# 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!

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.

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.

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.