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Let $f:\mathbb{R}\to\mathrm{S}$, where $\mathrm{S}\subseteq\mathbb{R}$, be a strictly increasing (decreasing) function. I wish to find out whether its inverse $f^{-1}$ is also strictly increasing (decreasing) or not. While a proof for this exists online, I will provide a counterexample that seems to contradict the conclusions in the proof.

The proof can be found in: https://proofwiki.org/wiki/Inverse_of_Strictly_Monotone_Function

I will for simplicity only consider a strictly increasing function. From the provided link above we have that: \begin{align*} x < y & \quad\Leftrightarrow\quad f(x) < f(y) \\ f^{-1}(x) < f^{-1}(y) & \quad\Leftrightarrow\quad f^{-1}(f(x)) < f^{-1}(f(y)) \\ f^{-1}(x) < f^{-1}(y) & \quad\Leftrightarrow\quad x < y \\ \end{align*} Thus, if $f$ is a strictly increasing function, then so is $f^{-1}$.

Now for the counterexample. With some loss of generality, let $f:\mathbb{R}\to\mathbb{R}_{++}$ be a strictly positive function, where $\mathbb{R}_{++}=\{ x\in\mathbb{R}: x>0\}$. Furthermore, for an inverse it follows that $f^{-1}(x)f(x)=f(x)f^{-1}(x)=1$. Then from \begin{align*} x < y & \quad\Leftrightarrow\quad f(x) < f(y) \end{align*} we can rewrite the right-hand side as \begin{align*} f^{-1}(y)=\frac{1}{f(y)}<\frac{1}{f(x)}=f^{-1}(x) \end{align*} which implies that if $f$ is a strictly increasing function, then its inverse $f^{-1}$ is strictly decreasing.

Please provide comments if and why my counterexample is invalid. Thank you in advance.

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    $\begingroup$ The reciprocal is different from the function inverse! A way to see the theorem heuristically is that the inverse is in some sense a reflection over the line $y=x$, and from there you can see the result with some doodling. $\endgroup$ Nov 2, 2016 at 14:20
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    $\begingroup$ Yes, I seem to have confused inverse with reciprocal. In the proof, how is one allowed to take $f^{−1}(\cdot)$ over $x<y$ on both sides without knowing if the "less than" sign should change direction? Doesn't this imply already that $f^{−1}$ is assumed to be strictly increasing? $\endgroup$
    – index
    Nov 2, 2016 at 14:36

2 Answers 2

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We may argue as follows:

Let $f:X \mapsto Y$ be a monotone increasing function ( hence injective ), and let $ y_1, y_2 \in Y$, such that $y_1 < y_2$. Then, $\exists \ x_1, x_2 \in X \ $ s.t. $x_1=f^{-1}(y_1), x_2=f^{-1}(y_2)$. Thus, we have $f(x_1)=y_1, f(x_2)=y_2$.

Now, we claim that $f^{-1}(y_1) < f^{-1}(y_2)$. Otherwise, $f^{-1}(y_1) \geq f^{-1}(y_2) \Rightarrow x_1 \geq x_2 \Rightarrow f(x_1) \geq f(x_2) \Rightarrow y_1 \geq y_2 $, which is obviously a contradiction, so this concludes our proof.

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It appears that you may be confusing the reciprocal function $\frac{1}{f(x)}$ with the inverse function (denoted $f^{-1}(x)$).

This latter notation, $f^{-1}$, is usually reserved for, and understood to be, the inverse function rather than the reciprocal function.


Edit:

In response to index's comment: I would prefer to structure the proof as:

We know $x<y \Leftrightarrow f(x)<f(y)$.

Hence $f^{-1}(x)<f^{-1}(y) \Leftrightarrow f(f^{-1}(x))<f(f^{-1}(y)) \Leftrightarrow x<y$.

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    $\begingroup$ Yes, I seem to have confused inverse with reciprocal. In the proof, how is one allowed to take $f^{-1}(\cdot)$ over $x<y$ on both sides without knowing if the "less than" sign should change direction? Doesn't this imply already that $f^{-1}$ is assumed to be strictly increasing? $\endgroup$
    – index
    Nov 2, 2016 at 14:08
  • $\begingroup$ Could you please comment a bit more on your edited part and your thinking? I don't see how your restructure of the proof answers my comment? $\endgroup$
    – index
    Nov 3, 2016 at 10:34

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