Exercise 17, § 1

Let $f$ be a real function defined on an open interval $I \subset \mathbf{R}$, and such that for every system of three points $x_1, x_2, x_3$ of $I$ satisfying $x_1 < x_2 < x_3 < x_1 + \pi$ one has

(1)

Show that:

a) $f$ is continuous at every point of $I$ and has a finite right and a finite left derivative at every point of $I$; further,

(2) for every pair of points $x, y$ of $I$ such that $|x - y| \leqslant \pi$; there is also an analogue of (2) where one replaces $f'_d$ by $f'_g$; finally, one has $f'_g(x) \leqslant f'_d(x)$ for every $x \in I$. (To show (2) let $x_2$ approach $x_1$ in (1), keeping $x_3$ fixed, and so obtain an upper bound for $\limsup_{x_2 \to x_1} \frac{f(x_2) - f(x_1)}{x_2 - x_1}$; then let $x_3$ approach $x_1$ in the inequality so obtained; deduce from this the existence of $f'_d(x)$ and the inequality (2) for $y > x$; proceed similarly for the other parts.)

b) Conversely, if (2) holds for every pair of points $x, y$ of $I$ such that $|x - y| \leqslant \pi$, then $f$ satisfies (1) on $I$ (consider the difference $\frac{f(x_2)}{\sin(x_3 - x_2)} - \frac{f(x_1)}{\sin(x_3 - x_1)}$).

c) If $f$ admits a second derivative on $I$, then (1) is equivalent to the condition

for every $x \in I$.

"Interpret these results, considering the plane curve defined by $x = \frac{1}{f(t)} \cos t,\ y = \frac{1}{f(t)} \sin t$ ("convexity with respect to the origin")."