Real Analysis/Inverse Functions

Real Analysis
Inverse Functions

In this chapter, we will be formalizing the definition and the intuitive behaviors of an inverse function. In earlier mathematics, you may have been taught a cursory amount on the subject, such as their reflection on the $y = x$ line or a list of functions and their inverses. You may have been taught the inverse trigonometric functions as a hand-wavy method to get an angle from the lengths of an imaginary triangle. You may have understood that the n^th root is the inverse of an n^th power. Here, we will formalize it—give it meaning and thus give it intrigue. Yet, many theorems related to inverse functions will not be immediately useful or understandable unless other concepts have been constructed beforehand. Thus, sections of this chapter may contain notices for any concepts needed beforehand.

Construction

The construction of the inverse function can be done in two methods. The first method is to simply define the inverse function outright. We will not define it in that manner in this wikibook. Instead, we will define it as a consequence of a new operation. We will define inverse functions in this manner because it assumes less concepts and gives us a new operator to work with (It will be used in a very important concept in this chapter). This section of this chapter will express how we will follow the second method in order to define it.

Inverse Operator

Definition of an Inverse Operator

An unary operation, taking the function

ƒ

, such that the set of all ordered pairs

(a, b)

is

(b, a)

in the function

ƒ

. Notated as

ƒ -1

.

The definition of an inverse is a straightforward modification to a function; simply reverse the ordering of each number in every ordered pair. But can you imagine if this definition is too simple? For example, would this definition work for $ƒ(x)=x 2$ ? Given that $(a, ƒ(a)) = (-a, ƒ(a))$ , the inversion will have the situation where there exists two ordered pairs such that $(ƒ(a), a) and (ƒ(a), -a)$ , which breaks the definition of a function. Given that this issue may arise when dealing with inverse functions, we can begin to craft a workaround to that problem immediately. Although the inverse of the function $ƒ(x)=x 2$ is not a function, we have only defined the definition of inverting a function. We have not defined an inverse function.

To prevent issues like $ƒ(x)=x 2$ , we will define an inverse function. However, we will not define an inverse function separately, but as a theorem. Why? so inverse functions become a resulting property of the inverse operator instead of being a separate class. Mathematics is built on principals that simple rules implies more complex ones. Shall we begin?

Inverse Function Theorem

Theorem defining Inverse Functions

ƒ -1

is a function means that

ƒ

is one-one and vice versa.

Definition of one-one

Given

a \neq b

, such that

ƒ(a) \neq ƒ(b)

\forall a, b \in domain of ƒ.

With this theorem, we can build our definition of an inverse function off the back of our initial definition of the inverse operator. However, did you notice that we slipped in another definition? We have this extra definition in place because this will allow us to sidestep a potential definition conflict.

What counts as a one-one function? Surprisingly, not all the functions you imagine when you imagine functions count. The identity function $ƒ = x$ is one-one, but not the square function $ƒ = x 2$ . Your favorite trigonometric functions $sin, cos, tan$ are not one-one functions. However, this piecewise function

$f(x)={\begin{cases}x&x\neq 0,1\\1&x=0\\0&x=1\end{cases}}$

is one-one. Now that we cleared ourselves from the need to find examples, let us prove this theorem.

Proof

The proof naturally follows from the definition, if set up correctly.

Proof that an Invertable Function must mean the Function is One-One
Assert that $ƒ -1, ƒ$ is a function and suppose any two pairs $(a, ƒ(a)), (b, ƒ(b)).$
Suppose that $ƒ(a) = ƒ(b).$ This implies that $(a, ƒ(a)) = (b, ƒ(a)).$
This means that the inverse function $ƒ -1$ will have $(ƒ(a), a), (ƒ(a), b).$
Our assertion that $ƒ -1$ is a function must mean $a = b .$
$a = b$ and $ƒ(a) = ƒ(b)$ is simply the negated definition for one-one.
$\blacksquare$

We also declared that the corollary, the biconditionality of the previous theorem, is true as well; let's prove that too.

Proof that a One-One Function, when inverted, is a Function
Assert that $ƒ$ is a one-one function.
Suppose two pairs $(a, b), (a, c)$ from the inverted function $ƒ -1$
This implies that the function $ƒ$ has the pairs $(b, a), (c, a).$
Our assertion that $ƒ$ is one-one must mean $b = c .$
The same input $a$ from the inverted function $ƒ$ gives the same output, i.e. $b = c$ , is simply the definition of a function.
$\blacksquare$

Notation

There is a small list of naming conventions associated with inverse functions that are designed to make things less confusing. Typical ordered pair names, such as $(x,y)$ or $(a,b)$ should be used to reflect the inverse function's nature of reversing ordered pairs. For example, $x = ƒ -1 (y)$ or $a = ƒ -1 (b)$ complements the naming convention of $y = ƒ(x)$ and $b = ƒ(a)$ respectively.

Theorems

The curious aspect of inverse functions is its ability to create new definitions of known functions and theorems by, in a sense, reversing known functions and theorems. We will first start with proving very simple consequences from the theorem defining inverse functions, then we will move to a curious proof relating to algebra, then we will advance into calculus by proving differentiation properties.

Basic Properties

If it was not obvious, there one seemingly necessary property that ought be easily derived from the definition.

Involution Law	$ƒ \circ ƒ -1 (y) = y$
Involution Law	$ƒ -1 \circ ƒ (x) = x$

Proof

The proof relies on an application of function and inverse function definitions.

Proof that Involution Applies to Inverse
$ƒ -1 \circ ƒ (x) = x$	$ƒ \circ ƒ -1 (y) = y$
The function $ƒ$ maps $x to y$	The function $ƒ$ maps $y to x$
The inverse function $g = ƒ -1$ maps $y to x$	The inverse function $g = ƒ -1$ maps $x to y$
$x{\stackrel {f}{\rightarrow }}y{\stackrel {g}{\rightarrow }}x$	$y{\stackrel {g}{\rightarrow }}x{\stackrel {f}{\rightarrow }}y$
$\blacksquare$	$\blacksquare$

The results of this proof seems to justify its notation; simply "cancel" out inverses of functions if the functions are composed on each other as if it was multiplication of a variable $a$ and its inverse $a -1 .$ It also proves our intuitive notion that we ought have a way to undo mathematical operations.

Algebra

Now that we have proven the intuitive understanding that the inverse function is the opposite of the function, we can work on the second part necessary to justify algebra, which is that of algebraic reversibility. Algebraic reversibility is the mathematical concept closely related to "if and only if"—the ability to apply an operation that can be reversed by applying the opposite operation. We will prove this concept now.

Theorem justifying Algebra, Part 2

If

f(a)=f(b)

and f is invertable, then you can apply an inverse function to both sides i.e. cancel out something.

Proof

This proof will rely only on applying the definition of an inverse function.

Proof of Algebra's Validity
Given that $f(a)=f(b)$ , we can use a variable x such that $x=f(a)=f(b)$ . This will simplify communication tremendously.
Given the definition of an inverse function, we can state that the variable x can map to the value at f^-1 of x, which is a as defined by $x=f(a)$ (why a? Use the Involution Law from the section just earler.	${\begin{aligned}f^{-1}&=\{\ldots ,\{\{x\},\{x,a\}\},\ldots \}\implies \\x&\rightarrow a\end{aligned}}$
Because f(b) = f(a) and that we are working with a function f, we can claim that b can also map to the value at f^-1 of x.	${\begin{aligned}f&=\{\ldots ,\{\{x\},\{x,b=a\}\},\ldots \}\implies \\x&\rightarrow b\end{aligned}}$
Combined together, the relationship becomes clear.	$f(a)=f(b)\implies x\rightarrow a=x\rightarrow b\implies a=b$ $\blacksquare$

Continuity

We will begin by proving that an inverse function is continuous if the original function is continuous. Why are we proving this as opposed to discontinuous functions? Well, on the chapter on limits, we discussed that discontinuity is the not special case. If it is not the special case, it usually has fewer properties to work with and does not need further expansion. THus, we will focus on continuity preservation. As a notice, this theorem requires proving some parts and combining it together. This will mean that the theorem itself will be very long. The theorem itself is written below.

Theorem

Given a function

ƒ

such that it is continuous and one-one, the inverse must also be continuous.

Note that from the inverse function definition we have already proved that all inverse functions $ƒ -1$ are one-one. From there, you might be a little daunted at the process you would have to undergo. However, this is a textbook so we will be your guide.

First, we will need to prove a separate theorem that will be instrumental in piecing the proof together. This theorem is below.

Theorem

Given a function

ƒ

such that it is continuous and one-one on some interval, it must either be increasing or decreasing on some interval.

To prove this, we will use the intermediate value theorem.

Proof of one-one → either increasing or decreasing.
	Increasing	Decreasing
Choose two values a, b, and c such that $a < b < c$ and that	$ƒ(a) < ƒ(c)$	$ƒ(a) > ƒ(c)$
Suppose	$ƒ(b) < ƒ(a)$	$ƒ(b) > ƒ(a)$
Then using the intermediate value theorem between $ƒ(b)$ and $ƒ(c)$ , you get another point that equals $ƒ(a)$ for a one-one function. Contradiction
Suppose	$ƒ(b) > ƒ(a)$	$ƒ(b) < ƒ(a)$
Also a contradiction using the same reasoning for the interval between $ƒ(a)$ and $ƒ(b)$ for the point $ƒ(c)$ .
$ƒ(b)$ must go in between.	$\blacksquare$	$\blacksquare$

From this theorem, we will use it to prove our initial theorem regarding continuity. As usual, we will focus on increasing in our proof, as we can emulate decreasing functions by negating the given function. The proof is an epsilon-delta proof.

Proof that continuous one-one functions imply inverse continuity
A refresher on the definition of continuity and its associated epsilon-delta meaning	${\begin{aligned}\lim _{x\rightarrow b}{f^{-1}(x)}&=f^{-1}(b)\\\|x-b\|<\delta &\rightarrow \|f^{-1}-f^{-1}(b)\|<\epsilon \end{aligned}}$
We first break down the absolute value notation of the epsilon-delta definition and apply some substitutions using our established naming conventions (In fact, we swap the inverse for the normal and the normal into the inverse).	${\begin{aligned}\|x-b\|<\delta &\rightarrow \|f^{-1}-f^{-1}(b)\|<\epsilon \\-\delta <x-b<\delta &\rightarrow -\epsilon <f^{-1}-f^{-1}(b)<\epsilon \\-\delta +b<x<\delta +b&\rightarrow -\epsilon +f^{-1}(b)<f^{-1}<\epsilon +f^{-1}(b)\\-\delta +f(a)<x<\delta +f(a)&\rightarrow -\epsilon +a<f^{-1}<\epsilon +a\end{aligned}}$
Given the epsilon inequality on the right side, we know that in between minus and plus ε must be the following value.	$-ε + a < a < ε + a$
Given that the function $ƒ$ is continuous and one-one, we can apply the theorem we just proved to imply that	$ƒ(-ε + a) < ƒ(a) < ƒ(ε + a)$
Let's suppose that	$δ = min(ƒ(a + ε) - ƒ(a), ƒ(a) - ƒ(a - ε))$
This relationship can then be verified [The proof is up to you]	$ƒ(a - ε) \leq ƒ(a) - δ$ and $ƒ(a) + δ \leq ƒ(ε + a)$
Let's suppose that, given a range, that some variable x exists between them.	$ƒ(a) - δ < x < ƒ(a) + δ$
We'll apply the relationship written above	$ƒ(a - ε) < x < ƒ(ε + a)$
Given that the increasing function $ƒ$ implies that $ƒ -1$ is increasing,	$ƒ -1 \circ ƒ(a - ε) < ƒ -1 (x) < ƒ -1 \circ ƒ(ε + a)$ $a - ε < ƒ -1 (x) < ε + a$
This shows that given a delta, you can imply an epsilon out of it.	$\blacksquare$

"The Reciprocal Definition" Property

This following definition that the concept of the inverse function provides is for a derivative definition, in which the derivative of the inverse is known.

Theorem

Given a differentiable function

ƒ -1

at

ƒ(a)

, it can be defined as

ƒ' = 1 / ƒ -1 \circ ƒ (a)

, provided

ƒ

is a function and

ƒ -1 \circ ƒ (a) \neq 0

.

Theorem

Given a differentiable function

ƒ

at

ƒ -1 (b)

, it can be defined as

ƒ -1' = 1 / ƒ \circ ƒ -1 (b)

, provided

ƒ -1

is a function and

ƒ \circ ƒ -1 (b) \neq 0

.

As always, this theorem works if you reverse whether the inverse function is inside or outside. We will not be discussing a proof for now. Why? The proof requires both continuity - something we haven't even verified to exist for inverses yet - and differentiation - something we also haven't verified to work for inverse functions. However, we will discuss the intuition behind this property, which we will show below if this property is not intuitive.

Intuitive Process of the Derivative Definition
	$ƒ' = 1 / ƒ -1' \circ ƒ (a)$	$ƒ -1' = 1 / ƒ' \circ ƒ -1 (a)$
Given the inverse function relationship	$ƒ -1 \circ ƒ (x) = x$	$ƒ \circ ƒ -1 (y) = y$
Derivative of x	$ƒ -1 \circ ƒ (x) \cdot ƒ'(x) = 1$	$ƒ \circ ƒ -1 (y) \cdot ƒ -1' = 1$
	$ƒ' = 1 / ƒ -1' \circ ƒ (a)$	$ƒ -1' = 1 / ƒ' \circ ƒ -1 (a)$
	$\blacksquare$	$\blacksquare$

Differentiation

Earlier on in the section on inverse functions, we have readily declared that the formula for calculating the derivative of an inverse function, in relationship to the derivative of the original function, as $ƒ' = 1 / ƒ -1' \circ ƒ (a)$ . However, it contained a caveat that $ƒ -1' \circ ƒ (a) \neq 0.$ The proof should be readily inferred from the formula itself (Hint: Substitute $ƒ -1' = ƒ (a) \neq 0$ ). We shown in that section that this can be derived from an application of Chain Rule with an appendix that it actually isn't a proof. We stated that it isn't because we have not worked out whether differentiation works for inverse functions. Well, since we proved that continuity works if the original function is continuous, we now need to check if differentiation also works (since differentiation requires continuity).

TODO	Editor's note Finish proof

Exercises

Prove that all periodic functions will not have an inverse function.
Prove that all non-restricted even functions $E(x) : E(a) = E(-a) \foralla\in R$ will not have an inverse function.