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The continuous path $γ (x) = (x, x \sin (1 / x))$ on $[0, 1]$ , with $γ (0) = (0, 0)$ , is not rectifiable

Counterexample

Define $f : [0, 1] \to R$ by $f (0) = 0$ and $f (x) = x \sin (1 / x)$ for $x > 0$ . Then $f$ is continuous, but the graph path $γ (x) = (x, f (x))$ is not rectifiable.

Facts & Assumptions

Given: The function $f$ and graph path $γ$ .

[L1]

The number $π$ is positive, and the shift formulas give $\sin (π / 2 + k π) = (- 1)^{k}$ for integers $k \geq 0$ (Pi as twice the smallest positive zero of cosine, Quarter-turn values and shifts by pi/2 and pi).

[L2]

The harmonic series diverges, the $p = 1$ case of the rational $p$ -series theorem (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ); a nonnegative series converges exactly when its partial sums are bounded above (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L3]

Bounded variation means that all partition variation sums are bounded above (Bounded variation and total variation on an interval).

[L4]

A path is rectifiable exactly when all of its coordinate functions have bounded variation (A path in $R^{n}$ is rectifiable exactly when every coordinate has bounded variation).

[L5]

Reciprocals of positive naturals tend below every positive bound (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

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

Verification

technique · divergent-variation

1.1

Since $∣ f (x) ∣ \leq x$ for $x > 0$ , $f (x) \to 0 = f (0)$ as $x ↓ 0$ ; away from zero it is continuous. Thus $γ$ is a path.

givenL6

1.2

Put $x_{k} = 2 / ((2 k + 1) π)$ . Positivity of $π$ and [L5] give $x_{k} ↓ 0$ , so choose $K$ with $x_{K} \leq 1$ ; and [L1] gives $f (x_{k}) = (- 1)^{k} x_{k}$ .

givenL1L5

2.1

For $N > K$ , take the partition whose points are $0, x_{N}, x_{N - 1}, \dots, x_{K}, 1$ , omitting a repeated endpoint if $x_{K} = 1$ . Its variation contribution from consecutive $x_{k}$ is $∣ f (x_{k}) - f (x_{k + 1}) ∣ = x_{k} + x_{k + 1} \geq x_{k}$ .

step 1.2L3

3.1

Since $x_{k} = 2 / ((2 k + 1) π) \geq 1 / (π (k + 1))$ , [L2] says the tails $\sum_{k = K}^{N - 1} x_{k}$ are unbounded. Hence the variation sums in step 2.1 are unbounded and $f$ is not of bounded variation.

step 2.1L2algebra

4.1

The first coordinate $x \mapsto x$ has bounded variation, but the second does not by step 3.1. Therefore [L4] says the graph path is not rectifiable.

step 3.1L4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

A. R. Shastri, Metric Spaces, Section 6 (standard reference, not scraped)

The continuous path γ(x)=(x,xsin⁡(1/x))γ(x)=(x,xsin(1/x)) on [0,1][0,1], with γ(0)=(0,0)γ(0)=(0,0), is not rectifiable