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I wish to use the Brouwer's fixed point theorem to show that continuous, bounded function $f$ from $[0, \infty) \times [0, \infty)$ to $[0, \infty) \times [0, \infty)$ has a fixed point.

However, my version of the Brouwer's fixed point theorem (BFPT) states that,

Let $A \subset \mathbb{R}^2$ be a nonempty convex compact set, then every continuous function $f: A \to A$ has a fixed point in $A$.

I need to use Brouwer's fixed point theorem for this problem (not homework by the way). The main difficulty is that the sets are closed but not compact.

In 1D, we have, suppose that $f$ is bounded, then let $M$ be the upper bound of $f$, this means $|f| \leq M, \forall x \in [0, \infty)$. So we can restrict the range of $f$ to $[0, M]$ which is compact. Then $f: [0, M] \to [0,M]$ has a fixed point by BFPT.

I am not sure how to upgrade this to 2D. I think the idea is the similar. But the notion of upper bound is strange. Can someone please help?

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  • $\begingroup$ Convexity is essential for BFPT. If $A$ is closed and convex and $f:A \to A$ is bounded, then we can replace $A$ by $A'=\operatorname{co} \overline{f(A)}$ which is convex and compact, hence the BFPT can be applied. $\endgroup$ – copper.hat Sep 14 '19 at 4:35
  • $\begingroup$ @copper.hat Is it true then "Let $A \subset \mathbb{R}^n$ be a nonempty convex closed set, then every continuous, bounded function $f: A \to A$ has a fixed point in $A$." $\endgroup$ – Sanjay Gupta Sep 14 '19 at 5:15
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Let $f=(f_1,f_2)$. Then $\|f(x)\| \leq M$ for all $x$ implies that $|f_1(x)| \leq M$ and $|f_2(x)| \leq M$. Hence $f$ maps $[0,M]\times [0,M]$ into itself. Since the rectangle $[0,M]\times [0,M]$ is compact and convex there exists $(x,y)$ in it with $f(x,y)=(x,y)$.

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  • $\begingroup$ Thank you. Is the following theorem true then: "Let $A \subset \mathbb{R}^n$ be a nonempty convex closed set, then every continuous, bounded function $f: A \to A$ has a fixed point in $A$." $\endgroup$ – Sanjay Gupta Sep 14 '19 at 5:14
  • $\begingroup$ Yes, that is true. Let $K$ be a closed ball with center at the origin. If the radius is large enough then $f$ maps $A\cap K$ into itself and $A\cap K$ is a compact convex set. Hence there is a fixed point in it. $\endgroup$ – Kavi Rama Murthy Sep 14 '19 at 5:22
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Since $\ f\ $ is bounded, all values of $\ f\ $ are contained in certain, say, square $[0;C]^2\subseteq[0;\infty)^2.\ $ You may consider a function which is like a chunk of $\ f,\ $ call it $\ g:[0;C]^2\rightarrow[0;C]^2\ $ where

$$ \forall_{z\in[0;C]^2}\quad g(z)\ :=\ f(z) $$

Then $\ g\ $ has a fixed point $\ z\in\ [0;C]^2.\ $ Of course this $\ z\ $ is a fixed point for $\ f\ $ as well.

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