Skip to main content
1 of 13
Robert Bryant
  • 108.4k
  • 8
  • 342
  • 453

Actually, there are lots of local solutions of this equation. For example, suppose that we start with a $C^2$ function $f$ on an interval $I\subset\mathbb{R}$ such that $f'$ is positive on $I$ and $f(I)$ is disjoint from $I$. Then an inverse $g:f(I)\to I$ of $f:I\to f(I)$ exists and is $C^2$. Now define $f$ on the interval $f(I)$ so that $f(y) = f''(g(y))$ for $y\in f(I)$. Then for $x\in I$, we will have $x = g(y)$ for some $y\in f(I)$ and, of course, $y = f(x)$. Then $f''(x) = f''(g(y)) = f(y) = f(f(x))$ for all $x\in I$.

Robert Bryant
  • 108.4k
  • 8
  • 342
  • 453