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.

  • $\begingroup$ The range of $f$ is certainly contained in $[f(a),f(b)]$, but why should it be the complete interval? $\endgroup$
    – Paul Frost
    Dec 10, 2018 at 22:35
  • $\begingroup$ Hi @PaulFrost. I'm a little unclear of your question. I think that knowing that the inverse of f is strictly increasing over the whole interval is how we can get the result that the inverse function is continuous at the end point. $\endgroup$
    – user624612
    Dec 10, 2018 at 23:08
  • $\begingroup$ The claim is that $f^{-1}$ is continuous over the range of original function. It is therefore not sufficient to consider only the end point. $\endgroup$
    – Paul Frost
    Dec 10, 2018 at 23:38
  • $\begingroup$ I see what you mean. In a similar problem (not posted here), we determined that a defined monotone strictly increasing function was differentiable, thus implying continuity. Basically here we just needed to further that proof for the more general case for a closed interval. Thank you for your answers by the way! @PaulFrost $\endgroup$
    – user624612
    Dec 11, 2018 at 15:56

1 Answer 1


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.

  • $\begingroup$ Well said. If it were me though, I wouldn't know where to go after finding that subsequence of $y_n$. $\endgroup$ Sep 30, 2020 at 8:47

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