RA Inverse function theorem

Section 4.4 Inverse function theorem

Note: less than 1 lecture (optional section, needed for Section 5.4, requires Section 3.6)

Subsection 4.4.1 Inverse function theorem

We start with a simple example. Consider the function \(f(x) \coloneqq a x\) for a number \(a \not= 0\text{.}\) Then \(f \colon \R \to \R\) is bijective, and the inverse is \(f^{-1}(y) = \frac{1}{a} y\text{.}\) In particular, \(f'(x) = a\) and \((f^{-1})'(y) = \frac{1}{a}\text{.}\) As differentiable functions are “infinitesimally like” linear functions, we expect the same behavior from the inverse function. The main idea of differentiating inverse functions is the following lemma.

Lemma 4.4.1.

Let \(I,J \subset \R\) be intervals. If \(f \colon I \to J\) is strictly monotone (hence one-to-one), onto (\(f(I) = J\)), differentiable at \(x_0 \in I\text{,}\) and \(f'(x_0) \not= 0\text{,}\) then the inverse \(f^{-1}\) is differentiable at \(y_0 = f(x_0)\) and

\begin{equation*} (f^{-1})'(y_0) = \frac{1}{f'\bigl( f^{-1}(y_0) \bigr)} = \frac{1}{f'(x_0)} . \end{equation*}

If \(f\) is continuously differentiable and \(f'\) is never zero, then \(f^{-1}\) is continuously differentiable.

Proof.

By Proposition 3.6.6, \(f\) has a continuous inverse. For convenience call the inverse \(g \colon J \to I\text{.}\) Let \(x_0,y_0\) be as in the statement. For \(x \in I\) write \(y \coloneqq f(x)\text{.}\) If \(x \not= x_0\) and so \(y \not= y_0\text{,}\) we find

\begin{equation*} \frac{g(y)-g(y_0)}{y-y_0} = \frac{g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)}{f(x)-f(x_0)} = \frac{x-x_0}{f(x)-f(x_0)} . \end{equation*}

See Figure 4.10 for the geometric idea.

Figure 4.10. Interpretation of the derivative of the inverse function.

Let

\begin{equation*} Q(x) \coloneqq \begin{cases} \frac{x-x_0}{f(x)-f(x_0)} & \text{if } x \neq x_0, \\ \frac{1}{f'(x_0)} & \text{if } x = x_0 \quad \text{(notice that } f'(x_0) \neq 0 \text{)}. \end{cases} \end{equation*}

As \(f\) is differentiable at \(x_0\text{,}\)

\begin{equation*} \lim_{x \to x_0} Q(x) = \lim_{x \to x_0} \frac{x-x_0}{f(x)-f(x_0)} = \frac{1}{f'(x_0)} = Q(x_0) , \end{equation*}

that is, \(Q\) is continuous at \(x_0\text{.}\) As \(g(y)\) is continuous at \(y_0\text{,}\) the composition \(Q\bigl(g(y)\bigr) = \frac{g(y)-g(y_0)}{y-y_0}\) is continuous at \(y_0\) by Proposition 3.2.7. Therefore,

\begin{equation*} \frac{1}{f'\bigl(g(y_0)\bigr)} = Q\bigl(g(y_0)\bigr) = \lim_{y \to y_0} Q\bigl(g(y)\bigr) = \lim_{y \to y_0} \frac{g(y)-g(y_0)}{y-y_0} . \end{equation*}

So \(g\) is differentiable at \(y_0\) and \(g'(y_0) = \frac{1}{f'\left(\vphantom{1^1_1}g(y_0)\right)}\text{.}\)

If \(f'\) is continuous and nonzero at all \(x \in I\text{,}\) then the lemma applies at all \(x \in I\text{.}\) As \(g\) is also continuous (it is differentiable), the derivative \(g'(y) = \frac{1}{f'\left(\vphantom{1^1_1}g(y)\right)}\) must be continuous.

What is usually called the inverse function theorem is the following result.

Theorem 4.4.2. Inverse function theorem.

Let \(f \colon (a,b) \to \R\) be a continuously differentiable function, \(x_0 \in (a,b)\) a point where \(f'(x_0) \not= 0\text{.}\) Then there exists an open interval \(I \subset (a,b)\) with \(x_0 \in I\text{,}\) the restriction \(f|_{I}\) is injective with a continuously differentiable inverse \(g \colon J \to I\) defined on an interval \(J \coloneqq f(I)\text{,}\) and

\begin{equation*} g'(y) = \frac{1}{f'\bigl( g(y) \bigr)} \qquad \text{for all } y \in J. \end{equation*}

Proof.

Without loss of generality, suppose \(f'(x_0) > 0\text{.}\) As \(f'\) is continuous, there must exist an open interval \(I = (x_0-\delta,x_0+\delta)\) such that \(f'(x) > 0\) for all \(x \in I\text{.}\) See Exercise 3.2.11.

By Proposition 4.2.8, \(f\) is strictly increasing on \(I\text{,}\) and hence the restriction \(f|_{I}\) is bijective onto \(J: = f(I)\text{.}\) As \(f\) is continuous, then by the Corollary 3.6.3 (or directly via the intermediate value theorem) \(f(I)\) is an interval. Now apply Lemma 4.4.1.

In Example 1.2.3 we saw how difficult an endeavor proving the existence of \(\sqrt{2}\text{.}\) With the intermediate value theorem we made the existence of roots almost trivial, and with the machinery of this section we will prove far more than mere existence.

Corollary 4.4.3.

Given \(n \in \N\) and \(x \geq 0\text{,}\) there exists a unique number \(y \geq 0\) (denoted \(x^{1/n} \coloneqq y\)), such that \(y^n = x\text{.}\) Furthermore, the function \(g \colon (0,\infty) \to (0,\infty)\) defined by \(g(x) \coloneqq x^{1/n}\) is continuously differentiable and

\begin{equation*} g'(x) = \frac{1}{nx^{(n-1)/n}} = \frac{1}{n} \, x^{(1-n)/n} , \end{equation*}

using the convention \(x^{m/n} \coloneqq {(x^{1/n})}^{m}\text{.}\)

Proof.

For \(x=0\) the existence of a unique root is trivial.

Let \(f \colon (0,\infty) \to (0,\infty)\) be defined by \(f(y) \coloneqq y^n\text{.}\) The function \(f\) is continuously differentiable and \(f'(y) = ny^{n-1}\text{,}\) see Exercise 4.1.3. For \(y > 0\) the derivative \(f'\) is strictly positive and so again by Proposition 4.2.8, \(f\) is strictly increasing (this can also be proved directly). Given any \(M > 1\text{,}\) \(f(M) = M^n \geq M\text{,}\) and given any \(1 > \epsilon > 0\text{,}\) \(f(\epsilon) = \epsilon^n \leq \epsilon\text{.}\) For every \(x\) with \(\epsilon < x < M\text{,}\) we have, by the intermediate value theorem, that \(x \in f\bigl( [\epsilon,M] \bigr) \subset f\bigl( (0,\infty) \bigr)\text{.}\) As \(M\) and \(\epsilon\) were arbitrary, \(f\) is onto \((0,\infty)\text{,}\) and hence \(f\) is bijective. Let \(g\) be the inverse of \(f\text{,}\) and we obtain the existence and uniqueness of positive \(n\)th roots. Lemma 4.4.1 says \(g\) has a continuous derivative and \(g'(x) = \frac{1}{f'\left(\vphantom{1^1_1}g(x)\right)} = \frac{1}{n {(x^{1/n})}^{n-1}}\text{.}\)

Example 4.4.4.

The corollary provides a good example of where the inverse function theorem gives us an interval smaller than \((a,b)\text{.}\) Take \(f \colon \R \to \R\) defined by \(f(x) \coloneqq x^2\text{.}\) Then \(f'(x) \not= 0\) as long as \(x \not= 0\text{.}\) If \(x_0 > 0\text{,}\) we can take \(I=(0,\infty)\text{,}\) but no larger.

Example 4.4.5.

Another useful example is \(f(x) \coloneqq x^3\text{.}\) The function \(f \colon \R \to \R\) is one-to-one and onto, so \(f^{-1}(y) = y^{1/3}\) exists on the entire real line including zero and negative \(y\text{.}\) The function \(f\) has a continuous derivative, but \(f^{-1}\) has no derivative at the origin. The point is that \(f'(0) = 0\text{.}\) See Figure 4.11 for a graph, notice the vertical tangent on the cube root at the origin. See also Exercise 4.4.4.

Figure 4.11. Graphs of \(x^3\) and \(x^{1/3}\text{.}\)

Subsection 4.4.2 Exercises

Exercise 4.4.1.

Suppose \(f \colon \R \to \R\) is continuously differentiable and \(f'(x) > 0\) for all \(x\text{.}\) Show that \(f\) is invertible on the interval \(J = f(\R)\text{,}\) the inverse is continuously differentiable, and \({(f^{-1})}'(y) > 0\) for all \(y \in f(\R)\text{.}\)

Exercise 4.4.2.

Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose you already know that both \(f\) and \(g\) are differentiable everywhere and \(f'\) is never zero. Using chain rule but not Lemma 4.4.1 prove the formula \(g'(y) = \frac{1}{f'\left(\vphantom{1_1^1}g(y)\right)}\text{.}\) Remark: This exercise is the same as Exercise 4.1.10, no need to do it again if you have solved it already..

Exercise 4.4.3.

Let \(n\in \N\) be even. Prove that every \(x > 0\) has a unique negative \(n\)th root. That is, there exists a negative number \(y\) such that \(y^n = x\text{.}\) Compute the derivative of the function \(g(x) \coloneqq y\text{.}\)

Exercise 4.4.4.

Let \(n \in \N\) be odd and \(n \geq 3\text{.}\) Prove that every \(x\) has a unique \(n\)th root. That is, there exists a number \(y\) such that \(y^n = x\text{.}\) Prove that the function defined by \(g(x) \coloneqq y\) is differentiable except at \(x=0\) and compute the derivative. Prove that \(g\) is not differentiable at \(x=0\text{.}\)

Exercise 4.4.5.

(requires Section 4.3) Show that if in the inverse function theorem \(f\) has \(k\) continuous derivatives, then the inverse function \(g\) also has \(k\) continuous derivatives.

Exercise 4.4.6.

Let \(f(x) \coloneqq x + 2 x^2 \sin(\nicefrac{1}{x})\) for \(x \not= 0\) and \(f(0) \coloneqq 0\text{.}\) Show that \(f\) is differentiable at all \(x\text{,}\) that \(f'(0) > 0\text{,}\) but that \(f\) is not invertible on any open interval containing the origin.

Exercise 4.4.7.

Let \(f \colon \R \to \R\) be a continuously differentiable function and \(k > 0\) be a number such that \(f'(x) \geq k\) for all \(x \in \R\text{.}\) Show \(f\) is one-to-one and onto, and has a continuously differentiable inverse \(f^{-1} \colon \R \to \R\text{.}\)
Find an example \(f \colon \R \to \R\) where \(f'(x) > 0\) for all \(x\text{,}\) but \(f\) is not onto.

Exercise 4.4.8.

Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose \(x \in I\) and \(y \coloneqq f(x) \in J\text{,}\) and that \(g\) is differentiable at \(y\text{.}\) Prove:

If \(g'(y) \not= 0\text{,}\) then \(f\) is differentiable at \(x\text{.}\)
If \(g'(y) = 0\text{,}\) then \(f\) is not differentiable at \(x\text{.}\)