```markdown
### Proof

- By the separating hyperplane theorem, there exists \((\tilde{\lambda}, \tilde{\nu}, \mu) \neq 0\) and \(\alpha\) such that:

  \[
  (u, v, t) \in \mathcal{A} \implies \tilde{\lambda}^T u + \tilde{\nu}^T v + \mu t \geq \alpha, \quad (1)
  \]

  and

  \[
  (u, v, t) \in \mathcal{B} \implies \tilde{\lambda}^T u + \tilde{\nu}^T v + \mu t \leq \alpha. \quad (2)
  \]

- From (1), we conclude that \(\tilde{\lambda} \succeq 0\) and \(\mu \geq 0\). (Otherwise, \(\tilde{\lambda}^T u + \mu t\) is unbounded below over \(\mathcal{A}\), contradicting (1).)

- The condition (2) simply means that \(\mu t \leq \alpha\) for all \(t < p^\star\), and hence, \(\mu p^\star \leq \alpha\).

Together with (1), we conclude that for any \(x \in \mathcal{D}\),

\[
\mu p^\star \leq \alpha \leq \mu f_0(x) + \sum_{i=1}^m \tilde{\lambda}_i f_i(x) + \tilde{\nu}^T(Ax - b) \quad (3)
\]
```
