Non-zero derivative does not imply the continuity of the inverse

June 3, 2022

Non-zero derivative does not imply the continuity of the inverse

Filed under: Uncategorized — Ngô Quốc Anh @ 0:20

A classical version of the inverse function theorem for 1D asserts the following.

Suppose that $f : U \to V$ is an invertible function with inverse $f^{-1} : V \to U$ . If $f$ is differentiable at $x_0$ with $f'(x_0) \ne 0$ and $f^{-1}$ is continuous at $y_0 = f(x_0)$ , then $f^{-1}$ is differentiable at $y_0$ and
$\displaystyle \big(f^{-1}\big)'(y_0) = \frac 1{f'(x_0)}.$

Very often, some of the above hypotheses for $f$ is replaced by, say, $f$ is continuous “in a neighborhood” and one-to-one. But in the proof, while the one-to-one property immediately gives the existence of the inverse function $f^{-1}$ , the “non-local” continuity yields the continuity of the inverse $f^{-1}$ . Here we are interested in the “local” continuity/differentiablity.

In this note, we show by constructing counter-examples that in general the continuity of $f^{-1}$ at $y_0$ cannot be dropped. In the sequel, we denote by $\mathbb N$ the set set of positive integers, namely $\mathbb N = \{1,2,3,...\}$ .

1. The case of continuity

We start with the simplest case, namely if we only assume that $f$ is continuous at $x_0$ . In this scenario, the following example due to R.E. Megginson is well-known. Consider the function $f : \mathbf R \to \mathbf R$ defined by

$\displaystyle f(x)=\left\{\begin{aligned} & \frac 1{2n} & & \text{if } x= \frac 1n \text{ for some } n \in \mathbb N \\ & \frac 1n & & \text{if } x= n \text{ for some odd } n \in \mathbb N \setminus \{1\} \\ & \frac n2 & & \text{if } x= n \text{ for some even } n \in \mathbb N \setminus \{1\} \\ & x & & \text{otherwise}.\end{aligned}\right.$

It is not hard to see that $f$ is well-defined and invertiable. Moreover, $f$ is continous at $x=0$ . However, as $f(0)=0$ , its inverse function $f^{-1}$ fails to be continous at $0$ because

$\displaystyle f^{-1} \Big( \frac 1{2n+1}\Big) = 2n+1 \nearrow +\infty$

as $n \to +\infty$ . This concludes the construction.

2. The case of zero derivative

We now modify the above example to handle the case of differentiable functions at the given point. A careful calculation shows that in the above example, the funtion $f$ is not differentiable at $x=0$ because

$\displaystyle \lim_{n \to +\infty} \frac{f (1/n) - f(0)}{1/n} = \frac 12 \ne 1 = \lim_{n \to +\infty}\frac{f (-1/n) - f(0)}{-1/n} .$

Toward a new construction, let us denote

$\phi : x \mapsto x^3,$

which is one-to-one. Having the function $\phi$ , we construct the desired function $f$ as follows.

$\displaystyle f(x)=\left\{\begin{aligned} & \phi (x/2) & & \text{if } x= 1/n \text{ for some } n \in \mathbb N \\ & \phi (1/x) & & \text{if } x= n \text{ for some odd } n \in \mathbb N \setminus \{1\} \\ & \phi (x/2) & & \text{if } x= n \text{ for some even } n \in \mathbb N \setminus \{1\} \\ & \phi (x) & & \text{otherwise}.\end{aligned}\right.$

It is not hard to see that $f$ is well-defined and invertiable. We now show that $f$ is differentiable at $0$ with $f'(0)=0$ . To see this, we just consider $f(x)$ for $|x|<1$ . From the definition of $f$ and the monotonicity of $\phi$ we note that

$\displaystyle |f(x)| \leq |\phi(x)| \leq x^2$

for any $|x|<1$ . Hence

$\displaystyle 0 \leq \lim_{x \to 0} \Big| \frac{f(x) - f(0)}{x}\Big| = \lim_{x \to 0} \Big| \frac{f(x) }{x}\Big| \leq \lim_{x \to 0} |x| = 0.$

This shows that $f$ is differentiable at $x=0$ with $f'(0)=0$ . Arguing as in the case of continuity, we know that

$\displaystyle f^{-1} \Big( \phi \big(\frac 1{2n+1} \big)\Big) = 2n+1 \nearrow +\infty$

and

$\displaystyle \phi \big(\frac 1{2n+1} \big) = \frac 1{(2n+1)^3} \searrow 0$

as $n \to +\infty$ . This shows that $f^{-1}$ is not continuous at $0$ .

3. The case of non-zero derivative

We are now arrive in the most delicated case. For simplicity, let us construct some bijective mapping $f : \mathbf R \to f(\mathbf R)$ . The following example is due to Mr. Nghia. The desired function $f$ is given as follows

$\displaystyle f(x)=\left\{\begin{aligned} & \frac 1{(n+1)^2 + k} & & \text{if } x= \frac 1{n^2+k} \text{ for some } n \in \mathbb N \text{ and some } k \in \{0,...,2n\} \\ & \frac 1x & & \text{if } x \in \{ (n+1)^2+2n+1, (n+1)^2 + 2n+2 \} \text{ for some } n \in \mathbb N \\ & x & & \text{otherwise}.\end{aligned}\right.$

It is now standard to verify that $f$ is well-defined and invertiable. Obviously, $f(\mathbf R)$ is not the whole $\mathbf R$ because $1/3 \notin f(\mathbf R)$ . However, there holds $(-1/4, 1/4) \subset f(\mathbf R)$ and $f(0)=0$ . We now examine $f$ at 0 and its inverse $f^{-1}$ at 0. We now show that $f$ is differentiable at $0$ with $f'(0)=1$ . Again, we just consider $f(x)$ for $|x|<1$ . Clearly, it suffices to show that

$\displaystyle \lim_{n \to +\infty} \frac{f (x_n) - f(0)}{x_n} = 1$

for any sequence $(x_n)$ converging to 0. This is done once we can show that both

$\displaystyle \lim_{n \to +\infty} \frac{f (y_n) - f(0)}{y_n} = 1 = \lim_{n \to +\infty} \frac{f (z_n) - f(0)}{z_n}$

hold true, where $z_n=1/n$ and $(y_n)$ is any sequence having no term $1/n$ . This is because given arbitrary $\varepsilon>0$ , there are two positive integers $N_1$ and $N_2$ such that

$\displaystyle \Big|\frac{f (y_n) - f(0)}{y_n} -1 \Big| < \varepsilon \quad \forall n \geq N_1$

and

$\displaystyle \Big|\frac{f (z_n) - f(0)}{z_n} -1 \Big| < \varepsilon \quad \forall n \geq N_2.$

Simply taking $N = \max\{N_1, N_2\}$ we conclude that the above estimate holds for all $n \geq N$ regardless of either $x_n \ne 1/n$ or $x_n=1/n$ . Going back to the limits while the case of $y_n$ is trivial by the definition of $f$ , the case $z_n = 1/n$ is also simple because in this scenario, we easily get

$\displaystyle \frac 1n = \frac 1{m^2+k}$

for some $m \geq 1$ and some $0 \leq k \leq 2m$ . Hence

$\displaystyle \frac{m^2}{(m+2)^2} \leq \frac{f (z_n) - f(0)}{z_n} = \frac{m^2+k}{(m+1)^2+k} \leq \frac{m^2+2m}{(m+1)^2}.$

As $n \to +\infty$ , we must have $m \to +\infty$ . Hence from the above two-sided estimate we must have

$\displaystyle \lim_{n \to +\infty} \frac{f (z_n) - f(0)}{z_n} = 1.$

Hence

$\displaystyle \lim_{n \to +\infty} \frac{f (x_n) - f(0)}{x_n} = 1$

as claimed. Thus, we have just shown that $f$ is differentiable at $0$ with $f'(0)=1$ . Finally, we show that the inverse function $f^{-1}$ is not continuous at $0$ . Indeed, this follows from

$\displaystyle f^{-1} \Big( \frac 1{(n+1)^2 + 2n+2}\Big) = (n+1)^2 + 2n+2 \nearrow +\infty$

as $n \to +\infty$ .

Our last observation concerns the case when $f$ is twice differentiable at $x_0$ . However, in this case $f$ is differentiable in a neighborhood of $x_0$ , which is enough to conclude that its inverse $f^{-1}$ must be continuous at $y_0$ .

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Ngô Quốc Anh

June 3, 2022