It is straightforward to see that $f \circ f$ is odd whenever $f$ is odd. Indeed, assuming $f(-x) = -f(x)$ for all $x$, we get

$$ f(f(-x)) = f(-f(x)) = -f(f(x)). $$

Hence, $f \circ f$ is an odd function as well.

My question is a converse of the above statement.

Suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous. If $f \circ f$ is an odd function. What can I say about $f$ itself? Is it odd?


Define $f(x)=\begin{cases}0&x\leq0\\-x&x>0\end{cases}$.

Then $f(f(-x))=-f(f(x))=0$ but $f(-x)\neq-f(x)$, except at $x=0$. Hence we have an odd $f(f(x))$ which doesn't imply an odd $f(x)$.

Note that $f(f(x))$ is in fact both even and odd. This answer was inspired in part by user @Henry_Lee.

  • 2
    $\begingroup$ That is a clever counterexample. If we replace $-x$ to $-x^2$ in the definition of $f$, then even differentiability of $f$ doesn't imply the converse statement. Thanks! $\endgroup$ – Juyoung Jeong Oct 9 '18 at 18:43
  • 1
    $\begingroup$ @JuyoungJeong Thanks and good point. We could even have $f(x)=\begin{cases}0&x\leq0\\-e^{-1/x}&x>0\end{cases}$ to have $f(x)$ smooth. $\endgroup$ – Jam Oct 9 '18 at 19:19

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