# Brouwer fixed poin in 1-D

In $\mathbb R_{+}$ (non-negative) I realize that any nonnegative valued continuous function $f\colon X \rightarrow Y$ where $X, Y \subseteq \mathbb R_{+}$ with a negative finite gradient has a fixed point.

But the Brouwer fixed point theorem tells us that the function should map the domain into codomain, i.e., $Y\subseteq X$. Is this required in the above case? I feel that even if the $Y \supset X$ the fixed point exists.

I mean if the function in the following figure didn't cut the red line and instead cut the $x$ axis say at 0.6 still the fixed point exists. http://en.wikipedia.org/wiki/File:Th%C3%A9or%C3%A8me-de-Brouwer-dim-1.svg http://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem

Edit: I forgot to mentions that:

The domain $X$ has left end $0$ and both $X, Y$ are closed. (like in the picture)

Thank you.

• If $X=[1,2]$ and $Y=[5,6]$ how can $f:X\to Y$ have a fixed point? So your "realization" isn't quite accurate. It's unclear what you are trying to say there. – Thomas Andrews Jun 23 '13 at 13:25
• I think I missed a point. I included that in the Edit. My confussion is $Y \supset X$ and still have a fixed point under conditions mentioned in the question. Thanks. – triomphe Jun 23 '13 at 13:40

Let $X=\{1,2\}$ and let $f\colon X\to X$ be $f(1)=2$ and $f(2)=1$. You probably want to assume $X$ and $Y$ are closed intervals.
Let $X= (1,2)$, $Y=(0,3)$. Then $f(x)=1-\frac x2$ has no fixed point.
• Even with the edit, Hagen's example still works; you can take $X= [0,1]$, $Y = [0,3]$, $f(x) = 3-x$. – fuglede Jun 23 '13 at 14:51
• Hello fuglede. In your example for the given $X=[0,1]$ $Y=[2,3]$. That is not what I asked. I mean when $Y \supset X$ there is a fixed point. But in your example $Y$ is not a super set of $X$. – triomphe Jun 24 '13 at 2:49