2.10 Monotone and Inverse functions

Definition 2.10.1. Let \(A\subseteq \mathbb {R}\). A function \(f: A\longrightarrow \mathbb {R}\) is increasing on \(A\) if \(f(x_1) \leq f(x_2)\) whenever \(x_1, x_2\in A\) and \(x_1 < x_2\). It is strictly increasing on \(A\) if \(f(x_1) < f(x_2)\) whenever \(x_1, x_2\in A\) and \(x_1 < x_2\).

Definition 2.10.2. Let \(A\subseteq \mathbb {R}\). A function \(f: A\longrightarrow \mathbb {R}\) is decreasing on \(A\) if \(f(x_1) \geq f(x_2)\) whenever \(x_1, x_2 \in A\) and \(x_1 < x_2\). It is strictly decreasing on \(A\) if \(f(x_1) > f(x_2)\) whenever \(x_1, x_2\in A\) and \(x_1 < x_2\).

Definition 2.10.3. A function which is either increasing or decreasing on \(A\) is said to be monotone on \(A\). It is strictly monotone on \(A\) if it is either strictly increasing or strictly decreasing on \(A\).

Theorem 2.10.4. Let \(I\subseteq \mathbb {R}\) be an interval and let \(f: I \longrightarrow \mathbb {R}\) be increasing on \(I\). Suppose that \(c\in I\) is not an end point of \(I\). Then

(i): \(\lim _{x\rightarrow c^-}f(x) = \sup \{f(x): x\in I,\ x<c\}\);
(ii): \(\lim _{x\rightarrow c^+}f(x) = \inf \{f(x): x\in I,\ x>c\}\).

Proof. (i) If \(x\in I\) and \(x<c\), then \(f(x) \leq f(c)\). So the set \(\{f(x): x\in I,\ x < c\}\) is bounded above by \(f(c)\), and it is non-empty since \(c\) is not an end point of \(I\). Let \(L\) be its supremum. Given \(\varepsilon > 0\), \(L- \varepsilon \) is not an upper bound of this set, so there exists \(y_{\varepsilon }\in I\) with \(y_{\varepsilon } < c\) and \(L - \varepsilon < f(y_{\varepsilon }) \leq L\). Put \(\delta = c- y_{\varepsilon }\). If \(0 < c - y < \delta \), then \(y_{\varepsilon } < y < c\), and since \(f\) is increasing, \[L-\varepsilon < f(y_{\varepsilon }) \leq f(y) \leq L .\] Therefore \(|f(y) - L| < \varepsilon \) whenever \(0 < c - y < \delta \), which proves (i).

(ii) The proof is similar, using the infimum of \(\{f(x): x\in I,\ x>c\}\), which is bounded below by \(f(c)\). □

Corollary 2.10.5. Let \(I\subseteq \mathbb {R}\) be an interval and let \(f: I \longrightarrow \mathbb {R}\) be increasing on \(I\). Suppose that \(c\in I\) is not an end point of \(I\). Then the following statements are equivalent:

1.: \(f\) is continuous at \(c\);
2.: \(\lim _{x\rightarrow c^-} f(x) = f(c) = \lim _{x\rightarrow c^+} f(x)\);
3.: \(\sup \{f(x): x\in I,\ x<c\} = f(c) = \inf \{f(x): x\in I,\ x>c\}\).

If \(a\in I\) is the left end point of \(I\), then \(f\) is continuous at \(a\) if and only if \(f(a) = \inf \{ f(x): x\in I,\ x>a\}\), that is, if and only if \(f(a) = \lim _{x\rightarrow a^+} f(x)\). Similarly for the right end point.

Proof. A function is continuous at an interior point \(c\) exactly when both one-sided limits at \(c\) exist and equal \(f(c)\), so 1 and 2 are equivalent. By the theorem, the one-sided limits are the supremum and infimum in 3, so 2 and 3 are equivalent. At an end point only one one-sided limit is involved, and the same reasoning applies. □

Definition 2.10.6. Let \(I\) be an interval, let \(f: I\longrightarrow \mathbb {R}\) be increasing on \(I\), and let \(c\in I\) be a point that is not an end point of \(I\). The jump of \(f\) at \(c\) is \[J_f(c) = \lim _{x\rightarrow c^+} f(x) - \lim _{x\rightarrow c^-} f(x) = \inf \{f(x): x\in I,\ x>c\} - \sup \{f(x): x\in I,\ x < c\}.\] If the left end point \(a\) of \(I\) belongs to \(I\), the jump of \(f\) at \(a\) is \(J_f(a) = \lim _{x\rightarrow a^+} f(x) - f(a)\). If the right end point \(b\) belongs to \(I\), the jump at \(b\) is \(J_f(b) = f(b) - \lim _{x\rightarrow b^-} f(x).\)

Theorem 2.10.7. Let \(I\subseteq \mathbb {R}\) be an interval and let \(f: I\longrightarrow \mathbb {R}\) be increasing on \(I\). If \(c\in I\), then \(f\) is continuous at \(c\) if and only if \(J_f(c) = 0\).

Proof. Let \(c\) be an interior point of \(I\), and write \(L^- = \lim _{x\rightarrow c^-}f(x)\) and \(L^+ = \lim _{x\rightarrow c^+}f(x)\). Since \(f\) is increasing, \(f(x)\leq f(c)\leq f(y)\) whenever \(x<c<y\), so \[L^-\leq f(c)\leq L^+ .\] Hence \(J_f(c) = L^+ - L^- = 0\) if and only if \(L^- = f(c) = L^+\), which by the corollary holds if and only if \(f\) is continuous at \(c\). At an end point, \(J_f\) is the difference between \(f\) and its one-sided limit there, and the corollary again gives the result. □

Remark. A monotone function therefore has one-sided limits everywhere, so its only possible discontinuities are jumps, where the two one-sided limits exist but differ. The open intervals \((L^-, L^+)\) at different jumps are disjoint, and each contains a rational number, so a monotone function has at most countably many discontinuities.

Recall that a function \(f: I\longrightarrow \mathbb {R}\) has an inverse function if and only if \(f\) is injective, that is, \(x, y\in I\) and \(x\neq y\) imply \(f(x) \neq f(y)\). A strictly monotone function is injective, and so has an inverse.

Theorem 2.10.8 (Continuous Inverse Theorem). Let \(I\subseteq \mathbb {R}\) be an interval and let \(f: I\longrightarrow \mathbb {R}\) be strictly monotone and continuous on \(I\). Then the inverse function \(g\) of \(f\) is strictly monotone and continuous on \(J = f(I)\).

Proof. We consider the case where \(f\) is strictly increasing. Since \(f\) is continuous and \(I\) is an interval, \(J = f(I)\) is an interval, and since \(f\) is injective, the inverse function \(g: J\longrightarrow I\) exists.

First, \(g\) is strictly increasing. Let \(y_1, y_2\in J\) with \(y_1 < y_2\), and write \(y_1 = f(x_1)\), \(y_2 = f(x_2)\) with \(x_1, x_2\in I\). If \(x_1\geq x_2\), then \(y_1 = f(x_1) \geq f(x_2) = y_2\), which is false. So \(x_1 < x_2\), that is, \(g(y_1) < g(y_2)\).

Next, \(g\) is continuous. Note that \(g(J) = I\) is an interval. Suppose \(g\) were discontinuous at an interior point \(c\) of \(J\). By the theorem above its jump there would be positive, \(L^- < L^+\), where \(L^\pm \) are the one-sided limits of \(g\) at \(c\). Since \(g\) is increasing, \(g(y)\leq L^-\) for \(y<c\) and \(g(y)\geq L^+\) for \(y>c\). Hence at most one point of the interval \((L^-, L^+)\), namely \(g(c)\), belongs to \(g(J)\). But \(g(J) = I\) contains points \(g(y)\leq L^-\) and \(g(y)\geq L^+\), so, being an interval, it contains all of \((L^-, L^+)\). This is a contradiction. The argument at an end point of \(J\) is the same, using one one-sided limit. Therefore \(g\) is continuous on \(J\). □

Example 2.10.9. Let \(f(x) = x + \lfloor x\rfloor \) on \([0,3]\), where \(\lfloor x\rfloor \) is the greatest integer not exceeding \(x\). Find the points of discontinuity and the jumps of \(f\).

Solution. The function is strictly increasing, and continuous except where \(\lfloor x\rfloor \) jumps, that is, at \(x = 1,2,3\). At \(c = 1\): \(\lim _{x\rightarrow 1^-}f(x) = 1+0 = 1\) and \(\lim _{x\rightarrow 1^+}f(x) = 1+1 = 2\), so \(J_f(1) = 1\). Similarly \(J_f(2) = 1\). At the right end point, \(J_f(3) = f(3) - \lim _{x\rightarrow 3^-}f(x) = 6 - 5 = 1\). At the left end point \(J_f(0) = 0\), so \(f\) is continuous there. The range of \(f\) is \([0,1)\cup [2,3)\cup [4,5)\cup \{6\}\), and its gaps have exactly the lengths of the jumps.

Example 2.10.10. Show that \(f(x) = x^3+x\) has a continuous inverse \(g\) on \(\mathbb {R}\), and compute \(g(2)\), \(g(10)\) and \(g(0.5)\) approximately.

Solution. If \(x<y\) then \(x^3<y^3\), so \(f(x)<f(y)\): \(f\) is strictly increasing and continuous. Also \(f(x)\longrightarrow \pm \infty \) as \(x\longrightarrow \pm \infty \), so \(f(\mathbb {R}) = \mathbb {R}\). By the Continuous Inverse Theorem the inverse \(g:\mathbb {R}\longrightarrow \mathbb {R}\) exists, is strictly increasing and is continuous.

Since \(f(1) = 2\) and \(f(2) = 10\), \(g(2) = 1\) and \(g(10) = 2\). For \(g(0.5)\) we solve \(x^3+x = 0.5\) by bisection: \(f(0.4) = 0.464\) and \(f(0.5) = 0.625\), then \(f(0.45) \approx 0.541\), \(f(0.425)\approx 0.502\), \(f(0.4125)\approx 0.483\), so \(g(0.5)\approx 0.424\).

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