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?

Note that $\psi$ is differentiable on $\mathbb{R}$. Then use the fact that a differentiable function is strictly convex if and only if its derivative is strictly increasing. — Frank Lu, Nov 14 '16 at 23:10

score 1 · Answer 1 · answered Nov 14 '16 at 23:17

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.

answered Nov 14 '16 at 23:17

W.314

348

1

Notice that every constant function, say $f(x) = c \ge 0$, is convex. But $f^{\alpha}$ isn't strictly convex. – user251257 Nov 15 '16 at 00:00
@W.314 why is $x^\alpha$ strictly convex? – roc11111111 Nov 15 '16 at 02:28
@roc11111111 $x^{\alpha}$ is strictly convex for $ x \geq 0$ since its derivative $\alpha x ^{\alpha - 1}$ is increasing – W.314 Nov 15 '16 at 09:34

score 0 · Answer 2 · answered Nov 15 '16 at 14:31

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: $$\begin{align} |\theta x_1 + (1-\theta) x_2 |^\alpha &\leq (|\theta x_1| + |(1-\theta) x_2 |)^\alpha \\ & < |\theta x_1|^\alpha + |(1-\theta) x_2 |^\alpha \\ & = \theta^\alpha | x_1|^\alpha + (1-\theta)^\alpha |x_2 |^\alpha \\ & < \theta| x_1|^\alpha + (1-\theta) |x_2 |^\alpha. \end{align}$$

Strict convexity of a function

2 Answers2