$\bproof$ For every $x \in \mcD$, the convexity of the function $f$ yields $$ f(x)-f(x^*) \geq \nabla f(x^*)^T (x-x^*). $$ It remains to prove $\nabla f(x^*)^T (x-x^*)\geq 0$. Using first $\nabla_x \mcl(x^*,\lambda^*)=0$ and then the convexity of the functions $g_i$, we get $$ \nabla f(x^*)^T (x-x^*)=-\sum_{i=1}^n \lambda^*_i \nabla_x g_i^T(x^*) (x-x^*) \geq -\sum_{i=1}^n \lambda^*_i (g_i(x)-g_i(x^*))=- \sum_{i=1}^n \lambda^*_i \underbrace{g_i(x)}_{\leq 0} \geq 0 $$ This finishes the proof. $\eproof$

$\bproof$ Using \eqref{eq:saddle} with $\lambda=\lambda^*$ first and then with $x=x^*$, we have for every $(x,\lambda) \in \Xset \times (\rset^+)^n$, \begin{equation*} \mcl(x^*,\lambda) \leq \mcl(x^*,\lambda^*) \leq \mcl(x,\lambda^*) \end{equation*} The existence of a saddle point implies the strong duality since $$ \inf_{x \in\Xset} \sup_{\lambda \geq 0} \mcl(x,\lambda) \leq \sup_{\lambda \geq 0} \mcl(x^*,\lambda) \leq \mcl(x^*,\lambda^*) \leq \inf_{x \in\Xset} \mcl(x,\lambda^*) \leq \sup_{\lambda \geq 0} \inf_{x \in\Xset} \mcl(x,\lambda) $$ which is the converse inequality of \eqref{eq:weak}. This finally implies: \begin{equation}\label{eq:all} \sup_{\lambda \geq 0} \inf_{x \in\Xset} \mcl(x,\lambda)=\mcl(x^*,\lambda^*)=\inf_{x \in\Xset} \sup_{\lambda \geq 0} \mcl(x,\lambda)=\inf_{x \in\mcD} f(x). \end{equation}

The upper bound in \eqref{eq:saddle} regardless the value of $\lambda\geq 0$ shows that $g_i(x^*)\leq 0$ for every $i \in [1:n]$, that is $x^* \in \mcD$. Now taking \eqref{eq:saddle} with $\lambda=0$ yields for $x\in \mcD$, $$ f(x^*)=\mcl(x^*,0) \leq \mcl(x,\lambda^*) \leq f(x) $$ which shows that $f(x^*)=\inf_{x \in\mcD} f(x)$. Combining with \eqref{eq:all} yields $\mcl(x^*,\lambda^*)=f(x^*)$ so that $\sum_{i=1}^n \lambda^*_i g_i(x^*_i)=0$ and since $\lambda^* \geq 0$ and $x^* \in \mcD$, this implies $\lambda^*_i g_i(x^*)=0$ for all $i \in [1:n]$. $\eproof$

$\bproof$ Set $$ U=\{u=u_{0:n} \in \rset^{n+1}; \exists\ x \in \mcD, f(x)<u_0 \ \mbox{and} \ g_i(x)\leq u_i \mbox{ for all } i \in [1:n]\}. $$ The condition $\{f<0\} \cap \mcD =\emptyset$ is equivalent to saying that $0 \notin U$ and since $U$ is clearly a convex set, by a separation argument, there exists a non-null vector $\phi \in \rset^{n+1}$ such that \begin{equation} \label{eq:ineg} \forall u \in U\,, \quad \phi^T u\geq 0 \end{equation} Take an arbitrary $i \in [0:n]$. If $u\in U$ then $u+t e_i \in U$ where $t>0$ and $e_i=(\indiacc{k=i})_{k \in [0:n]} \in \rset^{n+1}$. The previous inequality implies $\phi^T (u+t e_i)=\phi^T u + t\phi_i \geq 0$ for all $t>0$, therefore $\phi_i\geq 0$. Now, since $\phi^T u\geq 0$ for all $u \in U$, a simple limiting argument yields for all $x\in \mcD$, $$ \phi_0 f(x)+\sum_{i=1}^n \phi_i g_i(x) \geq 0 $$ To conclude, and since we already know that $\phi_i\geq 0$ for all $i \in [0:n]$, it only remains to prove that $\phi_0 \neq 0$ and to set in that case $\lambda^*_i=\frac{\phi_i}{\phi_0}$. The rest of the argument is by contradiction. If $\phi_0=0$, then the previous inequality on a Slater point $x=\tilde x$ yields $\sum_{i=1}^n \phi_i g_i(\tilde x) \geq 0$ but since $g_i(\tilde x)<0$ for all $i \in [1:n]$, we finally get $\phi_i=0$ for all $i \in [1:n]$. Finally all the components of $\phi$ are null and we are face to a contradiction. The proof is completed. $\eproof$

$\bproof$ Assume now that $f(x^*)=\inf_{x\in \mcD} f(x)$ for some $x^* \in\mcD$. Then, $\{f-f(x^*)<0\} \cap \mcD =\emptyset$ so that there exists $\lambda^* \geq 0$ satisfying for all $x\in \mcD$, $f(x)-f(x^*)+\sum_{i=1}^n \lambda^*_i g_i(x) \geq 0$. And this implies that $(x^*,\lambda^*)$ is a saddle point since: for all $\lambda \geq 0$ and $x \in \mcD$, $$ f(x^*)+\sum_{i=1}^n \lambda_i g_i(x^*) \leq f(x^*) \leq f(x)+\sum_{i=1}^n \lambda^*_i g_i(x) $$ so that $(x^*,\lambda^*)$ is a saddle point and this implies that the strong duality holds (as we have seen in the previous section). This concludes the proof. $\eproof$