Exercise 22, § 1

Let $f$ be a continuous increasing real function and $\geq 0$ on the interval $[0, 1]$ a) Show that there exists a convex function $g \geq 0$ on $[0, 1]$ such that $g \leq f$ and $\int_0^1 g(t)\,dt \geq \frac{1}{2} \int_0^1 f(t)\,dt$ (cf. I, p. 48, exerc. 23). b) Show that there exists a convex function $h$ on $[0, 1]$ such that $h \geq f$ and that

(For every $a$ such that $0 < a \leq 1$, let $f_a$ be the function equal to $f$ for $0 \leq t \leq a$ and to $f(a)$ for $a \leq t \leq 1$; let $A$ be the set of $a \in ]0, 1]$ for which there exists a convex function $h_a$ on $[0, 1]$ such that $h_a \geq f_a$ and $\int_0^1 h_a(t)\,dt \leq 2 \int_0^1 f_a(t)\,dt$. Show that the least upper bound $b$ of $A$ again belongs to $A$, using exerc. 1 of I, p. 45. Then prove that f_b = f, arguing by contradiction. For this, reduce to proving the following result: if $\varphi$ is continuous and increasing on $[0, 1]$, not constant, and such that $\varphi(0) = 0$. then there is a point $c \in ]0, 1[$ such that $\varphi(c) > 0$ and such that the linear function $\psi$ such that $\psi(c) = \varphi(c)$ and satisfies the relation $\psi(t) \geq \varphi(t)$ for $\max(0, 2c - 1) \leq t \leq c.$