Continuity of the inverse function – "Math for Non-Geeks"

In this chapter, we will introduce a theorem that gives a sufficient condition for the continuity of the inverse function of a bijective function. In mathematical literature, this theorem is often referred to as the "Inverse function Theorem". What is astounding about this result is that the inverse function of a discontinuous function can actually be continuous.

Motivation

The continuous and invertible function $f$ with the domain $[-1,0]\cup (1,2]$

First we want to consider the most general condition possible for when a bijective function $f:D\to W$ with $D,W\subseteq ℝ$ has a continuous inverse function. The first ansatz that we naturally want to investigate is the continuity of $f$ itself. We might spontaneously assume that the continuity of the inverse function $f^{-1}$ follows directly from the continuity of $f$. This is not necessarily the case, as is shown in the following example:

$\begin{array}{cc}f& :([-1,0]\cup (1,2])\to [-1,1]\\ & :x\mapsto f(x)=\{\begin{array}{cc}x& -1\le x\le 0\\ x-1& 1<x\le 2\end{array}\end{array}$

The discontinuous inverse function $g=f^{-1}$

The function is continuous since it's continuous on both $[-1,0]$ and $(1,2]$. Furthermore, it is strictly monotonically increasing, so it's injective. The image of $f$ is $[-1,1]$ and therefore $f:([-1,0]\cup (1,2])\to [-1,1]$ is surjective. All together, $f$ is bijective and therefore invertible. The inverse mapping $f^{-1}:[-1,1]\to ([-1,0]\cup (1,2])$ has functional values:

$f^{-1}(y)=\{\begin{array}{cc}y& -1\le y\le 0\\ y+1& 0<y\le 1\end{array}$

The inverse function is not continuous since it has a jump discontinuity at $y=0$. So it can definitely be the case that a continuous function has a discontinuous inverse.

But now another thing is apparent: the inverse function $f^{-1}=g$ is also invertible with the inverse function $g^{-1}={\left(f^{-1}\right)}^{-1}=f$. This means, however, that a discontinuous function (like $g$) can have a continuous inverse function. Hence, we posit:

The (dis)continuity of an invertible function has in no capacity any influence on the (dis)continuity of its inverse function.

The problem lies in the domain of $f$. Namely, this is $[-1,0]\cup (1,2]$, i.e. not an interval. The domain is not connected and has a "gap." How does this "gap" affect the inverse function $f^{-1}$?

Since the domain of $f$ corresponds to what should be the codomain of $f^{-1}$, the inverse function also has this "gap" in its codomain. Similarly, the codomain of $f$ corresponds to the domain of $f^{-1}$. It is for this reason that $f^{-1}$ can not be continuous. The domain of $f$ is an interval and therefore connected, while its codomain contains a gap. Since $f^{-1}$ is surjective and must attain every value in what should be its codomain, the graph of $f^{-1}$ must lend itself to a jump point somewhere. This is why $f^{-1}$ is not continuous.

Thus, we must require that, by assumption, the domain of $f$ is an interval to ensure that we don't encounter any problems. Usually this requirement is sufficient to guarantee the continuity of $f^{-1}$. We can prove this using the $ϵ$-$\delta$ characterization of continuity. With the help of the intermediate value theorem we can furthermore conclude that the codomain of $f$ - and therefore the domain of $f^{-1}$ - is an interval.

Theorem on the Continuity of the Inverse Function

Theorem (Continuity of the Inverse Function)

Let $D\subseteq ℝ$ be an interval and $f:D\to W$ a surjective, strictly monotone, and continuous mapping. Then $f$ is bijective, and the inverse function

$f^{-1}:W\to D$

is continuous and in the same sense as $f$ strictly monotone. Furthermore, $f(D)=W$ is an interval.

Proof (Continuity of the Inverse Function)

Proof step: $f$ is bijective

Let $x,y\in D$ with $x\ne y$. Then it either mus be the case $x<y$ or $x>y$. Since $f$ is strictly monotone, it follows $f(x)<f(y)$ or $f(x)>f(y)$. In either case it holds $f(x)\ne f(y)$. So $f$ is injective. Since $f$ is surjective by assumption, the mapping is bijective.

Proof step: $f^{-1}$ is strictly monotone

Without loss of generality let $f$ be strictly monotone increasing, i.e. $f(x)<f(y)$ follows from $x,y\in D$ with $x<y$. Now let $\tilde{x},\tilde{y}\in W$ with $\tilde{x}<\tilde{y}$. We must now show that $f^{-1}(\tilde{x})<f^{-1}(\tilde{y})$.

There exist some $x,y\in D$ with $f(x)=\tilde{x}$ and $f(y)=\tilde{y}$. Therefore, $f^{-1}(\tilde{x})=f^{-1}(f(x))=x$ and $f^{-1}(\tilde{y})=f^{-1}(f(y))=y$. Since $f^{-1}$ is injective, it holds $x\ne y$. It also can not be the case that $x>y$ holds. Otherwise, it must then hold $\tilde{x}=f(x)>f(y)=\tilde{y}$, which can not be the case because $\tilde{x}<\tilde{y}$.

Thus, $x<y$ and $f^{-1}(\tilde{x})<f^{-1}(\tilde{y})$. This proves that $f^{-1}$ is strictly monotone increasing like $f$. Proving that $f^{-1}$ is strictly monotone decreasing when we assume $f$ is strictly monotone decreasing is analogous.

Proof step: $f^{-1}$ is continuous

We will use the $ϵ$-$\delta$-characterization of continuity. To that end, let $ϵ>0$ and let $b\in W=f(D)$ be arbitrary. Further, let $a=f^{-1}(b)$ be the preimage of $b$ under $f$, i.e. $f(a)=b$. Now we have to show that there exists some $\delta >0$ such that $|f^{-1}(y)-f^{-1}(b)|<ϵ$ holds for all $y\in W$ with $|y-b|<\delta$. Then we will have achieved showing $f^{-1}$ is continuous in $b$. Since $b\in W$ can be chosen arbitrarily, from there we can conclude the continuity of the mapping $f^{-1}:W\to D$. We must, however, differentiate between the two cases of whether $a$ is a boundary point of $D$:

Case 1: $a$ is not a boundary point of $D$

Since $a$ is not a boundary point of $D$, there exists some $\widetilde{ϵ}$ with $0<\widetilde{ϵ}<ϵ$ and $]a-\widetilde{ϵ},a+\widetilde{ϵ}[\subseteq D$. We set $b_1=f(a-\widetilde{ϵ})$ and $b_2=f(a+\widetilde{ϵ})$. If $f$ is strictly monotone increasing, then it holds $b_1<b<b_2$. If $f$ is strictly monotone decreasing, then it holds $b_2<b<b_1$. Thus, $f$ maps the interval $[a-\widetilde{ϵ},a+\widetilde{ϵ}]$ bijectively to $[b_1,b_2]$ and $[b_2,b_1]$. If we set

$\delta =min\{|b-b_1|,|b-b_2|\}$

then it follows

$f^{-1}(]b-\delta ,b+\delta [)\subseteq ]a-\widetilde{ϵ},a+\widetilde{ϵ}[\subseteq ]a-ϵ,a+ϵ[$

Hence, the condition $|f^{-1}(y)-f^{-1}(b)|<ϵ$ for all $y\in W$ with $|y-b|<\delta$ is fulfilled and $f^{-1}$ is continuous in $b$.

Case 2: $a$ is a boundary point of $D$

We may assume without loss of generality that $a$ is the maximum of $D$. Now there exists some $\widetilde{ϵ}$ with $0<\widetilde{ϵ}<ϵ$ such that $a-\widetilde{ϵ}$ lies in $D$. Therefore, the value $f(a-\widetilde{ϵ})$ exists, and we can set $\delta =|f(a)-f(a-\widetilde{ϵ})|$.

If $f$ is strictly monotone increasing, then $f$ maps the interval $[a-\widetilde{ϵ},a]$ bijectively to $[f(a-\widetilde{ϵ}),f(a)]=[b-\delta ,b]$. Then it holds $f^{-1}([b-\delta ,b])=[a-\widetilde{ϵ},a]\subseteq (a-ϵ,a+ϵ)$. Because no $y\in W$ exists with $y>b$, it must therefore hold $|f^{-1}(y)-f^{-1}(b)|<ϵ$ for all $y\in W$ with $|y-b|<\delta$. This proves the continuity of $f^{-1}$ in $b$. The proof for the case where $f$ is monotone decreasing is analogous.

Examples

Root Functions

Example (Root Functions)

If $k\in \mathbb{N}$ with $k\ge 2$, then the $k$-th power function is defined as

$f:{ℝ}^{+}\to {ℝ}^{+}:x\mapsto f(x)=x^k$

This is a strictly monotonically increasing function, since for $x,y\in {ℝ}^{+}$ with $x<y$ it holds

$f(x)=x^k=\underset{k\text{ times}}{\underbrace{x\cdot \dots \cdot x}}<\underset{k\text{ times}}{\underbrace{y\cdot \dots \cdot y}}=y^k=f(y)$

Furthermore, this means $f$ is injective. The function $f$ is also surjective, since for every $y\in {ℝ}^{+}$ it holds for $x=\sqrt[k]{y}$:

$f(x)=f\left(\sqrt[k]{y}\right)={\left(\sqrt[k]{y}\right)}^k=y$

The inverse function is the $k$-th root function $f^{-1}:{ℝ}^{+}\to {ℝ}^{+}$ with $f(x)=\sqrt[k]{x}$. By the Inverse Function Theorem, this function is continuous and also strictly monotone increasing.

Natural Logarithm Functions

Example (Natural Logarithm Functions)

The exponential function

$f:ℝ\to {ℝ}^{+},f(x)=\exp (x)$

is strictly monotone increasing. Let $x,y\in {ℝ}^{+}$ with $x<y$. Then there exists some $t\in {ℝ}^{+}$ with $y=x+t$. It therefore holds

$\exp (y)=\exp (x+t)\underset{\text{equation}}{\stackrel{\text{functional}}{=}}\exp (x)\cdot \underset{>1}{\underbrace{\exp (t)}}>\exp (x)$

From the strict monotonicity it follows that $\exp$ is injective. Furthermore, the exponential function is surjective since:

$\begin{array}{ccc}\underset{x\to \mathrm{\infty }}{lim}\exp (x)& =\mathrm{\infty }\underset{x\to -\mathrm{\infty }}{lim}\exp (x)& =0\end{array}$

The surjectivity follows from the continuity of the exponential function and the Intermediate Value Theorem. The inverse function of the exponential function is the so-called natural logarithm function:

$f^{-1}:{ℝ}^{+}\to ℝ,f(x)=\ln x$

This function is also strictly monotone increasing (which is a result of the strict monotonicity of the exponential function). Furthermore, we can apply the Inverse Function Theorem and conclude that the logarithm function is also continuous.

Exercises

Exercise 1

Exercise (Continuity of the Inverse Function)

Let $f:ℝ\to ℝ$ be defined by the following values:

$f(x)=\frac{x}{1+|x|}$

Prove the following claims:

1. Show that $f$ is continuous, strictly monotone increasing, and injective on $ℝ$.
2. Show that $f:ℝ\to (-1,1)$ is surjective.
3. Give a justification for why the inverse function $f^{-1}:(-1,1)\to ℝ$ exists and determine the explicit functional values for $f^{-1}$. Show that $f^{-1}$ is continuous and strictly monotone increasing.

Solution (Continuity of the Inverse Function)

Solution sub-exercise 1:

$f$ is continuous since it is the quotient of continuous functions $a:ℝ\to ℝ:x\mapsto x$ and $b:ℝ\to ℝ:x\mapsto 1+|x|$. $1+|x|\ne 0$ thereby holds for all $x\in ℝ$. Let $x,y\in ℝ$ with $x<y$. To prove the strict monotonicity we will show $f(x)<f(y)$:

Case 1: $x,y\ge 0$

Then it holds

$x<y⟺\underset{=x(1+y)}{\underbrace{x+xy}}<\underset{=y(1+x)}{\underbrace{y+xy}}⟺\underset{=f(x)}{\underbrace{\frac{x}{1+x}}}<\underset{=f(y)}{\underbrace{\frac{y}{1+y}}}$

Case 2: $x\le 0<y$

Here it holds

$f(x)=\underset{\le 0}{\underbrace{\frac{x}{1+|x|}}}<\underset{>0}{\underbrace{\frac{y}{1+|y|}}}=f(y)$

Case 3: $x<0\le y$

Here it holds

$f(x)=\underset{<0}{\underbrace{\frac{x}{1+|x|}}}<\underset{\ge 0}{\underbrace{\frac{y}{1+|y|}}}=f(y)$

Case 4: $x,y<0$

It holds

$x<y⟺\underset{=x(1-y)}{\underbrace{x-xy}}<\underset{=y(1-x)}{\underbrace{y-xy}}⟺\underset{=f(x)}{\underbrace{\frac{x}{1-x}}}<\underset{=f(y)}{\underbrace{\frac{y}{1-y}}}$

So then $f$ is strictly monotone increasing on $ℝ$ and therefore also injective.

Solution sub-exercise 2:

It holds for all $x\in ℝ$:

$|f(x)|=\frac{|x|}{1+|x|}<\frac{|x|}{|x|}=1$

It follows immediately that $-1<f(x)<1$ holds. Furthermore,

$\begin{array}{cc}\underset{x\to \mathrm{\infty }}{lim}f(x)& =\underset{x\to \mathrm{\infty }}{lim}\frac{x}{1+|x|}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to \mathrm{\infty }}{x>0}}{lim}\frac{x}{1+|x|}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to \mathrm{\infty }}{x>0}}{lim}\frac{x}{1+x}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to \mathrm{\infty }}{x>0}}{lim}\frac{1}{\frac{1}{x}+1}=1\end{array}$

and

$\begin{array}{cc}\underset{x\to -\mathrm{\infty }}{lim}f(x)& =\underset{x\to -\mathrm{\infty }}{lim}\frac{x}{1+|x|}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to -\mathrm{\infty }}{x<0}}{lim}\frac{x}{1+|x|}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to -\mathrm{\infty }}{x<0}}{lim}\frac{x}{1-x}\\ & =\underset{\genfrac{}{}{0ex}{}{x\to -\mathrm{\infty }}{x<0}}{lim}\frac{1}{\frac{1}{x}-1}=-1\end{array}$

Since $f$ is continuous, by the Intermediate Value Theorem, for every $y\in (-1,1)$ there exists an $x\in ℝ$ with $f(x)=y$. So it holds for the image that $f(ℝ)=(-1,1)$, whereby $f:ℝ\to (-1,1)$ is surjective.

Solution sub-exercise 3:

Since $f$ is bijective, the inverse function $f^{-1}:(-1,1)\to ℝ$ exists. As an inverse function, $f^{-1}$ is also bijective. By the Theorem of Continuity of the Inverse Function, $f^{-1}$ is furthermore continuous and strictly monotone increasing. In order to calculate $f^{-1}$ we must first differentiate between two cases:

Case 1: $x\ge 0$

$\begin{array}{cc}& y=f(x)=\frac{x}{1+x}\\ ⟺& y+xy=x\\ ⟺& y=x-xy=x(1-y)\\ ⟺& x=\underset{=f^{-1}(y)}{\underbrace{\frac{y}{1-y}}}\end{array}$

Case 2: $x<0$

$\begin{array}{cc}& y=f(x)=\frac{x}{1-x}\\ ⟺& y-xy=x\\ ⟺& y=x+xy=x(1+y)\\ ⟺& x=\underset{=f^{-1}(y)}{\underbrace{\frac{y}{1+y}}}\end{array}$

If we now have $x\ge 0$, then $f(x)=y\ge 0$ and therefore $1-y=1-|y|$. On the other hand, if $x<0$, then so is $f(x)=y<0$ and therefore $1+y=1-|y|$. All together, we can conclude

$f^{-1}:(-1,1)\to ℝ,f^{-1}(y)=\frac{y}{1-|y|}$

Exercise 2

Exercise (Continuity of the Inverse Function 2)

Let $g:[0,\mathrm{\infty })\to ℝ$ with $g(x)=\exp (x)-\sin (\exp (-x))$

1. Show that $g$ is injective.
2. Determine the image $g([0,\mathrm{\infty }))$.
3. Justify why the inverse function $g^{-1}:g([0,\mathrm{\infty }))\to [0,\mathrm{\infty })$ is continuous.

Solution (Continuity of the Inverse Function 2)

Solution sub-exercise 1:

$g$ is continuous as it is the composition of the continuous functions $a:[0,\mathrm{\infty })\to ℝ:x\mapsto -x$, $b:[0,\mathrm{\infty })\to ℝ:x\mapsto \exp (x)$ and $c:[0,\mathrm{\infty })\to ℝ:x\mapsto \sin (x)$. Furthermore, for $x,y\in [0,\mathrm{\infty })$ with $x<y$ it holds:

$x<y⟹-x>-y⟹\exp (-x)>\exp (-y)$

Now it holds $\exp (-\tilde{x})\in (0,1]$ for $\tilde{x}\in [0,\mathrm{\infty })$. Since the sine function is strictly monotone increasing on the half-interval $(0,1]$, it follows

$\exp (-x)>\exp (-y)⟹\sin (\exp (-x))>\sin (\exp (-y))⟹-\sin (\exp (-x))<-\sin (\exp (-y))$

Because the exponential function is strictly monotone increasing, it furthermore holds that $\exp (x)<\exp (y)$. Then we have:

$\underset{=g(x)}{\underbrace{\exp (x)-\sin (\exp (-x))}}<\underset{=g(y)}{\underbrace{\exp (y)-\sin (\exp (-y))}}$

So $g$ is strictly monotone increasing and therefore injective.

Solution sub-exercise 2:

= First, we observe

$g(0)=\exp (0)-\sin (\exp (0))=1-\sin (1)$

Furthermore,

$\begin{array}{cc}\underset{x\to \mathrm{\infty }}{lim}\exp (x)& =\mathrm{\infty }\\ \underset{x\to \mathrm{\infty }}{lim}\exp (-x)& =0\end{array}$

and from there it follows that

$\underset{x\to \mathrm{\infty }}{lim}g(x)=\underset{x\to \mathrm{\infty }}{lim}[\underset{\to \mathrm{\infty }}{\underbrace{\exp (x)}}-\underset{\to \sin (0)=0}{\underbrace{\sin (\exp (-x))}}]=\mathrm{\infty }$

Since $g$ is continuous and $[0,\mathrm{\infty })$ is an interval, $g([0,\mathrm{\infty }))$ is also an interval (see the implications of the intermediate value theorem). Since $g$ is strictly monotone increasing and $\underset{x\to \mathrm{\infty }}{lim}g(x)=\mathrm{\infty }$, it follows

$g([0,\mathrm{\infty }))=[g(0),\mathrm{\infty })=[1-\sin (1),\mathrm{\infty })$

Solution sub-exercise 3:

Since $D=[0,\mathrm{\infty })$ is an interval and $g:[0,\mathrm{\infty })\to [1-\sin (1),\mathrm{\infty })$ is bijective, the inverse function $g^{-1}:[1-\sin (1),\mathrm{\infty })\to [0,\mathrm{\infty })$ is continuous by the Continuous Inverse Function Theorem.