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: $\| x\|_2 = \sqrt{\sum x_i^2}$.
 A: This answer gives a reply to the question if a space equipped with $L_2$ norm is a strictly convex space. The answer to which is "yes." This however appears not be what was asked for. I still leave this up for now as I think it might help to avoid potential confusions.
Intuitively, strictly convex means that if you have two points on the boundary on the unit ball then the line segment connecting these two points is in the interior of the unit ball (and not at the boundary). 
For the $L_2$ norm (say in $\mathbb{R}^3$ where we can visualize things) the unit ball is an "actual" ball, so if you have two points on the surface an you connect them with a straight line then the line goes though the interior of the ball. 
Contrast this with the $L_{\infty}$ norm where the unit ball has the form of a  cube. When you take two points on the same face and connect them you stay on the face that is on the boundary. 
Rigorously, it is a quite direct consequence of the characterization of the case of equality in the triangle inequality which in turn rest on the characterization of equality in the Cauchy-Schwarz inequality.
What you need to show is that for distinct $x,y $ with $\| x \| = \| y \| = 1 $ and for $0 < \lambda < 1 $ you have 
$$\|  \lambda x  + (1 - \lambda ) y \| < 1. $$ 
Note that $\lambda x  + (1 - \lambda ) y$ is a point on the line connecting $x$ and $y$. Further note that by the triangle inequality and the linearity of the norm you have right away
$$\|  \lambda x  + (1 - \lambda ) y \| \le  \|  \lambda x  \| +  \|(1 - \lambda ) y \| = \lambda \| x\| + (1 - \lambda) \| y \| = \lambda + (1 - \lambda)   = 1. $$ 
Thus, if the triangle inequality is strict then you would have what you want. 
Now, recall that you have equality in the triangle inequality if and only if $x = \mu y$ or $y = \mu x$  for a positive $\mu$. However, since $x$ and $y$ both have norm $1$ and are distinct this can never happen. 
To see this fact on the triangle inequality recall that for equality you need $\langle x , y \rangle = \|x \|  \| y \|$, which only holds under that condition. 
This arguments rests true for any norm that comes from an inner product. 
A: 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.
