2
$\begingroup$

I have proved the following statement and I would like to know if my proof is correct and/or/if/how it can be improved.

"Suppose $f:\mathbb{R}\to\mathbb{R}$ is a strictly increasing function.

Prove that the inverse function $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ is a continuous function."

My proof:

Let $f:\mathbb{R}\to\mathbb{R}$ be a strictly increasing function: then it is injective and as a function $f:\mathbb{R}\to f(\mathbb{R})$ it must be surjective so it has an inverse $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ which must be strictly increasing too$^{(1)}$.

Suppose now that $f^{-1}$ were discontinuous at a point $y_d\in f(\mathbb{R})$: then, being an increasing function, $y_d$ must be a jump discontinuity so the interval $I_{y_d}:=(\lim\limits_{y \to y_d^-,\ y\in f(\mathbb{R})\\}f^{-1}(y),\lim\limits_{y \to y_d^+,\ y\in f(\mathbb{R})}f^{-1}(y))=(\sup_{y<y_d,\ y\in f(\mathbb{R})}f^{-1}(y),\inf_{y>y_d,\ y\in f(\mathbb{R})} f^{-1}(y))$ must be nonempty and we can pick an element $\bar{x}\neq x_d=f^{-1}(y_d)$ in it so $f(\bar{x})=y_d$, but being $f$ strictly increasing by hypothesis it is also either $f(\bar{x})>y_d$ or $f(\bar{x})<y_d$, a contradiction.

So, $f^{-1}(\mathbb{R})\to\mathbb{R}$ cannot be discontinuous at any point i.e. it must be continuous on $f(\mathbb{R})$. $\square$


$^{(1)}$ let $y_1,y_2\in f(\mathbb{R})$ and suppose wlog $y_1<y_2$: then $f^{-1}(y_1)=f^{-1}(f(x_1))=x_1$ and $f^{-1}(y_2)=f^{-1}(f(x_2))=x_2$ and if $x_1\geq x_2$ then $f(x_1)=y_1\geq y_2=f(x_2)$ contradiction, so it must be $x_1=f^{-1}(y_1)<x_2=f^{-1}(y_2)$

$\endgroup$
6
  • $\begingroup$ Is the function f continuous? $\endgroup$ Jun 15, 2021 at 9:21
  • $\begingroup$ @EduardoMaza no, it is not continuous $\endgroup$
    – lorenzo
    Jun 15, 2021 at 9:22
  • 1
    $\begingroup$ See also math.stackexchange.com/q/3719441/42969 and math.stackexchange.com/q/4116908/42969. $\endgroup$
    – Martin R
    Jun 15, 2021 at 9:28
  • 1
    $\begingroup$ Note that the left-sided and/or the right-sided limit may not exist, e.g. if $y_d$ is an isolated point of $f(\Bbb R)$. Similarly, the sets $\{ y< y_d \mid y \in f(\Bbb R) \}$ and/or $\{ y> y_d \mid y \in f(\Bbb R) \}$ may be empty, so that their supremum resp. infimum is not defined. $\endgroup$
    – Martin R
    Jun 15, 2021 at 18:42
  • 1
    $\begingroup$ @Martin R thank you for your comment; regarding your first remark, being $f^{-1}$ increasing, shouldn't the left and right-sides limit always exist ( math.stackexchange.com/q/4171854 ) (even if $y_d$ is an isolated point, see math.stackexchange.com/questions/27429/…)? $\endgroup$
    – lorenzo
    Jun 15, 2021 at 20:14

4 Answers 4

4
$\begingroup$

Consider the function $g:\Bbb R\to\Bbb R$ given by $$g(y):=\begin{cases}y-1 & y<0\\y & y\ge 0.\end{cases}$$

Clearly, $g$ is strictly increasing, is a bijection from the range of its inverse to $\Bbb R,$ and has a jump disontinuity at $y=0.$ Letting $y_d=0,$ we see that $$\left(\lim_{y\to y_d^-,\ y\in\operatorname{dom}g}g(y),\lim_{y\to y_d^+,\ y\in\operatorname{dom}g}g(y)\right)=(-1,0)$$ is certainly nonempty, but contains no elements of the range of $g$--that is, no element of the domain of the inverse of $g.$

That's the flaw in your argument. Just because an interval is non-empty doesn't mean that it contains an element in the domain of an arbitrary, strictly increasing function. Since all you've concluded about $f^{-1}$ is that it is strictly increasing, is a bijection from the range of its inverse to $\Bbb R,$ and it has a jump discontinuity, then it is entirely possible that $f^{-1}=g,$ in which case your argument falls down.

In other words, you haven't actually justified the statement

we can pick an element $\bar{x}\neq x_d=f^{-1}(y_d)$ in [the nonempty interval] so $f(\bar{x})=y_d,$

so haven't obtained your contradiction.

Added: A better approach would be to proceed directly. Take an arbitrary $y_0\in f[\Bbb R],$ and let $x_0:=f^{-1}(y_0).$ Since $f$ is strictly increasing, then for $x<x_0$ (resp., for $x>x_0$) we have $f(x)<y_0$ (resp. $f(x)>y_0$).

Take an arbitrary $\varepsilon>0,$ let $y_m:=f(x_0-\varepsilon),$ and let $y_M:=f(x_0+\varepsilon),$ so that $y_m,y_M\in f[\Bbb R]$ and $y_m<y_0<y_M.$

Letting $\delta=\min\{y_0-y_m,y_M-y_0\},$ we have $\delta>0,$ and for all $y\in\Bbb R,$ if $|y-y_0|<\delta,$ then $y_m<y<y_M.$

In particular, take any $y\in f[\Bbb R]$ such that $|y-y_0|<\delta,$ and let $x=f^{-1}(y).$ Since $f$ is strictly increasing and $f(x_0-\varepsilon)=y_m<y=f(x),$ then $x_0-\varepsilon<x.$ Similarly, $x<x_0+\varepsilon,$ and so $|x-x_0|<\varepsilon,$ or equivalently, $$\bigl|f^{-1}(y)-f^{-1}(y_0)\bigr|<\varepsilon,$$ whence we have showed that $f^{-1}$ is continuous at $y_0,$ as desired.

$\endgroup$
0
2
$\begingroup$

Call $B=f(\mathbb{R})$ and let $g : B \to \mathbb{R}$ be $f$'s inverse. Note that $g$ is also stricly increasing. Suppose $g$ is discontinuous at a point $y \in B$. Let $\alpha = \lim_{w \in B, w \to y^{-}} g(w)$ and $\beta = \lim_{z \in B, z \to y^{+}} g(z)$. Since $g$ is increasing and discontinuous at $y$, we have $\alpha < \beta$.

Take $x_0 \in \left [\alpha + \frac{\beta - \alpha}{3}, \beta - \frac{\beta - \alpha}{3} \right ]$.

Then, for small enough $\delta > 0$, it happens that $g(y - \delta) < x_0 < g(y + \delta)$. Note that $\delta$ doesn't depend on $x_0$.

Applying $f$ to the above inequality yields $y - \delta < f(x_0) < y + \delta$. Letting $\delta \to 0$, we find that $f(x_0)=y$. So $f$ is actually constant at the nondegenerate interval $\left [\alpha + \frac{\beta - \alpha}{3}, \beta - \frac{\beta - \alpha}{3} \right ]$!!

Here I assumed $y$ can be approximated with points of $B$ from both sides. But the same argument can be used if $y$ is only a left limit point or right limit point of $B$, to do this you just need to use $g(y)$ in place of $\beta$ and $\alpha$, respectively.

$\endgroup$
1
  • $\begingroup$ Nice argument ! $\endgroup$
    – Medo
    Jul 31, 2021 at 19:44
2
$\begingroup$

First I would like to thank @Cameron Buie for his effort to point out the flaw in the OP proof.

Of course the statement is false. Monotonicity implies continuity almost everywhere. Even a strictly monotone function is not necessarily continuous. Neither is its inverse which is strictly monotone too.

Here is a simple counterexample:

$$ f(x)=\left\{ \begin{array}{ll} x+2, & \hbox{$x>0$;} \\ 1, & \hbox{$x=0$;} \\ x, & \hbox{$x<0$.} \end{array} \right.$$ The function $f:\mathbb{R}\rightarrow ]-\infty,0[\,\cup \,\{1\}\cup\,]2,\infty[\ $ is a strictly increasing bijection.

The strictly increasing inverse $f^{-1}: ]-\infty,0[\,\cup \,\{1\}\,\cup\,]2,\infty[ \rightarrow \mathbb{R} $

$$f^{-1}(x)=\left\{ \begin{array}{ll} x-2, & \hbox{$x>2$;} \\ 0, & \hbox{$x=1$;} \\ x, & \hbox{$x<0$.} \end{array} \right.$$ is discontinuous

$\endgroup$
7
  • $\begingroup$ It seems to me that $-2<0$ but $f(-2)=2>1=f(0)$ so $f$ is not strictly increasing $\endgroup$
    – lorenzo
    Jul 31, 2021 at 0:59
  • $\begingroup$ The typo $-x$ is corrected to $x$. $\endgroup$
    – Medo
    Jul 31, 2021 at 3:44
  • $\begingroup$ Let $g=f^{-1}$: then $g$ is clearly true for $x<0$ and also $\lim_{x\to 0^-}=0=\lim_{x\to 0^+}g(x)$ (it is vacuously true that the right-hand limit exists and is equal to $0$) so it is also continuous at $0$; it is also vacuously true that $\lim_{x\to 1^-}g(x)=\lim_{x\to 1^+}g(x)=g(1)=0$ (take any $\varepsilon >0$ and $\delta =\frac{1}{2}$); it is clearly continuous for $x>2$ and $\lim_{x\to 2^-}g(x)=\lim_{x\to 2^+}g(x)=g(2)=0$ so $g=f^{-1}$ is indeed continuous. $\endgroup$
    – lorenzo
    Jul 31, 2021 at 10:58
  • $\begingroup$ @ lorenzo. No, because the limits $\lim_{x\rightarrow 0^{+}}g$ and $\lim_{x\rightarrow 2^{-}}g$ do not exist. Let me prove discontinuity in another way for you. What is the inverse image of the open interval $]1/2,3/2[$ ? It is the set $\{1\}\cup\,]5/2,7/2[$ which is not an open set. Convinced ? $\endgroup$
    – Medo
    Jul 31, 2021 at 19:38
  • $\begingroup$ The reasoning @lorenzo used is not correct, as you say, but their conclusion is correct. There are some subtle points that you missed, Medo. First, the preimage of $(1/2,3/2)$ under $g$ is the open set $(5/2,7/2).$ On the other hand, the preimage of $(-1,1)$ under $g$ is $(-1,0)\cup\{1\}\cup(2,3),$ which is not open in $\Bbb R,$ but is open in $dom(g)$ taken as a subspace of $\Bbb R,$ since (for example) $$dom(g)\cap(-1,3)=(-1,0)\cup\{1\}\cup(2,3).$$ The kicker is that $g$ is (vacuously) continuous at $1$ even though the limits don't exist, since $1$ is isolated in $dom(g).$ $\endgroup$ Aug 21, 2023 at 18:16
0
$\begingroup$

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 23 on p.40 in Exercises 2B in this book.

Exercise 23
Suppose $f:\mathbb{R}\to\mathbb{R}$ is a strictly increasing function. Prove that the inverse function $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ is a continuous function.
[Note that this exercise does not have as a hypothesis that $f$ is continuous.]

My proof:

I use this Ramiro's idea.
Obviously, $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ is also a strictly increasing function.
Assume that $f^{-1}$ is not continuous at $y_0\in f(\mathbb{R})$.
Since $f^{-1}$ is not continuos at $y_0$, $y_0$ is a left limit point of $f(\mathbb{R})$ or a right limit point of $f(\mathbb{R})$.
If $y_0$ is a left limit point of $f(\mathbb{R})$, then $\lim_{y\to y_0-} f^{-1}(y)$ exists and $\lim_{y\to y_0-} f^{-1}(y)=\sup\{f^{-1}(y):y<y_0,y\in f(\mathbb{R})\}.$
If $y_0$ is a right limit point of $f(\mathbb{R})$, then $\lim_{y\to y_0+} f^{-1}(y)$ exists and $\lim_{y\to y_0+} f^{-1}(y)=\inf\{f^{-1}(y):y_0<y,y\in f(\mathbb{R})\}.$

(1) We consider the case in which $y_0$ is a left limit point of $f(\mathbb{R})$ and a right limit point of $f(\mathbb{R}).$
$\lim_{y\to y_0-} f^{-1}(y)\leq f^{-1}(y_0)\leq\lim_{y\to y_0+} f^{-1}(y)$ holds.
Since $f^{-1}$ is not continuous at $y_0$, $\lim_{y\to y_0-} f^{-1}(y)=\lim_{y\to y_0+} f^{-1}(y)=f^{-1}(y_0)$ does not hold.
So, $\lim_{y\to y_0-} f^{-1}(y)< f^{-1}(y_0)$ or $f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y)$ holds.

(2) We consider the case in which $y_0$ is a left limit point of $f(\mathbb{R})$, but is not a right limit point of $f(\mathbb{R}).$
$\lim_{y\to y_0-} f^{-1}(y)\leq f^{-1}(y_0)$ holds.
Since $f^{-1}$ is not continous at $y_0$, $\lim_{y\to y_0-} f^{-1}(y)=f^{-1}(y_0)$ does not hold.
So, $\lim_{y\to y_0-} f^{-1}(y)< f^{-1}(y_0)$ holds.

(3) We consider the case in which $y_0$ is a right limit point of $f(\mathbb{R})$, but is not a left limit point of $f(\mathbb{R}).$
$f^{-1}(y_0)\leq\lim_{y\to y_0+} f^{-1}(y)$ holds.
Since $f^{-1}$ is not continous at $y_0$, $f^{-1}(y_0)=\lim_{y\to y_0+} f^{-1}(y)$ does not hold.
So, $f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y)$ holds.

Therefore, at least, (a) or (b) holds:

(a) $y_0$ is a left limit point and $\lim_{y\to y_0-} f^{-1}(y)<f^{-1}(y_0).$
(b) $y_0$ is a right limit point and $f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y).$

First we consider the case (a):
Let $x_0\in (\lim_{y\to y_0-} f^{-1}(y), f^{-1}(y_0))$.
Then, since $x_0<f^{-1}(y_0)$, $f(x_0)<f(f^{-1}(y_0))=y_0.$
And $f(x_0)\in f(\mathbb{R})$.
So, $x_0=f^{-1}(f(x_0))\in\{f^{-1}(y):y<y_0,y\in f(\mathbb{R})\}.$
Note that $\lim_{y\to y_0-} f^{-1}(y)=\sup\{f^{-1}(y):y<y_0,y\in f(\mathbb{R})\}$ holds.
So, $x_0\leq\lim_{y\to y_0-} f^{-1}(y).$
This is a contradiction.

Next we consider the case (b):
Let $x_0\in (f^{-1}(y_0), \lim_{y\to y_0+} f^{-1}(y))$.
Then, since $f^{-1}(y_0)<x_0$, $y_0=f(f^{-1}(y_0))<f(x_0).$
And $f(x_0)\in f(\mathbb{R})$.
So, $x_0=f^{-1}(f(x_0))\in\{f^{-1}(y):y_0<y,y\in f(\mathbb{R})\}.$
Note that $\lim_{y\to y_0+} f^{-1}(y)=\inf\{f^{-1}(y):y_0<y,y\in f(\mathbb{R})\}$ holds.
So, $\lim_{y\to y_0+} f^{-1}(y)\leq x_0.$
This is a contradiction.

So, $f^{-1}$ is continuous at $y$ for any $y\in f(\mathbb{R}).$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .