Why L1 Norm is Great

1. Application

1.1. Loss Function

$$
\min {\omega}\left|y-\omega^{T} x\right|{1}
$$

1.2. Regularization

$$
\min {x} f(x)+|x|{1}
$$

2. Advantage

We can consider a easy 2D case, and show why L1 norm is better than L2 norm

2.1. Loss Function

L2 norm would strengthen the error

2.2. Regularization

The minimum is always located at one of the corners (Red)

• A corner is described as having 1 non-zero coordinate with the remaining coordinates being zero

So L1 norm would lead to better sparisity