Theorem 4.4.1.
Let \(I\) be an interval in \(\mathbb R\text{,}\) and \(f:I\longrightarrow {\mathbb R}\text{.}\)
- If \(f\) is strictly monotone, then \(f^{-1}\) exists.
- If \(f\) is continuous and \(f^{-1}\) exists, then \(f\) is strictly monotone.
Section 4.4 Invertible functions
Recall that given any sets \(D\) and \(R\text{,}\) a function \(f:D\longrightarrow R\) is one-to-one (or invertible) if \(f(x_1)=f(x_2)\) implies \(x_1=x_2\text{.}\) In this case, there is an inverse function \(f^{-1}:f(D)\longrightarrow D\) such that \(f^{-1}(f(x))=x\) for all \(x\in D\) and \(f(f^{-1}(y))=y\) for all \(y\in f(D)\text{.}\)
The next theorem shows that for functions defined on subsets of \(\mathbb R\text{,}\) invertibility is strictly related to monotonicity.
Hint.
- If \(f\) is strictly monotone, assume for a contradiction that it is not one-to-one.
- Assume there are numbers \(a\lt b\) in \(I\) with \(f(a)\lt f(b)\text{,}\) and (for a contradiction) there are numbers \(x_1\lt x_2\) such that \(f(x_1)\gt f(x_2)\text{.}\) Define \(g:[0,1]\longrightarrow {\mathbb R}\) by \(g(t)=f(tb+(1-t)x_2)-f(ta+(1-t)x_1)\)
Proof
- Suppose is strictly monotone, and \(f^{-1}\) does not exist. So \(f\) is not one-to-one and there are two elements \(x_1\neq x_2\) such that \(f(x_1)=f(x_2)\text{.}\) So either \(x_1\lt x_2\) or \(x_1\gt x_2\text{.}\) Since \(f\) is strictly monotone, either \(f(x_1)\lt f(x_2)\) or \(f(x_1)\gt f(x_2)\text{,}\) contradicting \(f(x_1)=f(x_2)\text{.}\)
-
Suppose \(f\) is continuous and \(f^{-1}\) exists. Let \(a,b\in I\) with \(a\lt b\text{,}\) and assume that \(f(a)\lt f(b)\text{.}\) We will show that \(f\) is strictly increasing (the case \(f(a)\gt f(b)\) is similar and would result in \(f\) being strictly decreasing.) If \(f\) is not strictly increasing, then there are numbers \(x_1\lt x_2\) such that \(f(x_1)\gt f(x_2)\text{.}\) Define \(g:[0,1]\longrightarrow {\mathbb R}\) by \(g(t)=f(tb+(1-t)x_2)-f(ta+(1-t)x_1)\text{.}\) Then \(g\) is continuous, and \(g(0)=f(x_2)-f(x_1)\lt 0\text{,}\) \(g(1)=f(b)-f(a) \gt 0\text{.}\) By the Intermediate Value Theorem there is some \(c\in (0,1)\) such that \(g(c)=0\text{,}\) so that \(f(cb+(1-c)x_2)=f(ca+(1-c)x_1)\text{.}\) Since \(f\) is one-to-one, \(cb+(1-c)x_2=ca+(1-c)x_1\text{,}\) or \(c(b-a)+(1-c)(x_2-x_1)=0\text{.}\) But this is impossible because all terms on the left are strictly positive.A more elegant and shorter proof can given by using the notion of connectedness (not defined in this course). Let \(D\) be the set in \({\mathbb R}^2\) defined by \(D=\{(x,y): x\lt y\}\text{.}\) Then \(D\) is connected. Intuitively, this means that \(D\) is “in one piece”, or that we can go from any point of \(D\) to any other point of \(D\) by following a path that never leaves \(D\text{.}\) The connected sets in \(\mathbb R\) are just the interval. There is a theorem that says that if the domain of a continuous function is connected, then the range is also connected. Continuity can be defined for functions of two variables, and if we define \(F:D\longrightarrow {\mathbb R}\) by \(F(x,y)=f(y)-f(x)\text{,}\) then it turns out that \(F\) is continuous and so its range must be an interval. But the range of \(F\) cannot contain zero, because \(f\) is one-to-one. An interval that does not contain zero must be either in \((0,\infty)\) or in \((-\infty,0)\text{,}\) and this means that if \((x,y)\in D\text{,}\) then either \(f(y)-f(x)\) is always positive (so \(f\) is strictly increasing) or \(f(y)-f(x)\) is always negative (so \(f\) is strictly decreasing).
Figure 4.4.2 shows that in Theorem 4.4.1, part 1, \(f\) need not be continuous. Figure 4.4.3 shows that Theorem 4.4.1, part 2, is false if the continuity assumption is dropped.
Note that in the example of Figure 4.4.2, the range of \(f\) is not an interval (and so the domain of \(f^{-1}\) will not be an interval). The next theorem shows that this is the only way that an invertible function can fail to be continuous.
Theorem 4.4.4.
Suppose that \(f:I\longrightarrow {\mathbb R}\) is strictly monotone, and the range \(f(I)\) is an interval. Then \(f\) is continuous.
Hint.
First use Definition 2.1.9 to show that (in case \(f\) is increasing) \(f([a,b])=[f(a),f(b)]\text{.}\) Then if \(c\in (a,b) \) and \(f(c)=y\text{,}\) given \(\epsilon \gt 0\text{,}\) by making \(\epsilon\) smaller if necessary we can make sure that \([y-\epsilon,y+\epsilon]\subset [f(a),f(b)]=f([a,b]).\)
Proof We prove the theorem in case \(I=[a,b]\) and \(f\) is strictly increasing. First we prove that \(f(I)=[f(a),f(b)]\text{.}\) If \(y\in f(I)\text{,}\) let \(x\in [a,b]\) such that \(f(x)=y\text{.}\) Then \(a\leq x \leq b \) and \(f(a)\leq f(x)\leq f(b)\text{,}\) so that \(y=f(x)\in [f(a),f(b)]\text{.}\) This proves that \(f(I)\subset [f(a),f(b)]\text{.}\) If \(y\in [f(a),f(b)]\text{,}\) then \(f(a)\leq y \leq f(b)\text{.}\) Since \(f(a), f(b)\in f(I)\) and \(f(I)\) is an interval, \(y\in f(I)\text{.}\) So \([f(a),f(b)]=f(I)\text{.}\)
Let now \(c\in (a,b)\text{.}\) We want to prove that \(f\) is continuous at \(x=c\text{.}\) Since \(a\lt c\lt b\text{,}\) we have \(f(a)\lt f(c) \lt f(b)\text{.}\) Suppose \(\epsilon \gt 0\) is given. Let \(\epsilon_1=\min\{\epsilon, f(b)-f(c), f(c)-f(a)\}\text{.}\) Then \(f(c)-\epsilon_1\geq f(c)-f(c)+f(a)=f(a)\) and \(f(c)+\epsilon_1\leq f(c)+f(b)-f(c)=f(b)\text{,}\) so \([f(c-\epsilon_1,f(c)+\epsilon_1]\subset [f(a), f(b)]=f([a,b])\text{.}\) So we can find \(x_1,x_2\in [a,b]\) such that \(f(x_1)=f(c)-\epsilon_1\) and \(f(x_2)=f(c)+\epsilon_1\text{.}\) If \(x_1\geq c\text{,}\) then \(f(c)-\epsilon_1=f(x_1)\geq f(c)\text{,}\) a contradiction. So \(x_1\lt c\text{.}\) In a similar way, \(c\lt x_2\text{.}\) So \(x_1\lt c\lt x_2\text{.}\) Let \(\delta =\min\{c-x_1,x_2-c\}\text{.}\) If \(|x-c|\lt \delta\text{,}\) then \(x_1\lt x \lt x_2\text{,}\) so \(f(c)-\epsilon_1=f(x_1)\lt f(x)\lt f(x_2)=f(c)+\epsilon_1\text{,}\) and since \(\epsilon_1\leq \epsilon\text{,}\) \(f(c)-\epsilon\lt f(x)\lt f(c)+\epsilon\text{,}\) or \(|f(x)-f(c)|\lt \epsilon\text{.}\) The cases \(c=a\) or \(c=b\) can be done in a similar way.
The previous theorem has an immediate and quite interesting corollary. It tells us that under the right conditions, taking the inverse function we get a continuous function even if the original function is not.
Corollary 4.4.5.
Suppose \(I\) is an interval, and \(f:I\longrightarrow {\mathbb R}\) is strictly monotone. Then the inverse function \(f^{-1}\) is continuous.
Hint.
Consider the range of \(f^{-1}\text{,}\) and apply the theorem.
It’s nice to know that we can count on \(f^{-1}\) being continuous even if \(f\) is not, as long as the domain of \(f\) is an interval. But note that in case \(f\) is not continuous, then its range will not be an interval (by Theorem 4.4.4 ), and this means that the domain of \(f^{-1}\) will not be an interval. So going from \(f\) to \(f^{-1}\) we gain continuity, but we lose a nice property of the domain. See the picture below.
