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I would like to prove that if $f:\left[0,1\right]\to\left[0,1\right]$ such that and $f\left(1-x\right)=1-f\left(x\right)$, then $f$ has a central symmetry at $\left(0.5,0.5\right)$.

This is actually an intuition but I can't prove it.

I did the small calculation below, but there is a mistake somewhere :

Starting with the classical central symmetry expression $$f\left(-x\right)=-f\left(x\right)$$ then translating it in $\left(0.5,0.5\right)$, it yields $$f\left(-x+0.5\right)+0.5=-f\left(x-0.5\right)-0.5$$ so $$f\left(-x+0.5\right)=-f\left(x-0.5\right)-1$$

A substitution with $x-0.5$, yields $$f\left(1-x\right)=-f\left(x-1\right)-1$$

There seems to be a sign mistake, but I cannot find it.

Thanks in advance for your help!

Gim
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1 Answers1

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Say, $f$ has cetral symmetry with center $(a,b)\neq(0,0)$. We define a new function $h$, with the help of $f$ as follows. $$h(x)=f(a-x)-b$$

Since $a$ is within the domain of $f$, we have that $f$ is defined on $a$ and $f(a)=b$. So, $h$ is just a relocation/transposition of $f$, such that $h(0)=f(a)-b=0$. Therefore, $h$ is also central symmetric and its center of symmetry now is $(0,0)$.

This also applies in the opposite direction: if $h$ has central symmetry with center $(0,0)$, then $f$ has central symmetry with center $(a,b)$. So, for $(a,b)=(0.5,0.5)$, if you show that $h(x)=-h(-x)$, then you're done. Well, indeed

\begin{align} h(x)&= f(0.5-x)-0.5\\ &=f(1-z)-b & \text{, where $z=x+0.5$}\\ &=1-f(z) -0.5\\ &=0.5-f(x+0.5)\\ &=0.5-f(0.5-(-x))\\ &=-(f(0.5-(-x))-0.5)\\ &=-h(-x) \end{align}

frabala
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