norms and sparsity Could anyone please elaborate on why $L^2$ norm moves toward the outliers compared to $L^1$ norm. I mean, what property/quantity in the mathematical expression of the norms makes it perform such way.
One more thing is, how $L^1$ norm introduces more sparsity in the solution?
Thank you.
Praveen 

I think the overall context you're referring to is the problem of $L^0$ minimization, i.e. compressed sensing. So, the goal of your question is to find the relationship between the following problems:
$(P_0)\qquad \min \|x\|_0 \;s.t.\;Ax=y\\
(P_1)\qquad \min \|x\|_1 \;s.t.\;Ax=y\\
(P_2)\qquad \min \|x\|_2 \;s.t.\;Ax=y$
Since our goal is ultimately to solve $P_0$, the problem with $L^2$ minimization is as follows:

As you can see, it is generally unlikely that a matrix $A$ will have a $P_2$ solution that lies on any of the axes (i.e. a sparse solution). However, because of the "diamond-shape" of the set of equal $L^1$ norm, the $L^1$ solution is more likely to be sparse
As for your second question, here's an exercise from my own coursework that might help you understand this better

Consider $P_1$ as described above, where $x\in\mathbb R^N, A\in \mathbb R^{m\times N},y\in \mathbb R^m$, with $m\ll N.$ Then $(P_1)$ has a solution with at most $m$ non-zero entries.  Accordingly, the solutions to the $P_1$ problem promote sparsity.

Actually proving this is an interesting exercise, and I can give you a hint there if you want it. The point is, we can guarantee a relatively sparse solution under $L^1$ minimization, which we can't generally do for $L^2$ minimization.  In fact, if $A$ has the null-space property, we find that there is a unique solution like this.

Here's the hint that came with the problem:

Hint: suppose $\overline{x}\in\mathbb R^N$ is a solution to $(P_1)$ and $\|\overline{x}\|=k$ where $m<k\leq N$.  It follows that $k$ columns of $A$ are linearly dependent.  As a result, there exists a nonzero vector $h$ in $\mathbb R^N$ such that $Ah=0$.  
Define $\widetilde{x}=\overline{x}+\epsilon h$ where $\epsilon\in\mathbb R$, then $A\widetilde{x}=y$, i.e., $\widetilde{x}$ is also a solution to $y=Ax$.  Since $\overline{x}$ is a solution to $(P_1)$, we have 
  $$\|\widetilde{x}\|_1=\|\overline{x}+\epsilon h\|_1 \geq \|\overline{x}\|_1$$
  Therefore, we can choose $\epsilon$ such that
$$\|\overline{x}+\epsilon h\|_1=\|\overline{x}\|_1 \quad \text{ and } \quad \|\overline{x}+\epsilon h\|_0<\|\overline{x}\|_0$$

