According to Brouwer's fixed point theorem, for compact convex $K\subset\mathbb{R}^n$, every continuous map $K\rightarrow K$ has a fixed point.

However, these fixed points cannot be chosen continuously, even for $K=[0,1]$, in the sense that there is no continuous map $fix:[0,1]^{[0,1]}\rightarrow[0,1]$ such that $\forall f:[0,1]\rightarrow[0,1]\;f(fix(f))=fix(f)$. To see this, consider a family of functions $f_x:[0,1]\rightarrow[0,1]$ ($x\in[0,1]$) such that $f_0(y)=\frac{1}{3}$; $f_{\frac{1}{2}}(y)=\frac{1}{3}$ for $y\leq\frac{1}{3}$, $f_{\frac{1}{2}}(y)=\frac{2}{3}$ for $y\geq\frac{2}{3}$, and linearly interpolates between those for $\frac{1}{3}<y<\frac{2}{3}$; $f_1(y)=\frac{2}{3}$; and $f_x$ linearly interpolates between $f_0$ and $f_\frac{1}{2}$ for $0<x<\frac{1}{2}$ and between $f_\frac{1}{2}$ and $f_1$ for $\frac{1}{2}<x<1$. In order for $fix$ to continuously select a fixed point, we would need $fix(f_x)=\frac{1}{3}$ for $x\leq\frac{1}{2}$ and $fix(f_x)=\frac{2}{3}$ for $x\geq\frac{1}{2}$, a contradiction.

But one could imagine gradually shifting probability from the lower fixed point of $f_x$ to the upper fixed point as $x$ increases. [Edit: actually, one couldn't do that; as Noam Elkies points out in the comments, this example answers my own question.]

Hence my question: For a compact convex $K\subset\mathbb{R}^n$, is there a continuous map $fix:K^K\rightarrow\Delta(K)$ such that $\forall f:K\rightarrow K$, $fix(f)$ is supported on fixed points of $f$? Here $K^K$ is given the compact-open topology and $\Delta(K)$ is the space of probability distributions over $K$, equipped with the weak topology.

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    $\begingroup$ Doesn't your counterexample exclude such a map as well? The support would have to be $\{1/3\}$ for $x < 1/2$ and $\{2/3\}$ for $x > 1/2$, making ${\rm fix}(f_x)$ one constant on $x \in [0,1/2)$ and another constant on $x \in (1/2,1]$; such a function of $x$ cannot be continuous at $x=1/2$. $\endgroup$ – Noam D. Elkies Aug 1 '18 at 3:02
  • $\begingroup$ In the range for which $f_x$ has multiple fixed points, you can gradually shift measure from the lowest fixed point to the highest fixed point as $x$ increases, which should be continuous. $\endgroup$ – Alex Mennen Aug 1 '18 at 3:17
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    $\begingroup$ but I thought the point of your example was that there is no "range" over which one can "gradually shift" a fixed point: the only $x$ for which $f_x$ has more than one fixed point is $x=1/2$. $\endgroup$ – Noam D. Elkies Aug 1 '18 at 3:59
  • $\begingroup$ Oops, I misvisualized my own example. I feel silly now. Thanks. $\endgroup$ – Alex Mennen Aug 1 '18 at 4:13
  • $\begingroup$ The example I had in mind, which I conveniently failed to accurately describe, is a similar family of functions in which $f_{\frac{1}{2}}$ has slope greater than $1$ in the middle, and there is a neighborhood around $\frac{1}{2}$ on which $f_x$ has 3 fixed points. $\endgroup$ – Alex Mennen Aug 1 '18 at 4:23

The problem here is not restricted to obtaining continuous choice functions, it already fails at the level of continuous multivalued functions.

Theorem The multivalued function $\mathrm{BFT}_k : \mathcal{C}([0,1]^k,[0,1]^k) \rightrightarrows [0,1]^k$ mapping a continuous function to some arbtirary fixed point is not continuous. (See https://arxiv.org/abs/1206.4809)

Theorem The multivalued function $\mathrm{SuppPoint} : \mathcal{PM}(\mathbf{X}) \rightrightarrows \mathbf{X}$ mapping a probability measure to an arbitrary point in its support is continuous (in fact, computable) for every "sufficiently complete" space $\mathbf{X}$, in particular it works for $\mathbf{X} = [0,1]^k$. (See http://logicandanalysis.org/index.php/jla/article/view/241/0)

By the latter, we see that getting probability distributions supported on the fixed points cannot be easier than getting a fixed point. The former tells us that this doesnt work.


There can be no such function even in the category of smooth functions.

Here an example for functions $f:[0,1]\to [0,1]$ with $f(0)=\epsilon$ and $f(1)=1-\epsilon$:

(1) Let $\operatorname{id}:x\mapsto x$ and choose a smooth function $f_0$ with $f_0=\operatorname{id}$ on $[\frac{1}{3},\frac{2}{3}]$, $f_0>\operatorname{id}$ on $[0,\frac{1}{3})$ and $f_0<\operatorname{id}$ $(\frac{2}{3},1]$.

(2) There is a sequence of smooth functions $f_n$ converging to $f_0$ with $f_n(\frac{1}{3})=\frac{1}{3}$, $f_n>\operatorname{id}$ on $[0,\frac{1}{3})$ and $f_n<\operatorname{id}$ on $(\frac{1}{3},1]$.

(3) There is a sequence of smooth functions $g_n$ converging to $f_0$ with $g_n(\frac{2}{3})=\frac{2}{3}$, $g_n>\operatorname{id}$ on $(0,\frac{2}{3})$ and $g_n<\operatorname{id}$ on $(\frac{2}{3},1]$.

Both sequences have each a unique fixed point equal to $\frac{1}{3}$ and resp. $\frac{2}{3}$. Thus there are exactly two distinct probability measures associated to the fixed point sets of $f_n$ and resp. $g_n$. Obviously the limits of those probability measures do not agree. In particular, any choice of $fix$ cannot be continuous at $f_0$.


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