This question concerns a subtle issue in the recovery of primal solution for support vector machine from the dual. None of the sources I read addresses it explicitly. In particular, it was missing from this paper (see p.15):
https://www.di.ens.fr/~mallat/papiers/svmtutorial.pdf
Recall the soft margin support vector machine problem \begin{align*} \text{minimize}_{w \in \mathbb{R}^p, b \in \mathbb{R}, \xi \in \mathbb{R}^n} & \quad \frac{1}{2}\|w\|^2 + C \sum_{i = 1}^n \xi_i\\ \text{subject to} & \quad y_i(b + w^T x_i) \geq 1 - \xi_i, \\ & \quad \xi_i \geq 0. \end{align*}
The dual is \begin{align*} \text{maximize} & \quad -\frac{1}{2}\Big\|\sum_{i = 1}^n \alpha_i y_i x_i\Big\|^2 + \mathbf{1}^T \alpha\\ \text{subject to} & \quad \alpha_i + \mu_i = C,\\ & \quad \alpha^T y = 0,\\ & \quad \alpha \succeq 0, \mu \succeq 0, \end{align*} where $\alpha_i$ is the Lagrange multiplier for constraint $y_i(b + w^T x_i) \geq 1 - \xi_i$, and $\mu_i$ is for the constraint $\xi_i \geq 0$.
Assume that optimal dual solution $\alpha^*, \mu^*$ has been found. We can recover $w^* = \sum_{i = 1}^n \alpha_i^* y_i x_i$ from KKT. To recover the optimal $b$, the paper says take any $i$ with $0 < \alpha_i^* < C$. Then complementary slackness forces $y_i(b + w^T x_i) = 1 - \xi_i$. Also, since $\alpha_i^* + \mu_i^* = C$, one obtains $\mu_i^* > 0$. Again, complementary slackness implies $\xi_i = 0$. Now one can recover $b$ by $y_i(b + w^T x_i) = 1$.
My question: How can you prove that there exists $i$ such that $0 < \alpha_i^* < C$? It's not clear to me at all.