{{page>:defs}}
{{tag>karush_kuhn_tucker kkt farkas_lemma}}



$$
\newcommand{\mcD}{\mathcal D}
\newcommand{\mcl}{\mathcal L}
$$

====== Weak duality ======
Let $f$ and $(h_i)_{1 \leq i \leq n}$ be convex differentiable functions on $\Xset=\rset^p$. Let $\mcD=\cap_{i=1}^n \{h_i \leq 0\}\neq \emptyset$ and note that by convexity of the functions $h_i$, the set $\mcD$ is actually a convex set. We are interested in the infimum of the convex function $f$ on the convex set $\mcD$. Define the Lagrange function
$$
\mcl(x,\lambda)=f(x)+\sum_{i=1}^n \lambda_i h_i(x)
$$
where $\lambda=(\lambda_1,\ldots,\lambda_n)^T \in \rset^n$. We first show the weak duality relation. Start with this simple remark: for all $(x,\lambda) \in \Xset \times \rset^n$, 
$$
\inf_{x \in \Xset} \mcl(x,\lambda) \leq \mcl(x,\lambda) 
$$
Taking the supremum wrt $\lambda \geq 0$ on both sides yields: 
$$
\sup_{\lambda\geq 0}\inf_{x \in \Xset} \mcl(x,\lambda) \leq \sup_{\lambda\geq 0} \mcl(x,\lambda)=\infty \indin{x\notin \mcD}+f(x)\indin{x\in \mcD}
$$
And taking now the infimum wrt $x \in \Xset$ on both sides, we finally get the <color red>**weak duality relation**</color>: 
\begin{equation} \label{eq:weak}
\sup_{\lambda\geq 0}\inf_{x \in \Xset} \mcl(x,\lambda) \leq \inf_{x \in \Xset} \sup_{\lambda\geq 0}\mcl(x,\lambda)=\inf_{x \in \mcD} f(x) 
\end{equation}
Note that in the lhs (left-hand side), the infimum is wrt $x\in \Xset$ and therefore, there is no constraint, which is nice... 
The rhs (right-hand side) is called the //primal problem// and the lhs the //dual problem//. Note, since $x\mapsto \mcl(x,\lambda)$ is convex, that the <color red>dual problem</color> $\sup_{\lambda\geq 0}\inf_{x \in \Xset} \mcl(x,\lambda)$ is equivalent to 
$$
\sup \set{\mcl(x,\lambda)}{\lambda \geq 0\mbox{ and }\nabla_x \mcl(x,\lambda)=0}
$$

To obtain the equality in \eqref{eq:weak} (which is named the <color red>**strong duality relation**</color>), we actually need some additional assumptions (for example the existence of a Slater point). But before that, let us check a very useful relation...
====== Karush-Kuhn-Tucker conditions ======

===== Sufficient conditions =====

<WRAP center round box 80%>
__**Lemma**__
Assume that there exist $(x^*,\lambda^*) \in \mcD\times (\rset^+)^n$ such that 
\begin{equation}
\nabla_x \mcl(x^*,\lambda^*)=0
\end{equation}
and for all $i \in [1:n]$, we have either $\lambda^*_i=0$ or $h_i(x^*)=0$ . 
Then, the strong duality holds and 
$$
f(x^*)=\inf_{x \in \mcD} f(x)=\mcl(x^*,\lambda^*)
$$
</WRAP>
<hidden Click to see the proof>
$\bproof$
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 $h_i$, we get
$$
\nabla f(x^*)^T (x-x^*)=-\sum_{i=1}^n \lambda^*_i \nabla_x h_i^T(x^*) (x-x^*) \geq -\sum_{i=1}^n \lambda^*_i (h_i(x)-h_i(x^*))=- \sum_{i=1}^n \lambda^*_i  \underbrace{h_i(x)}_{\leq 0} \geq 0
$$  
This finishes the proof. $\eproof$
</hidden>
===== Saddle points =====

<WRAP center round box 80%>
__**Definition**__
We say that $(x^*,\lambda^*) \in \Xset \times (\rset^+)^n$ is a saddle point of the Lagrange function $\mcl$ if for every $(x,\lambda) \in \Xset \times (\rset^+)^n$, 
\begin{equation}\label{eq:saddle}
\mcl(x^*,\lambda) \leq \mcl(x,\lambda^*) 
\end{equation}
</WRAP>
<WRAP center round box 80%>
__**Saddle point Lemma**__
If $(x^*,\lambda^*) \in \Xset \times (\rset^+)^n$ is a saddle point for $\mcl$ then  the strong duality holds, and the KKT conditions holds for $(x^*,\lambda^*)$.
</WRAP>
<hidden click here to see the proof>
$\bproof$ 
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)=\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} 

Note that $\mcl(x^*,\lambda^*)=\inf_{x \in\Xset} \mcl(x,\lambda^*)$ shows that $\nabla_x \mcl(x^*,\lambda^*)=0$. 
 
The upper bound in \eqref{eq:saddle} regardless the value of $\lambda\geq 0$ shows that $h_i(x^*)\leq 0$ 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 $\sum_{i=1}^n \lambda^*_i h_i(x^*_i)=0$ and since $\lambda^* \geq 0$ and $x^* \in \mcD$, this implies $\lambda^*_i h_i(x^*)=0$ for all $i \in [1:n]$. 
$\eproof$
</hidden>
===== Necessary conditions and strong duality =====
 
We now assume the existence of Slater points: there exists $\tilde x \in \mcD$ such that for all $i \in \{1,\ldots,n\}$, $h_i(\tilde x)<0$. 

<WRAP center round box 80%>
__**The convex Farkas lemma**__ 
Assume that there exists a Slater point. Then, $\{f<0\} \cap \mcD =\emptyset$ iff there exists $\lambda^* \geq 0$ such that for all $x\in \mcD$, $f(x)+\sum_{i=1}^n \lambda^*_i h_i(x) \geq 0$.
</WRAP>
<hidden Click here to see the proof> 
$\bproof$
Set 
$$
U=\{u=u_{0:n} \in \rset^{n+1}; \exists\ x \in \mcD, f(x)<u_0 \ \mbox{and} \ h_i(x)\leq u_i  \mbox{ for all } i \in [1:n]\}.
$$
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 $x\in \mcD$, 
$$
\phi_0 f(x)+\sum_{i=1}^n \phi_i h_i(x) \geq 0
$$
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 $\sum_{i=1}^n \phi_i h_i(\tilde x) \geq 0$ but since $h_i(\tilde x)<0$ 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.  
$\eproof$
</hidden>

The Farkas lemma will imply the strong duality under the existence of a Slater point. 
<WRAP center round box 80%>
__**The necessary condition**__ 
Assume that $f(x^*)=\inf_{x\in \mcD} f(x)$ for some $x^* \in\mcD$ and that there exists a Slater point. Then there exist $\lambda^* \geq 0$ such that $(x^*,\lambda^*)$ is a saddle point and by the Saddle point lemma, the strong duality holds. 
</WRAP>
<hidden Click here to see the proof>
$\bproof$
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 $x\in \mcD$, $f(x)-f(x^*)+\sum_{i=1}^n \lambda^*_i h_i(x) \geq 0$. And this implies that $(x^*,\lambda^*)$ is a saddle point since: for all $\lambda \geq 0$ and $x \in \mcD$, 
$$
f(x^*)+\sum_{i=1}^n \lambda_i h_i(x^*) \leq f(x^*) \leq f(x)+\sum_{i=1}^n \lambda^*_i h_i(x) 
$$
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. 
$\eproof$
</hidden>