Is there a $f\in C[0,1]$ such that $f(x)=\frac12 \sin (f(1-x))$

Question

Is there a $f\in (C[0,1],\lVert\cdot\rVert_{\infty})$ such that $f(x)=\frac12 \sin (f(1-x))$?

I feel like you need to apply Banach´s fixed point theorem, which means it suffices to prove that $T:C[0,1]\to C[0,1]: (Tf)(x)=\frac12 \sin (f(1-x))$ is a contraction mapping.

How can I prove the $T$ is indeed a contraction map?

Answer 1

Let $g(x)=\frac12\sin(x)$. Then $f(x)=g(f(1-x))$ for all $x\in[0,1]$, so in particular each point $f(x)$ is a period-2 fixpoint of $g$, and $f(1/2)$ is a period-1 fixpoint of $g$.

But note the following:

• $g(x)\le\frac x2$ for all $x>0$
• $g(x)>0$ for all $x\le1$
• $g(x)$ is odd.

So for any $x>0$, if $g(x)>0$ then $g(g(x))\le x/4$, and if $g(x)\le 0$ then $g(g(x))\le 0\le\frac x4$. Thus $g(x)$ has no period-2 fixpoints greater than $0$, and by the odd symmetry there are none less than $0$ either. Thus $f(x)=0$ for all $x\in[0,1]$.

PS: The Banach fixed-point theorem (on $\Bbb R$, not $C[0,1]$) is another way to see this, as $g'(x)\le1/2$, which implies that $|g(x)-g(y)|\le\frac12|x-y|$, and the fixpoint at $0$ is trivial to verify.