Support Vector machine : a simple question I think that a formulation of SVM for points x with label y is :
$$
\begin{align} 
\arg\min_{\substack{u,w,b}}  \frac{1}{2} \cdot |w|^2 + C \cdot \sum_{i} u_i  \\
 s.t.\ \  y_i\cdot (w \cdot x_i + b) &\geq 1-u_i \\\\
   u_i \geq  0 \\
\end{align}
$$
In that formulation, if we take C = 0, what prevents $u_i$ to go to infinity, so that the two constraints are always satisfied ?
 A: I don't think the $u_i$s will go to infinity.  If $C$ is set to zero this effectively disables the inequality constraints, so the optimisation problem is just to minimise the squared norm of the weights, which has a trivial solution at $w = 0$.  Substituting this into $y_i(w \cdot x_i + b) \geq 1 - u_i$ gives $y_i b \geq 1 - u_i$ which is satisfied by $u_i = 1 - y_i b$.  This means there will be a solution where the $u_i$ are finite, provided a sensible choice is made for $b$.  Whether the software actually finds this solution is another matter though; however as setting C=0 is effectively telling the SVM to completely ignore the training data, it is perhaps not too surprising that the programmer didn't consider this!
A: Actually, the slack variable $u_i$ means the tolerance of inconsistent labeling with $y_i$ by the linear function $(wx_i +b)$. The factor $C$ "adjusts" the weight of total tolerance. Therefore, setting $C$ to zero might be problematic in the case that the set $\{x_i,y_i\}$ is not linearly separable! 
