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.
The continuous and invertible function
with the domain
First we want to consider the most general condition possible for when a bijective function with has a continuous inverse function. The first ansatz that we naturally wan to investigate is the continuity of itself. We might spontaneously assume that the continuity of the inverse function follows directly from the continuity of . This is not necessarily the case, as is shown in the following example:
The discontinuous inverse function
The function is continuous since it's continuous on both and . Furthermore it is strictly monotonously increasing, so it's injective. The image of is and therefore is surjective. All together, is bijective and therefore invertible. The inverse mapping has functional values:
The inverse function is not continuous since it has a jump discontinuity at . So it can definitely be the case that a continuous function has a discontinuous inverse.
But now another thing is apparent: the inverse function is also invertible with the inverse function . This means, however, that a discontinuous function (like ) 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 . Namely, this is , i.e. not an interval. The domain is not connected and has a "gap." How does this "gap" affect the inverse function ?
Since the domain of corresponds to what should be the codomain of , the inverse function also has this "gap" in its codomain. Similarly, the codomain of corresponds to the domain of . It is for this reason that can not be continuous. The domain of is an interval and therefore connected, while its codomain contains a gap. Since is surjective and must attain every value in what should be its codomain, the graph of must lend itself to a jump point somewhere. This is why is not continuous.
Thus, we must require that, by assumption, the domain of is an interval to ensure that we don't encounter any problems. Usually this requirement is sufficient to guarantee the continuity of . We can prove this using the - characterization of continuity. With the help of Zwischenwertsatzes we can futhermore conclude that the codomain of - and therefore the domain of - is an interval.
Theorem on the Continuity of the Inverse Function [Bearbeiten]
Theorem (Continuity of the Inverse Function)
Let be an interval and a surjective, strictly monotone, and continuous mapping. Then is bijective, and the inverse function
is continuous and in the same sense as strictly monotone. Furthermore, is an interval.
Proof (Continuity of the Inverse Function)
Proof step: is bijective
Let with . Then it either mus be the case or . Since is strictly monotone, it follows or . In either case it holds . So therefore is injective. Since is surjective by assumption, the mapping is bijective.
Proof step: is strictly monotone
Without loss of generality let be strictly monotone increasing, i.e. follows from with . Now let with . Now we ust show that .
There exist some with and . Therefore, and . Since is injective, it holds . It also can not be the case that holds. Otherwise it must then hold , which can not be the case because .
Thus, and . This proves that is strictly monotone increasing like . Proving that is striclty monotone decreasing when we assume is strictly monotone decreasing is analogous.
Proof step: is continuous
We will use the --characterization of continuity. To that end, let and let be arbitrary. Further, let be the preimage of under , i.e. . Now we have to show that there exists some such that holds for all with . Then we will have achieved showing is continuous in . Since can be chosen arbitrarily, from there we can conclude the continuity of the mapping . We must, however, differentiate between the two cases of whether or not is a boundary point of :
Fall 1: is not a boundary point of
Since is not a boundary point of , there exists some with and . We set and . If is strictly monotone increasing, then it holds . If is strictly monotone decreasing, then it holds . Thus, maps the interval bijectively to and . If we set
then it follows
Hence, the condition for all with is fulfilled and is continuous in .
Fall 2: is a boundary point of
We may assume without loss of generality that is the maximum of . Now there exists some with such that lies in . Therefore the value exists and we can set.
If is strictly monotone increasing, then maps the interval bijectively to . Then it holds . Because no exists with , it must therefore hold for all with . This proves the continuity of in . The proof for the case where is monotone decreasing is analogous.
Example (Root Functions)
If with , then the -th power function is defined as
This is a strictly monotonously increasing function, since for with it holds
Furthermore, this means is injective. The function is also surjective, since for every it holds for :
The inverse function is the -th root function with .By the Inverse Function Theorem, this function is continuous and also strictly monotone increasing.
Natural Logarithm Functions[Bearbeiten]
Example (Natural Logarithm Functions)
The exponential function
is strictly monotone increasing. Let with . Then there exists some with . It therefore holds
From the strict monotonicity it follows that is injective. Furthermore, the exponential function is surjective since:
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:
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.
Exercise (Continuity of the Inverse Function)
Let be defined by the following values:
Prove the following claims:
- Show that is continuous, strictly monotone increasing, and injective on .
- Show that is surjective.
- Give a justification for why the inverse function exists and determine the explicit functional values for . Show that is continuous and strictly monotone increasing.
Solution (Continuity of the Inverse Function)
Solution sub-exercise 1:
is continuous since it is the quotient of continuous functions and . thereby holds for all . Let with . To prove the strict monotonicity we will show :
Then it holds
Here it holds
Here it holds
So then is strictly monotone increasing on and therefore also injective.
Solution sub-exercise 2:
It holds for all :
It follows immediately that holds. Furthermore,
Since is continuous, by the Intermediate Value Theorem, for every there exists an with . So it holds for the image that , whereby is surjective.
Solution sub-exercise 3:
Since is bijective, the inverse function exists. As an inverse function, is also bijective. By the Theorem of Continuity of the Inverse Function, is furthermore continuous and strictly monotone increasing. In order to calculate we must first differentiate between two cases:
If we now have , then and therefore . On the other hand, if , then so is and therefore . All together, we can conclude
Exercise (Continuity of the Inverse Function 2)
- Show that is injective.
- Determine the image .
- Justify why the inverse function is continuous.
Solution (Continuity of the Inverse Function 2)
Solution sub-exercise 1:
is continuous as it is the composition of the continuous functions , and . Furthermore, for with it holds:
Now it holds for . Since the sine function is strictly monotone increasing on the half-interval , it follows
Because the exponential function is strictly monotone increasing, it furthermore holds that . Then we have:
So is strictly monotone increasing and therefore injective.
Solution sub-exercise 2:
First, we observe
and from there it follows that
Since is continuous and is an interval, is also an interval (see „Folgerung zum Zwischenwertsatz“). Since is strictly monotone increasing and , it follows
Solution sub-exercise 3:
Since is an interval and is bijective, the inverse function is continuous by the Continuous Inverse Function Theorem.