# Properties of the Quadratic Transformation of Dual Variables
## 4 Qualitative Analysis of the Quadratic Transformation of Dual Variables  
Let assumption 1 be satisfied with respect to the primal problem of convex programming (1), (2). To reduce the problem (7) to an unconstrained optimization problem, we use the transformation of variables in the following form:  
\[y^{T}(u)=(u_{1}^{2},u_{2}^{2},\ldots,u_{m}^{2}),\ i=1,2,\ldots,m. \tag{20}\]  
The transformation method (20) was previously called the quadratic method. Let us denote \(\Phi(u)=\phi(y(u))\). The minimization problem (7) under such a change of variables becomes  
\[\min\ \Phi(u),\ u\in R^{m}. \tag{21}\]  
Problem (21) is multiextremal, with each minimum \(u^{*}\) giving the optimal value of the dual variables \(y(u^{*})\). Given the expressions (8) for the function \(\phi(y)\), the gradient components of the function are given by  
\[\
abla\Phi_{i}(u)=-2u_{i}g_{i}(\ x(y(u))\ ),\ i=1,2,\ldots,m. \tag{22}\]  
We want to show that the gradient of the function \(\Phi(u)\) satisfies the Lipschitz condition.  
**Theorem 1**.: _Let Assumption 1 hold for the primal convex programming problem (1), (2) and, then, in the bounded domain_  
\[||u||\leq R \tag{23}\]  
_the gradient of the function satisfies the Lipschitz condition._  
Proof of Theorem 1.: Denote \(a,\ \ b\in R^{m}\), \(y^{a}=y(a)\), \(y^{b}=y(b)\), \(g^{a}=g(y(a))\), \(g^{b}=g(y(b))\) and write the components of the gradient difference  
\[(\
abla\Phi(a)-\
abla\Phi(b))_{i}/2=b_{i}g_{i}^{b}-a_{i}g_{i}^{a}=a_{i}(g_{i}^ {b}-g_{i}^{a})+g_{i}^{b}(b_{i}-a_{i}),\ i=1,2,\ldots,m. \tag{24}\]  
Form vectors \(v,\ z\in R^{m}\) with components \(v_{i}=a_{i}(g_{i}^{b}-g_{i}^{a}),\ \ z_{i}=g_{i}^{b}(b_{i}-a_{i})\), \(i=1,2,\ldots,m\). Taking into account the notation and properties of the norm, we obtain from (24)  
\[||\
abla\Phi(a)-\
abla\Phi(b)||/2=||v+z||\leq||v||+||z||. \tag{25}\]  
Let us estimate \(||v||^{2}\) taking into account (9):  
\[\begin{array}{l}||v||^{2}=\sum\limits_{i=1}^{m}a_{i}^{2}(g_{i}^{b}-g_{i}^{a} )^{2}\leq\sum\limits_{i=1}^{m}a_{i}^{2}\sum\limits_{i=1}^{m}\left(g_{i}^{b}-g_ {i}^{a}\right)^{2}=||a||^{2}\cdot||g^{b}-g^{a}||^{2}\leq\\ L_{\phi}||a||^{2}\cdot||y^{b}-y^{a}||^{2}\leq L_{\phi}||a||^{2}\cdot||a-b||^{2 }\cdot||a+b||^{2}.\end{array}\]  
Here, the last inequality is derived on the basis of the estimate  
\[\begin{array}{l}||y^{b}-y^{a}||^{2}=\sum\limits_{i=1}^{m}\left(a_{i}^{2}-b_ {i}^{2}\right)^{2}=\sum\limits_{i=1}^{m}\left(a_{i}-b_{i}\right)^{2}(a_{i}+b _{i})^{2}\leq\\ \sum\limits_{i=1}^{m}\left(a_{i}-b_{i}\right)^{2}\sum\limits_{i=1}^{m}\left(a _{i}+b_{i}\right)^{2}=||a-b||^{2}\cdot||a+b||^{2}.\end{array}\]  
Get an estimate of \(||z||^{2}\):  
\[||z||^{2}=\sum\limits_{i=1}^{m}\left(g_{i}^{b}\right)^{2}(b_{i}-a_{i})^{2} \leq\sum\limits_{i=1}^{m}\left(g_{i}^{b}\right)^{2}\sum\limits_{i=1}^{m} \left(b_{i}-a_{i}\right)^{2}\leq||g^{b}||^{2}\cdot||b-a||^{2}.\]  
Finally, taking into account (23), we obtain the proof  
\[\begin{array}{l}||\
abla\Phi(a)-\
abla\Phi(b)||\leq 2(||v||+||z||)\leq\\ 2L_{\phi}||a||\cdot||a-b||\cdot||a+b||+2||g^{b}||\cdot||b-a||\leq\\ 4L_{\phi}R^{2}\cdot||a-b||+2C_{g}\cdot||b-a||\leq L_{\phi}||a-b||,\end{array}\]  
where \(C_{g}=\max\limits_{x\in Q}||g(x)||\).  
Gradient minimization methods are suitable for solving minimization problems with a Lipschitz gradient [34].