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It seems that the definition of convex hull is on a form $\sum a_nx_n$ in which coefficients sum up to 1. It implicitly implies that the combination is countable and discrete. I am interested in more abstract expression: $\int x da$ where $\int da =1$. Let me call it generalized convex hull. Are there some known results on the generalized convex hull? I believe it contains the convex hull but still a subset of closed convex hull. However, I cannot find a good example in $R^n$ to understand the concept. I think the generalized convex hull of open unit disk $\{(x,y):x^2+y^2<1\}$ is still the open unit disk? Am I right? If I am right, then we have Convex Hull$=$Generalized Convex Hull$\subset$Closure of Convex Hull in this case.

Another example I find interesting is suppose $A =\{\mathbf{1}_{[0,pt]}(x):pt\in [0,1],x\in [0,1]\} $. Then the convex combination of $A$ is just step functions and they are all discontinuous. However, generalized convex hull of $A$ contains functions which are continuous in $x\in(0,1)$. Am I right? If I am correct, then what is the closure of convex hull of $A$? It seems now that closure is just the generalized convex hull of $A$. Hence we have Convex Hull$\subset$Generalized Convex Hull$=$Closure of Convex Hull. As for topology in this example, we can use that induced by supremum norm on $[0,1]$.

When is that generalized convex hull and convex hull are equivalent and when is that closure of convex hull and generalized convex hull are equivalent?

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  • $\begingroup$ What does "$ \int x da $" mean ? also search the concept cs-convex sets $\endgroup$
    – Red shoes
    Jun 8, 2018 at 2:42
  • $\begingroup$ @Redshoes Hello, suppose you have set $A$,let each element in $A$ is indexed by $t\in T$, say $A = \{a(t):t \in T\}$ . For each $t$, we assign some non-negative value $q(t)$. All those negative values sum up to 1, i.e. $\int_T dq(t)=1$. We want to find out the generalized convex hull such that $\int_{t\in T} a(t)dq(t) $. Hope this will help. Btw, what is "cs-convex set"?? $\endgroup$
    – Ethanabc
    Jun 8, 2018 at 5:44

2 Answers 2

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Definitions I remember from 40 years ago.

Let $E$ be a set in an appropriate topological vector space.

$E$ is convex if all finite sums $$\sum_{i=1}^n a_i x_i$$ belong to $E$, where $n \in \{1,2,3,\dots\}$, $x_i \in E$, $a_i \in [0,1]$, $\sum_{i=1}^n a_i = 1$.

The convex hull of $E$ is the smallest convex set that contains $E$.

$E$ is $\sigma$-convex if all convergent series $$\sum_{i=1}^\infty a_i x_i$$ belong to $E$, where $x_i \in E$, $a_i \in [0,1]$, $\sum_{i=1}^\infty a_i = 1$.

$E$ is measure-convex if all Bochner integrals $$ \int_E x\;d\mu(x) \tag{3}$$ belong to $E$, where $\mu$ is a nonnegative Radon measure with $\mu(E) = 1$.

An enjoyable place to begin learning about such integrals as (3) is

Phelps, Robert R., Lectures on Choquet’s theorem, Lecture Notes in Mathematics. 1757. Berlin: Springer. 124 p. (2001). ZBL0997.46005.

More papers on such things may be found in:

Measure theory, Proc. Conf., Oberwolfach 1979, Lect. Notes Math. 794, (1980).

Measure theory, Proc. Conf., Oberwolfach 1981, Lect. Notes Math. 945, (1982).

Measure theory, Proc. Conf., Oberwolfach 1983, Lect. Notes Math. 1089, (1984).

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  • $\begingroup$ Convexity itself requires only a vector space (in fact, just an affine space). A set is convex if and only if for any two points in the set, the line segment connecting them is also in the set. The finite sum formula follows from that definition. $\sigma$-convex means convex + sequentially closed (every convergent sequence in the set has the limit also in the set). I think measure-convex is effectively convex + closed, but would have to look deeper to be sure. $\endgroup$ Dec 13, 2020 at 17:23
  • $\begingroup$ convex + sequentially closed implies $\sigma$-convex, but the converse is false. Convex+closed implies measure-convex, but the converse is false. $\endgroup$
    – GEdgar
    Dec 13, 2020 at 21:37
  • $\begingroup$ You are right - I fell once again into the trap of finite-dimensional thinking. $\endgroup$ Dec 13, 2020 at 22:21
  • $\begingroup$ Even in one dimension, the set $(0,1)$ is measure convex but not closed; also $(0,1)$ is $\sigma$-convex but not sequentially closed. $\endgroup$
    – GEdgar
    Dec 14, 2020 at 0:27
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Here is My thoughts:

Based your definition $T$ has to be a measure space. I assume the definition of Integral of vector valued functions is just component-wise integration. Let $f : T \to R^n$ and $q : T \to [0, +\infty)$. (for simplicity assume $f(t) \ge 0$ ) consider $$ \int_{t\in T} f(t)dq(t) := x \in \Bbb R^n $$

This is just integration of each component of $f$ Therefore every component of $x$ looks like $$x_i =\sum_{t \in T} f_i (t) q(t)$$ (for simplicity assume $f(t) \ge 0$ ) this is convergent if $\{t \in T ; ~ f(t) \neq 0\}$ is countable. So your definition coincides with the cs-closed set.

For **cs-closed ** set look at following paper in 1972

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0305004100050933

For infinite dimension case your definition is the generalization of cs-closed closed sets in infinite dimensions, which I think I has been used in optimal control.

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