find $v\in [a_n,b_n]$ such that $f(v)=u$. This implies that $\overline{\Omega}\subset f(\overline{\Omega})$

Edit: As Ayman pointed out, I just proved that $\overline{\Omega}\subset f(\overline{\Omega})$ and because I think this might be helpful to someone, I will not delete the answer.

Remark: When I answered the question, $\Omega$ was a subset of $\mathbb{R}$.

$\endgroup$
16
  • $\begingroup$ The question says $f : \overline{\Omega} \to \overline{\Omega}$. So equality holds after all. $\endgroup$ Feb 23, 2013 at 15:00
  • $\begingroup$ All you needed to prove was that $\overline{\Omega}\subseteq f\left(\overline{\Omega}\right),$ since we have $f:\overline{\Omega}\to\overline{\Omega}.$ However, the question has been edited so that $\Omega\subseteq\Bbb R^n$, so you'll need to take a different approach. $\endgroup$ Feb 23, 2013 at 15:02
  • $\begingroup$ It is true, but now I have to change it. $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:04
  • $\begingroup$ so you say that we have only $\overline{\Omega}\subset f(\overline{\Omega})$ ? $\endgroup$
    – Vrouvrou
    Feb 23, 2013 at 16:15
  • 1
    $\begingroup$ @KarimaMht You have $f(\overline{\Omega})\subseteq \overline{\Omega}$ by assumption. $\endgroup$
    – Julien
    Feb 23, 2013 at 16:21
0
$\begingroup$

Wrong Solution

If $\Omega = \emptyset$, then we are done.

The plan for nonempty $\Omega$ (as hinted at by JSchlather and Tomás) is to show that $f(\overline{U}) = \overline{U}$ for any connected component $U$ of $\Omega$. The result follows from this, since the connected components exhaust $\Omega$.

We already know $f(\overline{U}) \subset \overline{\Omega}$.

Now, $\overline{U}$ is connected (since $U$ is connected), so $f(\overline{U})$ is connected, since the continuous image of a connected set is connected.

Consequently, $f(\overline{U})$ lies entirely within one connected component of $\Omega$.

Since $f(x) = x$ on $\partial U \subset \overline{U}$, it must be that $f(\overline{U}) \subseteq \overline{U}$.

Therefore, all we need to show now is that $f(\overline{U}) \supseteq \overline{U}$.

Let $B \supset \overline{U}$ be a closed ball neighborhood of $\overline{U}$. Extend $f$ to a function $F: B \to B$ by the identity map on $B\smallsetminus\overline{U}$. Then $F$ fixes the boundary of $B$ (in addition to lots of other elements of $B$), and is continuous. Also $F$ takes $B\smallsetminus\overline{U}$ to itself, and $\overline{U}$ to itself. So it will suffice to show that $F$ is surjective for any map $F: B \to B$ from a ball to itself fixing the boundary.

Let $N_1 \supseteq N_2 \supseteq \cdots$ be a decreasing sequence of closed balls whose intersection is $x$. Let $K_i = F^{-1}(N_i)$. Then $K_1 \supseteq K_2 \supseteq \cdots$ is a decreasing sequence of closed sets such that $F^{-1}(\mbox{int }N_i) \subset K_i$; in particular, $K_i$ is nonempty. (The statement in bold is wrong, as pointed out by @G. Smith.)

Since $B$ is compact, the intersection of all the $K_i$ is nonempty; call it $\mathcal{K}$. Then $f(\mathcal{K}) \subseteq N_i$ for all $N_i$. Therefore $f(\mathcal{K}) = x$.

Now, $x$ was arbitrary, so $f$ is surjective, and we are done.

$\endgroup$
3
  • $\begingroup$ This is not correct because $K_i$ may be empty. Since $U$ is open and you are assuming $x\in U$, then eventually $\text{int}N_i\subset U$. But there is no guarantee that $F^{-1}(\text{int}N_i)\neq\emptyset$, which is exactly what we are trying to prove, i.e. there may be no points in $U$ that map to $\text{int}N_i$. $\endgroup$
    – mdg
    Mar 2, 2013 at 7:36
  • 1
    $\begingroup$ -blush- Well, I suppose I ought to leave this nonanswer up as a warning to those attempting a solution without using more heavy machinery. It's so tempting. $\endgroup$ Mar 6, 2013 at 20:17
  • 1
    $\begingroup$ It would be nice to be able to prove such a simple statement directly. Using degree theory seems to be shooting an ant with an atom bomb... $\endgroup$
    – mdg
    Mar 6, 2013 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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