The entire work here is to show that a continuous function from from the standard simplex Δd to itself has a fixed point. If that’s done, given S⊊Rd compact and convex and a continuous function ϕ on S, we can scale S to be a subset of Δd, take the continuous projection π:Δd→S, and ϕ∘π gives us a function from Δd to itself. Now, a fixed point of ϕ∘π is also a fixed point of ϕ.
For that, the intended way is presumably to mirror the step from Ex4 to Ex5. The problem is that the coloring of the disk isn’t drawn in a way that generalizes well. The nicer way to color it would be like this. One can see that these colors still work (i.e., a trichromatic triangle must contain the origin), and they’re subsets of the previous colors, so the conditions of sides not touching colors still hold. This way of coloring is analogous to what we do in d-dimension space.
Mathematically, one can describe these areas as follows:
The d areas of the kind Ck:={x∈Rd:∑dj=1xj<0∧xj≥0∀j≠k}
The area where all coordinates are non-negative, Cd+1:={x∈Rd:xj≥0∀j}
Given the d-dimensional standard simplex Δ and a continuous function f:Δ→Δ, the function g:Δ→Rd given by g(x)=f(x)−x does have the property that each face of the simplex has one color it can’t map into...
The first d faces can be characterized by the condition that xk=0. Then for a point x in this face, we have g(x)k=f(x)k−xk=f(x)k≥0, hence g(x)∉Ck.
The final face can be characterized by the condition that ∑xj=1. Then for a point x in this face, we have that ∑g(x)j=∑f(x)j−∑xj=∑f(x)j−1≤0, hence g(x)∉Cd+1. (Unless g(x)=O, in which case we’re also happy.)
We still have to show that the image points of the d+1 vertices of the simplex actually have all d+1 colors. This is not necessarily true, but as above we can show that either it is true or one of them maps directly into the origin.
The k-th vertex is the point x with xk=1 and x≠k=0. We have g(x)j=f(x)j−xj=f(x)j≥0 for all j≠k, and ∑g(x)j=∑f(x)j−∑xj=∑f(x)j−1≤0. Thus, either g(x)∈Ck or g(x)=O.
And for the origin, we have g(O)j=f(O)j−Oj=f(O)j≥0, so g(O)∈Cd+1
Now, either one of the first d vertices maps directly into the origin, or we can construct a simplex with all d+1 ‘colors’ for the map g in Rd. According to Ex9, this simplex has a d+1-chromatic component. It remains to show that the origin is always contained the span of such d+1 points (tedius but pretty clear from the 2-d case), then we can again construct a sequence of points that converges toward the origin, by making the simplex progressively more granular. This gives us a point x∗ such that g(x∗)=O=f(x∗)−x∗ and hence f(x∗)=x∗.
Ex10
The entire work here is to show that a continuous function from from the standard simplex Δd to itself has a fixed point. If that’s done, given S⊊Rd compact and convex and a continuous function ϕ on S, we can scale S to be a subset of Δd, take the continuous projection π:Δd→S, and ϕ∘π gives us a function from Δd to itself. Now, a fixed point of ϕ∘π is also a fixed point of ϕ.
For that, the intended way is presumably to mirror the step from Ex4 to Ex5. The problem is that the coloring of the disk isn’t drawn in a way that generalizes well. The nicer way to color it would be like this. One can see that these colors still work (i.e., a trichromatic triangle must contain the origin), and they’re subsets of the previous colors, so the conditions of sides not touching colors still hold. This way of coloring is analogous to what we do in d-dimension space.
Mathematically, one can describe these areas as follows:
The d areas of the kind Ck:={x∈Rd:∑dj=1xj<0∧xj≥0∀j≠k}
The area where all coordinates are non-negative, Cd+1:={x∈Rd:xj≥0∀j}
Given the d-dimensional standard simplex Δ and a continuous function f:Δ→Δ, the function g:Δ→Rd given by g(x)=f(x)−x does have the property that each face of the simplex has one color it can’t map into...
The first d faces can be characterized by the condition that xk=0. Then for a point x in this face, we have g(x)k=f(x)k−xk=f(x)k≥0, hence g(x)∉Ck.
The final face can be characterized by the condition that ∑xj=1. Then for a point x in this face, we have that ∑g(x)j=∑f(x)j−∑xj=∑f(x)j−1≤0, hence g(x)∉Cd+1. (Unless g(x)=O, in which case we’re also happy.)
We still have to show that the image points of the d+1 vertices of the simplex actually have all d+1 colors. This is not necessarily true, but as above we can show that either it is true or one of them maps directly into the origin.
The k-th vertex is the point x with xk=1 and x≠k=0. We have g(x)j=f(x)j−xj=f(x)j≥0 for all j≠k, and ∑g(x)j=∑f(x)j−∑xj=∑f(x)j−1≤0. Thus, either g(x)∈Ck or g(x)=O.
And for the origin, we have g(O)j=f(O)j−Oj=f(O)j≥0, so g(O)∈Cd+1
Now, either one of the first d vertices maps directly into the origin, or we can construct a simplex with all d+1 ‘colors’ for the map g in Rd. According to Ex9, this simplex has a d+1-chromatic component. It remains to show that the origin is always contained the span of such d+1 points (tedius but pretty clear from the 2-d case), then we can again construct a sequence of points that converges toward the origin, by making the simplex progressively more granular. This gives us a point x∗ such that g(x∗)=O=f(x∗)−x∗ and hence f(x∗)=x∗.