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$x \mapsto x^{3}$ is a $C^{1}$ bijection whose inverse is not differentiable at zero

Statement refuted

A $C^{1}$ bijection between open subsets of the real line must be a $C^{1}$ diffeomorphism.

The counterexample below establishes: The map $x \mapsto x^{3}$ is a smooth open bijection of $R$ with derivative zero at the origin, but its inverse is not differentiable there.

Facts & Assumptions

Given: The map $f (x) = x^{3}$ , the $C^{1}$ and diffeomorphism conventions (Continuously differentiable maps, local inverses, and local diffeomorphisms, $C^{k}$ Euclidean maps and diffeomorphisms), and the fact that a continuous strictly monotone function on an interval has a continuous inverse Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ .

[L1]

For every integer $d \geq 1$ , $p_{d} (x) = x^{d}$ is differentiable with $p_{d}^{'} (x) = d x^{d - 1}$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L2]

If differentiable functions are composable, then their composite is differentiable and its derivative is the product of the two derivatives (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L3]

A differentiable real function is continuous, and a continuous injective function on an interval has a continuous inverse onto its image (A function differentiable at $c$ is continuous at $c$ , Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[L4]

Sums and scalar multiples of differentiable real functions are differentiable with the expected derivatives (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L5]

The constant power $p_{0} (x) = 1$ has derivative zero (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L6]

Every nonnegative real has a unique nonnegative cube root (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L7]

The cube function is strictly increasing on the nonnegative reals (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

Counterexample

technique · contradiction

1.1

By [L1], [L4], and [L5], the successive derivatives of the cube map are $3 x^{2}$ , $6 x$ , $6$ , and then zero, so it is smooth. By [L7], the cube is strictly increasing on the nonnegative half-line; the identity $(- x)^{3} = - x^{3}$ then gives strict increase on the whole line. For $y \geq 0$ , [L6] supplies $x \geq 0$ with $x^{3} = y$ , while for $y < 0$ the negative of the cube root of $- y$ maps to $y$ . Thus the cube map is onto and hence bijective. By [L3] it and its inverse are continuous, so it is a homeomorphism and therefore open. Its derivative at zero is zero.

L1L3L4L5L6L7givenalgebra

2.1

Suppose its inverse $g$ were differentiable at zero. Applying [L2] to $g (x^{3}) = x$ at zero would give $g^{'} (0) f^{'} (0) = 1$ , but step 1.1 makes the left side zero. Therefore the inverse is not differentiable at zero. The map $x \mapsto x^{3}$ is a smooth open bijection of $R$ with derivative zero at the origin, but its inverse is not differentiable there.

step 1.1L2assume-contraalgebradischarge-contradiction∎

Depends on

Used by

FALSE: every C¹ bijection has a C¹ inverse False statement
FALSE: every open C¹ map has invertible derivative False statement

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, Exercise 8.5.4 (standard reference, not scraped)

x↦x3x↦x3 is a C1C1 bijection whose inverse is not differentiable at zero