4.10 The continuous inverse function theorem

To conclude the chapter, we consider inverses of continuous functions. We shall prove the inverse function theorem, which allows us to deduce information about the inverse $f^{-1}$ of a continuous function $f$ .

Continuity and injectivity

In order to define its inverse, we need a function $f$ to be injective. A simple example of an injective function is a function which is either increasing or decreasing (in other words, it is strictly monotone). However, there are examples of functions which are injective but are not strictly monotone.

Exercise 4.107.

Find an injective function $f\colon(-1,1)\to\mathbb{R}$ which is not strictly monotone.

The trick to Exercise 4.107 is to consider functions which have a jump or break in the graph. That is, your solution should be discontinuous. Indeed, if $f$ is both injective and continuous, then it is intuitively clear that this forces $f$ to be either increasing or decreasing. We make these ideas precise using the intermediate value theorem, leading to the following result.

Proposition 4.108.

Let $I\subseteq\mathbb{R}$ be an interval and $f\colon I\to\mathbb{R}$ be a continuous, injective function. Then $f$ is strictly monotone.

By the result of Exercise 4.107, we see that Proposition 4.108 may fail if we do not assume $f$ is continuous.

Proof (of Proposition 4.108).

We may assume that $I$ contains at least two points, otherwise the claim is trivial. Let $a$ , $b\in I$ with $a<b$ . Since $f$ is injective, this implies that either $f(a)<f(b)$ or $f(a)>f(b)$ . We assume $f(a)<f(b)$ ; the remaining case can be treated similarly or by reflection (replacing $f$ with $-f$ ).

We argue by contradiction, assuming that $f$ is not strictly monotone on $[a,b]$ . This means there must exist some $c\in(a,b)$ such that

\text{either}\quad f(c)\leq f(a)<f(b)\qquad\text{or}\qquad f(a)<f(b)\leq f(c).

In the first case, the intermediate value theorem implies that there is $x_{1}\in[c,b)$ such that $f(x_{1})=f(a)$ . In the second case, the same argument gives some $x_{2}\in(a,c]$ such that $f(x_{2})=f(b)$ . Since $x_{1}\neq a$ and $x_{2}\neq b$ , in either case this contradicts the injectivity of $f$ . Hence $f$ is strictly monotone. ∎

Recall from Corollary 4.95 that continuous functions map intervals to intervals. If we start with an open interval, however, then the image may not be open.

Exercise 4.109.

Consider the continuous function $p_{2}\colon(-1,1)\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ . Show that the image $\mathrm{Im}(p_{2}):=\{p_{2}(x):x\in(-1,1)\}$ is an interval, but not an open interval. Draw a figure to illustrate this.

However, if we work with injective continuous functions, then we can use Proposition 4.108 to show that the open property is preserved.

Corollary 4.110.

Let $I\subseteq\mathbb{R}$ be an open interval and $f\colon I\to\mathbb{R}$ be continuous and injective. Then the image $\mathrm{Im}(f):=\{f(x):x\in I\}$ is an open interval.

By the result of Exercise 4.109, we see that Corollary 4.110 may fail if we do not assume $f$ is injective.

To prove Corollary 4.110 we shall use a simple characterisation of open intervals.

Lemma 4.111.

Let $I\subseteq\mathbb{R}$ be an interval. Then the following are equivalent:

1

$I$ is an open interval.
2

For all $x\in I$ , there exists some $r>0$ such that $(x-r,x+r)\subseteq I$ .
3

For all $x\in I$ , there exists some $\alpha$ , $\beta\in\mathbb{R}$ with $\alpha<x<\beta$ such that $\alpha$ , $\beta\in I$ .

Roughly speaking, property 2 tells us that for any point $x\in I$ , we can move slightly to the left or slightly to the right of $x$ and still stay within $I$ . Property 3 is closely related and says that $I$ does not have any maximum or minimum.

Proof (of Lemma 4.111).

for Let $a$ , $b\in\mathbb{R}$ with $a<b$ .

1. $\Rightarrow$ 2.

Consider the case $I=(a,b)$ . Let $x\in I$ so that $a<x<b$ . It follows that $x-a>0$ and $b-x>0$ and so $r:=\min\{x-a,b-x\}>0$ . Here $r$ corresponds to the shortest distance from $x$ to one of the endpoints of $I$ . Observe that

$a=x-(x-a)\leq x-r<x<x+r\leq x+(b-x)=b.$

and so $(x-r,x+r)\subseteq(a,b)$ . This shows $I$ satisfies 2.

If $I=(a,\infty)$ , $I=(-\infty,b)$ , $I=\mathbb{R}$ or $I=\emptyset$ , then similar (but easier) arguments show that $I$ again satisfies 2. This shows that all open intervals satisfy 2.
2. $\Rightarrow$ 3.

Let $x\in I$ and assume 2 holds, so there exists some $r>0$ such that $(x-r,x+r)\subseteq I$ . If we define $\alpha:=x-r/2$ and $\beta:=x+r/2$ , then it follows that $\alpha<x<\beta$ and $\alpha$ , $\beta\in(x-r,x+r)\subseteq I$ , so $\alpha$ , $\beta\in I$ as required.
3. $\Rightarrow$ 1.

We prove the contrapositive. If $I=[a,b]$ , $I=(-\infty,b]$ or $I=(a,b]$ , then taking $x=b$ it is clear that there does not exist any $\beta>x$ such that $\beta\in I$ . If $I=[a,\infty)$ or $I=[a,b)$ , then taking $x=a$ it is clear that there does not exist any $\alpha<x$ such that $\alpha\in I$ . This shows that whenever $I$ is not open, property 3 fails.

∎

Proof (of Corollary 4.110).

We know from Corollary 4.95 that $\mathrm{Im}(f)$ is an interval, so it remains to show that $\mathrm{Im}(f)$ is an open interval.

By Proposition 4.108 we know that $f$ is strictly monotone. We shall assume that $f$ is increasing; the case where $f$ is decreasing can be treated similarly or by reflection.

Let $y\in\mathrm{Im}(f)$ so that $y=f(x)$ for some $x\in I$ . Since $I$ is an open interval, by Lemma 4.111 there exist $\alpha$ , $\beta\in\mathbb{R}$ with $\alpha<x<\beta$ such that $\alpha$ , $\beta\in I$ .

We now consider the points $f(\alpha)$ , $f(\beta)\in\mathrm{Im}(f)$ . Since $f$ is increasing, $f(\alpha)<f(x)=y<f(\beta)$ . However, this means $\mathrm{Im}(f)$ satisfies condition 3 of Lemma 4.111 and so it must be an open interval, as required. ∎

The continuous inverse function theorem

Let $I$ , $J\subseteq\mathbb{R}$ be intervals and $f\colon I\to J$ be a bijective function. The graph of the inverse function $f^{-1}\colon J\to I$ is obtained by reflecting the graph of $f$ across the diagonal $y=x$ : see Figure 4.28. Suppose $f$ is continuous. Intuitively, the graph of $f$ has no breaks, and reflecting the graph should not result in new breaks appearing. Thus, we expect that $f^{-1}$ should also be continuous. The following theorem confirms that this is indeed the case.

Figure 4.28: The graph of

f^{-1}\colon J\to I

is formed by reflecting the graph of

f\colon I\to J

Theorem 4.112 (Continuous inverse function theorem).

Let $I$ , $J\subseteq\mathbb{R}$ be intervals and $f\colon I\to J$ be a continuous, bijective function. Then the inverse $f^{-1}\colon J\to I$ is continuous.

Proof.

We shall only show the result when $I\subseteq\mathbb{R}$ is an open interval. The same arguments can be used to prove the general case, but there are additional technicalities when dealing with endpoints.

Fix $b\in J$ and let $\varepsilon>0$ be given. Our goal is to find some $\delta>0$ such that

(4.21) (4.21)

|f^{-1}(y)-f^{-1}(b)|<\varepsilon\qquad\text{for all $y\in J$ with $|y-b|<% \delta$.}

By the surjectivity of $f$ there exists some $a\in I$ such that $b=f(a)$ . Since $I$ is an open interval, by Lemma 4.111 there exists some $r>0$ such that $(a-r,a+r)\subseteq I$ . Thus, if we define $\rho:=\min\{\varepsilon,r\}$ , then $0<\rho\leq\varepsilon$ and $(a-\rho,a+\rho)\subseteq I$ . Since $f$ is bijective, it is injective. Since $f$ is both continuous and injective, by Corollary 4.110, it maps the open interval $(a-\varepsilon,a+\varepsilon)$ to an open interval $U$ containing $b=f(a)$ .

Since $U\subseteq\mathbb{R}$ is an open interval and $b\in U$ , by Lemma 4.111 there exists some $\delta>0$ such that $(b-\delta,b+\delta)\subseteq U$ . We claim that (4.21) holds for this choice of $\delta$ .

Proving the claim is just a matter of unpacking the definitions. Suppose $y\in J$ satisfies $|y-b|<\delta$ . Then $y\in U$ and so $f^{-1}(y)\in f^{-1}(U)=(a-\rho,a+\rho)$ , using the fact that $f^{-1}$ is the inverse to $f$ . Thus,

|f^{-1}(y)-f^{-1}(b)|=|f^{-1}(y)-a|<\rho\leq\varepsilon,

as required. ∎

We can use the continuous inverse function theorem to show that many familiar functions are continuous.

Example 4.113.

Given $n\in\mathbb{N}$ , let $q_{n}\colon[0,\infty)\to[0,\infty)$ be the function $q_{n}(x):=x^{1/n}$ . Then $q_{n}$ is continuous. Indeed, $q_{n}$ is the inverse of the bijective polynomial map $p_{n}\colon[0,\infty)\to[0,\infty)$ given by $p_{n}(x):=x^{n}$ . By Example 4.53, we know $p_{n}$ is continuous, and therefore $q_{n}$ is continuous by the continuous inverse function theorem.

Remark 4.114.

Whilst it is possible to argue directly from the definition to show the functions $q_{n}$ from Example 4.113 are continuous, this is quite an involved and messy exercise. It is much easier to let the continuous inverse function theorem do the work for us!

Example 4.115.

Since we know from Lemma 4.33 that $\sin\colon[-\pi/2,\pi/2]\to[-1,1]$ and $\cos\colon[0,\pi]\to[-1,1]$ are both continuous, it follows from the continuous inverse function theorem that the inverse trigonometric functions $\arcsin\colon[-1,1]\to[-\pi/2,\pi/2]$ and $\arccos\colon[-1,1]\to[0,\pi]$ are both continuous.

Example 4.116.

Suppose we know that $\exp\colon\mathbb{R}\to(0,\infty)$ is continuous. We can then apply the continuous inverse function theorem to deduce that $\log\colon(0,\infty)\to\mathbb{R}$ is continuous. Thus, in order to prove the ‘borrowed’ Lemma 4.35, we only need to show $\exp\colon\mathbb{R}\to(0,\infty)$ is continuous. We shall establish this fact in the next chapter.