Pareto Optimality for Multioptimization of Continuous Linear Operators

Abstract

This manuscript determines the set of Pareto optimal solutions of certain multiobjective-optimization 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 $max\parallel T\left(x\right)\parallel$, $min\parallel x\parallel$ 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.

1. Introduction

Multiobjective optimization problems (MOPs) appear quite often in all areas of pure and applied mathematics, for instance, in the geometry of Banach spaces [1,2,3], in operator theory [4,5,6,7], in lineability theory [8,9,10], in differential geometry [11,12,13,14], and in all areas of Experimental, Medical and Social Sciences [15,16,17,18,19,20]. 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 [18,19,20,21,22,23]. 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 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:

$$M:=\left\{\begin{array}{c}max{f}_i\left(x\right)i=1,\dots ,p,\hfill \\ min{g}_j\left(x\right)j=1,\dots ,q,\hfill \\ x\in \mathcal{R},\hfill \end{array}\right.$$

(1)

where $f_i,g_j:X\to \mathbb{R}$ are called objective functions, defined on a nonempty set X, and $\mathcal{R}$ 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 $\mathrm{sol}\left(M\right)$. In fact,

$$\mathrm{sol}\left(M\right):=\{x\in \mathcal{R}:\forall y\in \mathcal{R}\forall i\in \{1,\dots ,p\}\forall j\in \{1,\dots ,q\}{f}_i\left(x\right)\ge f_i\left(y\right),g_j\left(x\right)\le g_j\left(y\right)\}.$$

It is obvious that

$$\mathrm{sol}\left(M\right)=\mathrm{sol}\left(P_1\right)\cap \cdots \cap \mathrm{sol}\left(P_p\right)\cap \mathrm{sol}\left(Q_1\right)\cap \cdots \cap \mathrm{sol}\left(Q_q\right)$$

(2)

where

$$P_i:=\left\{\begin{array}{c}max{f}_i\left(x\right),\hfill \\ x\in \mathcal{R},\hfill \end{array}\right.\mathrm{and}{Q}_j:=\left\{\begin{array}{c}min{g}_j\left(x\right),\hfill \\ x\in \mathcal{R},\hfill \end{array}\right.$$

are single-objective optimization problems (SOPs) and $\mathrm{sol}\left(P_i\right)$, $\mathrm{sol}\left(Q_j\right)$ denote the set of general solutions of $P_i,Q_j$ for $i=1,\dots ,p$ and $j=1,\dots ,q$, respectively. The set of Pareto optimal solutions of MOP M is defined as

$$\begin{array}{ccc}\hfill \mathrm{Pos}\left(M\right):=\{x\in \mathcal{R}& :& \forall y\in \mathcal{R},\mathrm{if}\mathrm{there}\mathrm{exists}i\in \{1,\dots ,p\}\mathrm{with}{f}_i\left(y\right)>f_i\left(x\right)\hfill \\ & & \mathrm{or}j\in \{1,\dots ,q\}\mathrm{with}{g}_j\left(y\right)<g_j\left(x\right),\mathrm{then}\mathrm{there}\mathrm{exists}\hfill \\ & & i^{\prime }\in \{1,\dots ,p\}\mathrm{with}{f}_{i^{\prime }}\left(y\right)<f_{i^{\prime }}\left(x\right)\mathrm{or}{j}^{\prime }\in \{1,\dots ,q\}\hfill \\ & & \mathrm{with}{g}_{j^{\prime }}\left(x\right)<g_{j^{\prime }}\left(y\right)\}.\hfill \end{array}$$

To guarantee the existence of general solutions, it is usually asked for X to be a Hausdorff topological space, $\mathcal{R}$ 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 $P_i$s and $Q_j$s have at least one solution (Weierstrass extreme value theorem). Even more, solution sets $\mathrm{sol}\left(P_i\right)$ and $\mathrm{sol}\left(Q_j\right)$ are closed and thus compact, which makes $\mathrm{sol}\left(M\right)$ also compact. Nevertheless, even under these conditions, $\mathrm{sol}\left(M\right)$ 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, $f_i,g_j:X\to \mathbb{R}$ functions and $\mathcal{R}$ a nonempty subset of X. In $\mathcal{R}$, consider the equivalence relation given by

$$\mathcal{S}:=\left\{(x,y)\in {\mathcal{R}}^2:\forall i=1,\dots ,p\forall j=1,\dots ,q{f}_i\left(x\right)=f_i\left(y\right),g_j\left(x\right)=g_j\left(y\right)\right\}.$$

Next, in the quotient set of $\mathcal{R}$ by $\mathcal{S}$, $\frac{\mathcal{R}}{\mathcal{S}}$, consider the order relation given by

$${\left[x\right]}_{\mathcal{S}}\le {\left[y\right]}_{\mathcal{S}}\iff \forall i=1,\dots ,p\forall j=1,\dots ,q{f}_i\left(x\right)\le f_i\left(y\right),g_j\left(y\right)\le g_j\left(x\right).$$

Theorem 1. 

Consider MOP (1). Then,

$$\mathrm{Pos}\left(M\right)=\left\{x\in \mathcal{R}:{\left[x\right]}_{\mathcal{S}}isamaximalelementof\frac{\mathcal{R}}{\mathcal{S}}endowedwith\le \right\}$$

and

$$\mathrm{sol}\left(M\right):=\left\{x\in \mathcal{R}:{\left[x\right]}_{\mathcal{S}}isthemaximumof\frac{\mathcal{R}}{\mathcal{S}}endowedwith\le \right\}.$$

As a consequence, $\mathrm{sol}\left(M\right)\subseteq \mathrm{Pos}\left(M\right)$. If there exists $i_1\in \{1,\dots ,p\}$ or $j_1\in \{1,\dots ,q\}$ such that $\mathrm{sol}\left(P_{i_1}\right)$ or $\mathrm{sol}\left(Q_{j_1}\right)$ is a singleton, respectively, then $\mathrm{sol}\left(P_{i_1}\right)\subseteq \mathrm{Pos}\left(M\right)$ or $\mathrm{sol}\left(Q_{j_1}\right)\subseteq \mathrm{Pos}\left(M\right)$, respectively.

Proof. 

Fix an arbitrary $x_0\in \mathrm{Pos}\left(M\right)$. Let us assume that there is $y\in \mathcal{R}$, so that ${\left[x_0\right]}_{\mathcal{S}}<{\left[y\right]}_{\mathcal{S}}$. Then, $f_i\left(x_0\right)\le f_i\left(y\right)$ for all $i=1,\dots ,p$ and $g_j\left(x_0\right)\ge g_j\left(y\right)$ for all $j=1,\dots ,q$. However, ${\left[x_0\right]}_{\mathcal{S}}\ne {\left[y\right]}_{\mathcal{S}}$; therefore, there exists $i_0\in \{1,\dots ,p\}$ or $j_0\in \{1,\dots ,q\}$ such that $f_{i_0}\left(x_0\right)<f_{i_0}\left(y\right)$ or $g_{j_0}\left(x_0\right)<g_{j_0}\left(y\right)$, respectively. Since $x_0\in \mathrm{Pos}\left(M\right)$ by assumption, there exists $i_1\in \{1,\dots ,p\}$ or $j_1\in \{1,\dots ,q\}$, such that $f_{i_1}\left(x_0\right)>f_{i_1}\left(y\right)$ or $g_{j_1}\left(x_0\right)<g_{j_1}\left(y\right)$, respectively, which is a contradiction. Therefore, ${\left[x_0\right]}_{\mathcal{S}}$ is a maximal element of $\frac{\mathcal{R}}{\mathcal{S}}$ endowed with ≤. The arbitrariness of $x_0\in \mathrm{Pos}\left(M\right)$ shows that

$$\mathrm{Pos}\left(M\right)\subseteq \left\{x\in \mathcal{R}:{\left[x\right]}_{\mathcal{S}}is\mathrm{a}\mathrm{maximal}\mathrm{element}\mathrm{of}\frac{\mathcal{R}}{\mathcal{S}}\mathrm{endowed}\mathrm{with}\le \right\}.$$

Conversely, fix an arbitrary $x_0\in \mathcal{R}$, such that ${\left[x_0\right]}_{\mathcal{S}}$ is a maximal element of $\frac{\mathcal{R}}{\mathcal{S}}$ endowed with ≤. Take $y\in \mathcal{R}$ satisfying that there exists $i_0\in \{1,\dots ,p\}$ or $j_0\in \{1,\dots ,q\}$ with $f_{i_0}\left(y\right)>f_{i_0}\left(x_0\right)$ or $g_{j_0}\left(y\right)<g_{j_0}\left(x_0\right)$, respectively. If $f_i\left(x_0\right)\le f_i\left(y\right)$ for all $i\in \{1,\dots ,p\}\setminus \left\{i_0\right\}$ and $g_j\left(x_0\right)\ge g_j\left(y\right)$ for all $j\in \{1,\dots ,q\}\setminus \left\{j_0\right\}$, then ${\left[x_0\right]}_{\mathcal{S}}<{\left[y\right]}_{\mathcal{S}}$, reaching a contradiction with the maximality of ${\left[x_0\right]}_{\mathcal{S}}$ in $\frac{\mathcal{R}}{\mathcal{S}}$ endowed with ≤. This shows that

$$\mathrm{Pos}\left(M\right)=\left\{x\in \mathcal{R}:{\left[x\right]}_{\mathcal{S}}\mathrm{is}\mathrm{a}\mathrm{maximal}\mathrm{element}\mathrm{of}\frac{\mathcal{R}}{\mathcal{S}}\mathrm{endowed}\mathrm{with}\le \right\}.$$

Next, fix an arbitrary $x_0\in \mathrm{sol}\left(M\right)$. For every $y\in \mathcal{R}$, $f_i\left(x_0\right)\ge f_i\left(y\right)$ and $g_j\left(x_0\right)\le g_j\left(y\right)$ for all $i=1,\dots ,p$ and all $j=1,\dots ,q$. Then, ${\left[x_0\right]}_{\mathcal{S}}\ge {\left[y\right]}_{\mathcal{S}}$. The arbitrariness of $y\in \mathcal{R}$ ensures that ${\left[x_0\right]}_{\mathcal{S}}$ is a maximal element of $\frac{\mathcal{R}}{\mathcal{S}}$ endowed with ≤. Conversely, fix an arbitrary $x_0\in \mathcal{R}$, such that ${\left[x_0\right]}_{\mathcal{S}}$ is a maximal element of $\frac{\mathcal{R}}{\mathcal{S}}$ endowed with ≤. For every $y\in \mathcal{R}$, ${\left[y\right]}_{\mathcal{S}}\ge {\left[x_0\right]}_{\mathcal{S}}$; therefore, $f_i\left(x_0\right)\ge f_i\left(y\right)$ and $g_j\left(x_0\right)\le g_j\left(y\right)$ for all $i=1,\dots ,p$ and all $j=1,\dots ,q$. The arbitrariness of $y\in \mathcal{R}$ proves that $x_0\in \mathrm{sol}\left(M\right)$. We proved that

$$\mathrm{sol}\left(M\right):=\left\{x\in \mathcal{R}:{\left[x\right]}_{\mathcal{S}}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{of}\frac{\mathcal{R}}{\mathcal{S}}\mathrm{endowed}\mathrm{with}\le \right\}.$$

Lastly, suppose that $\mathrm{sol}\left(P_{i_1}\right)$ is a singleton for some $i_1\in \{1,\dots ,p\}$, and write $\mathrm{sol}\left(P_{i_1}\right)=\left\{x_0\right\}$. Take $y\in \mathcal{R}$ satisfying that there exists $i_0\in \{1,\dots ,p\}$ or $j_0\in \{1,\dots ,q\}$ with $f_{i_0}\left(y\right)>f_{i_0}\left(x_0\right)$ or $g_{j_0}\left(y\right)<g_{j_0}\left(x_0\right)$, respectively. If such $i_0$ exists, then $i_0\ne i_1$. By hypothesis, $f_{i_1}\left(x_0\right)>f_{i_1}\left(y\right)$ since $y\notin \mathrm{sol}\left(P_{i_1}\right)$. This shows that $x_0\in \mathrm{Pos}\left(M\right)$. Likewise, $\mathrm{sol}\left(Q_{j_1}\right)\subseteq \mathrm{Pos}\left(M\right)$ provides that $\mathrm{sol}\left(Q_{j_1}\right)$ is a singleton. □

Lemma 1. 

Consider MOP (1). Let $i_0\in \{1,\dots ,p\},j_0\in \{1,\dots ,q\}$. Then,

1. 

If there is $x_{i_0}\in \mathcal{R}$ so that ${\left[x_{i_0}\right]}_{\mathcal{S}}$ is a maximal element of $\left\{{\left[x\right]}_{\mathcal{S}}:x\in \arg {max}_{\mathcal{R}}{f}_{i_0}\right\}$, then ${\left[x_{i_0}\right]}_{\mathcal{S}}$ is a maximal element of $\mathcal{R}/\mathcal{S}$. Hence, $x_{i_0}\in \mathrm{Pos}\left(M\right)$.

2. 

If there is $x_{j_0}\in \mathcal{R}$ so that ${\left[x_{j_0}\right]}_{\mathcal{S}}$ is a maximal element of $\left\{{\left[x\right]}_{\mathcal{S}}:x\in \arg {min}_{\mathcal{R}}{g}_{j_0}\right\}$, then ${\left[x_{j_0}\right]}_{\mathcal{S}}$ is a maximal element of $\mathcal{R}/\mathcal{S}$. Hence, $x_{j_0}\in \mathrm{Pos}\left(M\right)$

Proof. 

We only prove the first item since the other follows a dual proof. Assume that ${\left[x_{i_0}\right]}_{\mathcal{S}}$ is not a maximal element of $\mathcal{R}/\mathcal{S}$. Then, we can find $y\in \mathcal{R}$ in such a way that ${\left[x_{i_0}\right]}_{\mathcal{S}}<{\left[y\right]}_{\mathcal{S}}$. In particular, $f_{i_0}\left(x_{i_0}\right)\le f_{i_0}\left(y\right)$; therefore, $f_{i_0}\left(y\right)={max}_{\mathcal{R}}{f}_{i_0}$; hence, $y\in \arg {max}_{\mathcal{R}}{f}_{i_0}$. As a consequence, ${\left[y\right]}_{\mathcal{S}}\in \left\{{\left[x\right]}_{\mathcal{S}}:x\in \arg {max}_{\mathcal{R}}{f}_{i_0}\right\}$, contradicting that ${\left[x_{i_0}\right]}_{\mathcal{S}}$ be a maximal element of $\left\{{\left[x\right]}_{\mathcal{S}}:x\in \arg {max}_{\mathcal{R}}{f}_{i_0}\right\}$. □

Theorem 2. 

Consider MOP (1). If X is a topological space, $\mathcal{R}$ is a compact Hausdorff subset of X and all the objective functions are continuous, then $\mathrm{Pos}\left(M\right)\ne ⌀$.

Proof. 

Fix $i_0\in \{1,\dots ,p\}$. In accordance with Lemma 1, it is only sufficient to find a maximal element of $A:=\left\{{\left[x\right]}_{\mathcal{S}}:x\in \arg {max}_{\mathcal{R}}{f}_{i_0}\right\}$. We rely on Zorn’s lemma. Consider a chain in A, that is, a totally ordered subset of elements ${\left[x_k\right]}_{\mathcal{S}}$, with k ranging a totally ordered set K in such a way that $k_1<k_2$ if and only if ${\left[x_{k_1}\right]}_{\mathcal{S}}<{\left[x_{k_2}\right]}_{\mathcal{S}}$. Since K is totally ordered, we have that ${\left(x_k\right)}_{k\in K}$ is a net in $\mathcal{R}$. The compactness of $\mathcal{R}$ allows for extracting a subnet ${\left(y_h\right)}_{h\in H}$ of ${\left(x_k\right)}_{k\in K}$ convergent to some $x_0\in \mathcal{R}$. Let us first show that $x_0\in \arg {max}_{\mathcal{R}}{f}_{i_0}$. The continuity of $f_{i_0}$ implies that ${\left(f_{i_0}\left(y_h\right)\right)}_{h\in H}$ converges to $f_{i_0}\left(x_0\right)$. Fix any $\epsilon >0$. There is $h_{\epsilon }\in H$ satisfying that, if $h\ge h_{\epsilon }$, then $\left|f_{i_0}\left(y_h\right)-f_{i_0}\left(x_0\right)\right|<\epsilon$. Fix any $k_0\in K$. There is $h_0\in H$, so that $\left\{y_h:h\ge h_0\right\}\subseteq \{x_k:k\ge k_0\}$. Since H is a directed set, we can find $h_1\in H$ with $h_1\ge h_{\epsilon }$ and $h_1\ge h_0$. There exists $k_1\in K$ with $k_1\ge k_0$ such that $y_{h_1}=x_{k_1}$. Next, $f_{i_0}\left(y_{h_1}\right)=f_{i_0}\left(x_{k_1}\right)={max}_{\mathcal{R}}{f}_{i_0}$. As a consequence,

$$\underset{\mathcal{R}}{max}{f}_{i_0}-f\left(x_0\right)=f_{i_0}\left(y_{h_1}\right)-f\left(x_0\right)<\epsilon .$$

The arbitrariness of $\epsilon$ shows that ${max}_{\mathcal{R}}{f}_{i_0}=f\left(x_0\right)$. Lastly, we prove that ${\left[x_0\right]}_{\mathcal{S}}$ is an upper bound for chain $\left\{{\left[x_k\right]}_{\mathcal{S}}:k\in K\right\}$. Fix an arbitrary $k_0\in K$. In order to prove that ${\left[x_{k_0}\right]}_{\mathcal{S}}\le {\left[x_0\right]}_{\mathcal{S}}$, we have to check that $f_i\left(x_{k_0}\right)\le f_i\left(x_0\right)$ for all $i\in \{1,\dots ,p\}$ and $g_j\left(x_{k_0}\right)\ge g_j\left(x_0\right)$ for all $j\in \{1,\dots ,q\}$. Indeed, fix $i\in \{1,\dots ,p\}$ and suppose to the contrary that $f_i\left(x_{k_0}\right)>f_i\left(x_0\right)$. Let $0<\epsilon <f_i\left(x_{k_0}\right)-f_i\left(x_0\right)$. There exists $h_{\epsilon }\in H$ such that, if $h\ge h_{\epsilon }$, then $\left|f_i\left(y_h\right)-f_i\left(x_0\right)\right|<\epsilon$. We can find $h_0\in H$, such that $\left\{y_h:h\ge h_0\right\}\subseteq \{x_k:k\ge k_0\}$. Since H is a directed set, we can find $h_1\in H$ with $h_1\ge h_{\epsilon }$ and $h_1\ge h_0$. There exists $k_1\in K$ with $k_1\ge k_0$ such that $y_{h_1}=x_{k_1}$. Since $k_0\le k_1$, we have that ${\left[x_{k_0}\right]}_{\mathcal{S}}\le {\left[x_{k_1}\right]}_{\mathcal{S}}$. Thus,

$$f_i\left(y_{h_1}\right)=f_i\left(x_{k_1}\right)\ge f_i\left(x_{k_0}\right)>f_i\left(x_0\right),$$

contradicting that $\left|f_i\left(y_{h_1}\right)-f_i\left(x_0\right)\right|<\epsilon$. In a similar way, it can be shown that $g_j\left(x_{k_0}\right)\ge g_j\left(x_0\right)$ for all $j\in \{1,\dots ,q\}$. As a consequence, ${\left[x_{k_0}\right]}_{\mathcal{S}}\le {\left[x_0\right]}_{\mathcal{S}}$. In other words, ${\left[x_0\right]}_{\mathcal{S}}$ is an upper bound for the chain $\left\{{\left[x_k\right]}_{\mathcal{S}}:k\in K\right\}$. 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, $\mathrm{sol}\left(M\right)=\varnothing$. This happens quite often with MOPs involving matrices. Even if the number of objective functions is short, we might still have $\mathrm{sol}\left(M\right)=\varnothing$. The following theorem [20], Theorem 2, is a very representative example of this situation of lack of general solutions.

Theorem 3. 

Let $T:X\to Y$ be a nonzero continuous linear operator, where $X,Y$ are normed spaces; then, the following max–min problem is free of general solutions:

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel ,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in X.\hfill \end{array}\right.$$

(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 $\mathrm{Pos}(3)$ (Theorem 6 and Corollary 1). If $X,Y$ are Hilbert spaces, say $H,K$, and $T_1,\dots ,T_k\in \mathcal{B}(H,K)$ are continuous linear operators, then the sets of Pareto optimal solutions of the MOPs

$$\left\{\begin{array}{c}max\parallel T_i\left(x\right)\parallel ,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(4)

for $i=1,\dots ,k$ are compared (Corollary 2) with the set of Pareto optimal solutions of MOP

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel ,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(5)

where

$$\begin{array}{cccc}\hfill T:& \hfill H& \to & K{\oplus }_2\stackrel{k}{\dots }{\oplus }_{2}K\hfill \\ & \hfill x& \mapsto & T\left(x\right)=\left(T_1\left(x\right),\dots ,T_k\left(x\right)\right).\hfill \end{array}$$

3.1. Formatting of Mathematical Components

Let $X,Y$ be normed spaces. Consider a nonzero continuous linear operator $T:X\to Y$. Then

$$\parallel T\parallel :=sup\{\parallel T\left(x\right)\parallel :x\in {\mathsf{B}}_X\}$$

is the norm of T. On the other hand,

$$\mathrm{suppv}\left(T\right):=\{x\in {\mathsf{S}}_X:\parallel T\left(x\right)\parallel =\parallel T\parallel \},$$

stands for the set of supporting vectors of T, where ${\mathsf{B}}_X:=\{x\in X:\parallel x\parallel \le 1\}$ is a (closed) unit ball, and ${\mathsf{S}}_X:=\{x\in X:\parallel x\parallel =1\}$ 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 $\mathcal{B}(X,Y)$.

Let H be a Hilbert space, and consider the dual map of H:

$$\begin{array}{cccc}\hfill J_H:& \hfill H& \to & H^{*}\hfill \\ & \hfill k& \mapsto & J_H\left(k\right):=k^{*}=(•|k).\hfill \end{array}$$

$J_H$ is a surjective linear isometry between H and $H^{*}$ (Riesz representation theorem). In the frame of the geometry of Banach spaces, $J_H$ is called duality mapping.

Consider $H,K$ Hilbert spaces, and let $T\in \mathcal{B}(H,K)$ be a bounded linear operator. We define the adjoint operator of T as $T\prime :={\left(J_H\right)}^{-1}\circ T^{*}\circ J_K\in \mathcal{B}(K,H)$, with $T^{*}:K^{*}\to H^{*}$ as the dual operator of T. The most representative property of the adjoint operator is that it is the unique operator in $\mathcal{B}(K,H)$ satisfying $\left(T\left(x\right)\right|y)=\left(x\right|T^{\prime }\left(y\right))$ for all $x\in H$ and all $y\in K$. It holds that $\parallel T^{\prime }\parallel =\parallel T\parallel$, ${\left(T^{\prime }\right)}^{\prime }=T$, ${(T+S)}^{\prime }=T^{\prime }+S^{\prime }$ and ${\left(\lambda T\right)}^{\prime }=\overline{\lambda }{T}^{\prime }$.

If $T\in \mathcal{B}\left(H\right)$ verifies $T=T^{\prime }$, then T is self-adjoint. This is equivalent to equality $\left(T\right(x\left)\right|y)=(x\left|T\right(y\left)\right)$ held for every $x,y\in H$. If T satisfies $\left(T\right(x\left)\right|x)\ge 0$ for each $x\in H$, then T is called positive. If H is complex, then $T\in \mathcal{B}\left(H\right)$ is self-adjoint if and only if $\left(T\right(x\left)\right|x)\in \mathbb{R}$ for each $x\in H$. Thus, in complex Hilbert spaces, positive operators are self-adjoint. T is strongly positive if there exists $S\in \mathcal{B}(H,K)$ with $T=S^{\prime }\circ S$. Typical examples of self-adjoint positive operators are strongly positive operators.

For each $T\in \mathcal{B}\left(H\right)$, the following set is the spectrum of T

$$\sigma \left(T\right):=\left\{\lambda \in \mathbb{C}:\lambda I-T\notin \mathcal{U}\left(\mathcal{B}\left(H\right)\right)\right\},$$

where $\mathcal{U}\left(\mathcal{B}\left(H\right)\right)$ is the multiplicative group of invertible operators on H. Among spectral properties, it is compact, nonempty, and $\parallel T\parallel \ge max\left|\sigma \right(T\left)\right|$. We work with a special subset of the spectrum:

$${\sigma }_p\left(T\right):=\left\{\lambda \in \mathbb{C}:ker(\lambda I-T)\ne \left\{0\right\}\right\},$$

called the point spectrum, whose elements are eigenvalues of T. It is clear that ${\sigma }_p\left(T\right)\subseteq \sigma \left(T\right)$. In addition, if $\lambda \in {\sigma }_p\left(T\right)$, the subspace of associated eigenvectors to $\lambda$ is

$$V\left(\lambda \right):=\left\{x\in H:T\left(x\right)=\lambda x\right\}.$$

If $\parallel T\parallel$ is an eigenvalue of T or, in other words, $\parallel T\parallel \in {\sigma }_p\left(T\right)$, then $\parallel T\parallel$ is the maximal element of $\left|\sigma \right(T\left)\right|$, i.e., $\parallel T\parallel =max\left|\sigma \right(T\left)\right|$. In this situation, we also write $\parallel T\parallel ={\lambda }_{max}\left(T\right)$.

Example 1. 

Let $T:H\to K$ be a continuous linear operator where $H,K$ are Hilbert spaces, such that $\parallel T\parallel \in {\sigma }_p\left(T\right)$; then, $V(\parallel T\parallel )\cap {\mathsf{S}}_X\subseteq \mathrm{suppv}\left(T\right)$. If $x\in V(\parallel T\parallel )\cap {\mathsf{S}}_X$, then $T\left(x\right)=\parallel T\parallel x$; therefore, $\parallel T\left(x\right)\parallel =\parallel T\parallel$ and hence $x\in \mathrm{suppv}\left(T\right)$.

In general, $\parallel T\parallel \notin {\sigma }_p\left(T\right)$, unless, for instance, T is compact, self-adjoint, and positive. This is why we have to rely on adjoint $T^{\prime }$ and strongly positive operator $T^{\prime }\circ T$. 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 $T^{\prime }\circ T$ 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 $T\in \mathcal{B}(H,K)$. Then,

1. 

${\parallel T\parallel }^2=\parallel T^{\prime }\circ T\parallel$.

2. 

$\mathrm{suppv}\left(T\right)\subseteq \mathrm{suppv}\left(T^{\prime }\circ T\right)$.

3. 

$\mathrm{supp}\left(T\right)\ne ⌀$ if and only if $\parallel T^{\prime }\circ T\parallel \in {\sigma }_p\left(T^{\prime }\circ T\right)$.

In this situation, $\parallel T\parallel =\sqrt{{\lambda }_{max}\left(T^{\prime }\circ T\right)}$ and $\mathrm{suppv}\left(T\right)=V\left({\lambda }_{max}(T^{\prime }\circ T)\right)\cap {\mathsf{S}}_H$.

Proof. 

• Fix an element $x\in H$, and the associated mapping $x^{*}:=(•|x)$. Then,

  $$\begin{array}{ccc}\hfill {\parallel T\left(x\right)\parallel }^2& =& \left(T\left(x\right)\left|T\right(x)\right)=\left(T^{\prime }\left(T\left(x\right)\right)|x\right)=x^{*}\left(\left(T^{\prime }\circ T\right)\left(x\right)\right)\hfill \end{array}$$

  (6)

  $$\begin{array}{ccc}& \le & \parallel x^{*}\parallel ∥T^{\prime }\left(T\left(x\right)\right)∥\le \parallel x^{*}\parallel ∥T^{\prime }\circ T∥\parallel x\parallel =∥T^{\prime }\circ T∥{\parallel x\parallel }^2\hfill \end{array}$$

  (7)

  $$\begin{array}{ccc}& \le & ∥T^{\prime }∥{\parallel T\parallel \parallel x\parallel }^2={\parallel T\parallel }^2{\parallel x\parallel }^2.\hfill \end{array}$$

  (8)
  If element x is taken in the unit sphere, i.e., $x\in {\mathsf{S}}_X$, and considering the previous inequalities, we concluded that ${\parallel T\parallel }^2=\parallel T^{\prime }\circ T\parallel$.
• Let $x\in \mathrm{suppv}\left(T\right)$ be an arbitrary element; then, Equation (6) implies that

  $$∥T^{\prime }\left(T\left(x\right)\right)∥={\parallel T\parallel }^2=\parallel T\parallel \parallel T\left(x\right)\parallel =∥T^{\prime }∥\parallel T\left(x\right)\parallel .$$
  Then, $x\in \mathrm{suppv}\left(T^{\prime }\circ T\right)$.
• Take $v\in \mathrm{suppv}\left(T\right)$. Before anything else, since $\mathrm{suppv}\left(T\right)\subseteq \mathrm{suppv}\left(T^{\prime }\circ T\right)$, we have that

  $$∥\frac{\left(T^{\prime }\circ T\right)\left(v\right)}{∥T^{\prime }\circ T∥}∥=\frac{∥\left(T^{\prime }\circ T\right)\left(v\right)∥}{∥T^{\prime }\circ T∥}=\frac{∥T^{\prime }\circ T∥}{∥T^{\prime }\circ T∥}=1.$$

  (9)
  Following chain of equalities (6),

  $$v^{*}\left(\frac{\left(T^{\prime }\circ T\right)\left(v\right)}{∥T^{\prime }\circ T∥}\right)=\frac{{\parallel T\left(v\right)\parallel }^2}{∥T^{\prime }\circ T∥}=\frac{{\parallel T\parallel }^2{\parallel v\parallel }^2}{∥T^{\prime }\circ T∥}=1.$$

  (10)
  Thanks to the strict convexity of space H,

  $$\frac{\left(T^{\prime }\circ T\right)\left(v\right)}{∥T^{\prime }\circ T∥}=v,$$

  that is,

  $$\left(T^{\prime }\circ T\right)\left(v\right)=∥T^{\prime }\circ T∥v$$

  and so $\parallel T^{\prime }\circ T\parallel \in {\sigma }_p\left(T^{\prime }\circ T\right)$. We implicitly proved that $\mathrm{suppv}\left(T\right)\subseteq V\left(∥T^{\prime }\circ T∥\right)\cap {\mathsf{S}}_H$.
  Conversely, let us suppose that $\parallel T^{\prime }\circ T\parallel \in {\sigma }_p\left(T^{\prime }\circ T\right)$. As we remarked before, $T^{\prime }\circ T$ is a strongly positive operator, so the eigenvalues of that operator are real and positive. Therefore, equality ${\lambda }_{max}\left(T^{\prime }\circ T\right)=∥T^{\prime }\circ T∥$ holds, which implies that

  $$\parallel T\parallel =\sqrt{{\parallel T\parallel }^2}=\sqrt{\parallel T^{\prime }\circ T\parallel }=\sqrt{{\lambda }_{max}\left(T^{\prime }\circ T\right)}.$$
  Take $w\in V\left({\lambda }_{max}\left(T^{\prime }\circ T\right)\right)\cap {\mathsf{S}}_H$. Then

  $$\begin{array}{ccc}\hfill {\parallel T\left(w\right)\parallel }^2& =& w^{*}\left(\left(T^{\prime }\circ T\right)\left(w\right)\right)\hfill \\ & =& w^{*}\left({\lambda }_{max}\left(T^{\prime }\circ T\right)w\right)\hfill \\ & =& {\lambda }_{max}\left(T^{\prime }\circ T\right)\hfill \\ & =& ∥T^{\prime }\circ T∥\hfill \\ & =& {\parallel T\parallel }^2.\hfill \end{array}$$
  This chain of equalities proves that $w\in \mathrm{suppv}\left(T\right)$. Consequently,

  $$\mathsf{V}\left({\lambda }_{max}\left(T^{\prime }\circ T\right)\right)\cap {\mathsf{S}}_H\subseteq \mathrm{suppv}\left(T\right).$$

 □

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 $T\in \mathcal{B}(H,K)$ defined between H and K, Hilbert spaces, then

$$V\left(T\right):=\bigcup _{\lambda \in {\sigma }_p\left(T\right)}V\left(\lambda \right).$$

Lemma 2. 

If we consider Hilbert spaces, $H,K$, and $T_1,\dots ,T_k\in \mathcal{B}(H,K)$, then, for every ${\alpha }_1,\dots ,{\alpha }_k\in \mathbb{C}$,

$$\bigcap _{i=1}^{k}V\left(T_i\right)\subseteq V\left(\sum _{i=1}^k{\alpha }_i{T}_i\right).$$

Proof. 

Take any $x\in {\bigcap }_{i=1}^{k}V\left(T_i\right)$. If $x=0$, there is nothing to prove, x is actually in $V\left({\sum }_{i=1}^k{\alpha }_i{T}_i\right)$. So, assume that $x\ne 0$. For every $i\in \{1,\dots ,k\}$, there exists ${\lambda }_i\in \sigma \left(T_i\right)$, such that $x\in V\left({\lambda }_i\right)$, that is, $T_i\left(x\right)={\lambda }_{i}x$. Then

$$\left(\sum _{i=1}^k{\alpha }_i{T}_i\right)\left(x\right)=\sum _{i=1}^k{\alpha }_i{T}_i\left(x\right)=\sum _{i=1}^k{\alpha }_i{\lambda }_{i}x=\left(\sum _{i=1}^k{\alpha }_i{\lambda }_i\right)x.$$

This shows that

$$x\in V\left(\sum _{i=1}^k{\alpha }_i{\lambda }_i\right)\subseteq V\left(\sum _{i=1}^k{\alpha }_i{T}_i\right).$$

 □

The hypothesis in Lemma 3 is, in fact, very restrictive.

Lemma 3. 

If $H,K$ are Hilbert spaces, and $T_1,\dots ,T_k\in \mathcal{B}(H,K)$, such that $V\left(T_i\right)\setminus ker\left(T_i\right)\subseteq ker\left(T_j\right)$ for all $i,j\in \{1,\dots ,k\}$ with $i\ne j$. For every $i\in \{1,\dots ,k\}$ and every $x_i\in V\left(T_i\right)\setminus ker\left(T_i\right)$, there are ${\alpha }_1,\dots ,{\alpha }_k\in \mathbb{C}$, such that

$$\sum _{i=1}^k{\beta }_i{x}_i\in V\left(\sum _{i=1}^k{\alpha }_i{T}_i\right)$$

for every ${\beta }_1,\dots ,{\beta }_k\in \mathbb{C}$.

Proof. 

For every $i\in \{1,\dots ,k\}$, there exists ${\lambda }_i\in \sigma \left(T_i\right)\setminus \left\{0\right\}$, such that $x_i\in V\left({\lambda }_i\right)$, that is, $T_i\left(x_i\right)={\lambda }_i{x}_i$. Define ${\alpha }_i:={\lambda }_i^{-1}$ for every $i\in \{1,\dots ,k\}$. Then

$$\left(\sum _{i=1}^k{\alpha }_i{T}_i\right)\left(\sum _{i=1}^k{\beta }_i{x}_i\right)=\sum _{i=1}^k{\alpha }_i{\beta }_i{T}_i\left(x_i\right)=\sum _{i=1}^k{\alpha }_i{\beta }_i{\lambda }_i{x}_i=\sum _{i=1}^k{\beta }_i{x}_i.$$

 □

If $H_1,\dots ,H_p$ are Hilbert spaces, then ${\left({⨁}_{i=1}^p{H}_i\right)}_2$ is a Hilbert space, considering the following scalar product and norm

$$\left({\left(h_i\right)}_{i=1}^p{\left(k_i\right)}_{i=1}^p\right):=\sum _{i=1}^p\left(h_i\right|k_i),∥\underset{i=1}{\overset{p}{\left(h_i\right)}}∥:=\sqrt{\sum _{i=1}^p{\parallel h_i\parallel }^2},$$

for all ${\left(h_i\right)}_{i=1}^p,{\left(k_i\right)}_{i=1}^p\in {\left({⨁}_{i=1}^p{H}_i\right)}_2$. If H is another Hilbert space and $T_i:H\to H_i$ is a continuous linear operator for each $i=1,\dots ,p$, then the direct sum of $T_1,\dots ,T_p$ is defined as

$$\begin{array}{cccc}\hfill {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2:& \hfill H& \to & {\left(\underset{i=1}{\overset{p}{⨁}}{H}_i\right)}_2\hfill \\ & \hfill x& \mapsto & {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2\left(x\right):={\left(T_i\left(x\right)\right)}_{i=1}^p.\hfill \end{array}$$

If $S_i:H_i\to H$ is a continuous linear operator for each $i=1,\dots ,p$, then the direct sum of $S_1,\dots ,S_p$ is now defined as

$$\begin{array}{cccc}\hfill {\left(\underset{i=1}{\overset{p}{⨁}}{S}_i\right)}_2:& \hfill {\left(\underset{i=1}{\overset{p}{⨁}}{H}_i\right)}_2& \to & H\hfill \\ & \hfill {\left(h_i\right)}_{i=1}^p& \mapsto & {\left(\underset{i=1}{\overset{p}{⨁}}{S}_i\right)}_2\left({\left(h_i\right)}_{i=1}^p\right):=\sum _{i=1}^p{S}_i\left(h_i\right).\hfill \end{array}$$

Theorem 5. 

Suppose that $H,H_1,\dots ,H_p$ are Hilbert spaces, and let $T_i:H\to H_i$ be a continuous linear operator for each $i=1,\dots ,p$. Then, ${\left({⨁}_{i=1}^p{T}_i\right)}_2^{\prime }={\left({⨁}_{i=1}^p{T}_i^{\prime }\right)}_2$ and

$${\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2^{\prime }\circ {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2=\sum _{i=1}^p{T}_i^{\prime }\circ T_i.$$

Proof. 

Fix arbitrary elements $x\in H$ and ${\left(h_i\right)}_{i=1}^p\in {\left({⨁}_{i=1}^p{H}_i\right)}_2$. Then,

$$\begin{array}{ccc}\hfill \left(x\left|{\left(\underset{i=1}{\overset{p}{⨁}}{T}_i^{\prime }\right)}_2\left({\left(h_i\right)}_{i=1}^p\right)\right)\right.& =& \left(x|\sum _{i=1}^p{T}_i^{\prime }\left(h_i\right)\right)\hfill \\ & =& \sum _{i=1}^p\left(x|T_i^{\prime }\left(h_i\right)\right)\hfill \\ & =& \sum _{i=1}^p\left(T_i\left(x\right)|h_i\right)\hfill \\ & =& \left({\left(T_i\left(x\right)\right)}_{i=1}^p|{\left(h_i\right)}_{i=1}^p\right)\hfill \\ & =& \left({\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2\left(x\right)|{\left(h_i\right)}_{i=1}^p\right).\hfill \end{array}$$

Lastly, for each $x\in H$,

$$\begin{array}{ccc}\hfill \left({\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2^{\prime }\circ {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2\right)\left(x\right)& =& {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2^{\prime }\left({\left(\underset{i=1}{\overset{p}{⨁}}{T}_i\right)}_2\left(x\right)\right)\hfill \\ & =& {\left(\underset{i=1}{\overset{p}{⨁}}{T}_i^{\prime }\right)}_2\left({\left(T_i\left(x\right)\right)}_{i=1}^p\right)\hfill \\ & =& \sum _{i=1}^p{T}_i^{\prime }\left(T_i\left(x\right)\right)\hfill \\ & =& \sum _{i=1}^p\left(T_i^{\prime }\circ T_i\right)\left(x\right).\hfill \end{array}$$

 □

3.2. Pareto Optimal Solutions of the MOP $max\parallel T\left(x\right)\parallel ,min\parallel x\parallel ,x\in X$

Under the settings of Theorem 3, $\arg {min}_{x\in X}\parallel x\parallel =\left\{0\right\}$; therefore, in view of Theorem 1, $0\in \mathrm{Pos}\left(3\right)$. This Pareto optimal solution is usually disregarded when it comes to a real-life problem.

Theorem 6. 

Let $X,Y$ be normed spaces, and $T:X\to Y$ be a nonzero continuous linear operator. Then, $\mathrm{Pos}\left(3\right)=\mathbb{R}\mathrm{suppv}\left(T\right)$.

Proof. 

Fix an arbitrary $x_0\in \mathrm{Pos}\left(3\right)$. Since $x_0=\parallel x_0\parallel \frac{x_0}{\parallel x_0\parallel }$, it is sufficient if we show that $\frac{x_0}{\parallel x_0\parallel }\in \mathrm{suppv}\left(T\right)$. Therefore, we may assume that $\parallel x_0\parallel =1$, so our aim was summed up to prove that $\parallel T\left(x_0\right)\parallel =\parallel T\parallel$. Since $x_0\in {\mathsf{S}}_X\subseteq {\mathsf{B}}_X$, $\parallel T\left(x_0\right)\parallel \le \parallel T\parallel$. Suppose that $\parallel T\left(x_0\right)\parallel <\parallel T\parallel$. By the definition of sup, there exists $y\in {\mathsf{B}}_X$, such that $\parallel T\left(x_0\right)\parallel <\parallel T\left(y\right)\parallel \le \parallel T\parallel$. $\parallel y\parallel \le 1=\parallel x_0\parallel$ and $\parallel T\left(x_0\right)\parallel <\parallel T\left(y\right)\parallel$, which contradicts that $x_0\in \mathrm{Pos}\left(3\right)$. As a consequence, $\parallel T\left(x_0\right)\parallel =\parallel T\parallel$; hence, $x_0\in \mathrm{suppv}\left(T\right)$. The arbitrariness of $x_0\in \mathrm{Pos}\left(3\right)$ shows that $\mathrm{Pos}\left(3\right)\subseteq \mathbb{R}\mathrm{suppv}\left(T\right)$. Conversely, fix an arbitrary $x_0\in \mathbb{R}\mathrm{suppv}\left(T\right)$. There exists $y_0\in \mathrm{suppv}\left(T\right)$ and $\alpha \in \mathbb{R}$, such that $x_0=\alpha y_0$. Observe that $\parallel x_0\parallel =\left|\alpha \right|\parallel y_0\parallel =\left|\alpha \right|$. We prove that $x_0\in \mathrm{Pos}\left(3\right)$. Let us consider an element $y\in X$ satisfying that $\parallel y\parallel <\parallel x_0\parallel =\left|\alpha \right|$, and we distinguish cases: if $y=0$, then $\parallel T\left(x_0\right)\parallel =\left|\alpha \right|\parallel T\left(y_0\right)\parallel =\left|\alpha \right|\parallel T\parallel >0=\parallel T\left(y\right)\parallel$. If $y\ne 0$, then

$$\parallel T\left(x_0\right)\parallel =\left|\alpha \right|\parallel T\left(y_0\right)\parallel =\left|\alpha \right|\parallel T\parallel \ge \left|\alpha \right|∥T\left(\frac{y}{\parallel y\parallel }\right)∥>\parallel T\left(y\right)\parallel .$$

Lastly, if there exists $y\in X$, such that $\parallel T\left(y\right)\parallel >\parallel T\left(x_0\right)\parallel$, then

$$\left|\alpha \right|\parallel T\parallel =\parallel T\left(x_0\right)\parallel <\parallel T\left(y\right)\parallel \le \parallel T\parallel \parallel y\parallel ,$$

which means that $\parallel x_0\parallel =\left|\alpha \right|<\parallel y\parallel$. □

When $X,Y$ are Hilbert spaces, the Pareto optimal solutions of (3) are directly obtained via combining Theorems 4 and 6.

Corollary 1. 

Let $T:H\to K$ be a continuous linear operator with $H,K$ Hilbert spaces. Then, $\mathrm{Pos}\left(3\right)=V\left({\lambda }_{max}\left(T^{\prime }\circ T\right)\right)$.

This last result allows for solving the following MOP (motivated in Section 4), given by

$$\left\{\begin{array}{c}max\parallel T_1{\left(x\right)\parallel }^2+\dots +{\parallel T_k\left(x\right)\parallel }^2,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H.\hfill \end{array}\right.$$

(11)

The Pareto optimal solutions of (11) are related to those of

$$\left\{\begin{array}{c}max\parallel T_i\left(x\right)\parallel ,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(12)

for $i=1,\dots ,k$.

Corollary 2. 

If $T_1,\dots ,T_k\in \mathcal{B}(H,K)$ are continuous linear operators between Hilbert spaces H and K, then:

1. 

$\mathrm{Pos}\left(11\right)=V\left({\lambda }_{max}\left({\sum }_{i=1}^k{T}_i^{\prime }\circ T_i\right)\right).$

2. 

${\bigcap }_{i=1}^k\mathrm{Pos}\left(12\right)\subseteq \mathrm{Pos}\left(11\right)$.

Proof. 

Consider bounded linear operator

$$\begin{array}{cccc}\hfill T_1{\oplus }_2\stackrel{k}{\dots }{\oplus }_2{T}_k:& \hfill H& \to & K{\oplus }_2\stackrel{k}{\dots }{\oplus }_{2}K\hfill \\ & \hfill x& \mapsto & \left(T_1{\oplus }_2\stackrel{k}{\dots }{\oplus }_2{T}_k\right)\left(x\right):=\left(T_1\left(x\right),\dots ,T_k\left(x\right)\right).\hfill \end{array}$$

The next equality trivially holds for every $x\in H$,

$${∥\left(T_1\left(x\right),\dots ,T_k\left(x\right)\right)∥}^2=\parallel T_1{\left(x\right)\parallel }^2+\dots +{\parallel T_k\left(x\right)\parallel }^2.$$

Since the square root is strictly increasing, (11) is equivalent to

$$\left\{\begin{array}{c}max∥\left(T_1{\oplus }_2\stackrel{k}{\dots }{\oplus }_2{T}_k\right)\left(x\right)∥,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(13)

which is an MOP of form (3).

• According to Corollary 1 and Theorem 5,

  $$\begin{array}{ccc}\hfill \mathrm{Pos}\left(11\right)& =& \mathrm{Pos}\left(13\right)\hfill \\ & =& V\left({\lambda }_{max}\left({\left(T_1{\oplus }_2\stackrel{k}{\dots }{\oplus }_2{T}_k\right)}^{\prime }\circ \left(T_1{\oplus }_2\stackrel{k}{\dots }{\oplus }_2{T}_k\right)\right)\right)\hfill \\ & =& V\left({\lambda }_{max}\left(\sum _{i=1}^k{T}_i^{\prime }\circ T_i\right)\right).\hfill \end{array}$$
• We rely on Theorem 6 and Corollary 1. Fix an arbitrary $x\in {\bigcap }_{i=1}^k\mathrm{Pos}\left(12\right)$. If $x=0$, then $x\in \mathbb{R}\mathrm{suppv}\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)=\mathrm{Pos}\left(13\right)=\mathrm{Pos}\left(11\right)$. Suppose that $x\ne 0$. In view of Theorem 6, $\frac{x}{\parallel x\parallel }\in {\bigcap }_{i=1}^k\mathrm{suppv}\left(T_i\right)$. We prove that $\frac{x}{\parallel x\parallel }\in \mathrm{suppv}\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)$. Take any $y\in {\mathsf{B}}_H$. Since $\frac{x}{\parallel x\parallel }\in {\bigcap }_{i=1}^k\mathrm{suppv}\left(T_i\right)$, for every $i\in \{1,\dots ,k\}$,

  $$∥T_i\left(\frac{x}{\parallel x\parallel }\right)∥\ge \parallel T_i\left(y\right)\parallel .$$
  As a consequence,

  $$\begin{array}{ccc}\hfill ∥\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)\left(\frac{x}{\parallel x\parallel }\right)∥& =& \sqrt{{∥T_1\left(\frac{x}{\parallel x\parallel }\right)∥}^2+\cdots +{∥T_k\left(\frac{x}{\parallel x\parallel }\right)∥}^2}\hfill \\ & \ge & \sqrt{{∥T_1\left(y\right)∥}^2+\cdots +{∥T_k\left(y\right)∥}^2}\hfill \\ & =& ∥\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)\left(y\right)∥.\hfill \end{array}$$
  This means that $\frac{x}{\parallel x\parallel }\in \mathrm{suppv}\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)$. In accordance with Theorem 6,

  $$x=\parallel x\parallel \frac{x}{\parallel x\parallel }\in \mathbb{R}\mathrm{suppv}\left(T_1{\oplus }_2\stackrel{k}{\cdots }{\oplus }_2{T}_k\right)=\mathrm{Pos}\left(13\right)=\mathrm{Pos}\left(11\right).$$

 □

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 [18,23,24,25]; then, MOPs like (3) come out:

$$\left\{\begin{array}{c}max\parallel E_x{\psi \parallel }_2,\hfill \\ min{\psi }^{T}L\psi ,\hfill \\ \psi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max\parallel E_y{\psi \parallel }_2,\hfill \\ min{\psi }^{T}L\psi ,\hfill \\ \psi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max\parallel E_z{\psi \parallel }_2,\hfill \\ min{\psi }^{T}L\psi ,\hfill \\ \psi \in {\mathbb{R}}^n,\hfill \end{array}\right.$$

(14)

where E is a matrix representing the electromagnetic field, $E_x,E_y,E_z$ 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,

$${\psi }^{T}L\psi ={\psi }^T{C}^{T}C\psi ={\left(C\psi \right)}^T\left(C\psi \right)={∥C\psi ∥}_2^2.$$

Next, we apply the following change of variables: $\phi :=C\psi$. Then, the previous problems can be rewritten as follows:

$$\left\{\begin{array}{c}max{∥\left(E_x{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2^2,\hfill \\ \phi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max{∥\left(E_y{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2^2,\hfill \\ \phi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max{∥\left(E_z{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2^2,\hfill \\ \phi \in {\mathbb{R}}^n.\hfill \end{array}\right.$$

(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):

$$\left\{\begin{array}{c}max{∥\left(E_x{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2,\hfill \\ \phi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max{∥\left(E_y{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2,\hfill \\ \phi \in {\mathbb{R}}^n,\hfill \end{array}\right.\left\{\begin{array}{c}max{∥\left(E_z{C}^{-1}\right)\phi ∥}_2,\hfill \\ {min\parallel \phi \parallel }_2,\hfill \\ \phi \in {\mathbb{R}}^n.\hfill \end{array}\right.$$

(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

$$\begin{array}{c}V\left({\lambda }_{max}\left({\left(E_x{C}^{-1}\right)}^T\left(E_x{C}^{-1}\right)\right)\right),\hfill \\ V\left({\lambda }_{max}\left({\left(E_y{C}^{-1}\right)}^T\left(E_y{C}^{-1}\right)\right)\right),\hfill \\ V\left({\lambda }_{max}\left({\left(E_z{C}^{-1}\right)}^T\left(E_z{C}^{-1}\right)\right)\right),\hfill \end{array}$$

respectively. On the other hand, we can consider the combined MOP, as in (11):

$$\left\{\begin{array}{c}max{∥\left(E_x{C}^{-1}\right)\phi ∥}_2^2+{∥\left(E_y{C}^{-1}\right)\phi ∥}_2^2+{∥\left(E_z{C}^{-1}\right)\phi ∥}_2^2,\hfill \\ {min\parallel \phi \parallel }_2,\hfill \\ \phi \in {\mathbb{R}}^n.\hfill \end{array}\right.$$

(17)

Let us define the following linear operator:

$$\begin{array}{cccc}\hfill T:& \hfill {\ell }_2^n& \to & {\ell }_2^n{\oplus }_2{\ell }_2^n{\oplus }_2{\ell }_2^n\hfill \\ \hfill & \hfill \phi & \mapsto & T\left(\phi \right)=\left(\left(E_x{C}^{-1}\right)\phi ,\left(E_y{C}^{-1}\right)\phi ,\left(E_z{C}^{-1}\right)\phi \right).\hfill \end{array}$$

The corresponding matrix to T is precisely

$$A:=\left(\begin{array}{c}{E}_x{C}^{-1}\\ E_y{C}^{-1}\\ E_z{C}^{-1}\end{array}\right).$$

For every $\phi \in {\mathbb{R}}^n$,

$${\parallel T\left(\phi \right)\parallel }_2={\left({∥\left(E_x{C}^{-1}\right)\phi ∥}_2^2+{∥\left(E_y{C}^{-1}\right)\phi ∥}_2^2+{∥\left(E_z{C}^{-1}\right)\phi ∥}_2^2\right)}^{\frac{1}{2}}.$$

Then (17) is the same as

$$\left\{\begin{array}{c}{max\parallel T\left(\phi \right)\parallel }_2,\hfill \\ {min\parallel \phi \parallel }_2,\hfill \\ \phi \in {\mathbb{R}}^n.\hfill \end{array}\right.$$

(18)

According to Corollary 1,

$$\mathrm{Pos}\left(18\right)=\mathbb{R}\mathrm{suppv}\left(T\right)=V\left({\lambda }_{max}\left(A^{T}A\right)\right).$$

Equivalently, according to Corollary 2,

$$\begin{array}{ccc}& & \mathrm{Pos}\left(18\right)=\mathrm{Pos}\left(17\right)\hfill \\ & =& V\left({\lambda }_{max}\left({\left(E_x{C}^{-1}\right)}^T\left(E_x{C}^{-1}\right)+{\left(E_y{C}^{-1}\right)}^T\left(E_y{C}^{-1}\right)+{\left(E_z{C}^{-1}\right)}^T\left(E_z{C}^{-1}\right)\right)\right).\hfill \end{array}$$

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 $T_1,\dots ,T_k\in \mathcal{B}(H,K)$ between H and K. Let ${\alpha }_1,\dots ,{\alpha }_k>0$. In bioengineering, it is common to assign weights ${\alpha }_i$ to different operators $T_i$ depending on the relevance of each $T_i$. Then, the following MOP comes into play:

$$\left\{\begin{array}{c}max{\alpha }_1\parallel T_1{\left(x\right)\parallel }^2+\dots +{\alpha }_k{\parallel T_k\left(x\right)\parallel }^2,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(19)

Nevertheless, the above MOP is, in fact, the same as the following:

$$\left\{\begin{array}{c}max\parallel S_1{\left(x\right)\parallel }^2+\dots +{\parallel S_k\left(x\right)\parallel }^2,\hfill \\ min\parallel x\parallel ,\hfill \\ x\in H,\hfill \end{array}\right.$$

(20)

where $S_i:=\sqrt{{\alpha }_i}{T}_i$ for each $i=1,\dots ,k$. $\mathrm{suppv}\left(T_i\right)=\mathrm{suppv}\left(\sqrt{{\alpha }_i}{T}_i\right)=\mathrm{suppv}\left(S_i\right)$ for every $i=1,\dots ,k$. By relying on Corollary 2, at least we can ensure that

$$\bigcap _{i=1}^k\mathrm{Pos}\left(12\right)\subseteq \mathrm{Pos}\left(20\right)=\mathrm{Pos}\left(19\right).$$

However, it is very unlikely that ${\bigcap }_{i=1}^k\mathrm{Pos}\left(12\right)\ne \left\{0\right\}$. Unless hypotheses similar to the ones employed in Lemma 2 or Lemma 3 are used, we cannot conclude any other relation between $\mathrm{Pos}\left(19\right)$ and $\mathrm{Pos}\left(12\right)$.