According to [Brouwer's fixed point theorem](https://en.wikipedia.org/wiki/Brouwer_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](https://en.wikipedia.org/wiki/Compact-open_topology) and $\Delta(K)$ is the space of probability distributions over $K$, equipped with the [weak topology](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures).



Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?
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.


