# Continuity of the inverse function – Serlo

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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 ${\displaystyle f}$ with the domain ${\displaystyle [-1,0]\cup (1,2]}$

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

{\displaystyle {\begin{aligned}f&:([-1,0]\cup (1,2])\to [-1,1]\\[1em]&:x\mapsto f(x)={\begin{cases}x&-1\leq x\leq 0\\x-1&1
The discontinuous inverse function ${\displaystyle g=f^{-1}}$

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

${\displaystyle f^{-1}(y)={\begin{cases}y&-1\leq y\leq 0\\y+1&0

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

But now another thing is apparent: the inverse function ${\displaystyle f^{-1}=g}$ is also invertible with the inverse function ${\displaystyle g^{-1}=\left(f^{-1}\right)^{-1}=f}$. This means, however, that a discontinuous function (like ${\displaystyle g}$) 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 ${\displaystyle f}$. Namely, this is ${\displaystyle [-1,0]\cup (1,2]}$, i.e. not an interval. The domain is not connected and has a "gap." How does this "gap" affect the inverse function ${\displaystyle f^{-1}}$?

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

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

## Theorem on the Continuity of the Inverse Function

Theorem (Continuity of the Inverse Function)

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

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

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

Proof (Continuity of the Inverse Function)

Proof step: ${\displaystyle f}$ is bijective

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

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

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

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

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

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

We will use the ${\displaystyle \epsilon }$-${\displaystyle \delta }$-characterization of continuity. To that end, let ${\displaystyle \epsilon >0}$ and let ${\displaystyle b\in W=f(D)}$ be arbitrary. Further, let ${\displaystyle a=f^{-1}(b)}$ be the preimage of ${\displaystyle b}$ under ${\displaystyle f}$, i.e. ${\displaystyle f(a)=b}$. Now we have to show that there exists some ${\displaystyle \delta >0}$ such that ${\displaystyle |f^{-1}(y)-f^{-1}(b)|<\epsilon }$ holds for all ${\displaystyle y\in W}$ with ${\displaystyle |y-b|<\delta }$. Then we will have achieved showing ${\displaystyle f^{-1}}$ is continuous in ${\displaystyle b}$. Since ${\displaystyle b\in W}$ can be chosen arbitrarily, from there we can conclude the continuity of the mapping ${\displaystyle f^{-1}:W\to D}$. We must, however, differentiate between the two cases of whether or not ${\displaystyle a}$ is a boundary point of ${\displaystyle D}$:

Fall 1: ${\displaystyle a}$ is not a boundary point of ${\displaystyle D}$

Since ${\displaystyle a}$ is not a boundary point of ${\displaystyle D}$, there exists some ${\displaystyle {\tilde {\epsilon }}}$ with ${\displaystyle 0<{\tilde {\epsilon }}<\epsilon }$ and ${\displaystyle ]a-{\tilde {\epsilon }},a+{\tilde {\epsilon }}[\subseteq D}$. We set ${\displaystyle b_{1}=f(a-{\tilde {\epsilon }})}$ and ${\displaystyle b_{2}=f(a+{\tilde {\epsilon }})}$. If ${\displaystyle f}$ is strictly monotone increasing, then it holds ${\displaystyle b_{1}. If ${\displaystyle f}$ is strictly monotone decreasing, then it holds ${\displaystyle b_{2}. Thus, ${\displaystyle f}$ maps the interval ${\displaystyle [a-{\tilde {\epsilon }},a+{\tilde {\epsilon }}]}$ bijectively to ${\displaystyle [b_{1},b_{2}]}$ and ${\displaystyle [b_{2},b_{1}]}$. If we set

${\displaystyle \delta =\min\{|b-b_{1}|,|b-b_{2}|\}}$

then it follows

${\displaystyle f^{-1}(]b-\delta ,b+\delta [)\subseteq ]a-{\tilde {\epsilon }},a+{\tilde {\epsilon }}[\subseteq ]a-\epsilon ,a+\epsilon [}$

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

Fall 2: ${\displaystyle a}$ is a boundary point of ${\displaystyle D}$

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

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

## Examples

### Root Functions

Example (Root Functions)

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

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

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

${\displaystyle f(x)=x^{k}=\underbrace {x\cdot \ldots \cdot x} _{k-{\text{mal}}}<\underbrace {y\cdot \ldots \cdot y} _{k-{\text{mal}}}=y^{k}=f(y)}$

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

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

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

### Natural Logarithm Functions

Example (Natural Logarithm Functions)

The exponential function

${\displaystyle f:\mathbb {R} \to \mathbb {R} ^{+},\ f(x)=\exp(x)}$

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

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

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

{\displaystyle {\begin{aligned}\lim _{x\to \infty }\exp(x)&=\infty \lim _{x\to -\infty }\exp(x)&=0\end{aligned}}}

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:

${\displaystyle f^{-1}:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)=\ln x}$

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 ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ be defined by the following values:

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

Prove the following claims:

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

Solution (Continuity of the Inverse Function)

Solution sub-exercise 1:

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

Fall 1: ${\displaystyle x,y\geq 0}$

Then it holds

${\displaystyle x

Fall 2: ${\displaystyle x\leq 0

Here it holds

${\displaystyle f(x)=\underbrace {\frac {x}{1+|x|}} _{\leq 0}<\underbrace {\frac {y}{1+|y|}} _{>0}=f(y)}$

Fall 3: ${\displaystyle x<0\leq y}$

Here it holds

${\displaystyle f(x)=\underbrace {\frac {x}{1+|x|}} _{<0}<\underbrace {\frac {y}{1+|y|}} _{\geq 0}=f(y)}$

Fall 4: ${\displaystyle x,y<0}$

It holds

${\displaystyle x

So then ${\displaystyle f}$ is strictly monotone increasing on ${\displaystyle \mathbb {R} }$ and therefore also injective.

Solution sub-exercise 2:

It holds for all ${\displaystyle x\in \mathbb {R} }$:

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

It follows immediately that ${\displaystyle -1 holds. Furthermore,

{\displaystyle {\begin{aligned}\lim _{x\to \infty }f(x)&=\lim _{x\to \infty }{\frac {x}{1+|x|}}\\[0.3em]&=\lim _{x\to \infty \atop x>0}{\frac {x}{1+|x|}}\\[0.3em]&=\lim _{x\to \infty \atop x>0}{\frac {x}{1+x}}\\[0.3em]&=\lim _{x\to \infty \atop x>0}{\frac {1}{{\frac {1}{x}}+1}}=1\end{aligned}}}

and

{\displaystyle {\begin{aligned}\lim _{x\to -\infty }f(x)&=\lim _{x\to -\infty }{\frac {x}{1+|x|}}\\[0.3em]&=\lim _{x\to -\infty \atop x<0}{\frac {x}{1+|x|}}\\[0.3em]&=\lim _{x\to -\infty \atop x<0}{\frac {x}{1-x}}\\[0.3em]&=\lim _{x\to -\infty \atop x<0}{\frac {1}{{\frac {1}{x}}-1}}=-1\end{aligned}}}

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

Solution sub-exercise 3:

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

Fall 1: ${\displaystyle x\geq 0}$

{\displaystyle {\begin{aligned}&y=f(x)={\frac {x}{1+x}}\\\iff {}&y+xy=x\\\iff {}&y=x-xy=x(1-y)\\\iff {}&x=\underbrace {\frac {y}{1-y}} _{=f^{-1}(y)}\end{aligned}}}

Fall 2: ${\displaystyle x<0}$

{\displaystyle {\begin{aligned}&y=f(x)={\frac {x}{1-x}}\\\iff {}&y-xy=x\\\iff {}&y=x+xy=x(1+y)\\\iff {}&x=\underbrace {\frac {y}{1+y}} _{=f^{-1}(y)}\end{aligned}}}

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

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

### Exercise 2

Exercise (Continuity of the Inverse Function 2)

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

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

Solution (Continuity of the Inverse Function 2)

Solution sub-exercise 1:

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

${\displaystyle x-y\implies \exp(-x)>\exp(-y)}$

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

${\displaystyle \exp(-x)>\exp(-y)\implies \sin(\exp(-x))>\sin(\exp(-y))\implies -\sin(\exp(-x))<-\sin(\exp(-y))}$

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

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

So ${\displaystyle g}$ is strictly monotone increasing and therefore injective.

Solution sub-exercise 2:

= First, we observe

${\displaystyle g(0)=\exp(0)-\sin(\exp(0))=1-\sin(1)}$

Furthermore,

{\displaystyle {\begin{aligned}\lim _{x\to \infty }\exp(x)&=\infty \\\lim _{x\to \infty }\exp(-x)&=0\end{aligned}}}

and from there it follows that

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

Since ${\displaystyle g}$ is continuous and ${\displaystyle [0,\infty )}$ is an interval, ${\displaystyle g([0,\infty ))}$ is also an interval (see „Folgerung zum Zwischenwertsatz“). Since ${\displaystyle g}$ is strictly monotone increasing and ${\displaystyle \lim _{x\to \infty }g(x)=\infty }$, it follows

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

Solution sub-exercise 3:

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