Let $f:[a,b] \rightarrow \Bbb R$ be a strictly monotone increasing. Then $f$ has an inverse function $g:[c,d]\rightarrow \Bbb R,$ where $[c,d]$ is the range of $f$. I'm trying to prove that $g$ is continuous at d.
My intial thoughts for an attempt of a proof:
Strictly monotone functions are injective. So if $\alpha, \beta \in [a,b]$ and $\alpha \not= \beta $ then $\alpha < \beta$. Since $f$ is strictly monotone increasing $f(\alpha) < f(\beta)$ and $f(\alpha) \not= f(\beta)$.
Since $f$ is strictly increasing, so is $f^{-1}$. So if $\alpha < \beta$ then $f^{-1}(f(\alpha)) < f^{-1}(f(\beta))$.
This is because if there exists $\alpha$ and $\beta $ $\in (a,b)$ with $\alpha < \beta$ such that $f^{-1}(\alpha)$ = $\alpha '$ and $f^{-1}(\beta)$ = $\beta '$ and $\alpha ' < \beta '$ then
$\beta = f^{-1}(\beta ') \le f^{-1}(\alpha ') = \alpha$
which is a contradiction if $f$ is strictly increasing.
The remainder of the proof is some form of an epsilon delta proof to show that the inverse function is continuous from the left at the right end point. My attempt:
Let $b$ be the upper limit $ \in [a,b]$ and define $d = f(b)$.
Next, I want to show that $\lim_{x\rightarrow d^{-}}f^{-1}(x) = b$ for any $\epsilon >0$ such that $(b-\epsilon) \subset [a,b]$.
So, $f(b-\epsilon) < f(b)$.
Let $\delta = 1/2 (f(b)-f(b-\epsilon))$
Then $f(x_0-\epsilon) < f(x_0)-\delta$
So if $|x-d| < \delta$, then $|f^{-1}(x)-f^{-1}(d)|<\epsilon$
then continuity holds at $f^{-1}(d)$, which is possible by the Archimedean principle. Currently, I'm having trouble with the epsilon-delta proof. I don't think the argument is strong enough.
Best Answer
Let $R = f([a,b])$ be the range of $f$. Since $f$ is strictly increasing, we have $R \subset [f(a),f(b)]$, but in general $R \ne [f(a),f(b)]$. For example, let $f : [0,2] \to \mathbb{R}, f(x) = x$ for $x \in [0,1)$, $f(1) = 2$, $f(x) = x + 2$ for $x \in (1,2]$.
But although $R$ is general no interval, the usual definition of continuity makes sense for $f^{-1} : R \to [a,b]$. Moreover, as you remarked in your question, $f^{-1}$ is strictly increasing, i.e. for $y,y'\in R$ with $y < y'$ we have $f^{-1}(y) < f^{-1}(y')$.
Let us assume that $f^{-1}$ is not continuous. This means that exist $y \in R$ and $\epsilon > 0$ such that for all $\delta > 0$ there exists $y_\delta \in R$ such that $\lvert y - y_\delta \rvert < \delta$ and $\lvert f^{-1}(y) - f^{-1}(y_\delta) \rvert \ge \epsilon$. We can therefore find a sequence $(y_n)$ in $R \setminus \{ y \}$ such that $y_n \to y$ and $\lvert f^{-1}(y) - f^{-1}(y_n) \rvert \ge \epsilon$. W.lo.g. we may assume that infinitely many $y_n < y$. Passing to a suitable subsequence, we may assume that all $y_n < y$ and that $(y_n)$ is strictly increasing. Write $x_n = f^{-1}(y_n)$, $x = f^{-1}(y)$. The sequence $(x_n)$ is strictly increasing such that $x_n < x$. It therefore converges to some $\xi \le x$. We have $y_n = f(x_n) < f(\xi) \le f(x) = y$, and this implies $f(\xi) = y$ because $y_n \to y$. Hence $\xi = f^{-1}(y) = x$. We conclude $x_n \to x$. But $\lvert x - x_n \rvert = \lvert f^{-1}(y) - f^{-1}(y_n) \rvert \ge \epsilon$ which is a contradiction.