Why "hinge" loss is equivalent to 0-1 loss in SVM? I'm reading a book named The Elements of Statistical Learning by Hastie et al. In $\S 12.3.2$ it introduced the SVM as a penalization method:
With $f(x)=h(x)^T \beta+\beta_0 $, the solution of the optimization problem
$$\min_{\beta_0,\beta} \sum_{i=1}^N [1-y_i f(x_i)]_+ +\frac{\lambda}{2}\|\beta\|^2 $$
with $\lambda=\frac{1}{C}$, is the same as that for
$$\min_{\beta_0,\beta} \frac{1}{2}\|\beta\|^2 +C \sum_{i=1}^N \xi_i$$
$$\text{subject to } \xi_i \geq 0,y_i f(x_i)\geq 1-\xi_i ~ \forall i, $$
Could anyone kindly give me some hint why they are equivalent, and what's the benefit of introducing the "hinge" loss?
Thanks a lot!
 A: Here is an intuitive illustration of difference between hinge loss and 0-1 loss:

(The image is from Pattern recognition and Machine learning)
As you can see in this image, the black line is the 0-1 loss, blue line is the hinge loss and red line is the logistic loss. The hinge loss, compared with 0-1 loss, is more smooth. The 0-1 loss have two inflection point and it have infinite slope at 0, which is too strict and not a good mathematical property. Thus, we soft this constraint to allow certain degree misclassificiton and provide convenient calculation.
A: I know this question old as dirt but I was just reading Hastie and had to figure it out.
From the constrains of the soft margin SVM formulation (the one at the bottom):
$1-y_if\left(x_i\right) \geq 1-\varepsilon_i$ we get
$\varepsilon_i \geq 1-y_if\left(x_i\right)$, since $\varepsilon_i$ must be greater than zero anyways, the constraint tells us
$\varepsilon_i \geq \left[ 1-y_if\left(x_i\right) \right]_+$
Now since we are minimizing, $\varepsilon_i$ will reach the lower bound
$\varepsilon_i = \left[ 1-y_if\left(x_i\right) \right]_+$
Substitute in the original formulation, multiply by $\lambda = \frac{1}{C} $ and you are good to go.
A: I know this question is very old but there is a more rigid way to get it：
Lagrangian:
    $$L = \|\omega\|^2_{2} + c\sum_{i=1}^{n}\zeta_i + \sum_{i = 1}^{n}r_i(\max(0,1-y_if(x_i))-\zeta_i)$$
    by KKT condition (with respect to $\zeta_i$):
    $$ \frac{\partial L}{\partial \zeta_i} = c - r_i = 0 \Rightarrow r_i = c\,, \forall i $$
    so
    $$ L = \|\omega\|^2_{2} + c\sum_{i=1}^{n}\max(0,1-y_if(x_i)) $$
thus it is equivalent to optimizing the hinge loss form.
