Exercise 8, § 1

Functions of a Real Variable, II, 80

It is set in small type as supplementary.

Show that there exists a continuous function on $[-1, +1]$ admitting a finite derivative at every point of this interval, this derivative being equal to $\sin \left( \frac{1}{\sin 1/x} \right)$ at points $x$ different from $1/n\pi$ ( $n$ an integer $\neq 0$ ) and from 0. (On a neighbourhood of $x = 1/n\pi$ make the change of variable $x = \frac{1}{n\pi + \operatorname{Arc}\sin u}$ and use exerc. 7; by means of the same change of variable, show that there exists a constant $a > 0$ independent of $n$ such that

\left| \int_{2/(2n-1)\pi}^{2/(2n+1)\pi} \sin \left( \frac{1}{\sin \frac{1}{x}} \right) dx \right| \leq \frac{a}{n^3};

and deduce that if one puts

g(x) = \lim_{\varepsilon \to 0} \int_x^1 \sin \left( \frac{1}{\sin \frac{1}{t}} \right) dt,

then $g$ has a derivative equal to 0 at the point $x = 0$ .)

Solution

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