Pareto Optimality for Multioptimization of Continuous Linear Operators

SS symmetry Article Pareto Optimality for Multioptimization of Continuous Linear Operators Clemente Cobos-Sánchez 1, *,† , José Antonio Vilchez-Membrilla 1,† , Almudena Campos-Jiménez 2,† Francisco Javier García-Pacheco 2,† 1 2 * †



 Citation: Cobos-Sánchez, C.; Vilchez-Membrilla, J.A.; Campos-Jiménez, A.; García-Pacheco, and Department of Electronics, College of Engineering, University of Cadiz, 11510 Puerto Real, Spain; [email protected] Department of Mathematics, College of Engineering, University of Cadiz, 11510 Puerto Real, Spain; [email protected] (A.C.-J.); [email protected] (F.J.G.-P.) Correspondence: [email protected] These authors contributed equally to this work. Abstract: This manuscript determines the set of Pareto optimal solutions of certain multiobjectiveoptimization problems involving continuous linear operators defined on Banach spaces and Hilbert spaces. These multioptimization problems typically arise in engineering. In order to accomplish our goals, we first characterize, in an abstract setting, the set of Pareto optimal solutions of any multiobjective optimization problem. We then provide sufficient topological conditions to ensure the existence of Pareto optimal solutions. Next, we determine the Pareto optimal solutions of convex max–min problems involving continuous linear operators defined on Banach spaces. We prove that the set of Pareto optimal solutions of a convex max–min of form max k T ( x )k, min k x k coincides with the set of multiples of supporting vectors of T. Lastly, we apply this result to convex max–min problems in the Hilbert space setting, which also applies to convex max–min problems that arise in the design of truly optimal coils in engineering. F.J. Pareto Optimality for Multioptimization of Continuous Linear Operators. Symmetry 2021, 13, Keywords: multioptimization; Pareto optimality; linear operators; adjoint operators; normed spaces; matrix norms 661. https://doi.org/10.3390/ sym13040661 MSC: 47L05; 47L90; 49J30; 90B50 Academic Editor: Juan Luis García Guirao 1. Introduction Received: 11 March 2021 Accepted: 8 April 2021 Published: 12 April 2021 Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. Copyright: © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/). Multiobjective optimization problems (MOPs) appear quite often in all areas of pure and applied mathematics, for instance, in the geometry of Banach spaces [1–3], in operator theory [4–7], in lineability theory [8–10], in differential geometry [11–14], and in all areas of Experimental, Medical and Social Sciences [15–20]. By means of MOPs, many real-life situations can be modeled accurately. However, the existence of a global solution that optimizes all the objective functions of an MOP at once is very unlikely. This is were Pareto optimal solutions (POS) come into play. Informally speaking, a POS is a feasible solution such that, if any other feasible solution is more optimal at one objective function, then it is less optimal at another objective function. Pareto optimal solutions are sometimes graphically displayed in Pareto charts (PC). In this manuscript, we prove a characterization of POS by relying on orderings and equivalence relations. We also provide a sufficient topological condition to guarantee the existence of Pareto optimal solutions. This work is mainly motivated by certain MOPs appearing in engineering, such as the design of truly optimal transcranial magnetic stimulation (TMS) coils [18–23]. The main goal of this manuscript is to characterize (Theorem 6) the set of Pareto optimal solutions of the MOPs that appear in the design of coils, such as (3). In the case of MOPs in which operators are defined on Hilbert spaces, this characterization is improved (Corollary 1). Under this Hilbert space setting, we also study the relationships between different MOPs Symmetry 2021, 13, 661. https://doi.org/10.3390/sym13040661 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 661 2 of 17 involving different operators, but which are defined on the same Hilbert space. These operators can be naturally combined to obtain a new MOP. The set of Pareto optimal solutions of this new MOP is compared (Corollary 2) to the set of Pareto optimal solutions of the initial MOPs. 2. Materials and Methods In this section, we compile all necessary tools to accomplish our results. We also develop new and original tools, such as Theorem 1 and Corollary 2, which contribute to enriching the literature on optimization theory. 2.1. Formal Description of MOPs A generic multiobjective optimization problem (MOP) has the following form:   max f i ( x ) i = 1, . . . , p, min g j ( x ) j = 1, . . . , q, M :=  x ∈ R, (1) where f i , g j : X → R are called objective functions, defined on a nonempty set X, and R is a nonempty subset of X called the feasible region or region of constraints/restrictions. The set of general solutions of the above MOP is denoted by sol( M). In fact, sol( M ) := { x ∈ R : ∀y ∈ R∀i ∈ {1, . . . , p}∀ j ∈ {1, . . . , q} f i ( x ) ≥ f i (y), g j ( x ) ≤ g j (y)}. It is obvious that sol( M ) = sol( P1 ) ∩ · · · ∩ sol( Pp ) ∩ sol( Q1 ) ∩ · · · ∩ sol( Qq ) where Pi :=  max f i ( x ), x ∈ R, and Q j :=  (2) min g j ( x ), x ∈ R, are single-objective optimization problems (SOPs) and sol( Pi ), sol( Q j ) denote the set of general solutions of Pi , Q j for i = 1, . . . , p and j = 1, . . . , q, respectively. The set of Pareto optimal solutions of MOP M is defined as Pos( M ) := { x ∈ R : ∀y ∈ R, if there exists i ∈ {1, . . . , p} with f i (y) > f i ( x ) or j ∈ {1, . . . , q} with g j (y) < g j ( x ), then there exists i′ ∈ {1, . . . , p} with f i′ (y) < f i′ ( x ) or j′ ∈ {1, . . . , q} with g j′ ( x ) < g j′ (y)}. To guarantee the existence of general solutions, it is usually asked for X to be a Hausdorff topological space, R is a compact subset of X, f i s are upper semicontinuous, and g j s are lower semicontinuous. This way, at least we make sure that the SOPs Pi s and Q j s have at least one solution (Weierstrass extreme value theorem). Even more, solution sets sol( Pi ) and sol( Q j ) are closed and thus compact, which makes sol( M ) also compact. Nevertheless, even under these conditions, sol( M ) might still be empty, as we can easily infer from Equation (2). 2.2. Characterizing Pareto Optimal Solutions A more abstract way to construct the set of Pareto optimal solutions follows. Let X be a nonempty set, f i , g j : X → R functions and R a nonempty subset of X. In R, consider the equivalence relation given by n o S := ( x, y) ∈ R2 : ∀i = 1, . . . , p ∀ j = 1, . . . , q f i ( x ) = f i (y), g j ( x ) = g j (y) . Symmetry 2021, 13, 661 3 of 17 Next, in the quotient set of R by S , R, S consider the order relation given by [ x ]S ≤ [y]S ⇔ ∀i = 1, . . . , p ∀ j = 1, . . . , q f i ( x ) ≤ f i (y), g j (y) ≤ g j ( x ). Theorem 1. Consider MOP (1). Then,   R endowed with ≤ Pos( M) = x ∈ R : [ x ]S is a maximal element of S and sol( M ) :=   R endowed with ≤ . x ∈ R : [ x ]S is the maximum of S As a consequence, sol( M ) ⊆ Pos( M). If there exists i1 ∈ {1, . . . , p} or j1 ∈ {1, . . . , q} such that sol( Pi1 ) or sol( Q j1 ) is a singleton, respectively, then sol( Pi1 ) ⊆ Pos( M) or sol( Q j1 ) ⊆ Pos( M), respectively. Proof. Fix an arbitrary x0 ∈ Pos( M). Let us assume that there is y ∈ R, so that [ x0 ]S < [y]S . Then, f i ( x0 ) ≤ f i (y) for all i = 1, . . . , p and g j ( x0 ) ≥ g j (y) for all j = 1, . . . , q. However, [ x0 ]S 6= [y]S ; therefore, there exists i0 ∈ {1, . . . , p} or j0 ∈ {1, . . . , q} such that f i0 ( x0 ) < f i0 (y) or g j0 ( x0 ) < g j0 (y), respectively. Since x0 ∈ Pos( M ) by assumption, there exists i1 ∈ {1, . . . , p} or j1 ∈ {1, . . . , q}, such that f i1 ( x0 ) > f i1 (y) or g j1 ( x0 ) < g j1 (y), respectively, which is a contradiction. Therefore, [ x0 ]S is a maximal element of R S endowed with ≤. The arbitrariness of x0 ∈ Pos( M ) shows that   R Pos( M ) ⊆ x ∈ R : [ x ]S is a maximal element of endowed with ≤ . S Conversely, fix an arbitrary x0 ∈ R, such that [ x0 ]S is a maximal element of R S endowed with ≤. Take y ∈ R satisfying that there exists i0 ∈ {1, . . . , p} or j0 ∈ {1, . . . , q} with f i0 (y) > f i0 ( x0 ) or g j0 (y) < g j0 ( x0 ), respectively. If f i ( x0 ) ≤ f i (y) for all i ∈ {1, . . . , p} \ {i0 } and g j ( x0 ) ≥ g j (y) for all j ∈ {1, . . . , q} \ { j0 }, then [ x0 ]S < [y]S , reaching a contradiction with the maximality of [ x0 ]S in R S endowed with ≤. This shows that Pos( M ) =  x ∈ R : [ x ]S is a maximal element of  R endowed with ≤ . S Next, fix an arbitrary x0 ∈ sol( M ). For every y ∈ R, f i ( x0 ) ≥ f i (y) and g j ( x0 ) ≤ g j (y) for all i = 1, . . . , p and all j = 1, . . . , q. Then, [ x0 ]S ≥ [y]S . The arbitrariness of y ∈ R ensures that [ x0 ]S is a maximal element of R S endowed with ≤. Conversely, fix an arbitrary x0 ∈ R, such that [ x0 ]S is a maximal element of R S endowed with ≤. For every y ∈ R, [y]S ≥ [ x0 ]S ; therefore, f i ( x0 ) ≥ f i (y) and g j ( x0 ) ≤ g j (y) for all i = 1, . . . , p and all j = 1, . . . , q. The arbitrariness of y ∈ R proves that x0 ∈ sol( M ). We proved that   R sol( M ) := x ∈ R : [ x ]S is the maximum of endowed with ≤ . S Lastly, suppose that sol( Pi1 ) is a singleton for some i1 ∈ {1, . . . , p}, and write sol( Pi1 ) = { x0 }. Take y ∈ R satisfying that there exists i0 ∈ {1, . . . , p} or j0 ∈ {1, . . . , q} with f i0 (y) > f i0 ( x0 ) or g j0 (y) < g j0 ( x0 ), respectively. If such i0 exists, then i0 6= i1 . By hypothesis, f i1 ( x0 ) > f i1 (y) since y ∈ / sol( Pi1 ). This shows that x0 ∈ Pos( M ). Likewise, sol( Q j1 ) ⊆ Pos( M) provides that sol( Q j1 ) is a singleton. Symmetry 2021, 13, 661 4 of 17 Lemma 1. Consider MOP (1). Let i0 ∈ {1, . . . , p}, j0 ∈ {1, . . . , q}. Then,    If 1.  there  is xi0 ∈ R so that xi0 S is a maximal element of [ x ]S : x ∈ arg maxR f i0 , then xi0 S is a maximal element of R/S . Hence, xi0 ∈ Pos( M).    If there is x j0 ∈ R so that x j0 S is a maximal element of [ x ]S : x ∈ arg minR g j0 , then 2.   x j0 S is a maximal element of R/S . Hence, x j0 ∈ Pos( M )   x i0 S Proof. We only prove the first item since the other follows a dual proof. Assume   that is not a maximal element of R / S . Then, we can find y ∈ R in such a way that < [y]S . x i0 S
In particular, f i0 xi0 ≤ f i0 (y); therefore, f i0 (y) = maxR f i0 ; hence, y ∈ arg maxR f i0 . As [y]S ∈ [ x ]S : x ∈ arg maxR f i0 , contradicting that xi0 S be a maximal a consequence,  element of [ x ]S : x ∈ arg maxR f i0 . Theorem 2. Consider MOP (1). If X is a topological space, R is a compact Hausdorff subset of X and all the objective functions are continuous, then Pos( M ) 6= ∅. Proof. Fix i0 ∈ {1, . . . , p} . In accordance with Lemma 1, it is only sufficient to find a maximal element of A := [ x ]S : x ∈ arg maxR f i0 . We rely on Zorn’s lemma. Consider a chain in A, that is, a totally ordered subset of elements [ x k ]S , with k ranging a totally  ordered set K in such a way that k1 < k2 if and only if xk1 S < xk2 S . Since K is totally ordered, we have that ( xk )k∈K is a net in R. The compactness of R allows for extracting a subnet (yh )h∈ H of ( xk )k∈K convergent to some x
0 ∈ R. Let us first show that x0 ∈ arg maxR f i0 . The continuity of f i0 implies that f i0 (yh ) h∈ H converges to f i0 ( x0 ). Fix any ε > 0. There is hε ∈ H satisfying that, if h ≥ hε , then f i0 (yh ) − f i0 ( x0 ) < ε. Fix any k0 ∈ K. There is h0 ∈ H, so that {yh : h ≥ h0 } ⊆ { xk : k ≥ k0 }. Since H is a directed set, we can find h1 ∈ H with
h1 ≥ h0 . There exists k1 ∈ K with k1 ≥ k0 such that
h1 ≥ hε and yh1 = xk1 . Next, f i0 yh1 = f i0 xk1 = maxR f i0 . As a consequence,
max f i0 − f ( x0 ) = f i0 yh1 − f ( x0 ) < ε. R The arbitrariness of ε shows that maxR f i0 = f ( x0 ). Lastly, we prove that [ x0 ]S is an upper   bound for chain {[ xk ]S : k ∈ K }. Fix an
arbitrary k0 ∈ K. In order to prove that x
k0 S ≤ [ x0 ]S , we have to check that f i xk0 ≤ f i ( x0 ) for all i ∈ {1, . . . , p} and g j xk0 ≥
g j ( x0 ) for all j ∈ {1, . . . , q}. Indeed, fix i ∈ {1, . . . , p} and suppose to the contrary
that f i xk0 > f i ( x0 ). Let 0 < ε < f i xk0 − f i ( x0 ). There exists hε ∈ H such that, if h ≥ hε , then | f i (yh ) − f i ( x0 )| < ε. We can find h0 ∈ H, such that {yh : h ≥ h0 } ⊆ { xk : k ≥ k0 }. Since H is a directed set, we can find h1 ∈ H with h1 ≥ hε andh1 ≥  h0 . There  exists k1 ∈ K with k1 ≥ k0 such that yh1 = xk1 . Since k0 ≤ k1 , we have that xk0 S ≤ xk1 S . Thus,


f i y h1 = f i x k 1 ≥ f i x k 0 > f i ( x 0 ),
contradicting that f i yh1 − f i ( x0 ) < ε. In a similar it can be shown that
 way,  g j xk0 ≥ g j ( x0 ) for all j ∈ {1, . . . , q}. As a consequence, xk0 S ≤ [ x0 ]S . In other words, [ x0 ]S is an upper bound for the chain {[ xk ]S : k ∈ K }. Since every chain of A has an upper bound, Zorn’s lemma ensures the existence of maximal elements in A. 2.3. MOPs in a Functional-Analysis Context A large number of objective functions in an MOP may cause a lack of general solutions, that is, sol( M ) = ∅. This happens quite often with MOPs involving matrices. Even if the number of objective functions is short, we might still have sol( M) = ∅. The following theorem [20], Theorem 2, is a very representative example of this situation of lack of general solutions. Symmetry 2021, 13, 661 5 of 17 Theorem 3. Let T : X → Y be a nonzero continuous linear operator, where X, Y are normed spaces; then, the following max–min problem is free of general solutions:   max k T ( x )k, min k x k,  x ∈ X. (3) Equation (3) describes an MOP that appears in bioengineering quite often after the linearization of forces or fields [18]. 3. Results We focus on MOPs similar to (3). In fact, we find Pos(3) (Theorem 6 and Corollary 1). If X, Y are Hilbert spaces, say H, K, and T1 , . . . , Tk ∈ B( H, K ) are continuous linear operators, then the sets of Pareto optimal solutions of the MOPs   max k Ti ( x )k, min k x k,  x ∈ H, (4) for i = 1, . . . , k are compared (Corollary 2) with the set of Pareto optimal solutions of MOP   max k T ( x )k, min k x k,  x ∈ H, where T: H x (5) k → K ⊕2 · · · ⊕2 K 7→ T ( x ) = ( T1 ( x ), . . . , Tk ( x )). 3.1. Formatting of Mathematical Components Let X, Y be normed spaces. Consider a nonzero continuous linear operator T : X → Y. Then k T k := sup{k T ( x )k : x ∈ BX } is the norm of T. On the other hand, suppv( T ) := { x ∈ SX : k T ( x )k = k T k}, stands for the set of supporting vectors of T, where BX := { x ∈ X : k x k ≤ 1} is a (closed) unit ball, and SX := { x ∈ X : k x k = 1} is the unit sphere. Continuous linear operators are also called bounded because they are bounded on the unit ball. The space of bounded linear operators from X to Y is denoted as B( X, Y ). Let H be a Hilbert space, and consider the dual map of H: JH : H k → 7→ H∗ JH (k) := k∗ = (•|k). JH is a surjective linear isometry between H and H ∗ (Riesz representation theorem). In the frame of the geometry of Banach spaces, JH is called duality mapping. Consider H, K Hilbert spaces, and let T ∈ B( H, K ) be a bounded linear operator. We define the adjoint operator of T as T ′ := ( JH )−1 ◦ T ∗ ◦ JK ∈ B(K, H ), with T ∗ : K ∗ → H ∗ as the dual operator of T. The most representative property of the adjoint operator is that it is the unique operator in B(K, H ) satisfying ( T ( x )|y) = ( x | T ′ (y)) for all x ∈ H and all y ∈ K. ′ It holds that k T ′ k = k T k, ( T ′ ) = T, ( T + S)′ = T ′ + S′ and (λT )′ = λT ′ . Symmetry 2021, 13, 661 6 of 17 If T ∈ B( H ) verifies T = T ′ , then T is self-adjoint. This is equivalent to equality ( T ( x )|y) = ( x | T (y)) held for every x, y ∈ H. If T satisfies ( T ( x )| x ) ≥ 0 for each x ∈ H, then T is called positive. If H is complex, then T ∈ B( H ) is self-adjoint if and only if ( T ( x )| x ) ∈ R for each x ∈ H. Thus, in complex Hilbert spaces, positive operators are self-adjoint. T is strongly positive if there exists S ∈ B( H, K ) with T = S′ ◦ S. Typical examples of self-adjoint positive operators are strongly positive operators. For each T ∈ B( H ), the following set is the spectrum of T σ ( T ) := {λ ∈ C : λI − T ∈ / U (B( H ))}, where U (B( H )) is the multiplicative group of invertible operators on H. Among spectral properties, it is compact, nonempty, and k T k ≥ max |σ( T )|. We work with a special subset of the spectrum: σp ( T ) := {λ ∈ C : ker(λI − T ) 6= {0}}, called the point spectrum, whose elements are eigenvalues of T. It is clear that σp ( T ) ⊆ σ( T ). In addition, if λ ∈ σp ( T ), the subspace of associated eigenvectors to λ is V (λ) := { x ∈ H : T ( x ) = λx }. If k T k is an eigenvalue of T or, in other words, k T k ∈ σp ( T ), then k T k is the maximal element of |σ( T )|, i.e., k T k = max |σ( T )|. In this situation, we also write k T k = λmax ( T ). Example 1. Let T : H → K be a continuous linear operator where H, K are Hilbert spaces, such that k T k ∈ σp ( T ); then, V (k T k) ∩ SX ⊆ suppv( T ). If x ∈ V (k T k) ∩ SX , then T ( x ) = k T k x; therefore, k T ( x )k = k T k and hence x ∈ suppv( T ). In general, k T k ∈ / σp ( T ), unless, for instance, T is compact, self-adjoint, and positive. This is why we have to rely on adjoint T ′ and strongly positive operator T ′ ◦ T. It is straightforward to verify that the eigenvalues of a positive operator are positive, and in the case of a self-adjoint operator, the eigenvalues are real. When T is compact, it holds that T ′ ◦ T is compact, self-adjoint, and positive. The next result was obtained by refining ([10] [Theorem 9]). In particular, we obtained the same conclusions with fewer hypotheses. Theorem 4. Consider H, K Hilbert spaces, and T ∈ B( H, K ). Then, 1. 2. 3. k T k2 = k T ′ ◦ T k. suppv( T ) ⊆ suppv( T ′ ◦ T ). supp( T ) 6= ∅ if and only if k T ′ ◦ T k ∈ σp ( T ′ ◦ T ). p In this situation, k T k = λmax ( T ′ ◦ T ) and suppv( T ) = V (λmax ( T ′ ◦ T )) ∩ S H . Proof. 1. Fix an element x ∈ H, and the associated mapping x ∗ := (•| x ). Then, k T ( x )k2
= ( T ( x )| T ( x )) = T ′ ( T ( x ))| x = x ∗ ∗ ′ ∗ ′

T ′ ◦ T (x) ′ ≤ k x k T ( T ( x )) ≤ k x k T ◦ T k x k = T ◦ T k x k ≤ T ′ k T kk x k2 = k T k2 k x k2 . (6) 2 (7) (8) If element x is taken in the unit sphere, i.e., x ∈ SX , and considering the previous inequalities, we concluded that k T k2 = k T ′ ◦ T k. Symmetry 2021, 13, 661 7 of 17 2. Let x ∈ suppv( T ) be an arbitrary element; then, Equation (6) implies that T ′ ( T ( x )) = k T k2 = k T kk T ( x )k = T ′ k T ( x )k. 3. Then, x ∈ suppv( T ′ ◦ T ). Take v ∈ suppv( T ). Before anything else, since suppv( T ) ⊆ suppv( T ′ ◦ T ), we have that ( T ′ ◦ T )(v) kT′ ◦ Tk k( T ′ ◦ T )(v)k = = 1. (9) = kT′ ◦ Tk kT′ ◦ Tk kT′ ◦ Tk Following chain of equalities (6), v∗  ( T ′ ◦ T )(v) kT′ ◦ Tk  = k T (v)k2 k T k2 k v k2 = = 1. ′ kT ◦ Tk kT′ ◦ Tk (10) Thanks to the strict convexity of space H, ( T ′ ◦ T )(v) = v, kT′ ◦ Tk that is,
T ′ ◦ T (v) = T ′ ◦ T v and so k T ′ ◦ T k ∈ σp ( T ′ ◦ T ). We implicitly proved that suppv( T ) ⊆ V (k T ′ ◦ T k) ∩ S H . Conversely, let us suppose that k T ′ ◦ T k ∈ σp ( T ′ ◦ T ). As we remarked before, T ′ ◦ T is a strongly positive operator, so the eigenvalues of that operator are real and positive. Therefore, equality λmax ( T ′ ◦ T ) = k T ′ ◦ T k holds, which implies that kTk = q k T k2 = q kT′ ◦ Tk = Take w ∈ V (λmax ( T ′ ◦ T )) ∩ S H . Then k T (w)k2 q λmax ( T ′ ◦ T ).

= w∗ T ′ ◦ T (w)

= w∗ λmax T ′ ◦ T w
= λmax T ′ ◦ T = T′ ◦ T = k T k2 . This chain of equalities proves that w ∈ suppv( T ). Consequently,

V λmax T ′ ◦ T ∩ S H ⊆ suppv( T ). The following technical lemma establishes the behavior of the point spectrum of a linear combination of operators. However, we first introduce some notation. Considering bounder linear operator T ∈ B( H, K ) defined between H and K, Hilbert spaces, then V ( T ) := [ V ( λ ). λ∈σp ( T ) Lemma 2. If we consider Hilbert spaces, H, K, and T1 , . . . , Tk ∈ B( H, K ), then, for every α 1 , . . . , α k ∈ C, ! k \ i =1 k V ( Ti ) ⊆ V ∑ αi Ti i =1 . Symmetry 2021, 13, 661 8 of 17 Proof. Takeany x ∈ ik=1 V ( Ti ). If x = 0, there is nothing to prove, x is actually in  V ∑ik=1 αi Ti . So, assume that x 6= 0. For every i ∈ {1, . . . , k }, there exists λi ∈ σ ( Ti ), such that x ∈ V (λi ), that is, Ti ( x ) = λi x. Then ! ! T k ∑ αi Ti k k k i =1 i =1 i =1 ∑ αi Ti (x) = ∑ αi λi x = ∑ αi λi (x) = i =1 This shows that k ∑ αi λi x∈V i =1 ! k ∑ αi Ti ⊆V i =1 ! x. . The hypothesis in Lemma 3 is, in fact, very restrictive. ∈ B( H, K ), such that Lemma 3. If H, K are Hilbert spaces, and T1 , . . . , Tk V ( Ti ) \ ker( Ti ) ⊆ ker( Tj ) for all i, j ∈ {1, . . . , k } with i 6= j. For every i ∈ {1, . . . , k} and every xi ∈ V ( Ti ) \ ker( Ti ), there are α1 , . . . , αk ∈ C, such that ! k k i =1 i =1 ∑ βi xi ∈ V ∑ αi Ti for every β 1 , . . . , β k ∈ C. Proof. For every i ∈ {1, . . . , k }, there exists λi ∈ σ ( Ti ) \ {0}, such that xi ∈ V (λi ), that is, Ti ( xi ) = λi xi . Define αi := λi−1 for every i ∈ {1, . . . , k}. Then k ∑ αi Ti i =1 ! k ∑ β i xi i =1 ! k = ∑ αi βi Ti (xi ) = i =1 If H1 , . . . , H p are Hilbert spaces, then following scalar product and norm  p p ( h i ) i =1 | ( k i ) i =1 : = p for all (hi )i=1 , (k i )i=1 ∈ L p i =1 p i =1 Hi  p  p L Hi  2 ∑ ( h i | k i ), i =1 k ∑ αi β i λi xi = i =1 2 k ∑ β i xi . i =1 is a Hilbert space, considering the p ( h i ) i =1 v u p u : = t ∑ k h i k2 , i =1 . If H is another Hilbert space and Ti : H → Hi is a continuous linear operator for each i = 1, . . . , p, then the direct sum of T1 , . . . , Tp is defined as L  L  p p T H : H → i i =1 i 2  L i =1  2 p p x 7→ i =1 Ti ( x ) : = ( Ti ( x ))i =1 . 2 If Si : Hi → H is a continuous linear operator for each i = 1, . . . , p, then the direct sum of S1 , . . . , S p is now defined as L p i =1 Si  2 : L p i =1 Hi  2 p ( h i ) i =1 → 7→ H L p i =1 Si   2  p p ( h i ) i = 1 : = ∑ i = 1 Si ( h i ) . Symmetry 2021, 13, 661 9 of 17 Theorem 5. Suppose that H, H1 , . . . , H p are Hilbert spaces, and let Ti : H → Hi be a continuous L  L ′ p p ′ T and T = linear operator for each i = 1, . . . , p. Then, i =1 i i =1 i 2 2 p M i =1 Ti !′ 2 p M ◦ Ti i =1 ! p i =1 2 p Proof. Fix arbitrary elements x ∈ H and (hi )i=1 ∈ x p M i =1 Ti′ ! 2  p ( h i ) i =1  ! ∑ Ti′ ◦ Ti . = L p i =1 p = ∑ x i =1 p  = ∑ = i =1 p  ∑ i =1 = =  Hi  2 Ti′ (hi ) x Ti′ (hi ) Ti ( x ) hi p . Then, !   p ( Ti ( x ))i=1 (hi )i=1 ! p M (x) Ti i =1  p ( h i ) i =1 2 ! . Lastly, for each x ∈ H, p M i =1 Ti !′ 2 ◦ p M i =1 Ti ! ! (x) = p M i =1 p 2 = M i =1 Ti !′ Ti′ ! 2  Ti ! p (x) 2 ( Ti ( x ))i=1 !  ∑ Ti′ (Ti (x)) i =1 p = i =1 2 p = p M ∑ i =1
Ti′ ◦ Ti ( x ). 3.2. Pareto Optimal Solutions of the MOP max k T ( x )k, min k x k, x ∈ X Under the settings of Theorem 3, arg minx∈ X k x k = {0}; therefore, in view of Theorem 1, 0 ∈ Pos(3). This Pareto optimal solution is usually disregarded when it comes to a real-life problem. Theorem 6. Let X, Y be normed spaces, and T : X → Y be a nonzero continuous linear operator. Then, Pos(3) = Rsuppv( T ). Proof. Fix an arbitrary x0 ∈ Pos(3). Since x0 = k x0 k k xx0 k , it is sufficient if we show that x0 k x0 k 0 ∈ suppv( T ). Therefore, we may assume that k x0 k = 1, so our aim was summed up to prove that k T ( x0 )k = k T k. Since x0 ∈ SX ⊆ BX , k T ( x0 )k ≤ k T k. Suppose that k T ( x0 )k < k T k. By the definition of sup,there exists y ∈ BX , such that k T ( x0 )k < k T (y)k ≤ k T k. kyk ≤ 1 = k x0 k and k T ( x0 )k < k T (y)k, which contradicts that x0 ∈ Pos(3). As a consequence, k T ( x0 )k = k T k; hence, x0 ∈ suppv( T ). The arbitrariness of x0 ∈ Pos(3) shows that Pos(3) ⊆ Rsuppv( T ). Conversely, fix an arbitrary x0 ∈ Rsuppv( T ). There exists y0 ∈ suppv( T ) and α ∈ R, such that x0 = αy0 . Observe that k x0 k = |α|ky0 k = |α|. We Symmetry 2021, 13, 661 10 of 17 prove that x0 ∈ Pos(3). Let us consider an element y ∈ X satisfying that kyk < k x0 k = |α|, and we distinguish cases: if y = 0, then k T ( x0 )k = |α|k T (y0 )k = |α|k T k > 0 = k T (y)k. If y 6= 0, then   y k T ( x0 )k = |α|k T (y0 )k = |α|k T k ≥ |α| T > k T (y)k. kyk Lastly, if there exists y ∈ X, such that k T (y)k > k T ( x0 )k, then |α|k T k = k T ( x0 )k < k T (y)k ≤ k T kkyk, which means that k x0 k = |α| < kyk. When X, Y are Hilbert spaces, the Pareto optimal solutions of (3) are directly obtained via combining Theorems 4 and 6. Corollary 1. Let T : H → K be a continuous linear operator with H, K Hilbert spaces. Then, Pos(3) = V (λmax ( T ′ ◦ T )). This last result allows for solving the following MOP (motivated in Section 4), given by   max k T1 ( x )k2 + · · · + k Tk ( x )k2 , (11) min k x k,  x ∈ H. The Pareto optimal solutions of (11) are related to those of   max k Ti ( x )k, min k x k,  x ∈ H, (12) for i = 1, . . . , k. Corollary 2. If T1 , . . . , Tk ∈ B( H, K ) are continuous linear operators between Hilbert spaces H and K, then:    1. Pos(11) = V λmax ∑ik=1 Ti′ ◦ Ti . 2. Tk i =1 Pos(12) ⊆ Pos(11). Proof. Consider bounded linear operator k T1 ⊕2 · · · ⊕2 Tk : k →  K ⊕2 · · · ⊕2 K  k x 7→ T1 ⊕2 · · · ⊕2 Tk ( x ) := ( T1 ( x ), . . . , Tk ( x )). H The next equality trivially holds for every x ∈ H, k( T1 ( x ), . . . , Tk ( x ))k2 = k T1 ( x )k2 + · · · + k Tk ( x )k2 . Since the square root is strictly increasing, (11) is equivalent to    k    max T1 ⊕2 · · · ⊕2 Tk ( x ) , min k x k,    x ∈ H, which is an MOP of form (3). (13) Symmetry 2021, 13, 661 11 of 17 1. According to Corollary 1 and Theorem 5, = Pos(13) Pos(11)  = V λmax k T1 ⊕2 · · · ⊕2 Tk k ∑ = V λmax i =1 2. Ti′ ◦ Ti !! ′  k ◦ T1 ⊕2 · · · ⊕2 Tk !! . We rely on Theorem 6 and Corollary 1. Fix an arbitrary x ∈ ik=1 Pos(12). If x = 0, then   k x ∈ Rsuppv T1 ⊕2 · · · ⊕2 Tk = Pos(13) = Pos(11). Suppose that x 6= 0. In view   T k of Theorem 6, k xxk ∈ ik=1 suppv( Ti ). We prove that k xxk ∈ suppv T1 ⊕2 · · · ⊕2 Tk . T Take any y ∈ B H . Since x kxk ∈ Tk i =1 suppv( Ti ), Ti  x kxk  for every i ∈ {1, . . . , k}, ≥ k Ti (y)k. As a consequence,  k T1 ⊕2 · · · ⊕2 Tk  x kxk  = s ≥ q = This means that x kxk x = kxk T1  x kxk  2 + · · · + Tk  k T1 (y)k2 + · · · + k Tk (y)k2   k T1 ⊕2 · · · ⊕2 Tk (y) . x kxk  2   k ∈ suppv T1 ⊕2 · · · ⊕2 Tk . In accordance with Theorem 6,   x k ∈ Rsuppv T1 ⊕2 · · · ⊕2 Tk = Pos(13) = Pos(11). kxk 4. Discussion In order to design truly optimal TMS coils, and depending on the nature and characteristics of the coil that we want to maximize or minimize, a linearization technique is applied to the electromagnetic field [18,23–25]; then, MOPs like (3) come out:     max k Ex ψk2 ,  max k Ey ψk2 ,  max k Ez ψk2 , (14) min ψ T Lψ, min ψ T Lψ, min ψ T Lψ,    n n ψ∈R , ψ∈R , ψ ∈ Rn , where E is a matrix representing the electromagnetic field, Ex , Ey , Ez are the components of E, and L represents inductance with a positive definite symmetric matrix. Using Cholesky decomposition, as L is positive definite and symmetric, the existence of an invertible matrix C, such that L = C T C, is guaranteed. Then, ψ T Lψ = ψ T C T Cψ = (Cψ) T (Cψ) = kCψk22 . Symmetry 2021, 13, 661 12 of 17 Next, we apply the following change of variables: ϕ := Cψ. Then, the previous problems can be rewritten as follows:
  max Ex C −1 ϕ min k ϕk22 ,  ϕ ∈ Rn , 2 ,
  max Ey C −1 ϕ min k ϕk22 ,  ϕ ∈ Rn , 2
  max Ez C −1 ϕ min k ϕk22 ,  ϕ ∈ Rn . , 2 , (15) Since the square root is strictly increasing, the previous MOPs are equivalent to the following (in the sense that they have the same set of global solutions and the same set of Pareto optimal solutions):
  max Ex C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn , 2 ,
  max Ey C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn , 2
  max Ez C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn . , 2 , (16) The three MOPs above are of the form (3). Therefore, in view of Corollary 1, the Pareto optimal solutions of each of them is determined by  
T
 V λmax Ex C −1 Ex C −1 ,   

T V λmax Ey C −1 Ey C −1 ,  

 T − 1 − 1 V λmax Ez C Ez C , respectively. On the other hand, we can consider the combined MOP, as in (11): 
2
2
2   max Ex C −1 ϕ 2 + Ey C −1 ϕ 2 + Ez C −1 ϕ 2 , min k ϕk2 ,   ϕ ∈ Rn . (17) Let us define the following linear operator: T: ℓ2n → ℓ2n ⊕2 ℓ2n ⊕2 ℓ2n



ϕ 7→ T ( ϕ) = Ex C −1 ϕ, Ey C −1 ϕ, Ez C −1 ϕ . The corresponding matrix to T is precisely  Ex C −1 A :=  Ey C −1 . Ez C −1  For every ϕ ∈ Rn , k T ( ϕ)k2 =   Ex C −1  ϕ 2 2 +  Ey C −1  ϕ 2 2 +  Ez C −1  ϕ 2 2 1 2 . Then (17) is the same as According to Corollary 1,   max k T ( ϕ)k2 , min k ϕk2 ,  ϕ ∈ Rn .   Pos(18) = Rsuppv( T ) = V λmax ( A T A) . (18) Symmetry 2021, 13, 661 13 of 17 Equivalently, according to Corollary 2, = Pos(18) = Pos(17)   T    T    T   −1 −1 −1 −1 −1 −1 V λmax Ex C Ex C + Ey C Ey C + Ez C . Ez C A very illustrative example of this situation is displayed in the Appendix A 5. Conclusions This section deals with linear combinations of MOPs of the form given in (3). Let H, K be Hilbert spaces. Consider continuous linear operators T1 , . . . , Tk ∈ B( H, K ) between H and K. Let α1 , . . . , αk > 0. In bioengineering, it is common to assign weights αi to different operators Ti depending on the relevance of each Ti . Then, the following MOP comes into play:   max α1 k T1 ( x )k2 + · · · + αk k Tk ( x )k2 , (19) min k x k,  x ∈ H, Nevertheless, the above MOP is, in fact, the same as the following:   max kS1 ( x )k2 + · · · + kSk ( x )k2 , min k x k,  x ∈ H, (20)
√ √ where Si := αi Ti for each i = 1, . . . , k. suppv( Ti ) = suppv αi Ti = suppv(Si ) for every i = 1, . . . , k. By relying on Corollary 2, at least we can ensure that k \ i =1 Pos(12) ⊆ Pos(20) = Pos(19). However, it is very unlikely that ik=1 Pos(12) 6= {0}. Unless hypotheses similar to the ones employed in Lemma 2 or Lemma 3 are used, we cannot conclude any other relation between Pos(19) and Pos(12). T Author Contributions: Conceptualization, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; methodology, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; formal analysis, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; investigation, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; writing—original-draft preparation, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; writing—review and editing, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; visualization, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; supervision, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; project administration, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P.; funding acquisition, C.C.-S., J.A.V.-M., A.C.-J. and F.J.G.-P. All authors have read and agreed to the published version of the manuscript. Funding: This research was funded by Ministry of Science, Innovation, and Universities of Spain, grant number PGC-101514-B-I00; and by the 2014-2020 ERDF Operational Programme and by the Department of Economy, Knowledge, Business, and University of the Regional Government of Andalusia, grant number FEDER-UCA18-105867. The APC was funded by FEDER-UCA18-105867. Acknowledgments: The authors would like to thank the reviewers for valuable comments and remarks that helped to improve the presentation and quality of the manuscript. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conflicts of Interest: The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results. Symmetry 2021, 13, 661 14 of 17 Abbreviations The following abbreviations are used in this manuscript: MDPI DOAJ MOP SOP POS PC TMS MRI ROI Multidisciplinary Digital Publishing Institute Directory of open access journals Multiobjective optimization problem Single-objective optimization problem Pareto optimal solution Pareto chart Transcranial magnetic stimulation Magnetic resonance imaging Region of interest Appendix A. Illustrative Example on Coil Design Appendix A.1. Coil Design in Engineering The use of coils that optimize one or more components of an electromagnetic field while minimizing power dissipation or stored magnetic energy is often required in bioengineering applications such as TMS [21,23] and magnetic resonance imaging (MRI) [24,25] or in high-precision magnetic measurement systems in space missions such as eLISA [26–29]. All these applications are characterized by the need of generating a prescribed and localized electromagnetic field in a specific region, and are subject to other performance requirements such as the minimization of stored magnetic energy or dissipated power. Therefore, the design of electromagnetic coils for these applications can be considered to be an MOP. MOPs from coil design are frequently expressed as a convex optimization and formulated in terms of the stream function of a quasistatic current [18]. Appendix A.2. Design of Maximal Bx and By Coil for Magnetic Measurement Systems in a Space Missions In the following, for the purpose of illustrating an application of the obtained theoretical results in this manuscript, we present the design of a planar coil over a (34 × 17 mm) PCBfor magnetic measurement systems in space missions. This coil was constructed with the aim of maximizing the magnetic field in a small and near region where a magnetic sensor capable of measuring the X and Y components of the B field is located. At the same time, resistance was minimized in order to avoid power dissipation. Hence, the initial requirement that the coil had to satisfy was that it had to produce a maximal magnetic field in a region of interest (ROI) that was located in the same position as that of the sensor (x = 22 mm, y = 6.5 mm, z = 1.25 mm), with its same dimensions (5.8 × 3.5 mm), formed by 200 points. Figure A1 illustrates the available surface for the coil design along with the ROI. In order to obtain stream function ϕ, which simultaneously maximizes Bx and By while minimizing power dissipation at the ROI, previously presented MOPs (16) and (17) were applied. Consequently, the current coil-design problem can be expressed as the following MOPs:
  max Bx C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn , 2 ,
  max By C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn , 
  max α Bx C −1 ϕ min k ϕk2 ,   ϕ ∈ Rn . 2 2 +β 2 ,
By C −1 ϕ 2 2
  max Bz C −1 ϕ min k ϕk2 ,  ϕ ∈ Rn . +γ
Bz C −1 ϕ 2 , 2 2 , (A1) (A2) where Bi ∈ Rm×n stands for the matrix of the magnetic field in the i-th direction (i = x, y, z); R ∈ Rn×n is the resistance matrix; n is the number of mesh points (n = 2000); m is the Symmetry 2021, 13, 661 15 of 17 number of ROI points (m = 200); and α, β, and γ are constants that provide specific weights for maximizing each component of the field (Bx , By , Bz ). Due to the fact that it is only necessary to maximize the Bx and By components in the current case, weights were chosen such that α = β < γ (in concrete α = β = 1 and γ = 10−2 ). 1.5 ROI z [mm] 1 0.5 0 16 14 rface 12 Su Coil 10 30 8 25 6 20 15 4 y [mm] 10 2 5 0 x [mm] 0 Figure A1. Representation of planar coil surface and region of interest (ROI) where optimal stream function ϕ is calculated. Figure A2 shows the stream function solution from the (A1) and (A2) MOPs (red and blue functions, respectively) computed by using the theoretical model developed in [4,18,19]. Three different optimal stream functions were obtained from (A1) MOP (ϕ x , ϕy , ϕz ). Consequently, the final ϕ1 solution was calculated as linear combination ϕ1 = αϕ x + βϕy + γϕz . However, stream function ϕ2 is the final solution obtained from the (A2) MOP. As expected from the conclusions of the manuscript, the stream functions were not equal. 1 Stream Function Coil 1 Stream Function Coil 2 0.9 0.8 0.7 Norm 0.6 0.5 0.4 0.3 0.2 0.1 0 0 200 400 600 800 1000 Mesh Points 1200 1400 1600 1800 2000 Figure A2. Stream functions obtained from Problems (A1) (Coil 1) and (A2) (Coil 2). Furthermore, stream function contours over the coil surface can be considered to be the current wire path [30,31]. Accordingly, coil wires were designed as the stream function contour, as is depicted in Figures A3 and A4, where the designed coils are different depending on the MOP. Symmetry 2021, 13, 661 16 of 17 y [mm] 15 10 5 0 0 5 10 15 20 x [mm] 25 30 Figure A3. 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