6. Properties of minimum functions

Let M1, <sup>1</sup>Γ � �, the leader decision affine manifold, and M2, <sup>2</sup>Γ � �, the follower decision affine manifold, be two connected affine manifolds of dimension n<sup>1</sup> and n2, respectively. Starting from a function with two vector variables

$$
\varphi: M\_1 \times M\_2 \to \mathbb{R}, (\mathfrak{x}, \mathfrak{y}) \to \varphi(\mathfrak{x}, \mathfrak{y})\_\prime
$$

and taking the infimum after one variable, let say y, we build a function

$$f(\mathbf{x}) = \inf\_{y} \left\{ \varphi(\mathbf{x}, y) : y \in a(\mathbf{x}) \right\},$$

which is called minimum function.

A minimum function is usually specified by a pointwise mapping a of the manifold M<sup>1</sup> in the subsets of a manifold M<sup>2</sup> and by a functional φð Þ x; y on M<sup>1</sup> � M2. In this context, some differential properties of such functions were previously examined in [4]. Now we add new properties related to increase and convexity ideas.

First we give a new proof to Brian White Theorem (see Mean Curvature Flow, p. 7, Internet 2017).

Theorem 6.1 Suppose that M<sup>1</sup> is compact, M<sup>2</sup> ¼ ½ � 0;T and f : M<sup>1</sup> � ½ �! 0;T R. Let ϕð Þ¼ t minxf xð Þ ;t . If, for each x with <sup>ϕ</sup>ðÞ¼ <sup>t</sup> f xð Þ ; <sup>t</sup> , we have <sup>∂</sup><sup>f</sup> <sup>∂</sup><sup>t</sup> ð Þ x; t ≥ 0, then ϕ is an increasing function.

ϕ 0 ðÞ¼ t ∂f ∂t

is convex.

g, in x, <sup>∂</sup><sup>g</sup>

f xð Þ<sup>0</sup> ≥ 0, as in Theorem 4.1.

x with <sup>ϕ</sup>ð Þ¼ <sup>y</sup> f xð Þ ; <sup>y</sup> , we have <sup>∂</sup><sup>f</sup>

∂ϕ <sup>∂</sup>y<sup>α</sup> <sup>¼</sup> <sup>∂</sup><sup>f</sup>

x with <sup>ϕ</sup>ð Þ¼ <sup>y</sup> f xð Þ ; <sup>y</sup> , we have d<sup>2</sup>

we must have

ð Þþ x0ð Þt ; t <

extends by continuity (see also the evolution of an extremum problem).

tf xð Þ<sup>0</sup> . On the other hand, in the minimum point, we have <sup>∂</sup><sup>g</sup>

point of M1. Since x0ð Þy is a critical point, we have

<sup>∂</sup>y<sup>α</sup> ð Þþ <sup>x</sup>0ð Þ<sup>y</sup> ; <sup>y</sup> <sup>&</sup>lt;

∂f ∂x

ð Þ <sup>x</sup>0ð Þ<sup>t</sup> ; <sup>t</sup> , x<sup>0</sup>

Consequently, ϕð Þt is an increasing function. If M<sup>1</sup> has a nonvoid boundary, then the monotony

Example 6.1 The single-time perspective of a function f : <sup>R</sup><sup>n</sup> ! <sup>R</sup> is the function g : <sup>R</sup><sup>n</sup> � <sup>R</sup><sup>þ</sup> ! <sup>R</sup>, g xð Þ¼ ; t tf xð Þ =t , dom g ¼ f g ð Þj x; t x=t∈ dom f ; t > 0 . The single-time perspective g is convex if f

The single-time perspective is an example verifying Theorem 7.1. Indeed, the critical point condition for

Theorem 6.2 Suppose that M<sup>1</sup> is compact and f : M<sup>1</sup> � M<sup>2</sup> ! R. Let ϕð Þ¼ y minxf xð Þ ; y . If, for each

Proof. Suppose that <sup>f</sup> is a <sup>C</sup><sup>2</sup> function and minxf xð Þ¼ ; <sup>y</sup> f xð Þ <sup>0</sup>ð Þ<sup>y</sup> ; <sup>y</sup> , where <sup>x</sup>0ð Þ<sup>y</sup> is an interior

ð Þ x0ð Þy ; y ,

Consequently, ϕð Þy is a partially increasing function. If M has a non-void boundary, then the monotony extends by continuity. □

Theorem 6.3 Suppose that M<sup>1</sup> is compact and f : M<sup>1</sup> � M<sup>2</sup> ! R. Let ϕð Þ¼ y minxf xð Þ ; y . If, for each

Proof. Without loss of generality, we work on Euclidean case. Suppose that f is a C<sup>2</sup> function and minxf xð Þ¼ ; y f xy ð Þ ð Þ; y , where x yð Þ is an interior point of M1. Since x yð Þ is a critical point,

<sup>∂</sup>xi ð Þ¼ x yð Þ; <sup>y</sup> <sup>0</sup>:

∂2 f ∂y<sup>α</sup>∂xi

∂xi <sup>∂</sup>y<sup>β</sup> <sup>¼</sup> <sup>0</sup>:

∂f

Taking the partial derivative with respect to y<sup>α</sup> and the scalar product with <sup>∂</sup>x<sup>i</sup>

∂xi ∂y<sup>β</sup> þ

∂2 f ∂xi ∂xj ∂xj ∂y<sup>α</sup>

∂f ∂x

<sup>∂</sup><sup>x</sup> ¼ 0, gives x ¼ tx0, where x<sup>0</sup> is a critical point of f . Consequently, ϕðÞ¼ t minxg xð Þ¼ ; t

<sup>∂</sup>y<sup>α</sup> ð Þ x; y ≥ 0, then ϕð Þy is a partially increasing function.

∂x<sup>0</sup> <sup>∂</sup>y<sup>α</sup> <sup>&</sup>gt;<sup>¼</sup> <sup>∂</sup><sup>f</sup>

<sup>y</sup> f xð Þ ; y ≤ 0, then ϕð Þy is an affine concave function.

<sup>0</sup>ð Þ<sup>t</sup> <sup>&</sup>gt;<sup>¼</sup> <sup>∂</sup><sup>f</sup> ∂t

ð Þ x0ð Þt ; t ≥ 0:

Bilevel Disjunctive Optimization on Affine Manifolds http://dx.doi.org/10.5772/intechopen.75643

<sup>∂</sup><sup>t</sup> ð Þ¼ x; t f xð Þ<sup>0</sup> . Then ϕð Þt is increasing if

<sup>∂</sup>y<sup>α</sup> ð Þ <sup>x</sup>0ð Þ<sup>y</sup> ; <sup>y</sup> <sup>≥</sup> <sup>0</sup>:

<sup>∂</sup>y<sup>β</sup> it follows

□

133

Proof. We shall prove the statement in three steps.

(1) If f is continuous, then ϕ is (uniformly) continuous.

Indeed, f is continuous on the compact M<sup>1</sup> � ½ � 0; 1 , hence uniformly continuous. So, for ε > 0 it exists δ > 0 such that if ∣t<sup>1</sup> � t2∣ < δ, then ∣f xð Þ� ; t<sup>1</sup> f xð Þ ; t<sup>2</sup> ∣ < ε, for any x∈ M1, or

$$-\varepsilon < f(\mathbf{x}, t\_1) - f(\mathbf{x}, t\_2) < \varepsilon$$

On one hand, if we put ϕð Þ¼ t<sup>1</sup> f xð Þ <sup>1</sup>; t<sup>1</sup> and ϕð Þ¼ t<sup>2</sup> f xð Þ <sup>2</sup>; t<sup>2</sup> , then we have

$$f(\mathbf{x}, t\_1) > f(\mathbf{x}, t\_2) - \varepsilon \ge f(\mathbf{x}\_2, t\_2) - \varepsilon.$$

Hence minxf xð Þ ; t<sup>1</sup> ≥ f xð Þ� <sup>2</sup>; t<sup>2</sup> ε, so is ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ≥ � ε.

On the other hand,

$$f(\mathbf{x}, t\_2) + \varepsilon > f(\mathbf{x}, t\_1) \ge f(\mathbf{x}\_1, t\_1).$$

Hence minxf xð Þþ ; t<sup>2</sup> ε ≥ f xð Þ <sup>1</sup>; t<sup>1</sup> , so is ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ≤ ε.

Finally, ∣ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ∣ ≤ ε, for ∣t<sup>1</sup> � t2∣ < δ, that is, ϕ is (uniformly) continuous.

(2) Let us fix t<sup>0</sup> <sup>∈</sup> ð � <sup>0</sup>; <sup>T</sup> . If <sup>ϕ</sup>ð Þ¼ <sup>t</sup><sup>0</sup> f xð Þ <sup>0</sup>; <sup>t</sup><sup>0</sup> and <sup>∂</sup><sup>f</sup> <sup>∂</sup><sup>t</sup> ð Þ x0; t<sup>0</sup> ≥ 0, then it exists δ > 0 such that ϕð Þt ≤ ϕð Þ t<sup>0</sup> , for any t∈ ð Þ t<sup>0</sup> � δ; t<sup>0</sup> .

Suppose <sup>∂</sup><sup>f</sup> <sup>∂</sup><sup>t</sup> ð Þ x0; t<sup>0</sup> > 0, it exists δ > 0 such that f xð Þ <sup>0</sup>; t ≤ f xð Þ <sup>0</sup>; t<sup>0</sup> , for each t ∈ð Þ t<sup>0</sup> � δ; t<sup>0</sup> . It follows minxf xð Þ ; t ≤ f xð Þ <sup>0</sup>; t ≤ f xð Þ <sup>0</sup>; t<sup>0</sup> , and so is ϕð Þt ≤ ϕð Þ t<sup>0</sup> .

If <sup>∂</sup><sup>f</sup> <sup>∂</sup><sup>t</sup> ð Þ¼ x0; t<sup>0</sup> 0, then we use f xð Þ¼ ; t f xð Þþ ; t εt, ε > 0. For f, the above proof holds, and we take ε ! 0.

(3) ϕ is an increasing function.

Let 0 <sup>≤</sup> <sup>a</sup> <sup>&</sup>lt; <sup>b</sup> <sup>≤</sup> <sup>T</sup> and note <sup>A</sup> <sup>¼</sup> <sup>t</sup> <sup>∈</sup>½ �j <sup>a</sup>; <sup>b</sup> <sup>ϕ</sup>ð Þ<sup>t</sup> <sup>≤</sup> <sup>ϕ</sup>ð Þ<sup>b</sup> . <sup>A</sup> is not empty. If <sup>α</sup> <sup>¼</sup> infA, then, by the step (2), α < b and, by the step (1), α∈ A. If α > a, we can use the step (2) for t<sup>0</sup> ¼ α and it would result that α was not the lower bound of A. Hence α ¼ a and ϕð Þa ≤ ϕð Þb .

Remark The third step shows that a function having the properties (1) and (2) is increasing. For this the continuity is essential. Only property (2) is not enough. For example, the function defined by ϕðÞ¼ t t on 0½ � ; 1 and ϕðÞ¼ t 1 � t on 1ð � ; 2 has only the property (2), but it is not increasing on 0½ � ; 2 .

Remark Suppose that <sup>f</sup> is a <sup>C</sup><sup>2</sup> function and minxf xð Þ¼ ; <sup>t</sup> f xð Þ <sup>0</sup>ð Þ<sup>t</sup> ; <sup>t</sup> , where <sup>x</sup>0ð Þ<sup>t</sup> is an interior point of M. Since x0ð Þt is a critical point, we have

Bilevel Disjunctive Optimization on Affine Manifolds http://dx.doi.org/10.5772/intechopen.75643 133

$$\boldsymbol{\phi}'(t) = \frac{\partial \boldsymbol{f}}{\partial t}(\boldsymbol{\chi}\_0(t), t) + < \frac{\partial \boldsymbol{f}}{\partial \boldsymbol{\chi}}(\boldsymbol{\chi}\_0(t), t), \boldsymbol{\chi}\_0'(t)> = \frac{\partial \boldsymbol{f}}{\partial t}(\boldsymbol{\chi}\_0(t), t) \succeq 0.$$

Theorem 6.1 Suppose that M<sup>1</sup> is compact, M<sup>2</sup> ¼ ½ � 0;T and f : M<sup>1</sup> � ½ �! 0;T R. Let ϕð Þ¼ t minxf xð Þ ;t .

Indeed, f is continuous on the compact M<sup>1</sup> � ½ � 0; 1 , hence uniformly continuous. So, for ε > 0 it

�ε < f xð Þ� ; t<sup>1</sup> f xð Þ ; t<sup>2</sup> < ε

f xð Þ ; t<sup>1</sup> > f xð Þ� ; t<sup>2</sup> ε ≥ f xð Þ� <sup>2</sup>; t<sup>2</sup> ε:

f xð Þþ ; t<sup>2</sup> ε > f xð Þ ; t<sup>1</sup> ≥ f xð Þ <sup>1</sup>; t<sup>1</sup> :

<sup>∂</sup><sup>t</sup> ð Þ x0; t<sup>0</sup> > 0, it exists δ > 0 such that f xð Þ <sup>0</sup>; t ≤ f xð Þ <sup>0</sup>; t<sup>0</sup> , for each t ∈ð Þ t<sup>0</sup> � δ; t<sup>0</sup> . It

<sup>∂</sup><sup>t</sup> ð Þ¼ x0; t<sup>0</sup> 0, then we use f xð Þ¼ ; t f xð Þþ ; t εt, ε > 0. For f, the above proof holds, and we

Let 0 <sup>≤</sup> <sup>a</sup> <sup>&</sup>lt; <sup>b</sup> <sup>≤</sup> <sup>T</sup> and note <sup>A</sup> <sup>¼</sup> <sup>t</sup> <sup>∈</sup>½ �j <sup>a</sup>; <sup>b</sup> <sup>ϕ</sup>ð Þ<sup>t</sup> <sup>≤</sup> <sup>ϕ</sup>ð Þ<sup>b</sup> . <sup>A</sup> is not empty. If <sup>α</sup> <sup>¼</sup> infA, then, by the step (2), α < b and, by the step (1), α∈ A. If α > a, we can use the step (2) for t<sup>0</sup> ¼ α and it

Remark The third step shows that a function having the properties (1) and (2) is increasing. For this the continuity is essential. Only property (2) is not enough. For example, the function defined by ϕðÞ¼ t t on 0½ � ; 1 and ϕðÞ¼ t 1 � t on 1ð � ; 2 has only the property (2), but it is not

Remark Suppose that <sup>f</sup> is a <sup>C</sup><sup>2</sup> function and minxf xð Þ¼ ; <sup>t</sup> f xð Þ <sup>0</sup>ð Þ<sup>t</sup> ; <sup>t</sup> , where <sup>x</sup>0ð Þ<sup>t</sup> is an interior

would result that α was not the lower bound of A. Hence α ¼ a and ϕð Þa ≤ ϕð Þb .

exists δ > 0 such that if ∣t<sup>1</sup> � t2∣ < δ, then ∣f xð Þ� ; t<sup>1</sup> f xð Þ ; t<sup>2</sup> ∣ < ε, for any x∈ M1, or

On one hand, if we put ϕð Þ¼ t<sup>1</sup> f xð Þ <sup>1</sup>; t<sup>1</sup> and ϕð Þ¼ t<sup>2</sup> f xð Þ <sup>2</sup>; t<sup>2</sup> , then we have

Finally, ∣ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ∣ ≤ ε, for ∣t<sup>1</sup> � t2∣ < δ, that is, ϕ is (uniformly) continuous.

Hence minxf xð Þ ; t<sup>1</sup> ≥ f xð Þ� <sup>2</sup>; t<sup>2</sup> ε, so is ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ≥ � ε.

Hence minxf xð Þþ ; t<sup>2</sup> ε ≥ f xð Þ <sup>1</sup>; t<sup>1</sup> , so is ϕð Þ� t<sup>1</sup> ϕð Þ t<sup>2</sup> ≤ ε.

(2) Let us fix t<sup>0</sup> <sup>∈</sup> ð � <sup>0</sup>; <sup>T</sup> . If <sup>ϕ</sup>ð Þ¼ <sup>t</sup><sup>0</sup> f xð Þ <sup>0</sup>; <sup>t</sup><sup>0</sup> and <sup>∂</sup><sup>f</sup>

point of M. Since x0ð Þt is a critical point, we have

follows minxf xð Þ ; t ≤ f xð Þ <sup>0</sup>; t ≤ f xð Þ <sup>0</sup>; t<sup>0</sup> , and so is ϕð Þt ≤ ϕð Þ t<sup>0</sup> .

ϕð Þt ≤ ϕð Þ t<sup>0</sup> , for any t∈ ð Þ t<sup>0</sup> � δ; t<sup>0</sup> .

(3) ϕ is an increasing function.

increasing on 0½ � ; 2 .

On the other hand,

Suppose <sup>∂</sup><sup>f</sup>

take ε ! 0.

If <sup>∂</sup><sup>f</sup>

<sup>∂</sup><sup>t</sup> ð Þ x; t ≥ 0, then ϕ is an increasing function.

<sup>∂</sup><sup>t</sup> ð Þ x0; t<sup>0</sup> ≥ 0, then it exists δ > 0 such that

If, for each x with <sup>ϕ</sup>ðÞ¼ <sup>t</sup> f xð Þ ; <sup>t</sup> , we have <sup>∂</sup><sup>f</sup>

132 Optimization Algorithms - Examples

Proof. We shall prove the statement in three steps. (1) If f is continuous, then ϕ is (uniformly) continuous. Consequently, ϕð Þt is an increasing function. If M<sup>1</sup> has a nonvoid boundary, then the monotony extends by continuity (see also the evolution of an extremum problem).

Example 6.1 The single-time perspective of a function f : <sup>R</sup><sup>n</sup> ! <sup>R</sup> is the function g : <sup>R</sup><sup>n</sup> � <sup>R</sup><sup>þ</sup> ! <sup>R</sup>, g xð Þ¼ ; t tf xð Þ =t , dom g ¼ f g ð Þj x; t x=t∈ dom f ; t > 0 . The single-time perspective g is convex if f is convex.

The single-time perspective is an example verifying Theorem 7.1. Indeed, the critical point condition for g, in x, <sup>∂</sup><sup>g</sup> <sup>∂</sup><sup>x</sup> ¼ 0, gives x ¼ tx0, where x<sup>0</sup> is a critical point of f . Consequently, ϕðÞ¼ t minxg xð Þ¼ ; t tf xð Þ<sup>0</sup> . On the other hand, in the minimum point, we have <sup>∂</sup><sup>g</sup> <sup>∂</sup><sup>t</sup> ð Þ¼ x; t f xð Þ<sup>0</sup> . Then ϕð Þt is increasing if f xð Þ<sup>0</sup> ≥ 0, as in Theorem 4.1.

Theorem 6.2 Suppose that M<sup>1</sup> is compact and f : M<sup>1</sup> � M<sup>2</sup> ! R. Let ϕð Þ¼ y minxf xð Þ ; y . If, for each x with <sup>ϕ</sup>ð Þ¼ <sup>y</sup> f xð Þ ; <sup>y</sup> , we have <sup>∂</sup><sup>f</sup> <sup>∂</sup>y<sup>α</sup> ð Þ x; y ≥ 0, then ϕð Þy is a partially increasing function.

Proof. Suppose that <sup>f</sup> is a <sup>C</sup><sup>2</sup> function and minxf xð Þ¼ ; <sup>y</sup> f xð Þ <sup>0</sup>ð Þ<sup>y</sup> ; <sup>y</sup> , where <sup>x</sup>0ð Þ<sup>y</sup> is an interior point of M1. Since x0ð Þy is a critical point, we have

$$\frac{\partial \phi}{\partial y^{\alpha}} = \frac{\partial f}{\partial y^{\alpha}}(\mathbf{x}\_{0}(y), y) + < \frac{\partial f}{\partial \mathbf{x}}(\mathbf{x}\_{0}(y), y), \frac{\partial \mathbf{x}\_{0}}{\partial y^{\alpha}} > = \frac{\partial f}{\partial y^{\alpha}}(\mathbf{x}\_{0}(y), y) \ge 0.$$

Consequently, ϕð Þy is a partially increasing function. If M has a non-void boundary, then the monotony extends by continuity. □

Theorem 6.3 Suppose that M<sup>1</sup> is compact and f : M<sup>1</sup> � M<sup>2</sup> ! R. Let ϕð Þ¼ y minxf xð Þ ; y . If, for each x with <sup>ϕ</sup>ð Þ¼ <sup>y</sup> f xð Þ ; <sup>y</sup> , we have d<sup>2</sup> <sup>y</sup> f xð Þ ; y ≤ 0, then ϕð Þy is an affine concave function.

Proof. Without loss of generality, we work on Euclidean case. Suppose that f is a C<sup>2</sup> function and minxf xð Þ¼ ; y f xy ð Þ ð Þ; y , where x yð Þ is an interior point of M1. Since x yð Þ is a critical point, we must have

$$\frac{\partial f}{\partial x^i}(x(y), y) = 0.$$

Taking the partial derivative with respect to y<sup>α</sup> and the scalar product with <sup>∂</sup>x<sup>i</sup> <sup>∂</sup>y<sup>β</sup> it follows

$$
\frac{\partial^2 f}{\partial x^i \partial x^j} \frac{\partial x^i}{\partial y^\alpha} \frac{\partial x^i}{\partial y^\beta} + \frac{\partial^2 f}{\partial y^\alpha \partial x^i} \frac{\partial x^i}{\partial y^\beta} = 0.
$$

□

On the other hand

$$d\_y \phi(y) = d\_y f(\mathbf{x}(y), y) = \left(\frac{\partial f}{\partial \mathbf{x}^i} \frac{\partial \mathbf{x}^i}{\partial y^\alpha} + \frac{\partial f}{\partial y^\alpha}\right) dy^\alpha = \frac{\partial f}{\partial y^\alpha} dy^\alpha$$

$$d\_y^2 \phi(y) = \left(\frac{\partial^2 f}{\partial y^\alpha \partial \mathbf{x}^i} \frac{\partial \mathbf{x}^i}{\partial y^\beta} + \frac{\partial^2 f}{\partial y^\alpha \partial y^\beta}\right) dy^\alpha dy^\beta$$

$$= \left(-\frac{\partial^2 f}{\partial \mathbf{x}^i \partial \mathbf{x}^i} \frac{\partial \mathbf{x}^i}{\partial y^\delta} \frac{\partial \mathbf{x}^j}{\partial y^\alpha} + \frac{\partial^2 f}{\partial y^\alpha \partial y^\beta}\right) dy^\alpha dy^\beta \le 0.$$

Theorem 6.4 Let f : <sup>M</sup><sup>1</sup> � <sup>M</sup><sup>2</sup> ! <sup>R</sup> be a C<sup>2</sup> function and

$$\phi(y) = \min\_{\mathfrak{x}} f(\mathfrak{x}, y) = f(\mathfrak{x}(y), y).$$

□

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If the set A ¼ f g ð Þ x yð Þ; y : y∈ M<sup>2</sup> is affine convex and f j <sup>A</sup> is affine convex, then ϕð Þy is affine convex. Proof. Suppose <sup>f</sup> is a <sup>C</sup><sup>2</sup> function. At points ð Þ x yð Þ; <sup>y</sup> , we have

$$0 \le d^2 f(\mathbf{x}(y), y) = \left(\frac{\partial^2 f}{\partial \mathbf{x}^i \partial \mathbf{x}^j} \frac{\partial \mathbf{x}^i}{\partial y^\alpha} \frac{\partial \mathbf{x}^j}{\partial y^\delta} + 2 \frac{\partial^2 f}{\partial \mathbf{x}^i \partial y^\alpha} \frac{\partial \mathbf{x}^i}{\partial y^\delta} + \frac{\partial^2 f}{\partial y^\alpha \partial y^\delta} \right) dy^\alpha dy^\beta$$

$$= \left(\frac{\partial^2 f}{\partial \mathbf{x}^i \partial y^\alpha} \frac{\partial \mathbf{x}^i}{\partial y^\delta} + \frac{\partial^2 f}{\partial y^\alpha \partial y^\delta} \right) dy^\alpha dy^\beta = d^2 \phi(y).$$
