(From Rudin Principles of Mathematical Analysis, 5.2)
Suppose $f'(x) > 0$ in ($a, b$). Prove that $f$ is strictly increasing in ($a, b$), and let
$g$ be its inverse function.
Prove that $g$ is differentiable, and that $g'(f(x)) = \frac{1}{f′(x)} \quad
(a < x < b)$
Here's an answer I found online:
Let $g : f(a, b) → (a, b) $ be the inverse function of $f$, i.e., $ g(f(x)) = x $ for all $x ∈ (a, b)$.
We now show that $g′(y) = \lim\limits_{z→y} \frac {g(z) − g(y)}{z − y}$ exists for all $y ∈ f(a, b)$.
Put $y = f(x)$ and $z = f(t)$, where $x, t ∈ (a, b)$, then
since $f$ is continuous (by Theorem 5.2), so is $g$ (by Theorem 4.17), and $z → y$
implies $t → x$.
It follows that
$$\begin{align*}
\lim_{z→y} \frac{g(z) − g(y)}{z − y} &= \lim_{t→x}\frac{g(f(t)) − g(f(x))}{f(t) − f(x)} \\
&= \lim_{t→x}\frac{t − x}{f(t) − f(x)} \\
&= \lim_{t→x}\frac{1}{\frac{f(t) − f(x)}{t − x}} \\
&= \frac{1}{f′(x)}
\end{align*}$$
Question:
Why it is necessary to for g to be continuous?
The only step that uses continuity is the changing of the limit values ( $z → y$
implies $t → x$), but that comes from $f$ I think?
Best Answer
You do not need to assume that the inverse function $g\colon f([a,b])\to [a,b]$ is continuous because that actually does come for free. To see this, it suffices to show that if $U\subset [a,b]$ is an relatively open interval, then $g^{-1}(U)$ is open in $f([a,b])$. Now, $g^{-1}(U) = f(U)$, and since $f$ is strictly increasing, the image of a relatively open interval in $[a,b]$ under $f$ is another relatively open interval in $f([a,b])$.