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$x^{2} \sin (1 / x^{2})$ has an unbounded, non-Riemann-integrable derivative

Example

Define $F : R \to R$ by

$F (0) : = 0, F (x) : = x^{2} \sin (1 / x^{2}) (x \neq 0) .$

The function $F$ is differentiable on $R$ , with $F^{'} (0) = 0$ , and $F^{'}$ is unbounded on every neighbourhood of zero. Consequently no extension of $F^{'} ∣_{(0, 1]}$ to $[0, 1]$ is Riemann integrable under the Darboux convention.

Facts & Assumptions

Given: The function $F$ in the Example.

[L1]

For every real $u$ , $∣ \sin u ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L2]

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

[L3]

The chain rule computes the derivative of a composite (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L4]

Products and scalar multiples of differentiable functions are differentiable with the usual derivative formulas (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 power rule gives the derivatives of $x^{2}$ and $x^{- 2}$ on their natural domains (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]

Sine vanishes and cosine equals $1$ at every integer multiple of $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi, Quarter-turn values and shifts by pi/2 and pi, The derivatives of sine and cosine are cosine and minus sine).

[L7]

Every positive real has a unique positive square root (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , case $n = 2$ ).

[L8]

For every real $ε > 0$ , there is a positive integer $N$ with $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L9]

Darboux integrability on $[a, b]$ is defined for bounded real functions on that interval (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L10]

The number $π = 2 γ$ is positive because $γ \in (0, 2)$ (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).

[L11]

The derivative at zero is the limit of $(F (h) - F (0)) / h$ as $h \to 0$ through nonzero reals (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).

Verification

technique · direct

1.1

For $h \neq 0$ , the difference quotient at zero from [L11] is $h \sin (1 / h^{2})$ , whose absolute value is at most $∣ h ∣$ by [L1]. Hence $F^{'} (0) = 0$ .

L1L11algebra

1.2

For $x \neq 0$ , [L2] to [L5] give $F^{'} (x) = 2 x \sin (1 / x^{2}) - \frac{2 \cos (1 / x^{2})}{x} .$

L2L3L4L5algebra

2.1

For $k \in N$ , let $x_{k}$ be the positive square root of $1 / (2 π (k + 1))$ . It exists by [L7] and [L10], and $1 / x_{k}^{2} = 2 π (k + 1)$ , so [L6] and step 1.2 give $F^{'} (x_{k}) = - 2 / x_{k}$ .

step 1.2L6L7L10constructalgebra

3.1

Let $η > 0$ . Applying [L8] below the positive real $2 π η^{2}$ shows that $1 / (2 π (k + 1)) < η^{2}$ for all sufficiently large $k$ , hence $0 < x_{k} < η$ by uniqueness and order of the positive square root. Thus $x_{k} \to 0$ . Given a real $M > 0$ , the same argument with $η = 2 / M$ gives $2 / x_{k} > M$ eventually, so $1 / x_{k} \to + \infty$ . Therefore the values $∣ F^{'} (x_{k}) ∣ = 2 / x_{k}$ exceed every real bound arbitrarily close to zero.

step 2.1L7L8L10algebra

4.1

Steps 1.1, 1.2, and 3.1 show that $F$ is differentiable on $R$ , with $F^{'} (0) = 0$ , while $F^{'}$ is unbounded on every neighbourhood of zero.

step 1.1step 1.2step 3.1

5.1

Every extension of $F^{'} ∣_{(0, 1]}$ to $[0, 1]$ retains the unbounded values from step 3.1, but [L9] requires boundedness for Darboux integrability. No such extension is Riemann integrable on $[0, 1]$ .

step 3.1step 4.1L9∎

Depends on

Used by

FALSE: every differentiable function has a continuous derivative False statement

Dependency tree · two levels

63 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

Adam Coffman, Yifei Pan, and Yuan Zhang, Continuous Solutions of Nonlinear Cauchy–Riemann Equations and Pseudoholomorphic Curves in Normal Coordinates, Remark 3.6 (standard reference, not scraped)

x2sin⁡(1/x2)x2sin(1/x2) has an unbounded, non-Riemann-integrable derivative