# Properties of the Quadratic Transformation of Dual Variables
## 3 Qualitative Analysis of the Modulus Transformation of Dual Variables  
Assume that Assumption 1 holds for the primal convex programming problem (1), (2). To reduce problem (7) to an unconstrained optimization problem, we use the following transformation of variables [32]:  
\[y^{T}(u)=(|u_{1}|,|u_{2}|,\ldots,|u_{m}|),\ i=1,2,\ldots,m. \tag{10}\]  
The transformation method (10) was previously called the modulus method. Let us denote \(\theta(u)=\phi(y(u))\). The minimization problem (7) under such a variables substitution takes the form  
\[\min\,\theta(u),\ u\in R^{m}. \tag{11}\]  
The problem (11) is multi-extremal, but in each of the quadrants determined by the sign constancy of the components of the variable \(u\), the function is convex, and the components of its gradient are given by the expression  
\[\
abla\theta_{i}(u)=-sign(u_{i})\cdot g_{i}(\ x(y(u))\ ),\ i=1,2,\ldots,m. \tag{12}\]  
In each of the quadrants, the gradient satisfies the Lipschitz condition. Note that, because of the convexity of the function, the gradients in the corresponding quadrants are also subgradients. At points on the boundaries of adjacent quadrants, we have gradients of adjacent areas. For this reason, to solve the problem of minimizing the function (11), we can use relaxation subgradient methods, in which the descent direction is chosen so that leaving the neighborhood of the current minimum is possible by one-dimensional minimization along this direction. Using a two-dimensional problem as an example, let us consider the problem of increasing the degree of conditionality of the problem at the extremum area. Let us assume that the number of constraints in the problem (1), (2) is \(m=2\), and the minimum in the corresponding problem (11) is at the point  
\[(u*)^{T}=(\ u_{1}^{*}>0,\ \ u_{2}^{*}=0). \tag{13}\]Here, we assume that only the first condition is active. The gradients (12) at the point (13) at \(u_{2}=0\), depending on whether they belong to the area \(u_{2}\geq 0\) or \(u_{2}\leq 0\), respectively, have the form  
\[\left(g_{+}^{*}\right)^{T}=(0,\ \ -g_{2}^{*}),\ \left(g_{-}^{*}\right)^{T}=(0,\ \ g_{2}^{*}). \tag{14}\]  
Since constraint \(g_{2}(x)\) is inactive, \(g_{2}^{*}<0\). For a small \(\delta_{1}>0\), let us consider the gradients of problem (11) on the boundary of the domains at \(u_{2}=0\) in the point  
\[\left(u^{0}\right)^{T}=(u_{1}^{*}-\delta_{1}>0\,\ \ \ u_{2}^{0}=0), \tag{15}\]  
determined by belonging to half-planes \(u_{2}\geq 0\) and \(u_{2}\leq 0\), respectively:  
\[\left(g_{+}^{0}\right)^{T}=(\varepsilon_{1},\ \ -g_{2}^{*}+\varepsilon_{2}),\ \left(g_{-}^{0}\right)^{T}=( \varepsilon_{1},\ \ g_{2}^{*}-\varepsilon_{2}). \tag{16}\]  
Since the gradients in the half-planes \(u_{2}\geq 0\) and \(u_{2}\leq 0\) satisfy the Lipschitz condition and \(||u^{0}-u^{*}||=\delta_{1}\), the value \(\delta_{1}\) determines the smallness of the values \(\varepsilon_{1}\) and \(\varepsilon_{2}\). Here \(\varepsilon_{1}<0\), since the gradients (16) are subgradients of the function \(\theta(u)\) that is convex in the corresponding half-planes. Therefore, each subgradient represents a normal on the tangential hyperplane to the level surface, and the reverse direction of the subgradients shows the half-plane of the minimum point for this hyperplane. In plane \(u_{1}\), \(u_{2}\), Figure 1 shows points \(u^{*}\), \(u^{0}\), gradients \(g_{+}^{0}\), \(g_{-}^{0}\) and the line segments \(L_{+}\), \(L_{-}\), whose normals are the corresponding gradients \(g_{+}^{0}\), \(g_{-}^{0}\). The set \(G\) is the convex hull of the gradients \(g_{+}^{0}\), \(g_{-}^{0}\). To reduce the function from the point \(u^{0}\), we need to find the direction falling into the interior of the cone with the opening angle \(\alpha\), bounded by the lines \(L_{+}\), \(L_{-}\). The value of the angle \(\alpha\) is proportional to the component \(\varepsilon_{1}\).