A strictly separable finite sample has a unique hard-margin vector

Strict separability gives $u$ with $m_i(u)>0$ for every $i$. Since the sample is finite, $a:={\min }_i{m}_i(u)>0$. The scaled vector $u\mathrm{/}a$ satisfies $m_i(u\mathrm{/}a)=m_i(u)\mathrm{/}a\ge 1$, so the feasible set in [F1] is nonempty.

Put $R=\mathrm{\parallel }u\mathrm{/}a{\mathrm{\parallel }}_2$ and $K=\mathcal{F}\cap \{w:\mathrm{\parallel }w{\mathrm{\parallel }}_2\le R\}$. The set $K$ is nonempty by step 1.1 and is closed and bounded by [F1], so [F2] makes it compact and gives a minimizer $w_{\ast }$ of $\frac{1}{2}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$ on $K$. Every feasible vector outside the radius-$R$ ball has norm greater than $R$, while $u\mathrm{/}a\in K$ has norm $R$; therefore no such vector improves on $w_{\ast }$. Hence $w_{\ast }$ is a global minimizer on $\mathcal{F}$.

If two distinct feasible minimizers $w,z$ existed, their midpoint would remain feasible because each $m_i$ is linear. But ${\parallel \frac{w+z}{2}\parallel }_2^2=\frac{1}{2}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2+\frac{1}{2}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2-\frac{1}{4}\mathrm{\parallel }w-z{\mathrm{\parallel }}_2^2$ is strictly smaller than their common minimum value, a contradiction. Therefore the minimizer is unique.