This question came to me as part of a larger problem in game theory I was trying to figure out.
$F$ is a family of continuous functions defined in the range $[0,1]$. Initially $F$ contains only the constant functions $0$ and $1$.
Perform the following steps on repeat:
- Pick any two distinct functions in $F$
- Check if they intersect in the range $(0,1)$. If they do, stop.
- If not, add $$f(x) = xf_1(x) + (1-x)f_2(x)$$ to $F$ where $f_1 (x)$ is the strictly bigger of the two functions and $f_2(x)$ is the smaller.
Will this process ever generate functions that intersect in the range (0,1)?
Using induction we can show that $f(0) = 0$, $f(1) = 1$ for all functions generated besides $0$ and $1$. We'll have to take three cases to do so: $f_1 = 1$, $f_2 = 0$ and $f_1,f_2 \neq 1,0$ Also we can show that all the functions are polynomials.
If we differentiate we get
$$f'(x) = f_1(x) - f_2(x) + xf_1'(x) + (1-x)f_2'(x)$$
Since $f_1(x) > f_2(x)$, assuming $f_1'(x) \geq 0$, $f_2'(x) \geq 0$ gives $f'(x) > 0$. This proves that all the functions are strictly increasing.
It seems intuitive (to me atleast) that two such functions won't ever intersect. I tried calculating second derivative, which still didn't help
First, we have the constant function $f_0(x)= 0$ and $f_1(x)=1$.
We then generate the line $f_2(x)=x$.
Let's pick $f_0$ and $f_2$, we get $f_3(x)=x(x)=x^2$.
Let's pick $f_1$ and $f_2$, we get $f_4(x)=x(1)+(1-x)x=2x-x^2$
Let's pick $f_3$ and $f_4$, we get $f_5(x)=x(2x-x^2)+(1-x)x^2=-2x^3+3x^2$
Notice that $(0.5,0.5)$ is a point on both $f_5$ and $f_2$.