3

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.

Bob
  • 55
  • 1
    First you say that f is continuous followed by requiring that $f'(x)\ge 0$ which indicates that $f$ is also supposed to be differentiable. Is that the case? – Friedrich Philipp Mar 04 '16 at 19:10
  • $f(0)$ cannot be determined from $xf(x)+(1-x)f(1-x)=1$. However, since $f$ is non-decreasing, and it can be shown that for all $x\ne0$, $f(x)=1$, we must have $f(0)=1$. – robjohn Mar 04 '16 at 19:37

2 Answers2

5

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 $x\in\mathbb{R}$.

robjohn
  • 345,667
2

How about $f(x) = 1$? $\,\,\,\,\,$

Cameron Williams
  • 29,432