Prove convexity of squared Euclidean norm

First we show that any norm $|| \cdot ||: \mathbb{R}^n \to \mathbb{R_+}$ is a convex function:

It's clear the domain $\mathbb{R}^n$ is a convex set. Then by properties of the norm, $|| k x || =|k| || x ||$ and $||x+y|| \leq ||x|| + ||y|| $, for any $x,y \in \mathbb{R}^n$ and $0 \leq \theta \leq 1$,

$$|| \theta x + (1-\theta)y || \leq ||\theta x || + ||(1-\theta)y || = \theta || x || + (1-\theta)||y || $$

Now that $f(x)=x^2$ and $g(x)=||x||_2$ are both convex, and $f(x)=x^2$ is non-decreasing on $[0, \infty)$, the range of $g$ , therefore the composition $ f \circ g=||\cdot ||_2^2$ is convex.

Notice that $f(x)=x^2$ is convex ($2=f''(x)>0$.) Thus it is convex, which means that it satisfies $f\big(\theta\|x\|+(1-\theta)\|y\|\big)\leq \theta f(\|x\|)+(1-\theta)f(\|y\|)$ for $\theta\in [0, 1]$.

The Hessian of the squared Euclidean norm is everywhere (edit: twice) the identity matrix, which is positive definite.