Absorbing set - WikiHQ

In functional analysis and related areas of mathematics an absorbing set in a vector space is a set <math>S</math> which can be "inflated" or "scaled up" to eventually always include any given point of the vector space.

Alternative terms are radial or absorbent set.

Every neighborhood of the origin in every topological vector space is an absorbing subset.

Definition

Notation for scalars

Suppose that <math>X</math> is a vector space over the field <math>\mathbb{K}</math> of real numbers <math>\R</math> or complex numbers <math>\Complex,</math> and for any <math>-\infty \leq r \leq \infty,</math> let

<math display=block>B_r = \{a \in \mathbb{K} : |a| < r\} \quad \text{ and } \quad B_{\leq r} = \{a \in \mathbb{K} : |a| \leq r\}</math>

denote the open ball (respectively, the closed ball) of radius <math>r</math> in <math>\mathbb{K}</math> centered at <math>0.</math> <!-- For any <math>-\infty \leq r \leq R \leq \infty</math> and subset <math>A \subseteq X,</math> let

Similarly, if <math>K \subseteq \mathbb{K}</math> and <math>k</math> is a scalar then let <math>K A = \{k a : k \in K, a \in A\},</math> <math>K x = \{k x : k \in K\},</math> <math>k A = \{k a : a \in A\},</math> and <math>\mathbb{K} x = \{k x : k \in \mathbb{K}\} = \operatorname{span} \{x\}.</math> -->

Define the product of a set <math>K \subseteq \mathbb{K}</math> of scalars with a set <math>A</math> of vectors as <math>K A = \{k a : k \in K, a \in A\},</math> and define the product of <math>K \subseteq \mathbb{K}</math> with a single vector <math>x</math> as <math>K x = \{k x : k \in K\}.</math>

Preliminaries

Balanced core and balanced hull

A subset <math>S</math> of <math>X</math> is said to be Balanced set| if <math>a s \in S</math> for all <math>s \in S</math> and all scalars <math>a</math> satisfying <math>|a| \leq 1;</math> this condition may be written more succinctly as <math>B_{\leq 1} S \subseteq S,</math> and it holds if and only if <math>B_{\leq 1} S = S.</math>

Given a set <math>T,</math> the smallest balanced set containing <math>T,</math> denoted by <math>\operatorname{bal} T,</math> is called the of <math>T</math> while the largest balanced set contained within <math>T,</math> denoted by <math>\operatorname{balcore} T,</math> is called the of <math>T.</math>

These sets are given by the formulas

<math display=block>\operatorname{bal} T ~=~ {\textstyle\bigcup\limits_{|c| \leq 1 c \, T = B_{\leq 1} T</math>

and

<math display=block>\operatorname{balcore} T ~=~ \begin{cases}

{\textstyle\bigcap\limits_{|c| \geq 1 c \, T & \text{ if } 0 \in T \\

\varnothing & \text{ if } 0 \not\in T, \\

\end{cases}</math>

(these formulas show that the balanced hull and the balanced core always exist and are unique).

A set <math>T</math> is balanced if and only if it is equal to its balanced hull (<math>T = \operatorname{bal} T</math>) or to its balanced core (<math>T = \operatorname{balcore} T</math>), in which case all three of these sets are equal: <math>T = \operatorname{bal} T = \operatorname{balcore} T.</math>

If <math>c</math> is any scalar then

<math display=block>\operatorname{bal} (c \, T) = c \, \operatorname{bal} T = |c| \, \operatorname{bal} T</math>

while if <math>c \neq 0</math> is non-zero or if <math>0 \in T</math> then also

<math display=block>\operatorname{balcore} (c \, T) = c \, \operatorname{balcore} T = |c| \, \operatorname{balcore} T.</math>

One set absorbing another

If <math>S</math> and <math>A</math> are subsets of <math>X,</math> then <math>A</math> is said to <math>S</math> if it satisfies any of the following equivalent conditions:

Definition: There exists a real <math>r > 0</math> such that <math>S \, \subseteq \, c \, A</math> for every scalar <math>c</math> satisfying <math>|c| \geq r.</math> Or stated more succinctly, <math>S \; \subseteq \; {\textstyle\bigcap\limits_{|c| \geq r c \, A</math> for some <math>r > 0.</math>
If the scalar field is <math>\R</math> then intuitively, "<math>A</math> absorbs <math>S</math>" means that if <math>A</math> is perpetually "scaled up" or "inflated" (referring to <math>t A</math> as <math>t \to \infty</math>) then (for all positive <math>t > 0</math> sufficiently large), all <math>t A</math> will contain <math>S;</math> and similarly, <math>t A</math> must also eventually contain <math>S</math> for all negative <math>t < 0</math> sufficiently large in magnitude.
This definition depends on the underlying scalar field's canonical norm (that is, on the absolute value <math>|\cdot|</math>), which thus ties this definition to the usual Euclidean topology on the scalar field. Consequently, the definition of an absorbing set (given below) is also tied to this topology.
There exists a real <math>r > 0</math> such that <math>c \, S \, \subseteq \, A</math> for every non-zero scalar <math>c \neq 0</math> satisfying <math>|c| \leq r.</math> Or stated more succinctly, <math>{\textstyle\bigcup\limits_{0 < |c| \leq r c \, S \, \subseteq \, A</math> for some <math>r > 0.</math>
Because this union is equal to <math>\left(B_{\leq r} \setminus \{0\}\right) S,</math> where <math>B_{\leq r} \setminus \{0\} = \{c \in \mathbb{K} : 0 < |c| \leq r\}</math> is the closed ball with the origin removed, this condition may be restated as: <math>\left(B_{\leq r} \setminus \{0\}\right) S \, \subseteq \, A</math> for some <math>r > 0.</math>
The non-strict inequality <math>\,\leq\,</math> can be replaced with the strict inequality <math>\,<\,,</math> which is the next characterization.
There exists a real <math>r > 0</math> such that <math>c \, S \, \subseteq \, A</math> for every non-zero and because <math>Y</math> is 1-dimensional, the only vector topologies on it are the Hausdorff Euclidean topology and the trivial topology, which is a subset of the Euclidean topology.

So regardless of which of these vector topologies is on <math>Y,</math> the set <math>U \cap Y</math> will be a neighborhood of the origin in <math>Y</math> with respect to its unique Hausdorff vector topology (the Euclidean topology).

Thus <math>U</math> is absorbing.</li>

</ul>

</li>

<li><math>A</math> contains the origin and for every 1-dimensional vector subspace <math>Y</math> of <math>X,</math> <math>A \cap Y</math> is absorbing in <math>Y</math> (according to any defining condition of "absorbing" other than this one).

This characterization shows that the property of being absorbing in <math>X</math> depends on how <math>A</math> behaves with respect to 1 (or 0) dimensional vector subspaces of <math>X.</math> In contrast, if a finite-dimensional vector subspace <math>Z</math> of <math>X</math> has dimension <math>n > 1</math> and is endowed with its unique Hausdorff TVS topology, then <math>A \cap Z</math> being absorbing in <math>Z</math> is no longer sufficient to guarantee that <math>A \cap Z</math> is a neighborhood of the origin in <math>Z</math> (although it will still be a necessary condition). For this to happen, it suffices for <math>A \cap Z</math> to be an absorbing set that is also convex, balanced, and closed in <math>Z</math> (such a set is called a and it will be a neighborhood of the origin in <math>Z</math> because every finite-dimensional Euclidean space, including <math>Z,</math> is a barrelled space).

</li>

</ol>

If <math>\mathbb{K} = \Reals</math> then to this list can be appended:

<li value="9">The algebraic interior of <math>A</math> contains the origin (that is, <math>0 \in {}^{i}A</math>).</li>

If <math>A</math> is balanced then to this list can be appended:

<li value="10"> For every <math>x \in X,</math> there exists a scalar <math>c \neq 0</math> such that <math>x \in c A</math> (or equivalently, such that <math>c x \in A</math>).</li>
For every <math>x \in X,</math> there exists a scalar <math>c</math> such that <math>x \in c A.</math>

If <math>A</math> is convex or balanced then to this list can be appended:

<li value="12"> For every <math>x \in X,</math> there exists a positive real <math>r > 0</math> such that <math>r x \in A.</math>
The proof that a balanced set <math>A</math> satisfying this condition is necessarily absorbing in <math>X</math> follows immediately from condition (10) above and the fact that <math>c A = |c| A</math> for all scalars <math>c \neq 0</math> (where <math>r := |c| > 0</math> is real).
The proof that a convex set <math>A</math> satisfying this condition is necessarily absorbing in <math>X</math> is less trivial (but not difficult). A detailed proof is given in this footnote

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