SVM optimality criterion in Bottou, Lin (2006) My question relates to an alternative optimality criterion for an SVM dual solution derived in Bottou, Lin (2006) in pages 8 and 9. 
Let:


*

*$\alpha^* = (\alpha_1^*,\dots,\alpha_n^*)$ be a dual solution to the SVM. Recall that any solution $\alpha^*$ must satisfy: $\forall i \in \{1,...,n\}, 0 \leq \alpha_i^* \leq C$; 

*$g_i^*$ the derivative of the dual objective function $\mathcal{D}$ with respect to $\alpha_i^*$; 

*$B_i$ equal to $C \times 1_{\{y_i=1\}}$ and $A_i$ equal to $-C \times 1_{\{y_i=-1\}}$ such that $\forall i \in \{1,...,n\},\; y_i\alpha_i \in [A_i,B_i]$ $-$ i.e. $[A_i,B_i]$ might be equal to $[0,C]$ or $[-C,0]$;

*$I_{up} = \{i:y_i\alpha_i < B_i\}$ $-$ i.e.  $\alpha_i < C$ or $\alpha_i < 0$;

*$I_{down} = \{j:y_j\alpha_j > A_j\}$  $-$ i.e.  $\alpha_j > 0$ or $\alpha_j > -C$.


Then Bottou and Lin's necessary and sufficient optimality criterion $-$ let call it $\mathcal{O}_{SVM}(\alpha^*)$ $-$ is:
$$
\exists \: \rho \in \mathbb{R} : \max_{i \, \in \, I_{up}} y_ig_i^* \leq \rho \leq \min_{j \, \in \, I_{down}} y_jg_j^*
$$
I understand how $\alpha^*$ being a solution implies $\mathcal{O}_{SVM}(\alpha^*)$ $-$ it is shown in the paper $-$ however I am not sure how $\mathcal{O}_{SVM}(\alpha^*)$ implies $\alpha^*$ is a solution. 
Does anybody know a reference where the reciprocal is proved? If not, could someone shed more light on this? As noted, details can be found in the above link in pages 8 and 9; the criterion corresponds to equation $(1.11)$.

[Addendum - edited]
My issue relates primarily to $\varepsilon$ $-$ see paper: the inequality $\mathcal{D}(\alpha^{\varepsilon}) \leq \mathcal{D}(\alpha^*)$ should hold for all $\alpha^{\varepsilon}$ for $\alpha^*$ to be an optimum, however $\varepsilon$ should be sufficiently small for $\alpha^{\varepsilon}$ to satisfy its constraints, hence the inequality does not hold for all $\alpha^{\varepsilon}$, does it? Except if we are implicitly restricting ourselves to the feasible region by considering only the following set of epsilons: $S_{\varepsilon} = \{\varepsilon:\forall i \in \{1,\dots,n\}, \alpha^{\varepsilon} = (\alpha_1^{\varepsilon},\dots,\alpha_n^{\varepsilon}), 0\leq\alpha_i^{\varepsilon}\leq C\}$ ?
 A: If we can find $\mathbf{\alpha^*}$ and $(\mathbf{w^*}, b^*, \mathbf{\xi^*})$ such that the optimality criterion holds:
$$\exists \: \rho \in \mathbb{R} : \max_{i \, \in \, I_{up}} y_ig_i^* \leq \rho \leq \min_{j \, \in \, I_{down}} y_jg_j^*$$
with $I_{up} = \{i \; | \; y_i\alpha_i < B_i\}$ and $I_{down} = \{j \; | \; y_j \alpha_j > A_j \}$
then we have:
$$
\exists \: \rho \in \mathbb{R} \; \forall k, \left\{ \begin{array} \\
\text{if} \;\; g_k^* \; > \; y_k \rho \;\;\; \text{then} \;\;\; \alpha_k^* = C \\
\text{if} \;\; g_k^* \; < \; y_k \rho \;\;\; \text{then} \;\;\; \alpha_k^* = 0
\end{array} \right\}
$$
Using this we can prove that (PROOF 1 at the end): $$C \xi_k^* - \alpha_k^* g_k^* = -y_k\alpha_k^*\rho$$
And since we know that 
$$\mathcal{P}(\mathbf{w}^*, b^*, \mathbf{\xi}^*) - \mathcal{D}(\mathbf{\alpha}^*) = \sum_{k=1}^{n} (C \xi_k^* - \alpha_k^* g_k^*) $$
we get:
$$ \begin{split} \mathcal{P}(\mathbf{w}^*, b^*, \mathbf{\xi}^*) - \mathcal{D}(\mathbf{\alpha}^*) & = \sum_{k=1}^n (-y_k \alpha_k^* \rho) \\
& = -\rho \sum_{i=1}^n y_i \alpha_i^* \\
& = -\rho·0 \\
& = 0 \end{split}$$
According to the $\textit{strong duality}$, if we can find $\mathbf{\alpha^*}$ and $(\mathbf{w^*}, b^*, \mathbf{\xi^*})$ such that $\mathcal{D}(\mathbf{\alpha^*}) = \mathcal{P}(\mathbf{w^*}, b^*, \mathbf{\xi^*})$, then $(\mathbf{w^*}, b^*, \mathbf{\xi^*})$ and $\mathbf{\alpha^*}$ are solutions of the primal and dual problems.
PROOF 1:
Either $g_k^* > y_k \rho \;\;$ or $\;\;g_k^* < y_k \rho$, we prove that in both cases: $\;\;C \xi_k^* - \alpha_k^* g_k^* = - y_k \alpha_k^* \rho$.
If $g_k^* > y_k \rho$:
$C = \alpha_k^* \;\;\;$ and $\;\;\; \xi_k^* = max\{0, g_k^* - y_k \rho\} = g_k^* - y_k \rho$
Thus
\begin{align}
C \xi_k^* - \alpha_k^* g_k^* & = C \;\; max\{0, g_k^* - y_k \rho\} - \alpha_k^* g_k^* \\
& = \alpha_k^* \; (g_k^* - y_k \rho) - \alpha_k^* g_k^* \\
& = - y_k \alpha_k^* \rho
\end{align}
If $g_k^* < y_k \rho$:
$\alpha_k^* = 0 \;\;\;$ and $\;\;\; \xi_k^* = max\{0, g_k^* - y_k \rho\} = 0$
Thus
$$C \xi_k^* - \alpha_k^* g_k^* = - y_k \alpha_k^* \rho = 0$$
