Exact Solutions to the Maxmin Problem max‖Ax‖ Subject to ‖Bx‖≤1

Abstract

In this manuscript we provide an exact solution to the maxmin problem $max\parallel Ax\parallel$ subject to $\parallel Bx\parallel \le 1$, where A and B are real matrices. This problem comes from a remodeling of $max\parallel Ax\parallel$ subject to $min\parallel Bx\parallel$, because the latter problem has no solution. Our mathematical method comes from the Abstract Operator Theory, whose strong machinery allows us to reduce the first problem to $max\parallel Cx\parallel$ subject to $\parallel x\parallel \le 1$, which can be solved exactly by relying on supporting vectors. Finally, as appendices, we provide two applications of our solution: first, we construct a truly optimal minimum stored-energy Transcranian Magnetic Stimulation (TMS) coil, and second, we find an optimal geolocation involving statistical variables.

1. Introduction

1.1. Scope

Different scientific fields, such as Physics, Statistics, Economics, or Engineering, deal with real-life problems that are usually modelled by the experts on those fields using matrices and their norms (see [1,2,3,4,5,6]). A typical modelling is the following original maxmin problem

$$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ min\parallel Bx\parallel .\hfill \end{array}\right.$$

One of the most iconic results in this manuscript (Theorem 2) shows that the previous problem, regarded strictly as a multiple optimization problem, has no solutions. To save this obstacle we provide a different model, such as

$$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel Bx\parallel \le 1.\hfill \end{array}\right.$$

Here in this article we justify the remodelling of the original maxmin problem and we solve it by making use of supporting vectors. This concept comes from the Theory of Banach Spaces and Operator Theory. Given a matrix A, a supporting vector is a unit vector x such that A attains its norm at x, that is, x is a solution of the following single optimization problem:

$$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel x\parallel =1.\hfill \end{array}\right.$$

The geometric and topological structure of supporting vectors can be consulted in [7,8,9]. On the other hand, generalized supporting vectors are defined and studied in [7,8]. The generalized supporting vectors of a finite sequence of matrices $A_1,\cdots ,A_n$, for the Euclidean norm ${\parallel •\parallel }_2$, are the solutions of

$$\left\{\begin{array}{c}max\parallel A_1{x\parallel }_2^2+\cdots +{\parallel A_{n}x\parallel }_2^2\hfill \\ {\parallel x\parallel }_2=1.\hfill \end{array}\right.$$

This optimization problem clearly generalizes the previous one.

Supporting vectors were originally applied in [10] to truly optimally design a TMS coil, because until that moment TMS coils had only been designed by means of heuristic methods, which were never proved to be convergent. In [10] a three-component TMS coil problem is posed but only the one-component case was resolved. The three-component case was stated and solved by means of the generalized supporting vectors in [8]. In this manuscript, we model a TMS coil with a maxmin problem and solve it exactly with our method.

A second application of supporting vectors was given in [8], where an optimal location situation using Principal Component Analysis (PCA) was solved. In this manuscript, we model a more complex PCA problem as an optimal maxmin geolocation involving statistical variables.

For other perspective on supporting vectors and generalized supporting vectors, we refer the reader to [9].

1.2. Background

In the first place, we refer the reader to [8] (Preliminaries) for a general review of multiobjective optimization problems and their reformulations to avoid the lack of solutions (generally caused by the existence of many objective functions).

The original maxmin optimization problem has the form

$$M:=\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ minf\left(x\right)\hfill \end{array}\right.$$

where $f,g:X\to (0,\infty )$ are real-valued functions and X is a nonempty set. Notice that

$$\mathrm{sol}\left(M\right)=\arg maxg\left(x\right)\cap \arg minf\left(x\right).$$

Many real-life problems can be mathematically model, such as a maxmin. However, this kind of multiobjective optimization problems may have the inconvenience of lacking a solution. If this occurs, then we are in need of remodeling the real-life problem with another mathematical optimization problem that has a solution and still models the real-life problem very accurately.

According to [10] (Theorem 5.1), one can realize that, in case $\mathrm{sol}\left(M\right)=⌀$, the following optimization problems are good alternatives to keep modeling the real-life problem accurately:

• $\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ minf\left(x\right)\hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}min\frac{f\left(x\right)}{g\left(x\right)}\hfill \\ g\left(x\right)\ne 0\hfill \end{array}\right.$.
• $\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ minf\left(x\right)\hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}max\frac{g\left(x\right)}{f\left(x\right)}\hfill \\ f\left(x\right)\ne 0\hfill \end{array}\right.$.
• $\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ minf\left(x\right)\hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ f\left(x\right)\le a\hfill \end{array}\right.$.
• $\left\{\begin{array}{c}maxg\left(x\right)\hfill \\ minf\left(x\right)\hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}minf\left(x\right)\hfill \\ g\left(x\right)\ge b\hfill \end{array}\right.$.

We will prove in the third section that all four previous reformulations are equivalent for the original maxmin $\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ min\parallel Bx\parallel \hfill \end{array}\right.$. In the fourth section, we will solve the reformulation $\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel Bx\parallel \le 1\hfill \end{array}\right.$.

2. Characterizations of Operators with Null Kernel

Kernels will play a fundamental role towards solving the general reformulated maxmin (2) as shown in the next section. This is why we first study the operators with null kernel.

Throughout this section, all monoid actions considered will be left, all rings will be associative, all rings will be unitary rngs, all absolute semi-values and all semi-norms will be non-zero, all modules over rings will be unital, all normed spaces will be real or complex and all algebras will be unitary and complex.

Given a rng R and an element $s\in R$, we will denote by $\ell \mathit{d}\left(s\right)$ to the set of left divisors of s, that is,

$$\ell \mathit{d}\left(s\right):=\{r\in R:\exists t\in R\setminus \{0\}\mathrm{with}rt=s\}.$$

Similarly, $\mathit{rd}\left(s\right)$ stands for the set of right divisors of s. If R is a ring, then the set of its invertibles is usually denoted by $\mathcal{U}\left(R\right)$. Notice that $\ell \mathit{d}\left(1\right)$ ($\mathit{rd}\left(1\right)$) is precisely the subset of elements of R which are right-(left) invertible. As a consequence, $\mathcal{U}\left(R\right)=\ell \mathit{d}\left(1\right)\cap \mathit{rd}\left(1\right)$. Observe also that $\ell \mathit{d}\left(0\right)\cap \mathit{rd}\left(1\right)=⌀=\mathit{rd}\left(0\right)\cap \ell \mathit{d}\left(1\right)$. In general we have that $\ell \mathit{d}\left(0\right)\cap \ell \mathit{d}\left(1\right)\ne ⌀$ and $\mathit{rd}\left(0\right)\cap \mathit{rd}\left(1\right)\ne ⌀$. Later on in Example 1 we will provide an example of a ring where $\mathit{rd}\left(0\right)\cap \mathit{rd}\left(1\right)\ne ⌀$.

Recall that an element p of a monoid is called involutive if $p^2=1$. Given a rng R, an involution is an additive, antimultiplicative, composition-involutive map $\ast :R\to R$. A $\ast$-rng is a rng endowed with an involution.

The categorical concept of monomorphism will play an important role in this manuscript. A morphism $f\in {hom}_{\mathcal{C}}(A,B)$ between objects A and B in a category $\mathcal{C}$ is said to be a monomorphism provided that $f\circ g=f\circ h$ implies $g=h$ for all $C\in \mathrm{ob}\left(\mathcal{C}\right)$ and all $g,h\in {hom}_{\mathcal{C}}(C,A)$. Once can check that if $f\in {hom}_{\mathcal{C}}(A,B)$ and there exist $C_0\in \mathrm{ob}\left(\mathcal{C}\right)$ and $g_0\in {hom}_{\mathcal{C}}(B,C_0)$ such that $g_0\circ f$ is a monomorphism, then f is also a monomorphism. In particular, if $f\in {hom}_{\mathcal{C}}(A,B)$ is a section, that is, exists $g\in {hom}_{\mathcal{C}}(B,A)$ such that $g\circ f=I_A$, then f is a monomorphism. As a consequence, the elements of ${hom}_{\mathcal{C}}(A,A)$ that have a left inverse are monomorphisms. In some categories, the last condition suffices to characterize monomorphisms. This is the case, for instance, of the category of vector spaces over a division ring.

Recall that $\mathcal{CL}(X,Y)$ denotes the space of continuous linear operators from a topological vector space X to another topological vector space Y.

Proposition 1.

A continuous linear operator $T:X\to Y$ between locally convex Hausdorff topological vector spaces X and Y verifies that $ker\left(T\right)\ne \left\{0\right\}$ if and only if exists $S\in \mathcal{CL}(Y,X)\setminus \left\{0\right\}$ with $T\circ S=0$. In particular, if $X=Y$, then $ker\left(T\right)\ne \left\{0\right\}$ if and only if $T\in \ell \mathit{d}\left(0\right)$ in $\mathcal{CL}\left(X\right)$.

Proof. 

Let $S\in \mathcal{CL}(Y,X)\setminus \left\{0\right\}$ such that $T\circ S=0$. Fix any $y\in Y\setminus ker\left(S\right)$, then $S\left(y\right)\ne 0$ and $T\left(S\right(y\left)\right)=0$ so $S\left(y\right)\in ker\left(T\right)\setminus \left\{0\right\}$. Conversely, if $ker\left(T\right)\ne \left\{0\right\}$, then fix $x_0\in ker\left(T\right)\setminus \left\{0\right\}$ and $y_0^{\ast }\in Y^{\ast }\setminus \left\{0\right\}$ (the existence of $y^{\ast }$ is guaranteed by the Hahn-Banach Theorem on the Hausdorff locally convex topological vector space Y). Next, consider

$$\begin{array}{cccc}\hfill S:& \hfill Y& \to & X\hfill \\ & \hfill y& \mapsto & S\left(y\right):=y_0^{\ast }\left(y\right)x_0.\hfill \end{array}$$

Notice that $S\in \mathcal{CL}(Y,X)\setminus \left\{0\right\}$ and $T\circ S=0$. □

Theorem 1.

Let $T:X\to Y$ be a continuous linear operator between locally convex Hausdorff topological vector spaces X and Y. Then:

1. 

If T is a section, then $ker\left(T\right)=\left\{0\right\}$

2. 

In case X and Y are Banach spaces, $T\left(X\right)$ is topologically complemented in Y and $ker\left(T\right)=\left\{0\right\}$, then T is a section.

Proof. 

• Trivial since sections are monomorphisms.
• Consider $T:X\to T\left(X\right)$. Since $T\left(X\right)$ is topologically complemented in Y we have that $T\left(X\right)$ is closed in Y, thus it is a Banach space. Therefore, the Open Mapping Theorem assures that $T:X\to T\left(X\right)$ is an isomorphism. Let $T^{-1}:T\left(X\right)\to X$ be the inverse of $T:X\to T\left(X\right)$. Now consider $P:Y\to Y$ to be a continuous linear projection such that $P\left(Y\right)=T\left(X\right)$. Finally, it suffices to define $S:=T^{-1}\circ P$ since $S\circ T=I_X$.

 □

We will finalize this section with a trivial example of a matrix $A\in {\mathbb{R}}^{3\times 2}$ such that $A\in \mathit{rd}\left(I\right)\cap \mathit{rd}\left(0\right)$.

Example 1.

Consider

$$A=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right).$$

It is not hard to check that $ker\left(A\right)=\left\{\right(0,0\left)\right\}$ thus A is left-invertible by Theorem 1(2) and so $A\in \mathit{rd}\left(I\right)$. In fact,

$$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).$$

Finally,

$$\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$$

3. Remodeling the Original Maxmin Problem $max\parallel \mathbf{T}\left(\mathbf{x}\right)\parallel$ Subject to $min\parallel \mathbf{S}\left(\mathbf{x}\right)\parallel$

3.1. The Original Maxmin Problem Has No Solutions

This subsection begins with the following theorem:

Theorem 2.

Let $T,S:X\to Y$ be nonzero continuous linear operators between Banach spaces X and Y. Then the original maxmin problem

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

(1)

has trivially no solution.

Proof. 

Observe that $\arg min\parallel S\left(x\right)\parallel =ker\left(S\right)$ and $\arg max\parallel T\left(x\right)\parallel =⌀$ because $T\ne \left\{0\right\}$. Then the set of solutions of Problem (1) is

$$\arg min\parallel S\left(x\right)\parallel \cap \arg max\parallel T\left(x\right)\parallel =ker\left(S\right)\cap ⌀=⌀.$$

 □

As a consequence, Problem (1) must be reformulated or remodeled.

3.2. Equivalent Reformulations for the Original Maxmin Problem

According to the Background section, we begin with the following reformulation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ \parallel S\left(x\right)\parallel \le 1\hfill \end{array}\right.\end{array}$$

(2)

Please note that ${\displaystyle \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel }$ is a $\mathbb{K}$-symmetric set, where $\mathbb{K}:=\mathbb{R}\mathrm{or}\mathbb{C}$, in other words, if $\lambda \in \mathbb{K}$ and $\left|\lambda \right|=1$, then ${\displaystyle \lambda x\in \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel }$ for every ${\displaystyle x\in \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel }$. The finite dimensional version of the previous reformulation is

$$\begin{array}{c}\hfill \left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel Bx\parallel \le 1\hfill \end{array}\right.\end{array}$$

(3)

where $A,B\in {\mathbb{R}}^{m\times n}$.

Recall that $\mathcal{B}(X,Y)$ denotes the space of bounded operators from X to Y.

Lemma 1.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. If the general reformulated maxmin problem

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ \parallel S\left(x\right)\parallel \le 1\hfill \end{array}\right.$$

has a solution, then $ker\left(S\right)\subseteq ker\left(T\right)$.

Proof. 

If $ker\left(S\right)\setminus ker\left(T\right)\ne ⌀$, then it suffices to consider the sequence ${\left(n{x}_0\right)}_{n\in \mathbb{N}}$ for $x_0\in ker\left(S\right)\setminus ker\left(T\right)$, since $\parallel S\left(n{x}_0\right)\parallel =0\le 1$ for all $n\in \mathbb{N}$ and $\parallel T\left(n{x}_0\right)\parallel =n\parallel T\left(x_0\right)\parallel \to \infty$ as $n\to \infty$. □

The general maxmin (1) can also be reformulated as

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ min\parallel S\left(x\right)\parallel \hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}max\frac{\parallel T\left(x\right)\parallel }{\parallel S\left(x\right)\parallel }\hfill \\ \parallel S\left(x\right)\parallel \ne 0\hfill \end{array}\right.$$

Lemma 2.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. If the second general reformulated maxmin problem

$$\left\{\begin{array}{c}max\frac{\parallel T\left(x\right)\parallel }{\parallel S\left(x\right)\parallel }\hfill \\ \parallel S\left(x\right)\parallel \ne 0\hfill \end{array}\right.$$

has a solution, then $ker\left(S\right)\subseteq ker\left(T\right)$.

Proof. 

Suppose there exists $x_0\in ker\left(S\right)\setminus ker\left(T\right)$. Then fix an arbitrary $x_1\in X\setminus ker\left(S\right)$. Notice that

$$\frac{\parallel T(n{x}_0+x_1)\parallel }{\parallel S(n{x}_0+x_1)\parallel }\ge \frac{n\parallel T\left(x_0\right)\parallel -\parallel T\left(x_1\right)\parallel }{\parallel S\left(x_1\right)\parallel }\to \infty$$

as $n\to \infty$. □

The next theorem shows that the previous two reformulations are in fact equivalent.

Theorem 3.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. Then

$$\bigcup _{t>0}t\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel =\arg \underset{\parallel S\left(x\right)\parallel \ne 0}{max}\frac{\parallel T\left(x\right)\parallel }{\parallel S\left(x\right)\parallel }.$$

Proof. 

Let $x_0\in \arg {max}_{\parallel S\left(x\right)\parallel \le 1}\parallel T\left(x\right)\parallel$ and $t_0>0$. Fix an arbitrary $y\in X\setminus ker\left(S\right)$. Notice that $x_0\notin ker\left(S\right)$ in virtue of Theorem 1. Then

$$\parallel T\left(x_0\right)\parallel \ge ∥T\left(\frac{y}{\parallel S\left(y\right)\parallel }\right)∥,$$

therefore

$$\frac{\parallel T\left(t{x}_0\right)\parallel }{\parallel S\left(t{x}_0\right)\parallel }=\frac{\parallel T\left(x_0\right)\parallel }{\parallel S\left(x_0\right)\parallel }\ge \parallel T\left(x_0\right)\parallel \ge ∥T\left(\frac{y}{\parallel S\left(y\right)\parallel }\right)∥.$$

Conversely, let $x_0\in \arg {max}_{\parallel S\left(x\right)\parallel \ne 0}\frac{\parallel T\left(x\right)\parallel }{\parallel S\left(x\right)\parallel }$. Fix an arbitrary $y\in X$ with $\parallel S\left(y\right)\parallel \le 1$. Then

$$∥T\left(\frac{x_0}{\parallel S\left(x_0\right)\parallel }\right)∥=\frac{\parallel T\left(x_0\right)\parallel }{\parallel S\left(x_0\right)\parallel }\ge \frac{\parallel T\left(y\right)\parallel }{\parallel S\left(y\right)\parallel }\ge \parallel T\left(y\right)\parallel$$

which means that

$$\frac{x_0}{\parallel S\left(x_0\right)\parallel }\in \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel$$

and thus

$$x_0\in \parallel S\left(x_0\right)\parallel \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel \subseteq \bigcup _{t>0}t\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel .$$

 □

The reformulation

$$\left\{\begin{array}{c}min\frac{\parallel S\left(x\right)\parallel }{\parallel T\left(x\right)\parallel }\hfill \\ \parallel T\left(x\right)\parallel \ne 0\hfill \end{array}\right.$$

is slightly different from the previous two reformulations. In fact, if $ker\left(S\right)\setminus ker\left(T\right)\ne ⌀$, then $\arg {min}_{\parallel T\left(x\right)\parallel \ne 0}\frac{\parallel S\left(x\right)\parallel }{\parallel T\left(x\right)\parallel }=ker\left(S\right)\setminus ker\left(T\right)$. The previous reformulation is equivalent to the following one as shown in the next theorem:

$$\left\{\begin{array}{c}min\parallel S\left(x\right)\parallel \hfill \\ \parallel T\left(x\right)\parallel \ge 1\hfill \end{array}\right.$$

Theorem 4.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. Then

$$\bigcup _{t>0}t\arg \underset{\parallel T\left(x\right)\parallel \ge 1}{min}\parallel S\left(x\right)\parallel =\arg \underset{\parallel T\left(x\right)\parallel \ne 0}{min}\frac{\parallel S\left(x\right)\parallel }{\parallel T\left(x\right)\parallel }.$$

We spare of the details of the proof of the previous theorem to the reader. Notice that if $ker\left(S\right)\setminus ker\left(T\right)\ne ⌀$, then $\arg {min}_{\parallel T\left(x\right)\parallel \ge 1}\parallel S\left(x\right)\parallel =ker\left(S\right)\setminus \{x\in :\parallel T\left(x\right)\parallel <1\}$. However, if $ker\left(S\right)\subseteq ker\left(T\right)$, then all four reformulations are equivalent, as shown in the next theorem, whose proof’s details we spare again to the reader.

Theorem 5.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. If $ker\left(S\right)\subseteq ker\left(T\right)$, then

$$\arg \underset{\parallel S\left(x\right)\parallel \ne 0}{max}\frac{\parallel T\left(x\right)\parallel }{\parallel S\left(x\right)\parallel }=\arg \underset{\parallel T\left(x\right)\parallel \ne 0}{min}\frac{\parallel S\left(x\right)\parallel }{\parallel T\left(x\right)\parallel }.$$

4. Solving the Maxmin Problem $max\parallel \mathbf{T}\left(\mathbf{x}\right)\parallel$ Subject to $\parallel \mathbf{S}\left(\mathbf{x}\right)\parallel \le \mathbf{1}$

We will distinguish between two cases.

4.1. First Case: S Is an Isomorphism Over Its Image

By bearing in mind Theorem 5, we can focus on the first reformulation proposed at the beginning of the previous section:

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ min\parallel S\left(x\right)\parallel \hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ \parallel S\left(x\right)\parallel \le 1\hfill \end{array}\right.$$

The idea we propose to solve the previous reformulation is to make use of supporting vectors (see [7,8,9,10]). Recall that if $R:X\to Y$ is a continuous linear operator between Banach spaces, then the set of supporting vectors of R is defined by

$$\mathrm{suppv}\left(R\right):=\arg \underset{\parallel x\parallel \le 1}{max}\parallel R\left(x\right)\parallel .$$

The idea of using supporting vectors is that the optimization problem

$$\left\{\begin{array}{c}max\parallel R\left(x\right)\parallel \hfill \\ \parallel x\parallel \le 1\hfill \end{array}\right.$$

whose solutions are by definition the supporting vectors of R, can be easily solved theoretically and computationally (see [8]).

Our first result towards this direction considers the case where S is an isomorphism over its image.

Theorem 6.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. Suppose that S is an isomorphism over its image and $S^{-1}:S\left(X\right)\to X$ denotes its inverse. Suppose also that $S\left(X\right)$ is complemented in Y, being $p:Y\to Y$ a continuous linear projection onto $S\left(X\right)$. Then

$$S^{-1}\left(S\left(X\right)\cap \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥\right)\subseteq \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel .$$

If, in addition, $\parallel p\parallel =1$, then

$$\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel =S^{-1}\left(S\left(X\right)\cap \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥\right).$$

Proof. 

We will show first that

$$S\left(X\right)\cap \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥\subseteq S\left(\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel \right).$$

Let ${\displaystyle y_0=S\left(x_0\right)\in \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥}$. We will show that ${\displaystyle x_0\in \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel }$. Indeed, let $x\in X$ with $\parallel S\left(x\right)\parallel \le 1$. Since $\parallel S\left(x_0\right)\parallel =\parallel y_0\parallel \le 1$, by assumption we obtain

$$\begin{array}{ccc}\hfill \parallel T\left(x\right)\parallel & =& ∥\left(T\circ S^{-1}\circ p\right)\left(S\left(x\right)\right)∥\hfill \\ & \le & ∥\left(T\circ S^{-1}\circ p\right)\left(y_0\right)∥\hfill \\ & =& ∥\left(T\circ S^{-1}\circ p\right)\left(S\left(x_0\right)\right)∥\hfill \\ & =& ∥T\left(x_0\right)∥.\hfill \end{array}$$

Now assume that $\parallel p\parallel =1$. We will show that

$$S\left(\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel \right)\subseteq S\left(X\right)\cap \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥.$$

Let ${\displaystyle x_0\in \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel }$, we will show that ${\displaystyle S\left(x_0\right)\in \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥}$. Indeed, let $y\in {\mathsf{B}}_Y$. Observe that

$$∥S\left(S^{-1}\left(p\left(y\right)\right)\right)∥=\parallel p\left(y\right)\parallel \le \parallel y\parallel \le 1$$

so by assumption

$$\begin{array}{ccc}\hfill ∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥& =& ∥T\left(S^{-1}\left(p\left(y\right)\right)\right)∥\hfill \\ & \le & \parallel T\left(x_0\right)\parallel \hfill \\ & =& ∥T\left(S^{-1}\left(p\left(S\left(x_0\right)\right)\right)\right)∥\hfill \\ & =& ∥\left(T\circ S^{-1}\circ p\right)\left(S\left(x_0\right)\right)∥.\hfill \end{array}$$

 □

Notice that, in the settings of Theorem 6, $S^{-1}\circ p$ is a left-inverse of S, in other words, S is a section, as in Theorem 1(2).

Taking into consideration that every closed subspace of a Hilbert space is 1-complemented (see [11,12] to realize that this fact characterizes Hilbert spaces of dimension $\ge 3$), we directly obtain the following corollary.

Corollary 1.

Let $T,S\in \mathcal{B}(X,Y)$ where X is a Banach space and Y a Hilbert space. Suppose that S is an isomorphism over its image and let $S^{-1}:S\left(X\right)\to X$ be its inverse. Then

$$\begin{array}{ccc}\hfill \arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel & =& S^{-1}\left(S\left(X\right)\cap \arg \underset{\parallel y\parallel \le 1}{max}∥\left(T\circ S^{-1}\circ p\right)\left(y\right)∥\right)\hfill \\ & =& S^{-1}\left(S\left(X\right)\cap \mathrm{suppv}\left(T\circ S^{-1}\circ p\right)\right)\hfill \end{array}$$

where $p:Y\to Y$ is the orthogonal projection on $S\left(X\right)$.

4.2. The Moore–Penrose Inverse

If $B\in {\mathbb{K}}^{m\times n}$, then the Moore–Penrose inverse of B, denoted by $B^{+}$, is the only matrix $B^{+}\in {\mathbb{K}}^{n\times m}$ which verifies the following:

• $B=B{B}^{+}B$.
• $B^{+}=B^{+}B{B}^{+}$.
• $B{B}^{+}={\left(B{B}^{+}\right)}^{\ast }$.
• $B^{+}B={\left(B^{+}B\right)}^{\ast }$.

If $ker\left(B\right)=0$, then $B^{+}$ is a left-inverse of B. Even more, $B{B}^{+}$ is the orthogonal projection onto the range of B, thus we have the following result from Corollary 1.

Corollary 2.

Let $A,B\in {\mathbb{R}}^{m\times n}$ such that $ker\left(B\right)=\left\{0\right\}$. Then

$$\begin{array}{ccc}\hfill B\left(\arg \underset{{\parallel Bx\parallel }_2\le 1}{max}{\parallel Ax\parallel }_2\right)& =& B{\mathbb{R}}^n\cap \arg \underset{{\parallel y\parallel }_2\le 1}{max}{∥A{B}^{+}y∥}_2\hfill \\ & =& B{\mathbb{R}}^n\cap \mathrm{suppv}\left(A{B}^{+}\right)\hfill \end{array}$$

According to the previous Corollary, in its settings, if $y_0\in \arg {max}_{{\parallel y\parallel }_2\le 1}{∥A{B}^{+}y∥}_2$ and there exists $x_0\in {\mathbb{R}}^n$ such that $y_0=B{x}_0$, then $x_0\in \arg {max}_{{\parallel Bx\parallel }_2\le 1}{\parallel Ax\parallel }_2$ and $x_0$ can be computed as

$$x_0=B^{+}B{x}_0=B^{+}{y}_0.$$

4.3. Second Case: S Is Not an Isomorphism Over Its Image

What happens if S is not an isomorphism over its image? Next theorem answers this question.

Theorem 7.

Let $T,S\in \mathcal{B}(X,Y)$ where X and Y are Banach spaces. Suppose that $ker\left(S\right)\subseteq ker\left(T\right)$. If

$$\begin{array}{cccc}\hfill \pi :& \hfill X& \to & X/ker\left(S\right)\hfill \\ & \hfill x& \mapsto & \pi \left(x\right):=x+ker\left(S\right)\hfill \end{array}$$

denotes the quotient map, then

$$\arg \underset{\parallel S\left(x\right)\parallel \le 1}{max}\parallel T\left(x\right)\parallel ={\pi }^{-1}\left(\arg \underset{\parallel \overline{S}\left(\pi \left(x\right)\right)\parallel \le 1}{max}\parallel \overline{T}\left(\pi \left(x\right)\right)\parallel \right),$$

where

$$\begin{array}{cccc}\hfill \overline{T}:& \hfill \frac{X}{ker\left(S\right)}& \to & Y\hfill \\ & \hfill \pi \left(x\right)& \mapsto & \overline{T}\left(\pi \left(x\right)\right):=T\left(x\right)\hfill \end{array}$$

and

$$\begin{array}{cccc}\hfill \overline{S}:& \hfill \frac{X}{ker\left(S\right)}& \to & Y\hfill \\ & \hfill \pi \left(x\right)& \mapsto & \overline{S}\left(\pi \left(x\right)\right):=S\left(x\right).\hfill \end{array}$$

Proof. 

Let $x_0\in \arg {max}_{\parallel S\left(x\right)\parallel \le 1}\parallel T\left(x\right)\parallel$. Fix an arbitrary $y\in X$ with $\parallel \overline{S}\left(\pi \left(y\right)\right)\parallel \le 1$. Then $\parallel S\left(y\right)\parallel =\parallel \overline{S}\left(\pi \left(y\right)\right)\parallel \le 1$ therefore

$$\parallel \overline{T}(\pi \left(x_0\right)\parallel =\parallel T\left(x_0\right)\parallel \ge \parallel T\left(y\right)\parallel =\parallel \overline{T}\left(\pi \left(y\right)\right)\parallel .$$

This shows that $\pi \left(x_0\right)\in \arg {max}_{\parallel \overline{S}\left(\pi \left(x\right)\right)\parallel \le 1}\parallel \overline{T}\left(\pi \left(x\right)\right)\parallel$. Conversely, let

$$\pi \left(x_0\right)\in \arg \underset{\parallel \overline{S}\left(\pi \left(x\right)\right)\parallel \le 1}{max}\parallel \overline{T}\left(\pi \left(x\right)\right)\parallel .$$

Fix an arbitrary $y\in X$ with $\parallel S\left(y\right)\parallel \le 1$. Then $\parallel \overline{S}\left(\pi \left(y\right)\right)\parallel =\parallel S\left(y\right)\parallel \le 1$ therefore

$$\parallel T\left(x_0\right)\parallel =\parallel \overline{T}\left(\pi \left(x_0\right)\right)\parallel \ge \parallel \overline{T}\left(\pi \left(y\right)\right)\parallel =\parallel T\left(y\right)\parallel .$$

This shows that $x_0\in \arg {max}_{\parallel S\left(x\right)\parallel \le 1}\parallel T\left(x\right)\parallel$. □

Please note that in the settings of Theorem 7, if $S\left(X\right)$ is closed in Y, then $\overline{S}$ is an isomorphism over its image $S\left(X\right)$, and thus in this case Theorem 7 reduces the reformulated maxmin to Theorem 6.

4.4. Characterizing When the Finite Dimensional Reformulated Maxmin Has a Solution

The final part of this section is aimed at characterizing when the finite dimensional reformulated maxmin has a solution.

Lemma 3.

Let $S:X\to Y$ be a bounded operator between finite dimensional Banach spaces X and Y. If ${\left(x_n\right)}_{n\in \mathbb{N}}$ is a sequence in $\{x\in X:\parallel S(x)\parallel \le 1\}$, then there is a sequence ${\left(z_n\right)}_{n\in \mathbb{N}}$ in $ker\left(S\right)$ so that ${(x_n+z_n)}_{n\in \mathbb{N}}$ is bounded.

Proof. 

Consider the linear operator

$$\begin{array}{cccc}\hfill \overline{S}:& \hfill \frac{X}{ker\left(S\right)}& \to & Y\hfill \\ & \hfill x+ker\left(S\right)& \mapsto & \overline{S}(x+ker\left(S\right))=S\left(x\right).\hfill \end{array}$$

Please note that

$$∥\overline{S}(x_n+ker\left(S\right))∥=\parallel S\left(x_n\right)\parallel \le 1$$

for all $n\in \mathbb{N}$, therefore the sequence ${(x_n+ker\left(S\right))}_{n\in \mathbb{N}}$ is bounded in $\frac{X}{ker\left(S\right)}$ because $\frac{X}{ker\left(S\right)}$ is finite dimensional and $\overline{S}$ has null kernel so its inverse is continuous. Finally, choose $z_n\in ker\left(S\right)$ such that $\parallel x_n+z_n\parallel <\parallel x_n+ker\left(S\right)\parallel +\frac{1}{n}$ for all $n\in \mathbb{N}$. □

Lemma 4.

Let $A,B\in {\mathbb{R}}^{m\times n}$. If $ker\left(B\right)\subseteq ker\left(A\right)$, then A is bounded on $\{x\in {\mathbb{R}}^n:\parallel Bx\parallel \le 1\}$ and attains its maximum on that set.

Proof. 

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a sequence in $\{x\in {\mathbb{R}}^n:\parallel Bx\parallel \le 1\}$. In accordance with Lemma 3, there exists a sequence ${\left(z_n\right)}_{n\in \mathbb{N}}$ in $ker\left(B\right)$ such that ${(x_n+z_n)}_{n\in \mathbb{N}}$ is bounded. Since $A\left(x_n\right)=A(x_n+z_n)$ by hypothesis (recall that $ker\left(B\right)\subseteq ker\left(A\right)$), we conclude that A is bounded on $\{x\in {\mathbb{R}}^n:\parallel Bx\parallel \le 1\}$. Finally, let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a sequence in $\{x\in {\mathbb{R}}^n:\parallel Bx\parallel \le 1\}$ such that ${\displaystyle \parallel A{x}_n\parallel \to \underset{\parallel Bx\parallel \le 1}{max}\parallel Ax\parallel }$ as $n\to \infty$. Please note that $∥\overline{A}(x_n+ker\left(B\right))∥=\parallel A{x}_n\parallel$ for all $n\in \mathbb{N}$, so ${\left(\overline{A}(x_n+ker\left(B\right))\right)}_{n\in \mathbb{N}}$ is bounded in ${\mathbb{R}}^m$ and so is ${\left(\overline{A}(x_n+ker\left(B\right))\right)}_{n\in \mathbb{N}}$ in $\frac{{\mathbb{R}}^n}{ker\left(B\right)}$. Fix $b_n\in ker\left(B\right)$ such that $\parallel x_n+b_n\parallel <\parallel x_n+ker\left(B\right)\parallel +\frac{1}{n}$ for all $n\in \mathbb{N}$. This means that ${(x_n+b_n)}_{n\in \mathbb{N}}$ is a bounded sequence in ${\mathbb{R}}^n$ so we can extract a convergent subsequence ${\left(x_{n_k}+b_{n_k}\right)}_{k\in \mathbb{N}}$ to some $x_0\in X$. At this stage, notice that $∥B\left(x_{n_k}+b_{n_k}\right)∥=∥B{x}_{n_k}∥\le 1$ for all $k\in \mathbb{N}$ and ${\left(B\left(x_{n_k}+b_{n_k}\right)\right)}_{k\in \mathbb{N}}$ converges to $B{x}_0$, so $\parallel B{x}_0\parallel \le 1$. Note also that, since $ker\left(B\right)\subseteq ker\left(A\right)$, ${\left(∥A{x}_{n_k}∥\right)}_{n\in \mathbb{N}}$ converges to $\parallel A{x}_0\parallel$, which implies that

$$x_0\in \arg \underset{\parallel Bx\parallel \le 1}{max}\parallel Ax\parallel .$$

 □

Theorem 8.

Let $A,B\in {\mathbb{R}}^{m\times n}$. The reformulated maxmin problem

$$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel Bx\parallel \le 1\hfill \end{array}\right.$$

has a solution if and only if $ker\left(B\right)\subseteq ker\left(A\right)$.

Proof. 

If $ker\left(B\right)\subseteq ker\left(A\right)$, then we just need to call on Lemma 4. Conversely, if $ker\left(B\right)\setminus ker\left(A\right)\ne ⌀$, then it suffices to consider the sequence ${\left(n{x}_0\right)}_{n\in \mathbb{N}}$ for $x_0\in ker\left(B\right)\setminus ker\left(A\right)$, since $\parallel B\left(n{x}_0\right)\parallel =0\le 1$ for all $n\in \mathbb{N}$ and $\parallel A\left(n{x}_0\right)\parallel =n\parallel A\left(x_0\right)\parallel \to \infty$ as $n\to \infty$. □

4.5. Matrices on Quotient Spaces

Consider the maxmin

$$\left\{\begin{array}{c}max\parallel T\left(x\right)\parallel \hfill \\ \parallel S\left(x\right)\parallel \le 1\hfill \end{array}\right.$$

being X and Y Banach spaces and $T,S\in \mathcal{B}(X,Y)$ with $ker\left(S\right)\subseteq ker\left(T\right)$. Notice that if ${\left(e_i\right)}_{i\in I}$ is a Hamel basis of X, then ${\left(e_i+ker\left(S\right)\right)}_{i\in I}$ is a generator system of $\frac{X}{ker\left(S\right)}$. By making use of the Zorn’s Lemma, it can be shown that ${\left(e_i+ker\left(S\right)\right)}_{i\in I}$ contains a Hamel basis of $\frac{X}{ker\left(S\right)}$. Observe that a subset C of $\frac{X}{ker\left(S\right)}$ is linearly independent if and only if $S\left(C\right)$ is a linearly independent subset of Y.

In the finite dimensional case, we have

$$\begin{array}{cccc}\hfill \overline{B}:& \hfill \frac{{\mathbb{R}}^n}{ker\left(B\right)}& \to & {\mathbb{R}}^m\hfill \\ & \hfill x+ker\left(B\right)& \mapsto & \overline{B}(x+ker\left(B\right)):=Bx.\hfill \end{array}$$

and

$$\begin{array}{cccc}\hfill \overline{A}:& \hfill \frac{{\mathbb{R}}^n}{ker\left(B\right)}& \to & {\mathbb{R}}^m\hfill \\ & \hfill x+ker\left(B\right)& \mapsto & \overline{A}(x+ker\left(B\right)):=Ax.\hfill \end{array}$$

If $\{e_1,\cdots ,e_n\}$ denotes the canonical basis of ${\mathbb{R}}^n$, then $\{e_1+ker\left(B\right),\cdots ,e_n+ker\left(B\right)\}$ is a generator system of $\frac{{\mathbb{R}}^n}{ker\left(B\right)}$. This generator system contains a basis of $\frac{{\mathbb{R}}^n}{ker\left(B\right)}$ so let $\{e_{j_1}+ker\left(B\right),\cdots ,e_{j_l}+ker\left(B\right)\}$ be a basis of $\frac{{\mathbb{R}}^n}{ker\left(B\right)}$. Please note that $\overline{A}\left(e_{j_k}+ker\left(B\right)\right)=A{e}_{j_k}$ and $\overline{B}\left(e_{j_k}+ker\left(B\right)\right)=B{e}_{j_k}$ for every $k\in \{1,\cdots ,l\}$. Therefore, the matrix associated with the linear map defined by $\overline{B}$ can be obtained from the matrix B by removing the columns corresponding to the indices $\{1,\cdots ,n\}\setminus \{j_1,\cdots ,j_l\}$, in other words, the matrix associated with $\overline{B}$ is $\left[B{e}_{j_1}|\cdots |B{e}_{j_l}\right]$. Similarly, the matrix associated with the linear map defined by $\overline{A}$ is $\left[A{e}_{j_1}|\cdots |A{e}_{j_l}\right]$. As we mentioned above, recall that a subset C of $\frac{{\mathbb{R}}^n}{ker\left(B\right)}$ is linearly independent if and only if $B\left(C\right)$ is a linearly independent subset of ${\mathbb{R}}^m$. As a consequence, in order to obtain the basis $\{e_{j_1}+ker\left(B\right),\cdots ,e_{j_l}+ker\left(B\right)\}$, it suffices to look at the rank of B and consider the columns of B that allow such rank, which automatically gives us the matrix associated with $\overline{B}$, that is, $\left[B{e}_{j_1}|\cdots |B{e}_{j_l}\right]$.

Finally, let

$$\begin{array}{cccc}\hfill \pi :& \hfill {\mathbb{R}}^n& \to & \frac{{\mathbb{R}}^n}{ker\left(B\right)}\hfill \\ & \hfill x& \mapsto & \pi \left(x\right):x+ker\left(B\right)\hfill \end{array}$$

denote the quotient map. Let $l:=\mathrm{rank}\left(B\right)=dim\left(\frac{{\mathbb{R}}^n}{ker\left(B\right)}\right)$. If $x=(x_1,\cdots ,x_l)\in {\mathbb{R}}^l$, then ${\sum }_{k=1}^l{x}_k\left(e_{j_k}+ker\left(B\right)\right)\in \frac{{\mathbb{R}}^n}{ker\left(B\right)}$. The vector $z\in {\mathbb{R}}^n$ defined by

$$z_p:=\left\{\begin{array}{cc}\hfill x_k& p=j_k\hfill \\ \hfill 0& p\notin \{j_1,\cdots ,j_l\}\hfill \end{array}\right.$$

verifies that

$$p\left(z\right)=\sum _{k=1}^l{x}_k\left(e_{j_k}+ker\left(B\right)\right).$$

To simplify the notation, we can define the map

$$\begin{array}{cccc}\hfill \alpha :& \hfill {\mathbb{R}}^l& \to & {\mathbb{R}}^n\hfill \\ & \hfill x& \mapsto & \alpha \left(x\right):=z\hfill \end{array}$$

where z is the vector described right above.

5. Discussion

Here we compile all the results from the previous subsections and define the structure of the algorithm that solves the maxmin (3).

Let $A,B\in {\mathbb{R}}^{m\times n}$ with $ker\left(B\right)\subseteq ker\left(A\right)$. Then

$$\left\{\begin{array}{c}{max\parallel Ax\parallel }_2\hfill \\ {min\parallel Bx\parallel }_2\hfill \end{array}\right.\stackrel{\mathrm{reform}}{⟶}\left\{\begin{array}{c}{max\parallel Ax\parallel }_2\hfill \\ {\parallel Bx\parallel }_2\le 1\hfill \end{array}\right.$$

Case 1:

$ker\left(B\right)=\left\{0\right\}$. $B^{+}$ denotes the Moore–Penrose inverse of B.

$$\left\{\begin{array}{c}{max\parallel Ax\parallel }_2\hfill \\ {\parallel Bx\parallel }_2\le 1\hfill \end{array}\right.\stackrel{\mathrm{supp}.\mathrm{vec}.}{⟶}\left\{\begin{array}{c}max\parallel A{B}^{+}{y\parallel }_2\hfill \\ {\parallel y\parallel }_2\le 1\hfill \end{array}\right.\stackrel{\mathrm{solution}}{⟶}\left\{\begin{array}{c}{\displaystyle y_0\in \arg \underset{{\parallel y\parallel }_2\le 1}{max}{\parallel A{B}^{+}y\parallel }_2}\hfill \\ \mathrm{rank}\left(B\right)=\mathrm{rank}\left(\left[B\right|y_0]\right)\hfill \end{array}\right.\stackrel{\mathrm{final}\mathrm{sol}.}{⟶}{x}_0:=B^{+}{y}_0$$

Case 2:

$ker\left(B\right)\ne \left\{0\right\}$. $\overline{B}=\left[B{e}_{j_1}|\cdots |B{e}_{j_l}\right]$ where $\mathrm{rank}\left(B\right)=l=\mathrm{rank}\left(\overline{B}\right)$ and $\overline{A}=\left[A{e}_{j_1}|\cdots |A{e}_{j_l}\right]$.

$$\left\{\begin{array}{c}{max\parallel Ax\parallel }_2\hfill \\ {\parallel Bx\parallel }_2\le 1\hfill \end{array}\right.\stackrel{\mathrm{case}1}{⟶}\left\{\begin{array}{c}max\parallel \overline{A}{y\parallel }_2\hfill \\ \parallel \overline{B}{y\parallel }_2\le 1\hfill \end{array}\right.\stackrel{\mathrm{solution}}{⟶}{y}_0\stackrel{\mathrm{final}\mathrm{sol}.}{⟶}{x}_0:=\alpha \left(y_0\right)$$

In case a real-life problem is modeled like a maxmin involving more operators, we proceed as the following remark establishes in accordance with the preliminaries of this manuscript (reducing the number of multiobjective functions to avoid the lack of solutions):

Remark 1.

Let ${\left(T_n\right)}_{n\in \mathbb{N}}$ and ${\left(S_n\right)}_{n\in \mathbb{N}}$ be sequences of continuous linear operators between Banach spaces X and Y. The maxmin

$$\left\{\begin{array}{c}max\parallel T_n\left(x\right)\parallel n\in \mathbb{N}\hfill \\ min\parallel S_n\left(x\right)\parallel n\in \mathbb{N}\hfill \end{array}\right.$$

(4)

can be reformulated as (recall the second typical reformulation)

$$\left\{\begin{array}{c}max\sum _{n=1}^{\infty }{\parallel T_n\left(x\right)\parallel }^2\hfill \\ min\sum _{n=1}^{\infty }{\parallel S_n\left(x\right)\parallel }^2\hfill \end{array}\right.$$

(5)

which can be transformed into a regular maxmin as in (1) by considering the operators

$$\begin{array}{cccc}\hfill T:& \hfill X& \to & {\ell }_2\left(Y\right)\hfill \\ & \hfill x& \mapsto & T\left(x\right):={\left(T_n\left(x\right)\right)}_{n\in \mathbb{N}}\hfill \end{array}$$

and

$$\begin{array}{cccc}\hfill S:& \hfill X& \to & {\ell }_2\left(Y\right)\hfill \\ & \hfill x& \mapsto & S\left(x\right):={\left(S_n\left(x\right)\right)}_{n\in \mathbb{N}}\hfill \end{array}$$

obtaining then

$$\left\{\begin{array}{c}{max\parallel T\left(x\right)\parallel }^2\hfill \\ {min\parallel S\left(x\right)\parallel }^2\hfill \end{array}\right.$$

which is equivalent to

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

Observe that for the operators T and S to be well defined it is sufficient that $(\parallel T_n{\parallel )}_{n\in \mathbb{N}}$ and $(\parallel S_n{\parallel )}_{n\in \mathbb{N}}$ be in ${\ell }_2$.

6. Materials and Methods

The initial methodology employed in this research work is the Mathematical Modelling of real-life problems. The subsequent methodology followed is given by the Axiomatic-Deductive Method framed in the First-Order Mathematical language. Inside this framework, we deal with the Category Theory (the main category involved is the Category of Banach spaces with the Bounded Operators). The final methodology used is the implementation of our mathematical results in the MATLAB programming language.

7. Conclusions

We finally enumerate the novelties provided in this work, which serve as conclusions for our research:

• We prove that the original maxmin problem

  $$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ min\parallel Bx\parallel \hfill \end{array}\right.$$

  (6)

  has no solution (Theorem 2).
• We then rewrite (6) as

  $$\left\{\begin{array}{c}max\parallel Ax\parallel \hfill \\ \parallel Bx\parallel \le 1\hfill \end{array}\right.$$

  (7)

  which still models the real-life problem very accurately and has a solution if and only if $ker\left(B\right)\subseteq ker\left(A\right)$ (Theorem 8).
• We provide an exact solution of (7) assuming $ker\left(B\right)\subseteq ker\left(A\right)$, not an heuristic method for approaching it. See Section 5.
• A MATLAB code is provided for computing the solution to the maxmin problem. See Appendix C.
• Our solution applies to design truly optimal minimum stored-energy TMS coils and to find more complex optimal geolocations involving statistical variables. See Appendixes Appendix A and Appendix B.
• This article represents an interdisciplinary work involving pure abstract nontrivial proven theorems and programming codes that can be directly applied to different situations in the real world.