How to show that $x\mapsto|x^3| $ is convex?

Question

how to show that $x\mapsto|x^3| $ is convex .

I tried to use the definition $$f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda) f(y)$$

But how can I get the inequality ?

Answer 1

For $x\in\mathbb{R}$, $|x^3|=|x|^3$.

Note that $h(x)=|x|$ is convex and that $g(x)= x^3$ on $x\in[0,\infty$) is convex and nondecresing.

First apply the convexity of $h$ and then the monotonicity of $g$. Finally, use the convexity of $g$ on $[0,\infty)$.

$$g(\lambda|x|+(1-\lambda)|y|)\leq\lambda g(|x|)+(1-\lambda)g(|y|)$$

Answer 2

Using the fact that: Second derivative positive $\implies$ convex

Using the chain rule:

$$(|x^3|)^{''}=(3x\left | x \right |)'=\frac{6x^{2}}{\left | x \right |}\geq 0$$

This proves that $x\mapsto|x^3|$ is convex.