Supporting hyperplane

Illustration 2.1.7 Figure 2.6 provides a few examples of supporting hyperplanes for convex and nonconvex sets. 

The dual function at (fa, fa) corresponds to determining the lowest plane with slope (-fa,—fa) which intersects the image set /. This corresponds to the supporting hyperplane h which is tangent to the image set I at the point P, as shown in Figure 4.1. 

The minimum value of this problem is the value of z3 where the supporting hyperplane h intersects the ordinate, denoted as z3 in Figure 4.1. 

Determine the value of which defines the slope of a supporting hyperplane to the... 

Remark 1 The value of (/Zi, fi2) that intersects the ordinate at the maximum possible value in Figure 4.1 is the supporting hyperplane of I that goes through the point P, which is the optimal solution to the primal problem (P). 

Remark 6 The geometrical interpretation of the primal and dual problems clarifies the weak and strong duality theorems. More specifically, in the vicinity of y — 0, the perturbation function v(y) becomes the 23-ordinate of the image set I when zi and z2 equal y. In Figure 4.1, this ordinate does not decrease infinitely steeply as y deviates from zero. The slope of the supporting hyperplane to the image set I at the point P, (-pi, -p2), corresponds to the subgradient of the perturbation function u(y) at y = 0. 

Remark 7 An instance of unstable problem (P) is shown in Figure 4.2 The image set I is tangent to the ordinate 23 at the point P. In this case, the supporting hyperplane is perpendicular, and the value of the perturbation function v(y) decreases infinitely steeply as y begins to increase above zero. Hence, there does not exist a subgradient at y = 0. In this case, the strong duality theorem does not hold, while the weak duality theorem holds. 

Remark 1 The difference in the optimal values of the primal and dual problems can be due to a lack of continuity of the perturbation function v(y) at y = 0. This lack of continuity does not allow the existence of supporting hyperplanes described in the geometrical interpretation section. 

Figure 5.3 A convex set S of final states with a supporting hyperplane P. A hyperplane is a tangent line in two dimensions...

Now for a convex set, there exists a supporting hyperplane at any boundary point of the set (see Appendix 5.B, p. 149). We apply this result to our convex set of final states, in which the final optimal state... 

We will now show that if 5 is a convex set, then at each of its boundary point we have a supporting hyperplane, or equivalently p (y — y) <0 for aU y in S. In the following two steps we show that ... 

However, a supporting hyperplane in the transformed space will generally not correspond to a supporting hyperplane in the original space, and this may introduce biases in the failure probability estimates in either direction. See (Huseby et al.,... 

A hyperplane H is a supporting hyperplane of a convex set B if B is entirely contained in one of the two closed half-spaces determined by H and B has at least one boundary-point on H. 

Figure 1. The set e, part of the environmental contour dB and the supporting hyperplane H ) ...

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