Strictly convex

The previous considerations can be specified. Namely, if J is a strictly convex functional then... 

The existence of a unique duality mapping could easily be proved. Indeed, let 1 = u G y I u = 1 be a unit sphere in V. According to the Hahn-Banach theorem, for every fixed u G E there exists a unique element u G V such that u = 1, u, u) = 1 due to the strict convexity of V. Let us define... 

Let K c V he a convex closed set. We assume that y is a strictly convex reflexive Banach space. For given u G V an element Pu G K is called a projection of u onto the set K if... 

It is easily seen that Theorem 1.11 guarantees the solvability of problem (1.94). This solution is unique due to the strict convexity of V. 

If the matrix Q is positive semidefinite (positive definite) when projected into the null space of the active constraints, then (3-98) is (strictly) convex and the QP is a global (and unique) minimum. Otherwise, local solutions exist for (3-98), and more extensive global optimization methods are needed to obtain the global solution. Like LPs, convex QPs can be solved in a finite number of steps. However, as seen in Fig. 3-57, these optimal solutions can lie on a vertex, on a constraint boundary, or in the interior. A number of active set strategies have been created that solve the KKT conditions of the QP and incorporate efficient updates of active constraints. Popular methods include null space algorithms, range space methods, and Schur complement methods. As with LPs, QP problems can also be solved with interior point methods [see Wright (1996)]. 

It can be shown from a Taylor series expansion that if/(x) has continuous second partial derivatives, /(x) is concave if and only if its Hessian matrix is negative-semidefinite. For/(x) to be strictly concave, H must be negative-definite. For /(x) to be convex H(x) must be positive-semidefinite and for/(x) to be strictly convex, H(x) must be positive-definite. 

The eigenvalues of H are 7 and 1, hence H(x) is positive-definite. Consequently, fix) is strictly convex (as well as convex). 

Show that f=eXi + eXl is convex. Is it also strictly convex ... 

If fix) is convex, H(x) is positive-semidefinite at all points x and is usually positive-definite. Hence Newton s method, using a line search, converges. If fix) is not strictly convex (as is often the case in regions far from the optimum), H(x) may not be positive-definite everywhere, so one approach to forcing convergence is to replace H(x) by another positive-definite matrix. The Marquardt-Levenberg method is one way of doing this, as discussed in the next section. 

The quantity slope2 is the slope of the line search objective function gk(a) at a = ak (see Figure 6.8) and slopel is its slope at a = 0, so (dk)Tyk > 0 if and only if slope2 > slopel. This condition is always satisfied if /is strictly convex. A good line search routine attempts to meet this condition if it is not met, then H is not updated. 

Exercise. Actually a function / is called convex when f"(x) 0, and strictly convex when f"(x)>0, as was required in (5.1). Show by a counterexample that mere convexity is not enough for our purpose. 

Remark 1 A strictly convex function on a subset S of 3 n is convex on S. The converse, however, is not true. For instance, a linear function is convex but not strictly convex. 

Remark 5 The above two theorems can be directly extended to strictly convex and strictly concave functions by replacing the inequalities > and < with strict inequalities > and <. 

Here, for convenience, we omit in eqn. (84) all items that are linear with respect to N containing In Nf8 and also convex function of IN/5. ) Let us make use of one more characteristic property of smooth, strictly convex functions in any direction a second derivative of G0 in D must be rigorously positive. To check it, let us write... 

The equality to zero is obtained only in the case where, for any i,j = 1,.. ., n we have dJNfii = 8jlN, i.e. when the vector 3 (with components 8t) is proportional to that with components N t, in other words there exists a value of X such that 6, = XNf)t. But this is possible only in the case in which all the components 8t are simultaneously either positive or negative. Since, at some non-zero value of x, the vectors with components N0t and Noi + 3t must lie in the same reaction polyhedron, the simultaneous positivity or negativity for all the Si values is forbidden by, for example, the law of conservation of the overall (taking into account its adsorption) gas mass = Zm,-(iV0i + x3t) 1.171 1 = 0, for any Af we have m, > 0, hence <5 cannot have the same signs. Consequently, in the reaction polyhedron, G is a strictly convex function since the sum of a strictly convex G0 with a linear function of Df and a strictly convex function of IV5 is strictly convex in this polyhedron. 

The strict convexity of the function G in the reaction polyhedron D results in the following important property. In this polyhedron G has the unique local minimum. At the same time this local minimum is a global one. 

It is clearly seen that, at a vertex of the reaction polyhedron, G achieves its local maximum value (due to the strict convexity of G and the fact that its minimum point is positive). Therefore near each vertex, as well as in the vicinity of some faces, the G function can be used to construct a region that is unattainable from outside. Let us consider the case of one vertex and then a more awkward general situation. 

Extrema (minima or maxima) of a function can be examined by checking the eigenvalues of the Hessian at its stationary points. If all the eigenvalues of the Hessian are positive (negative) at a stationary point, then the function / is at a local minimum (maximum). Likewise, if all the eigenvalues of the Hessian are positive for all x, then /(x) is said to be strictly convex, with a global minimum at the stationary point. 

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