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4.5 Invertible functions

Let $f:[a,b]\to[c,d]$ be a function. We say that $f$ in invertible if it is injective and surjective. That is

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If $x\neq y\in[a,b]$ , then $f(a)\neq f(b)$ . [Injectivity]
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For any $z\in[c,d]$ there exists $x\in[a,b]$ such that $f(x)=z$ . [Surjectivity]

The inverse of $f$ , $f^{-1}:[c,d]\to[a,b]$ is defined as $f^{-1}(x)=y$ , where $y$ is the unique element such that $f(y)=x$ .

Example 4.5.1

The following functions are invertible.

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$f:[0,1]\to[0,1]$ $f(x)=x^{k}$ , where $k\geq 1$ .
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$f:[0,\frac{\pi}{2}]\to[0,1]$ , $f(x)=sin(x)$ .
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$f:[0,1]\to[0,1]$ $f(x)=x$ if $x\neq\frac{1}{3}$ or $\frac{1}{2}$ . On the other hand, $f(\frac{1}{3})=\frac{1}{2}$ and $f(\frac{1}{2})=\frac{1}{3}$ .

Proposition 4.5.1

Let $f:[a,b]\to[c,d]$ be a continuous invertible function. Then its inverse function $f^{-1}$ is continuous as well.

Proof: Let $\{y_{n}\}^{\infty}_{n=1}$ be a convergent sequence in $[c,d]$ tending to $y$ . We need to show that $\{f^{-1}(y_{n})\}^{\infty}_{n=1}$ is also convergent and tends to $f^{-1}(y)$ . Suppose that it is not true. Then, there exists a subsequence $\{f^{-1}(y_{n_{k}})\}^{\infty}_{k=1}$ converging to $z$ such that $z\neq f^{-1}(y)$ .

Remember that in Question 2.2.1 we have seen that if $\{x_{n}\}^{\infty}_{n=1}$ do not converge to $x$ , then there exists a subsequence $\{x_{n_{k}}\}^{\infty}_{k=1}$ and $\varepsilon>0$ such that for any $k\geq 1$ $|x_{n_{k}}-x|\geq\varepsilon$ . By the Bolzano-Weierstrass Theorem, we have a convergent subsequence of this subsequence. The limit of this subsequence it at least $\varepsilon$ -far from $x$ !

Since $f$ is continuous, $\{f(f^{-1}(y_{n_{k}}))\}^{\infty}_{k=1}$ must converge to $f(z)$ . That is $\{y_{n_{k}}\}^{\infty}_{k=1}$ must converge to $f(z)\neq y$ . This leads to a contradiction. $\square$

Definition 4.5.1

A function $f:[a,b]\to[c,d]$ is strictly monotonic if for any $x<y$ $f(x)<f(y)$ [this is the functional version of strictly increasing].

Example 4.5.2

If a function is differentiable and has positive derivative, then it is strictly monotonic. This is the reason you find the local maximum of a function by searching for the zeros of its derivative $f^{\prime}$ . This fact will be proven rigorously in Math114. On the other hand, there are strictly monotonic functions that are not even continuous at a certain point. Let $f:[0,1]\to[0,1]$ be a function defined the following way:

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$f(x)=x$ , whenever $x\leq\frac{1}{2}$ .
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$f(x)=x+1$ , if $x>\frac{1}{2}$ .

Exercise 4.5.1

If $g:[a,b]\to[c,d]$ is strictly monotonic and $f:[c,d]\to[e,f]$ is strictly monotonic, then $f\circ g:[a,b]\to[e,f]$ is strictly monotonic.

Proposition 4.5.2

Let $f:[a,b]\to[c,d]$ be strictly monotonic and continuous, such that $f(a)=c,f(b)=d$ . Then $f$ is invertible. The inverse is again strictly monotonic.

Proof: Injectivity follows from the definition of strict monotonicity. If $x<y$ , then $f(x)\neq f(y)$ . So, we need to prove that for any $z\in[c,d]$ there exists $t\in[a,b]$ such that $f(t)=z$ . However, this statement follows immediately from the Intermediate Value Theorem. $\square$

Recall that $f:x\to x^{n}$ is a strictly monotonic function mapping the non-negative halfline $[0,\infty)$ to itself. Hence its inverse $f^{-1}$ exists and it is a continuous function by Proposition 4.5.1 and Proposition 4.5.2. This strictly monotonic function is called the $n$ -th root function $g:x\to x^{\frac{1}{n}}$ . As I promised at the beginning, we see know that there exists a unique positive number $x$ such that $x^{2}=2$ .