# Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II

## Section4.4Inverse function theorem

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

### Subsection4.4.1Inverse 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.

#### 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.

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.

#### 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.

#### 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 $$g'(x) = \frac{1}{f'\left(\vphantom{1^1_1}g(x)\right)} = \frac{1}{n {(x^{1/n})}^{n-1}}\text{.}$$

#### Example4.4.4.

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.

#### Example4.4.5.

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.

### Subsection4.4.2Exercises

#### Exercise4.4.1.

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 $$y \in f(\R)\text{.}$$

#### Exercise4.4.2.

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..

#### Exercise4.4.3.

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 $$g(x) \coloneqq y\text{.}$$

#### Exercise4.4.4.

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 $$x=0\text{.}$$

#### Exercise4.4.5.

(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.

#### Exercise4.4.6.

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.

#### Exercise4.4.7.

1. 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 $$f^{-1} \colon \R \to \R\text{.}$$
2. Find an example $$f \colon \R \to \R$$ where $$f'(x) > 0$$ for all $$x\text{,}$$ but $$f$$ is not onto.

#### Exercise4.4.8.

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