Exercise 3, § 1

a) Show that $(1 + x)^\alpha \geq 1 + \alpha x$ for $\alpha \geq 1$ and $x \geq 0$, and that $(1 + x)^\alpha \leq 1 + \alpha x$ for $0 \leq \alpha \leq 1$ and $x \geq 0$.

b) Show that $(1 - x)^\alpha \leq \frac{1}{1 + \alpha x}$ for $0 \leq x \leq 1$ and $\alpha \geq 0$.

c) For $0 \leq \alpha \leq 1$ show that for $x \geq 0$ and $y \geq 0$ one has $x^\alpha y^{1-\alpha} \leq \alpha x + b y$ for every pair of numbers $> 0$ for which $a^\alpha b^{1-\alpha} = \alpha(1 - \alpha)^{1-\alpha}$.

d) Deduce from c) that if $f, g$ are two regulated $\geq 0$ functions on an interval $I$ such that the integrals $\int_I f(t)\,dt$ and $\int_I g(t)\,dt$ converge, then the integral $\int_I (f(t))^\alpha (g(t))^{1-\alpha}\,dt$ converges, and

From this deduce the Hölder inequality (which is called the "Cauchy-Buniakowski-Schwarz inequality" when $\alpha = \frac{1}{2}$).