Exercise 8.3

If $h$ is the optimal hyperplane, its $w_h$ minimizes $w^{T}w$. Also, for a separating hyperplane, the data point that is nearest to the hyperplance has distance $\frac{1}{\left|w_h\right|}$. Assume ${\rho }_{\text{+}}>{\rho }_{\text{−}}$, so we should have ${\rho }_{\text{−}}=\frac{1}{\left|w_h\right|}$.

Let $l=\frac{{\rho }_{\text{+}}-{\rho }_{\text{−}}}{2}$, and ${\rho }_{\text{+}}^{\prime }={\rho }_{\text{+}}-l=\frac{{\rho }_{\text{+}}+{\rho }_{\text{−}}}{2}>\frac{1}{\left|w_h\right|}$, ${\rho }_{\text{−}}^{\prime }={\rho }_{\text{−}}+l=\frac{{\rho }_{\text{+}}+{\rho }_{\text{−}}}{2}>\frac{1}{\left|w_h\right|}$.

If we let $w^{\prime }={\rho }_{\text{+}}^{\prime }={\rho }_{\text{−}}^{\prime }$, then we have $\left|w^{\prime }\right|<\left|w_h\right|$, which contradicts the fact that $w_h$ minimizes $w^{T}w$.