Is $L_2$-norm a strictly convex function? I am new to convex analysis, and just wondering whether there is a simple check to see whether $L_2$-norm is strictly convex. How to mathematically prove/disprove this?
$L_2$-norm: $\| x\|_2 = \sqrt{\sum x_i^2}$.

For $x$ and $y=2x\ne x$ we have for any $\lambda\in(0,1)$
$$
\|\lambda x+(1-\lambda)y\|=\|(\lambda+(1-\lambda)2)x\|=\lambda \|x\|+(1-\lambda)2\|x\|=\lambda \|x\|+(1-\lambda)\|y\|
$$
which means no, $f(x)=\|x\|$ is not a strictly convex function. If you think of the function graph (the cone) you will see that it becomes flat on the radial rays, i.e. the Hessian (second derivative) degenerates. However, the function $\|x\|^2$ is strictly convex.
