Let $f:\mathbb R\longrightarrow \mathbb R$ a differentiable function s.t. $f'(x)\neq 0$ for all $x$. Suppose $f$ is bijective and that it's inverse $g:=f^{-1}$ is continuous. Prove that $g$ is also $\mathcal C^1$. I tried as follow :

Let $b\in \mathbb R$ and set $g(b)=a$, i.e. $f(a)=b$.

  • Since $g$ is continuous, there is $\varepsilon$ s.t. $\varepsilon(h)\to 0$ when $h\to 0$ s.t. $$g(b+h)-g(b)=\varepsilon(h)\implies g(b+h)=a+\varepsilon(h)\implies b+h=f(a+\varepsilon(h))=f(a)+f'(a)\varepsilon(h)+o(\varepsilon(h))\implies h=f'(a)\varepsilon(h)+o(\varepsilon(h)).$$

  • Therefore, $$g(b+h)-g(b)-\frac{1}{f'(g(b))}h=\varepsilon(h)-\frac{1}{f'(a)}(f'(a)\varepsilon(h)+o(\varepsilon(h))=\varepsilon(h)-\varepsilon(h)+o(\varepsilon(h))=o(\varepsilon(h)).$$

Therefore $$\lim_{h\to 0}g(b+h)-g(b)-\frac{1}{f'(g(b))}h=0,$$ and thus $g$ is differentiable.

Question : Is it correct ? Because I'm note very confortable with this type of proof.

  • $\begingroup$ For me it's not enough since by continuity of $g$, you have that $\lim_{h\to 0}g(b+h)-g(b)-\frac{1}{f'(g(b))}h=0$ anyway. What you have to prove is that $\lim_{h\to 0}g(b+h)-g(b)-\frac{1}{f'(g(b))}=0$. $\endgroup$ – Surb Aug 10 '18 at 13:10

No, it's not enough since by continuity of $g$, you have that $$\lim_{h\to 0}g(b+h)-g(b)-\frac{1}{f'(g(b))h}=0\tag{E}$$ anyway.

Remark that (E) gives you $g(b+h)=g(b)+o(1)$ and what you need is $$g(b+h)=g(b)+ch+o(h).$$

What you have to prove is that $$\lim_{h\to 0}\frac{g(b+h)-g(b)}{h}-\frac{1}{f'(g(b))}=0.$$

Until $h=f'(a)\varepsilon(h)+o(\varepsilon(h))$ it's fine. Now, remark that there is $R(h)\to 0$ s.t. $$h=f'(a)\varepsilon(h)+\varepsilon(h)R(h).$$ Therefore $$|h|\geq |f'(a)\varepsilon(h)|-|\varepsilon(h)||R(h)|=|\varepsilon(h)|(|f'(a)|-|R(h)|).$$ Let $\delta>0$ s.t. for all $|h|<\delta$ small enough s.t. $|R(h)|\leq |f'(a)|/2$ and thus $$|\varepsilon(h)|\leq 2h/|f'(a)|.$$ Therefore $\varepsilon(h)=\mathcal O(h).$

Finally, we get $$\left|g(b+h)-g(b)-\frac{1}{f'(g(b))}h\right|\leq |\varepsilon(h)R(h)|=\frac{|\varepsilon(h)|}{|h|}|h|R(h)|\leq C|h| |R(h)|,$$ with $R(h)\to 0$, and thus, the claim follow.

  • $\begingroup$ Why do we need to prove $\lim_{h\to 0}g(b+h)-g(b)-\frac{1}{f'(g(b))}=0.$ ? $\endgroup$ – onurcanbektas Aug 10 '18 at 13:43
  • $\begingroup$ isn't the definition of $g'(b) = \lim_{h\to 0}(g(b+h)-g(b))/h -\frac{1}{f'(g(b))}=0.$ $\endgroup$ – onurcanbektas Aug 10 '18 at 13:43
  • 1
    $\begingroup$ @onurcanbektas: Thank you. It's a typo, I corrected it. $\endgroup$ – Surb Aug 10 '18 at 13:45


Use inverse function theorem.

About your proof; I think it is OK.

See @Surb's answer.

  • 1
    $\begingroup$ No, it's not OK for reason I mentioned. $\endgroup$ – Surb Aug 10 '18 at 13:33

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