Let $f:[0,1]^n \rightarrow [0,1]^n$ be a continuous mapping. Brouwer's fixed point theorem says that $f$ has a fixed point, i.e., some $x$ such that $f(x) = x$.

Suppose we have a continuous family, i.e., a continuous function $f:[0,1]^n \times [0,1] \rightarrow [0,1]^n$. Then for each $r \in [0,1]$ we have that there is a point $x_r$ such that $f(x_r,r) = x_r$. It is known that, in general, there is no continuous mapping $r \mapsto x_r$. The following counter-example exists in the case $n=1$: take $f(x,r) = 2rx$ if $r \leq 1/2$ and $f(x,r) = (2r-1) + (2-2r)x$ - for $r < 1/2$ we have the only fixed point as 0, and $r > 1/2$ we have the only fixed point as $1$.

My question is "what happens if we demand there are unique fixed points?" i.e., if every $f(\cdot,r)$ has a unique fixed point, is the mapping $r \mapsto x_r$ continuous?

In the case $n=1$ my intuition (which is very possibly wrong!) seems to indicate that if each of the functions $f(\cdot,r)$ has a unique fixed point, then the mapping sending such a function to its fixed point is continuous.

I'm sure this has been studied before by someone, however searching for it is a little difficult. Does anyone have any references, or know what happens for other $n$? I have little to no geometric insight as to what happens for $n \geq 2$.


2 Answers 2


The answer is yes. Fixed points of $f$ are zeros of the continuous $g(x)=f(x)-x$. So one has to prove the following: If $g:[0,1]^n\times[0,1]\to[0,1]^n$ is continuos, and for each $r\in[0,1]$, $g(.,r)$ has exactly one zero $x_r$ then $r\to x_r$ is continuous. Consider the set $$E=\{ (x,r):g(x,r)=0\}\subset [0,1]^{n+1}.$$ this set is closed. On the other hand, this set is the graph of the function $r\mapsto x_r$. A function on $[0,1]$ whose graph is closed is continuous. The proof is straightforward. Let $r_n\to r^*$, choose a subsequence such that $x_{r_n}\to x^*$, then $(r^*,x^*)\in E$ because $E$ is closed, and $x_{r^*}=x^*$ because there is a unique $x^*$ such that $(r^*,x^*)\in E$.


Yes, it is indeed continuous. Consider the correspondence (set-valued map) $F:[0,1]\to 2^{[0,1]^n}$ given by $$F(x)=\{y\in[0,1]^n:f(y,x)=y\}.$$ Obviously, $F$ has closed values in a compact Hausdorff space and has a closed graph. Therefore, $F$ is upper hemicontinuous. Now if $F$ is single-valued, and therefore essentally a function, being upper hemicontinuous coincides with being continuous.


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