Rudin claims that the exponential function has a differentiable inverse on all of $\mathbb{R}$ because it is strictly increasing and differentiable. I am failing to see which theorem proves this for all $\mathbb{R}$ not just some interval.

How does rudin intend us to prove that this is true? One of the exercises seems to prove something similar however, it only works for the open interval $(a,b)$ but I feel this does not apply because all of $\mathbb{R}$ is not an open interval.

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  • $\begingroup$ I guess the point is that there is a valid inverse for an open interval and you can patch all of these intervals together to get the whole of the real line. $\endgroup$
    – daruma
    Jun 19, 2021 at 1:07
  • $\begingroup$ @daruma What if the intervals are overlapping? How can I ensure that the functions are differentiable between intervlas? I was thinking that myself but I feel a bit skeptical about the validity of this method $\endgroup$ Jun 19, 2021 at 1:09
  • $\begingroup$ See for example math.stackexchange.com/q/2226089/42969. That results holds also for intervals with infinite endpoints, i.e. if $a=-\infty$ or $b=\infty$. $\endgroup$
    – Martin R
    Jun 19, 2021 at 3:40
  • $\begingroup$ Strictly increasing and differentiable does not imply that the inverse is differentiable. A counterexample is $f(x)=x^3$ (or any other function with derivative vanishing at any point). $\endgroup$ Aug 25, 2022 at 22:56
  • $\begingroup$ However, here it still works as the derivative of the exponential function never vanishes. $\endgroup$ Aug 25, 2022 at 23:12

1 Answer 1


Let $E$ denote the Exponential Function. $E$ is then a bijective, strictly increasing and 1-differentiable function mapping $\Bbb R$ onto $\Bbb R_+$, therefore it has an inverse function, which we shall denote by $L$, mapping $\Bbb R_+$ onto $\Bbb R$.

Let us now prove the following claims.

Claim 1: If $f: A \to B$, where $A,B \subseteq \Bbb R$, is a bijective and striclty increasing function, then its inverse $g \equiv f^{-1}: B \to A$ is also strictly increasing.

Proof: Suppose $b_1, b_2 \in B$ and $b_1 < b_2$, then there exists $a_1, a_2 \in A$ such that $$f(a_1) = b_1 < b_2 = f(a_2)$$ hence, it must be the case that $a_1 < a_2$ by hypothesis. However, by the definition of $g$ $$g(b_1) = a_1 < a_2 = g(b_2)$$ thus completing the proof by the arbitrariety of $b_1$ and $b_2$.

Claim 2: If $f: ]a,b[ \to B$, where $B$ is the range of $f$, is 1-differentiable, strictly increasing and such that $\forall x \in ]a,b[, f'(x) \neq 0$, then its inverse function is also 1-differentiable on $B$ and: $$\forall y_0 \in B, g'(y_0) = 1/f'(x_0)$$ where $y_0$ is the unique element of $B$ such that $f(x_0) = y_0$ or, equivalently, $g(y_0) = x_0$.

Proof: Let $y_0 \in B$ be arbitrary. Then, notice that, by the intermediate value property and the fact that $f$ is strictly increasing, $B = ]\text{inf} f, \text{sup}f[$, hence $g$ is defined on an open interval containing $y_0$ and it makes sense to study its differentiability at $y_0$. Moreover there exist points $c, d \in ]a,b[$ such that: $$c < d \hspace{5mm} \text{ and } \hspace{5mm} \text{inf}f < f(c) < y_0 < f(d) < \text{sup}f$$ therefore $f|_{[c,d]}$ is a bijective and continuous function mapping the compact set $[c,d]$ into $[f(c), f(d)]$. It follows that its inverse $g|_{[f(c), f(d)]}$ is also continuous (here compacteness of the domain of $f|_{[a,b]}$ is needed). The conclusion is that $g$ is continuous at $y_0$.

Finally, let $\phi: ]a,b[-\{x_0\} \to \Bbb R$ be defined by the equation:

$$ \phi(x) = \frac{x - x_0}{f(x) - f(x_0)} $$

then consider the following compositions: $$ y \mapsto g(y) \mapsto \phi(g(y)) $$ Since $\lim \limits_{y \to y_0} g(y) = g(y_0) = x_0$ by the continuity previously proved, $\lim \limits_{x \to x_0} \phi (x) = 1/f'(x_0)$ by hypothesis and $g$ is strictly increasing (hence, there is a punctured neighbourohood of $y_0$ in which $g$ never attains the value $g(y_0)$) we can apply the Theorem on the Composition of limits to conclude that: $$\lim \limits_{y \to y_0} \phi(g(y)) = \lim \limits_{y \to y_0} \frac{g(y) - g(y_0)}{y - y_0} = 1/f'(x_0) = g'(y_0)$$ The proof is thus complete by the arbitrariety of $y_0 \in B$.

By CLAIM 1 it follows immediately that $L$ is strictly increasing on $\Bbb R_+$. By CLAIM 2, instead, it follows that, for every $y_0$ in $\Bbb R_+$, $L|_{]y_0/2,y_0+1[}$ is differentiable at $y_0$ (just restrict $E$ to $]L(y_0/2), L(y_0 +1)[$ and apply the claim). However, by the local character of a limit, it is clear that:

$$\lim \limits_{y \to y_0} \frac{L(y) - L(y_0)}{y - y_0} = \lim \limits_{y \to y_0} \frac{L|_{]y_0/2,y_0+1[}(y) - L|_{]y_0/2,y_0+1[}(y_0)}{y - y_0}$$ so that $L$ is also differentiable at $y_0$ and it has the same first derivative as $L|_{]y_0/2,y_0+1[}$.


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