Strict convexity of a function

Question

How to prove that for each $\alpha > 1$ the function $\psi(t)=|t|^\alpha$ is strictly convex on the real line?

Answer 1

You can see $\psi$ as a composition of two functions, namely $x \mapsto |x|$ and $x \mapsto x^{\alpha}$ the first one is convex, and the second one is strictly convex , and increasing. Hence the result.

Answer 2

Here is a direct proof. Let $x_1$, $x_2 \in \mathbb{R}$ ($x_1 \neq x_2$), and let $0<\theta<1$. We have: $$\begin{align} |\theta x_1 + (1-\theta) x_2 |^\alpha &\leq (|\theta x_1| + |(1-\theta) x_2 |)^\alpha \\ & < |\theta x_1|^\alpha + |(1-\theta) x_2 |^\alpha \\ & = \theta^\alpha | x_1|^\alpha + (1-\theta)^\alpha |x_2 |^\alpha \\ & < \theta| x_1|^\alpha + (1-\theta) |x_2 |^\alpha. \end{align}$$