Why is lasso not strictly convex I know a nonmonotonic convex function which attains its minimum value at a unique point only is strictly convex. I didn't get how lasso is not strictly convex. For eg if I consider two dimensional case.
$||x||_1$ attains its minimum value at (0,0) which is unique
 A: 
I thought that a strictly convex function is a convex function which has a unique minimizer. I am saying this was the wrong definition. 

Indeed, this was quite wrong, and the source of confusion here. What is true is that strict convexity is related to uniqueness of minimizer: if a strictly convex function attains its minimum, then it does so at exactly one point. However, having unique minimizer does not imply strict convexity, or any convexity at all. 
It's important to remember that there are two different notions of strict convexity. Namely, we have: 


*

*Strictly convex functions: $u(ta+(1-t)b)<tu(a)+(1-t)u(b)$ for $0<t<1$

*Strictly convex norms: $\|ta+(1-t)b\|<t\|a\|+(1-t)\|b\|$ for $0<t<1$, unless $a$ and $b$ are parallel vectors.  


A norm can  never be strictly convex function in the sense of definition 1, because for any nonzero vector $x$ we have $$\|2x\|=\frac12({\|x\|+\|3x\|}),\quad \text{where }\ 2x=\frac12(x+3x)$$
This is why the definition of a strictly convex norm requires non-parallel vectors. But $\|\cdot\|_1$ fails the second definition too: for example, 
$$\|e_1+e_2\|=\frac12({\|2e_1\|+\|2e_2\|}),\quad \text{where }\ e_1+e_2=\frac12(2e_1+2e_2)$$
and $e_1,e_2$ are standard basis vectors.
