Can a "continuous" convex combination not be element of the convex hull?

Question

Short version of question: can a "continuous" convex combination not be element of the convex hull?

I am not a mathematician, so please excuse me if I am not precise. I consider first, e.g., 4 dimensional real valued vectors $a \in \mathbb{R}^4$. Now consider a set of $n$ vectors $a_i, i={1,2,...,n}$ and the set containing all convex combinations of these vectors \begin{equation} C=\left\{\sum_{i=1}^k \hat{w}_i a_i | k\in\{1,2,...,n\}, i\in\{1,2,...,n\}, \sum_{i=1}^k \hat{w}_i = 1, \hat{w}_i \geq 0 \forall i\right\} \ . \end{equation} As far as I understand the definition of the convex hull, see 3.definition in Wikipedia, the set $C$ is the convex hull of these vectors and trivially any convex combination of vectors lies in $C$.

Now, I am taking a look at the following problem over a non-convex region $\Omega \subset \mathbb{R}^2$ for vector valued vector functions $a(x) \in \mathbb{R}^4$ and $x \in \Omega$ \begin{equation} \lambda = \int_\Omega w(x) a(x) dx \in \mathbb{R}^4 \end{equation} with real valued $w(x) \in \mathbb{R}$ with the following properties \begin{equation} \int_\Omega w(x) dx = 1 , \quad w(x) \geq 0 \quad \forall x \in \Omega \end{equation} at what $w(x)$ is a distribution. Due to the properties of $w(x)$, I interpret for any $w(x)$ the integral $\lambda$ to be a "continuous" convex combination of the values of $a(x)$ over $\Omega$. The set of all possible $\lambda$ for all distributions $w(x)$ having the properties mentioned above will be denoted as \begin{equation} \Lambda = \left\{\lambda | \lambda = \int_\Omega w(x) a(x) dx , \int_\Omega w(x) dx = 1 , w(x) \geq 0 \quad \forall x \in \Omega\right\} \end{equation} and the convex hull of all values of $a(x)$ as \begin{equation} \Gamma = \left\{\sum_{i=1}^k \hat{w}_i a(x_i) | k\in\mathbb{N}, x_i \in \Omega, \sum_{i=1}^k \hat{w}_i = 1, \hat{w}_i \geq 0 \forall i\right\} \ . \end{equation}

Question: are the sets $\Lambda$ and $\Gamma$ the same or can I find a $w(x)$ such that the resulting $\lambda \not\in \Gamma$? This would be somehow very unintuitive for me, but I am not a mathematician. I keep thinking of this with Dirac distributions defined for $\Omega$ and the $n$ going to infinity in the case of simple vectors, as sketched at the beginning. Therefore, I can not imagine any case for which I should be able to combine values of $a(x)$ and end outside of $\Gamma$. But the more I read about distributions, the more weird things are possible! Any help is very appreciated. Thanks a lot!

Answer 1

As others have said the two sets are the same. The fact that $\Gamma\subset \Lambda$ essentially follows from the fact that a convex combination $\sum_1^k w_i a_i$ is equal to $\int a(x)w(x)\, dx$ with $w=\sum_1^k w_i\delta_{a_i}$ (Dirac deltas concentrated at $a_i$).

The opposite inclusion follows from Jensen's inequality. Consider the function (this is called characteristic (or indicator) function in convex analysis) $$I_\Gamma(x)=\begin{cases} 0 & x\in \Gamma \\ +\infty & x\notin \Gamma\end{cases}$$ This function is convex and $\Gamma=\{x\ :\ I_\Gamma(x)=0\}$. Now let $\lambda = \int a(x)w(x)\, dx\in\Lambda$. By Jensen's inequality $$ I_\Gamma(\lambda)\le \int I_\Gamma(a(x))w(x)\, dx=0, $$ so $I_\Gamma(\lambda)=0$, which means that $\lambda \in \Gamma$.

Answer 2

One solution is to consider the following characterization of convex sets:

A closed set is convex if and only if it is the intersection of closed half-spaces.

So, all half-spaces are of the form $x\cdot v \geq c$, so if we want to prove this in the case where $\Gamma$ is closed, all that we need is the following: If $a(x)\cdot v\leq c$ everywhere, then $\int_{\Omega}w(x)a(x)\cdot v\,dx\leq c$ for any distribution $w$. However, this is trivial by monotonicity: That integrand cannot exceed $w(x)c$ so the integral cannot exceed $\int_{\Omega}w(x)c\,dx = c$, as desired. This tells us that $\Lambda$ is a subset of $\text{cl}(\Gamma)$.

To work out the boundary, you just note that for any point $p$ on the boundary, there is at least one plane through it not intersecting the interior. Moreover, if $w(x)$ assigns positive weight a portion of $a(x)$ not on this plane, then the integral will not be on this plane (since it will be strictly below it). Otherwise, $w(x)$ only assigns weight on that plane, in which case we're dealing with the two dimensional analog of the problem (since the intersection of that plane and $\Omega$ and $\Gamma$ acts as expected). So, we can work out an induction proof on dimension that will make sure the boundary works out.