What is wrong with this reasoning in proving that $||\textbf{x}||^3$ is convex?

Question

I am trying to prove that $||\textbf{x}||^3$ is convex by showing that $$||\lambda \textbf{x}_1+(1-\lambda)\textbf{x}_2||^3\leq \lambda||\textbf{x}_1||^3+(1-\lambda)||\textbf{x}_2||^3$$ where $0\leq \lambda\leq1$. On the left hand side I use $$||\lambda \textbf{x}_1+(1-\lambda)\textbf{x}_2||\cdot||\lambda \textbf{x}_1+(1-\lambda)\textbf{x}_2||\cdot||\lambda \textbf{x}_1+(1-\lambda)\textbf{x}_2||\leq (\lambda ||\textbf{x}_1||+(1-\lambda)||\textbf{x}_2||)^3$$ After this if I could prove that $$\lambda||\textbf{x}_1||^3+(1-\lambda)||\textbf{x}_2||^3-(\lambda ||\textbf{x}_1||+(1-\lambda)||\textbf{x}_2||)^3\geq 0$$ then it is obvious that $||\textbf{x}||$ is convex but unfortunately I am unable to show this (the result I get is given below) $$(1-\lambda)\left[\lambda ||\textbf{x}_1||^3(1-\lambda)+||\textbf{x}_2||^3(1-(1-\lambda)^2)-3\lambda^2||\textbf{x}_1||^2||\textbf{x}_2||-3\lambda ||\textbf{x}_1||(1-\lambda)||\textbf{x}_2||^2\right]$$ (which is not obvious to be positive). Is there any other way through which I could show that $||\textbf{x}||^3$ is convex? Thanks in advance.

Answer 1

More generally, if $f$ is convex and $g$ is non-decreasing and convex then the composition $g\circ f$ is convex: $$ g\left(f\left(\lambda \textbf{x}_1+(1-\lambda)\textbf{x}_2\right)\right) \le g\left(\lambda f(\textbf{x}_1) + (1 - \lambda) f(\textbf{x}_2)\right) \le \lambda g(f(\textbf{x}_1)) + (1 - \lambda)g(f(\textbf{x}_2)).$$ Take $f(x)=\|\textbf{x}\|$ and $g(x)=x^3$ (which is convex in $[0,+\infty)$).

Answer 2

Which theorems can you use? For onedimensional twice differentiable functions you can use

$$ f\text{ is convex }\Leftrightarrow f''\geq 0. $$

This implies that $f:[0,\infty)\to[0,\infty)$, $f(x)=x^3$ is convex. You can conclude \begin{align*} \lambda||\textbf{x}_1||^3+(1-\lambda)||\textbf{x}_2||^3 &=\lambda f(\|\textbf{x}_1\|)+(1-\lambda)f(\|\textbf{x}_2\|)\\ &\geq f(\lambda\|\textbf{x}_1\|+(1-\lambda)\|\textbf{x}_2\|)\\ &=(\lambda ||\textbf{x}_1||+(1-\lambda)||\textbf{x}_2||)^3 \end{align*}