# Closed form for functional equation $g(x)=\frac{1}{2}(g(x^2)+g((1-x)^2))$ on $[0,1]$

See bottom for edit to question:

Before providing background, my question is if there is a closed-form solution to

$$g(x)=\frac{1}{2}(g(x^2)+g((1-x)^2))$$

with $$g(0)=g(1)=0$$ and $$g(1/2)=1$$. If this can not be found, can it at least be proven that the solution is unique. A further question I have that I have not been able to show is that $$g(x)$$ is increasing on $$[0,1/2)$$ and decreasing on $$(1/2,1]$$.

This is an offshoot of the question here. It was initially asked if the functions defined

$$f_0(x)=x(1-x)$$

$$f_n(x)=\frac{1}{2}(f_{n-1}(x^2)+f_{n-1}((x-1)^2))$$

are increasing on $$[0,1/2]$$ and decreasing on $$[1/2,0]$$. For example, $$\frac{f_{10}(x)}{f_{10}(1/2)}$$ is given by the graph

However, it seems that whatever function $$f_n(x)$$ is approaching is universal up to a constant regardless of $$f_0(x)$$ (provided $$f_0(x)$$ is continuous on $$[0,1]$$). For example, if $$f_0(x)=\sin(2\pi x)$$, then $$f_{15}(x)$$ looks like

This then led to a conjecture: That for any continuous $$f_0(x)$$ on $$[0,1]$$, $$f_n(x)$$ approaches

$$a_ng(x)+b_n$$

where $$a_n,b_n$$ are constants and $$g(x)$$ is the solution to

$$g(x)=\frac{1}{2}(g(x^2)+g((1-x)^2)).$$

with $$g(0)=g(1)=0$$ and $$g(1/2)=1$$. Note that I do not have a proof that these conditions on $$g(x)$$ produce a unique solution, but it seems likely from testing. The reason that the constants do not matter in $$ag(x)+b$$ is that for $$h(x)=ag(x)+b$$

$$g(x)=\frac{1}{2}(g(x^2)+g((1-x)^2))$$

$$ag(x)+b=a\frac{1}{2}(g(x^2)+g((1-x)^2))+b$$

$$h(x)=\frac{1}{2}(ag(x^2)+b+ag((1-x)^2)+b)$$

$$h(x)=\frac{1}{2}(h(x^2)+h((1-x)^2))$$

and therefore $$h(x)$$ also satisfies the functional relation.

Some properties of a solution $$g(x)$$ are:

$$1)\ g(x)=g(1-x)$$

$$2)\ g^{(n)}(1)=(-1)^n g^{(n)}(0)$$

$$3)\ g^{(n)}(0)=g^{(n)}(0)=0\text{ for }n>1$$

This is found differentiating $$g(x)$$ and the functional equation and then plugging in $$x=1$$. For example, for $$n=2$$, we get

$$g''(1)=2 g''(1)+g'(0)+g'(1)$$

However, since we know $$-g'(0)=g'(1)$$, this simplifies to

$$g''(1)=2g''(1)\Rightarrow g''(1)=g''(0)=0$$

This can be extended by induction to all $$n>1$$.

EDIT: As was pointed out in the comments, the only reason the parabola-like curve appears is that $$f_n(x)$$ was normalized by $$f_n(1/2)$$ with that in mind, what can we say about the limit of

$$\frac{f_n(x)}{f_n(1/2)}$$

as $$n$$ goes to infinity. Does this curve have a shape? If so, does this shape have a closed form solution?

• The solution certainly cannot be unique without some additional assumption like continuity, since each value of $g$ is only related to countably many others using the functional equation. Nov 12, 2019 at 20:51
• Also, doesn't the functional equation immediately give $g(1/2)=g(1/4)$? That would seem to contradict the behavior your graphs suggest. Nov 12, 2019 at 20:58
• I guess what's going on is that while $f_n$ is approaching a solution to the functional equation (namely, $0$), the normalization where you divide by $f_n(1/2)$ is not because $f_n(1/2)$ is going to $0$ fast. Nov 12, 2019 at 21:01
• Yep, that seems to be the case. Still makes me wonder what that vaguely parabola-like curve is though Nov 12, 2019 at 21:18

There is no continuous solution to this functional equation. Indeed, suppose such a $$g$$ was continuous. Then it achieves a maximum value at some $$t\in[0,1]$$. We then must have $$g(t)=g(t^2)$$ since $$g((1-t)^2)\leq g(t)$$, and then similarly $$g(t^2)=g(t^4)=g(t^8)=\dots$$. By continuity we conclude $$g(t)=g(0)=0$$ which is a contradiction since $$g(1/2)=1$$.
Note that because of your normalization where you divide by $$f_n(1/2)$$, the function that your graphs approach would instead be a solution to $$cg(x)=\frac{1}{2}(g(x^2)+g((1-x)^2))$$ where the constant $$c$$ is the limiting value of $$f_{n+1}(1/2)/f_n(1/2)$$.
• Ahh, that constant $c$ is what I was missing when I normalized. Do you have any insight on what the shape of that graph the functions would approach might be? Nov 12, 2019 at 21:26
• Not particularly. It looks like it has vertical tangents at $0$ and $1$, though, so the derivatives are undefined rather than the higher derivatives vanishing there. Nov 12, 2019 at 21:37