What are the conditions to apply Brouwer fixed-point theorem (one dimentional case)?

Question

Does this theorem work for a continuous function $f : ]a,b[ \rightarrow ]a,b[$ with $a, b \in \mathbb{\bar{R}}$ ?

Thanks in advance.

Answer 1

No, but it does hold for a continuous function $f:[a,b]\to [a,b]$ Here's the proof for the special case when $[a,b]=[0,1]$

Case 1: $f(0)=0$. In which case, done, that's our fixed point. Case 2: $f(1)=1$. Again, we're done, that's our fixed point.

Case 3: $f(0)>0,f(1)<1$. Then define a new function $g(x)=f(x)-x$ on $[0,1]$. We have $g(0)=f(0)-0>0$, $g(1)=f(1)-1<0$, so by the intermediate value theorem, for some point $c\in (0,1)$, $g(c)=f(c)-c=0$, hence $f(c)=c$.

The generalization to $[a,b]$ comes from either redoing the g function or the ultra simple proof that "All continuous functions on the topological space X have a fixed point" is a topological property, and $[a,b]$ is homeomorphic to $[0,1]$.

Answer 2

The set is not compact, so we can't assume that Brouwer's fixed point theorem will hold, as the assumptions are not completely satisfied.

Answer 3

No, the theorem does not hold. For example, define $f:(0,1)\to(0,1)$ to be $f(x) = x^2$. This function has no fixed points on its domain; you need the endpoints for this to work out, which is why you require a compact set.