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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