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Changed wording as per Matt F.'s suggestion
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Kacper Kurowski
Kacper Kurowski
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Let $(V, \lVert \cdot \rVert)$ be a normed space. LetFor any $A \subseteq V$, let $O(A)$ be the intersection of all closed balls containing $A$, or more precisely, let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one can allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to one of at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Let $(V, \lVert \cdot \rVert)$ be a normed space. Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one can allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Let $(V, \lVert \cdot \rVert)$ be a normed space. For any $A \subseteq V$, let $O(A)$ be the intersection of all closed balls containing $A$, or more precisely, let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one can allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to one of at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property
Deleted spurious 'For'
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LSpice
LSpice
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Which convex subsets of a normed space are intersections of balls?

Let $(V, \lVert \cdot \rVert)$ be a normed space. For Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one callcan allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Which convex subsets of a normed space are intersections of balls

Let $(V, \lVert \cdot \rVert)$ be a normed space. For Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one call allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Which convex subsets of a normed space are intersections of balls?

Let $(V, \lVert \cdot \rVert)$ be a normed space. Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one can allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property
small fixes
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Kacper Kurowski
Kacper Kurowski
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Let $(V, \lVert \cdot \rVert)$ be a normed space. For Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one call allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that directionhasdirection has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Let $(V, \lVert \cdot \rVert)$ be a normed space. For Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one call allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line is unique, then that directionhas the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

Let $(V, \lVert \cdot \rVert)$ be a normed space. For Let $O \colon 2^V \to 2^V$ be defined by the formula: \begin{equation} \forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)), \end{equation} where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one call allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:

I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to at least two noncollinear directions. Here, the reason being that:

  • if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,
  • if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property
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Kacper Kurowski
Kacper Kurowski
  • 820
  • 5
  • 10