I've been wracking my brain for a couple of days without making much progress trying to find a function $f(x):\mathbb{R} \rightarrow \mathbb{R}$ with the following properties.
- $f(0)=0$
- $f(1)=1$
- When $0\leq x\leq 1$:
- $f(x)$ is continuous
- $f'(x)\geq 0$
- $xf(x)+(1-x)f(1-x)=1$
Any help would be greatly appreciated.
Edit: Sorry everyone! I made a bad typo. $f(0)=0$, not 1.
Plugging in $x=0$ or $x=1$, we get $f(1)=1$. Plugging in $x=\frac12$, we get $f\left(\frac12\right)=1$. So if $f$ is non-decreasing, we have that $f(x)=1$ for $x\in\left[\frac12,1\right]$. Now, for $x\in\left[0,\frac12\right]$, $$ \begin{align} f(x) &=\frac{1-(1-x)f(1-x)}x\\ &=\frac{1-(1-x)\cdot1}x\\[3pt] &=1\tag{1} \end{align} $$ Thus, we must have $f(x)=1$ for $x\in[0,1]$.
For $x\not\in[0,1]$, suppose $x\gt1$ and $f(x)\gt1$, then $1-x\lt0$ and $$ \begin{align} f(1-x) &=\frac{1-xf(x)}{1-x}\\ &=\frac{1-x-x(f(x)-1)}{1-x}\\ &=1+\frac{x}{x-1}(f(x)-1)\\[3pt] &\gt1\tag{2} \end{align} $$ Thus, $f$ must have decreased for $x\lt0$. Therefore, $f(x)=1$ for all $x\gt1$. We can then apply $(1)$ to get that $f(x)=1$ for all $x\lt0$. Since $f(x)$ is non-decreasing, we get that $f(0)=1$, as well.
Therefore, $f(x)=1$ for all $x\in\mathbb{R}$.