2
$\begingroup$

Suppose I have some function $f$: $\mathbb{R}_+\rightarrow\mathbb{R}_+$ that is bijective, continuous and strictly increasing. Moreover, it is (at least) twice continuously differentiable everywhere. Since $f$ is bijective, its inverse $f^{-1}$ exists. What conditions need to be satisfied so that $f^{-1}$ is also (at least) twice continuously differentiable everywhere?


From wikipedia's Inverse FT article: when $f$ is a continuously differentiable function with nonzero derivative at the point $a$, then $f^{-1}$ is continuously differentiable. Given that $f$ in my case is strictly increasing $\implies$ $f'>0$ this condition seems satisfied for all $a \in \mathbb{R}_+$.


Counterexamples in comments assume that $f'(x)=0$ at some $a$, but this is ruled out by the assumption that $f'(x)>0$.

$\endgroup$
  • 2
    $\begingroup$ You might want to look at $f(x) = x + c \sin x$ for $0 < c < 1$ to get an idea of what might go wrong $\endgroup$ – John Hughes Dec 16 '17 at 22:25
  • 3
    $\begingroup$ That function is not strictly increasing and is not bijective. $\endgroup$ – Mark Fischler Dec 16 '17 at 22:32
  • $\begingroup$ every injection $\mathbb R \to \mathbb R$ is already strictly monotone. $\endgroup$ – Andres Mejia Dec 16 '17 at 22:38
  • $\begingroup$ the derivative of $f^{-1}$ is not defined at every point where $f'(x)=0$. $\endgroup$ – Masacroso Dec 16 '17 at 22:39
  • $\begingroup$ @MarkFischler: Did you notice that I restricted to the case where $0 < c < 1$? That guarantees positive derivative. $\endgroup$ – John Hughes Dec 16 '17 at 22:40
1
$\begingroup$

The condition is exactly that $f'(x)$ is nonzero (or equivalently, positive) for all $x$. Indeed, if $f^{-1}$ is differentiable everywhere, then differentiating the identity $f^{-1}(f(x))=x$ gives $(f^{-1})'(f(x))f'(x)=1$ so $f'(x)$ can never be zero.

Conversely, if $y=f(x)$ and $f'(x)\neq 0$, then $f^{-1}$ is differentiable at $y$ with $(f^{-1})'(y)=1/f'(x)$ (indeed, you can prove this directly from the definition, and the difference quotients to compute $(f^{-1})'(y)$ are the reciprocals of the difference quotients for $f'(x)$). So if $f'$ is always nonzero, then $f^{-1}$ is differentiable everywhere with $(f^{-1})'(y)=1/f'(f^{-1}(y))$. If $f$ is twice continuously differentiable, we can then differentiate $(f^{-1})'$ by the chain rule to find that $f^{-1}$ is twice continuously differentiable.


Note, however, that merely assuming $f$ is differentiable and strictly increasing does not imply $f'$ is positive everywhere. For instance, consider $f(x)=x^3$, which is strictly increasing and infinitely differentiable but $f'(0)=0$ and $f^{-1}(x)=\sqrt[3]{x}$ is not differentiable at $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.