Convex Optimization. Mikhail Moklyachuk

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      The matrix –A₁ is non-negative definite. Therefore, the point (0, 0) satisfies the second-order necessary conditions for a maximum. However, a direct verification of the behavior of the function f in a neighborhood of the point (0, 0) shows that (0, 0) ∉ locextr f. The matrix A₂ is positive definite. Thus, by theorem 1.13, the local minimum of the problem is attained at the points (1, 1), ( –1, –1).

      Answer. (0, 0) ∉ locextr; (1, 1), ( –1, –1) ∈ locmin.

1.4. Constrained optimization problems

      1.4.1. Problems with equality constraints

      Let fk : ℝⁿ → ℝ, k = 0,1, … , m, be differentiable functions of n real variables. The constrained optimization problem (with equality constraints) is the problem

[1.2]

The points x ∈ ℝⁿ, which satisfy the equation fk(x) = 0, , are called admissible in the problem [1.2]. An admissible point gives a local minimum (maximum) of the problem [1.2] if there exists a number δ > 0 such that for all admissible x ∈ ℝⁿ that satisfy the conditions fk(x) = 0, k = 1, 2, … , m, and the condition , the following inequality

      holds true.

      The main method of solving constrained optimization problems is the method of indeterminate Lagrange multipliers. It is based on the fact that the solution of the constrained optimization problem [1.2] is attained at points that are critical in the unconstrained optimization problem

      where is the Lagrange function, and λ₀, … , λ_m are the Lagrange multipliers.

      THEOREM 1.15.– (Lagrange theorem) Let be a point of local extremum of the problem [1.2], and let the functions fi(x), i = 0, 1, … , m, be continuously differentiable in a neighborhood U of the point . Then there will exist the Lagrange multipliers λ₀, … , λ_m, not all equal to zero, such that the stationary condition on x of the Lagrange function is fulfilled

      In order for λ₀ ≠ 0, it is sufficient that the vectors be linearly independent.

      To prove the theorem, we use the inverse function theorem in a finite-dimensional space.

      THEOREM 1.16.– (Inversefunction theorem) Let F₁(x_, … , xs), … , Fs(x₁, ..., xs) be s continuously differentiable in a neighborhood U of the point functions of s variables and let the Jacobian

      be not equal to zero. Then there exist numbers ε > 0, δ > 0, K > 0 such that for any y = (y₁, … , ys), ∥y∥ ≤ ε we can find x = (x₁, … , xs), which satisfies conditions ∥x∥ < δ, , ∥x∥ ≤ K ∥y∥.

      PROOF.– We prove the Lagrange theorem by contradiction. Suppose that the stationarity condition

      does not hold, and vectors , i = 0, 1, … , m, are linearly independent, which means that the rank of the matrix

      is equal to m +1. Therefore, there СКАЧАТЬ