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FALSE: an invertible derivative at one point gives a local inverse

Statement

False claim: if a real function is differentiable at $a$ and $f^{'} (a) \neq 0$ , then it has a local inverse at $a$ in the sense of Continuously differentiable maps, local inverses, and local diffeomorphisms.

Facts & Assumptions

Given: Define $f (0) = 0$ and $f (x) = x + 2 x^{2} \sin (1 / x)$ for $x \neq 0$ . We use derivative algebra, the chain and power rules (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 $g (c) \neq 0$ , 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , 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), boundedness of sine Signs, monotonicity intervals, and ranges of sine and cosine, and differentiability implying continuity A function differentiable at $c$ is continuous at $c$ .

[L1]

A real function is differentiable at $c$ when its relative difference quotient has a finite limit (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L2]

The derivatives of sine and cosine satisfy $(\sin x)^{'} = \cos x$ and $(\cos x)^{'} = - \sin x$ (The derivatives of sine and cosine are cosine and minus sine).

[L3]

Both sine and cosine have period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[L4]

The quarter-turn values include $\sin (π / 2) = 1$ , $\cos (π / 2) = 0$ , $\sin π = 0$ , and $\cos π = - 1$ (Quarter-turn values and shifts by pi/2 and pi).

[L5]

A continuous injective real function on an interval is strictly monotone (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 $f$ ).

Refutation

technique · contradiction

1.1L1L2givenalgebra

By [L1], $(f (h) - f (0)) / h = 1 + 2 h \sin (1 / h) \to 1$ , so $f^{'} (0) = 1$ . For $x \neq 0$ , the algebra, chain, and power rules with [L2] give $f^{'} (x) = 1 + 4 x \sin (1 / x) - 2 \cos (1 / x) .$

2.1step 1.1L3L4algebra

Put $x_{n} = 1 / (2 π n)$ and $y_{n} = 1 / ((2 n + 1) π)$ for $n \geq 1$ . By [L3] and [L4], step 1.1 gives $f^{'} (x_{n}) = - 1$ and $f^{'} (y_{n}) = 3$ . Both sequences tend to zero, so derivatives of both signs occur in every neighbourhood of zero.

3.1step 2.1L1L5givenassume-contradischarge-contradiction∎

Suppose $f$ were injective on an interval about zero. It is continuous there, so [L5] would make it strictly increasing or strictly decreasing. Difference quotients show that the derivative of an increasing differentiable function is nonnegative and that of a decreasing one is nonpositive, contradicting step 2.1. Thus $f^{'} (0) = 1$ is invertible but no local inverse exists.

Depends on

Used by

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Sources

J. Lebl, Basic Analysis II, Exercise 8.5.7 (standard reference, not scraped)