 Solution

Use Lagrange multipliers ${\alpha }_i\ge 0$ for $1-{\xi }_i-y_i(w\cdot x_i+b)\le 0$ and ${\mu }_i\ge 0$ for $-{\xi }_i\le 0$. The optimization Lagrangian isThe Lagrange dual function is the infimum over the now unconstrained primal variables. The first-order optimality conditions in $w$ and $b$ giveIf the latter equality fails, the infimum over $b$ is $-\infty$. The first-order optimality condition in ${\xi }_i$ givesSubstitution, or completing the square separately in $w$ and each ${\xi }_i$, producesprovided ${\sum }_i{\alpha }_i{y}_i=0$. For fixed nonnegative ${\alpha }_i$, this expression is maximized over ${\mu }_i\ge 0$ at ${\mu }_i=0$. Thus the simplest Lagrangian dual problem isIn particular, there is no upper box constraint ${\alpha }_i\le C$. That constraint belongs to the linear-slack penalty; squared slack instead contributes the displayed diagonal quadratic penalty. With $Y=\mathrm{diag}(y_1,\dots ,y_m)$ and Gram matrix $G_{ij}=x_i\cdot x_j$, the objective can also be writtenSince $G$ is positive semidefinite and $C>0$, the matrix in parentheses is positive-definite. The objective of the Lagrangian dual problem is therefore strictly concave, and the feasible maximizer is unique.

The primal problem is a convex optimization problem, and $w=0,b=0,{\xi }_i=2$ is strictly feasible for both families of inequalities. Slater's condition gives strong duality, so an optimum of the Lagrangian dual problem yields a primal optimum. The Karush-Kuhn-Tucker conditions supplyFor any $i$ with ${\alpha }_i>0$, the intercept is consequentlyAll such choices agree at an exact optimum. If all labels have one sign, the optimal coefficients can all be zero: choose $w=0$, $\xi =0$, and any constant intercept of that sign with magnitude at least one.