Abstract
In the paper we present the generalized Minkowski functionals. We also establish some useful properties of the Minkowski functionals, criterium of the continuity of such functionals, and a generalization of a Kolmogorov result.
In Honor of Constantin Carathéodory
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1 Introduction
We shall introduce basic ideas, which will be used in the paper.
Let X be a linear topological (Hausdorff) space over the set of real numbers \(\mathbb{R}\). Denote \(\mathbb{R}_{+} = [0,\infty )\), \(\mathbb{R}^{+} = [0,\infty ]\). Also \(0 \cdot \infty = \infty \cdot 0 = 0\). Let A be a subset of X. As usual for \(\alpha \in \mathbb{R}\),
We shall call A a symmetric, provided that \(A = -A\). Moreover, a set \(A \subset X\) is said to be bounded (sequentially) (see [6]) iff for every sequence \(\{t_{n}\} \subset \mathbb{R}\), t n → 0 as \(n \rightarrow \infty\) and every sequence \(\{x_{n}\} \subset A\), the sequence \(\{t_{n} \cdot x_{n}\} \subset X\) satisfies \(t_{n} \cdot x_{n} \rightarrow 0\) as \(n \rightarrow \infty\).
We also recall the idea of generalized metric space (briefly gms) introduced by Luxemburg (see [5] and also [2]). Let X be a set. A function
is called a generalized metric on X, provided that for all x, y, z ∈ X,
-
(i)
d(x, y) = 0 if and only if x = y,
-
(ii)
d(x, y) = d(y, x),
-
(iii)
d(x, y) ≤ d(x, z) + d(z, y),
A pair (X, d) is called a generalized metric space.
Clearly, every metric space is a generalized metric space.
Analogously, for a linear space X, we can define a generalized norm and a generalized normed space.
Let’s note that any generalized metric d is a continuous function.
For if \(x_{n},x,y_{n},y \in (X,d)\) for \(n \in \mathbb{N}\) (the set of all natural numbers) and
i.e. d(x n , x) → 0 and d(y n , y) → 0 as \(n \rightarrow \infty\), then in the case \(d(x,y) <\infty\), we can prove, in standard way, that
But if \(d(x,y) = \infty\), we have for \(\varepsilon> 0\)
and, for n > n 0, \(n,n_{0} \in \mathbb{N}\)
i.e. \(d(x_{n},y_{n}) = \infty\) for n > n 0 and consequently
as claimed.
2 Generalized Minkowski Functionals
Now we shall prove the following basic result.
Theorem 1.
Let X be a linear topological (Hausdorff) space over \(\mathbb{R}\) and let a subset U of X satisfy the conditions:
-
(i)
U is a convex (nonempty) set,
-
(ii)
U is a symmetric set.
Then the function \(p: X \rightarrow \mathbb{R}^{+}\) defined by the formula
where
has the properties:
Proof.
Clearly, \(p(x) \in [0,\infty ]\) for x ∈ X. Since 0 ∈ U, by the definition (1) we get (3). To prove (4), consider at first the case α > 0 (if α = 0, the property (4) is obvious). Assume that \(p(x) <\infty\) for x ∈ X. Then we have
If \(p(x) = \infty\), then {t > 0: x ∈ tU} = ∅. Therefore,
and consequently \(p(\alpha x) = \infty\), i.e. (4) holds true.
Now consider the case α < 0. Taking into account that (ii) implies that also tU for \(t \in \mathbb{R}\) is a symmetric, one gets for x ∈ X and \(p(x) <\infty\)
If \(p(x) = \infty\), then
which implies also \(p(-x) = \infty\), and consequently
Thus, for α < 0, x ∈ X and in view of the first part of the proof,
i.e. (4) has been verified.
Finally, if \(p(x) = \infty\) or \(p(y) = \infty\), then (5) is satisfied. So assume that x, y ∈ X and
Take an \(\varepsilon> 0\). From the definition (1), there exist numbers t 1 ≥ p(x) and t 2 ≥ p(y), \(t_{1} \in A_{x}\), \(t_{2} \in A_{y}\) such that
The convexity of U implies that
and consequently
which means that \(t_{1} + t_{2} \in A_{x+y}\).
Hence
i.e.
and since \(\varepsilon\) is arbitrarily chosen, this concludes the proof. □
Example 1.
Consider \(X = \mathbb{R} \times \mathbb{R}\), \(U = (-1,1)\).
Then
because \(\{t> 0: (x_{1},y_{1}) \in tU\} =\emptyset\) for y 1 ≠ 0.
We see that p is a generalized norm in \(\mathbb{R}^{2} = \mathbb{R} \times \mathbb{R}\) (p takes values in \([0,\infty ]\)).
Remark 1.
The function \(p: X \rightarrow \mathbb{R}^{+}\) defined by (1) we shall call the generalized Minkowski functional of U (also a generalized seminorm).
Remark 2.
Under some stronger assumptions (see e.g. [3]), the function p is called the Minkowski functional of U.
The next basic property of the functional p is given in
Theorem 2.
Suppose that the assumptions of Theorem 1 are satisfied. If, moreover, U is bounded (sequentially), then
Proof.
Assume that p(x) = 0 for x ∈ X. Suppose that x ≠ 0. From the definition of p(x) for every \(\varepsilon _{n} = \frac{1} {n}\), there exists a t n > 0 such that x ∈ t n U, \(n \in \mathbb{N}\) and \(t_{n} <\frac{1} {n}\). Hence, \(x = t_{n}x_{n}\), x n ∈ U for \(n \in \mathbb{N}\) and by the boundedness of U, \(x = t_{n}x_{n} \rightarrow 0\) as \(n \rightarrow \infty\). But clearly x → x, whence x = 0, which is a contradiction and completes the proof. □
Remark 3.
Under the assumptions of Theorem 2, the generalized Minkowski functional is a generalized norm in X.
Let’s note the following useful
Lemma 1.
Let \(U \subset X\) be a convex set and 0 ∈ U. Then
for all 0 ≤α ≤ 1.
The simple proof of this Lemma is omitted here.
Next we prove
Lemma 2.
Let U be as in Theorem 1 . If, moreover, U does not contain half-lines, then
Proof.
For the contrary, suppose that x ≠ 0. By the definition of p(x) for every \(\varepsilon> 0\), there exists a \(0 <t <\varepsilon\) such that x ∈ tU. Take r > 0 and \(\varepsilon <\frac{1} {r}\). Clearly, \(\frac{x} {t} \in U\). Furthermore,
By Lemma 1, \(rx =\alpha \frac{x} {t} \in \alpha U \subset U\), which means that there exists an x ≠ 0 such that for every r > 0, rx ∈ U what contradicts the assumptions on U. This yields our statement. □
We have also
Lemma 3.
Let U be as in Theorem 1 . Then
Proof.
For the contrary, suppose that there exists an x ≠ 0 such that for every r > 0 we have rx ∈ U. Hence
and therefore
which implies that p(x) = 0. From (9) we get x = 0, which is a contradiction. Eventually, one gets the implication (9) and this ends the proof. □
Therefore, Lemmas 2 and 3, we can rewrite as the following
Proposition 1.
Let the assumptions of Theorem 1 be satisfied. Then the generalized Minkowski functional p for U is a generalized norm iff U does not contain half-lines.
3 Properties of the Generalized Minkowski Functionals
In this part we start with the following
Theorem 3.
Let X be a linear topological (Hausdorff) space over \(\mathbb{R}\) and let \(f: X \rightarrow \mathbb{R}^{+}\) be any function with properties:
Define
Then
-
a)
U is a symmetric set,
-
b)
U is a convex (nonempty) set,
-
c)
f = p, i.e. f is the generalized Minkowski functional of U.
Proof.
The conditions a) and b) follow directly from the definition (12) and properties (10) and (11), respectively. To prove c), assume that x ∈ U, thus \(f(x) <\infty\). Therefore, for t > 0
i.e. \(x \in tU \Leftrightarrow f(x) \in [0,t)\) for t > 0.
Thus
whence
Now let \(f(x) = \infty\). For the contrary, assume that \(p(x) <\infty\). Then by the definition of p,
which implies that there exists t > 0 such that x ∈ tU, thus also
and finally f(x) ∈ [0, t) which is impossible. This completes the proof. □
The next result reads as follows.
Theorem 4.
Let X, f, U be as in the Theorem 3 . If U is sequentially bounded, then f = p is a generalized norm.
Proof.
Assume that \(f(x) = p(x) = 0\). One has
therefore, for every \(\varepsilon _{n} = \frac{1} {n}\), \(n \in \mathbb{N}\), there exists \(0 <t_{n} <\frac{1} {n}\) such that x ∈ t n U, i.e. \(x = t_{n}u_{n}\), where u n ∈ U for \(n \in \mathbb{N}\). Since U is bounded
thus x = 0, as claimed. □
4 Continuity of the Generalized Minkowski Functionals
Let’s note the following
Theorem 5.
Let the assumptions of Theorem 1 be satisfied. Then
Proof.
From the assumption, for \(0 <\varepsilon <1\), there exists a neighbourhood V of zero such that
But p(u) < 1 for u ∈ V, whence by Lemma 1 u ∈ U, and therefore, \(V \subset U\), which proves the implication (13). □
We have also
Theorem 6.
Let the assumptions of Theorem 1 be satisfied. Then
Proof.
Let U 0 be a neighbourhood of zero such that \(U_{0} \subset U\). For the contrary, suppose that there exists an \(\varepsilon _{0}> 0\) such that for every neighbourhood V of zero there exists an x ∈ V with \(p(x) \geq \varepsilon _{0}\). Take \(V = V _{n} = \frac{1} {n}U_{0}\), \(n \in \mathbb{N}\) (clearly V n is a neighbourhood of zero). Then there exists an \(x_{n} \in \frac{1} {n}U_{0} \subset \frac{1} {n}U\), such that
Take n such that \(\frac{1} {n} <\varepsilon _{0}\). Thus, one has
i.e.
which contradicts the inequality (15) and completes the proof. □
We have even more.
Theorem 7.
Let the assumptions of Theorem 1 be satisfied. Then
Proof.
First of all, observe that since there exists a neighbourhood V of zero, contained in U, then for x ∈ X
and hence
Therefore, we have \(p(x) <\infty\) for any x ∈ X. Since p is also convex and, by Theorem 6, p is continuous at zero, then by the famous theorem of Bernstein–Doetsch (see e.g. [1]), p is continuous in X, which ends the proof. □
Remark 4.
To see that the condition 0 ∈ int U is essential in Theorems 5 and 6, the reader is referred to Example 1.
Eventually, taking into account Theorems 5 and 7, we can state the following useful result about the continuity of the generalized Minkowski functionals.
Proposition 2.
Under the assumptions of Theorem 1 , the equivalence
holds true.
5 Kolmogorov Type Result
Let X be a linear space (over \(\mathbb{R}\) or \(\mathbb{C}\)—the set of all complex numbers) and a generalized metric space. We say that X is a generalized linear-metric space, if the operations of addition and multiplication by constant are continuous, i.e. if x n → x and y n → y, then \(x_{n} + y_{n} \rightarrow x + y\) and tx n → tx (with respect to a generalized metric in X).
For example, if generalized metric is introduced by a generalized norm, then we get a generalized linear-metric space.
We shall prove the following.
Theorem 8.
Let (X,ϱ) be a generalized linear-metric space over \(\mathbb{R}\) . Suppose that \(U \subset X\) is an open, convex and sequentially bounded set. Then there exists a generalized norm \(\|\cdot \|\) such that the generalized metric induced by this norm is equivalent to a generalized metric ϱ.
Proof.
Take a point x 0 ∈ U, then
is an open, convex, symmetric and sequentially bounded subset of X (the details we omit here).
Define
By Theorem 2 we see that this function is a generalized norm.
At first we shall show the implication:
To this end, take \(\varepsilon> 0\). Then the set \(\varepsilon V\) is also open: for it, because \(f(x) = \frac{1} {\varepsilon } x\), x ∈ X, is a continuous function and
we see that also \(\varepsilon V\) is open. Therefore, \(x_{n} \in \varepsilon V\) for n > n 0 and consequently
i.e. (19) is satisfied.
Conversely, assume that \(\|x_{n}\| \rightarrow 0\) as \(n \rightarrow \infty\). By the definition (18) for every \(\varepsilon _{n} = \frac{1} {n}\), n > n 0, there exists t n > 0 such that
Let \(\varepsilon _{n} \rightarrow 0\), then t n → 0 as \(n \rightarrow \infty\). Also \(\frac{x_{n}} {t_{n}} \in V\), but since V is bounded, then
in the generalized metric ϱ thus ϱ(x n , 0) → 0, which ends the proof. □
Remark 5.
If ϱ is a metric, from Theorem 7, we get the Kolmogorov result (see [4]).
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Czerwik, S., Król, K. (2016). Generalized Minkowski Functionals. In: Pardalos, P., Rassias, T. (eds) Contributions in Mathematics and Engineering. Springer, Cham. https://doi.org/10.1007/978-3-319-31317-7_4
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