Let $X$ be a normed vector space and $A$ be a subset of $X$. $\operatorname{conv}(A)$ is called the intersection of all convex subsets of $X$ that contain $A$

a) Show that $\operatorname{conv}(A)$ is a convex set

b) Show that

$$\operatorname{conv}(A) = \bigg\{\sum_{i=1}^n\lambda _i x_i : \sum_{i=1}^n \lambda_i =1, \lambda_i\ge 0, x_i\in A, i=1,\cdots,n\bigg\}$$

c) If $A$ is compact then is $\operatorname{conv}(A)$ compact?

d) Show that if $A\subseteq \mathbb{R}^n$ is compact then $\operatorname{conv}(A)$ is compact

a) Take two elements $a$ and $b$ at the intersection of all convex subsets of $X$ that contain $A$. Now take $\lambda a + (1-\lambda)b$. It is contained in every of the subsets of $X$ that contain $A$, therefore it is contained in the intersection. Q.E.D.


[$\Leftarrow$] Suppose by induction that $\sum_{i=1}^n\lambda_i x_i $ belongs to $\operatorname{conv}(A)$. We must prove that $\sum_{i=1}^{k+1}\lambda_i x_i $, with $\sum_{i=1}^{k+1}=1$ also belongs.

$$\sum_{i=1}^{k+1}\lambda_i x_i = \sum_{i=1}^{k}\lambda_i x_i + \lambda_{k+1}x_{k+1} = \sum_{i=1}^{k}\lambda_i x_i + (1-\sum_{i=1}^n\lambda_i)x_{k+1}$$

Now choose $\delta$ such that $\delta\sum_{i=1}^{k}\lambda_i=1$, then

$$\frac{\delta}{\delta} \left(\sum_{i=1}^{k}\lambda_i x_i + (1-\sum_{i=1}^n\lambda_i)x_{k+1}\right)= \left(\frac{1}{\delta}\sum_{i=1}^{k}(\delta\lambda_i) x_i + \frac{1}{\delta}(\delta-1)x_{k+1}\right) =\\ \frac{1}{\delta}x + \left(1-\frac{1}{\delta}\right)x_{k+1}$$

which is a combination of elements of $A$ that sums to $1$

[$\Rightarrow$] Can I just say that $x\in\operatorname{conv}(A)$, then $x = 1x + 0\cdot everything$ therefore it is a combination that sums to $1$ of elements of $A$?

c) A hint is provided: show that $\operatorname{conv}(A\cup B)$ is the image by a continuous function of the compact $\{(\alpha,\beta; \alpha,\beta\ge 0, \alpha + \beta = 1)\}\times A\times B$. I don't understand how to show this.

d) Any hints?

  • 1
    $\begingroup$ Your part $b$ is correct. For part $c$, recall that the image of a compact set under a continuous function is also compact. I am not sure what part $d$ is asking (i.e it seems the same as part $c$. $\endgroup$ – rubikscube09 Mar 2 '19 at 22:11
  • $\begingroup$ Part b is not correct; you cannot conclude on the basis that $x = 1x$ that $x$ belongs to the proposed convex hull, as $x$ may not belong to $A$. Instead, try proving that the proposed convex hull is convex. If it is, then it is a convex set containing $A$, so if $x$ is not in this set, it must not be in the convex hull of $A$. $\endgroup$ – Theo Bendit Mar 2 '19 at 23:07
  • $\begingroup$ d) is proved in Rudin's Functional Analysis. $\endgroup$ – Kavi Rama Murthy Mar 2 '19 at 23:48
  • $\begingroup$ An interesting side note on part (d): it does not hold true in more general vector spaces--see this SE post. $\endgroup$ – David M. Mar 8 '19 at 3:14
  • $\begingroup$ @rubikscube09 Part (c) is asking about a general normed vector space $X$, while part (d) is specific to the case $X=\mathbb{R}^n$. $\endgroup$ – David M. Mar 8 '19 at 13:34

Part a and the first implication of part b are correct.

For the converse of part $b$, note that writing $x = 1x+0$ does not make it a member of the RHS of the second conclusion, which requires that each element of the combination be in $A$, while $x \notin A$ may be possible.

Therefore, since you have shown that $\mbox{conv}(A)$ contains the LHS, showing that the LHS is convex and contains $A$ will show that it contains $\mbox{conv}(A)$.

That the RHS contains $A$ follows from $x = 1x$, where $\color{blue}{x \in A}$. Now we need to show that the RHS is convex.

So pick two elements, $\sum_{i=1}^n \lambda_i x_i$ and $\sum_{j=1}^m \mu_j y_j$, and a $t \in [0,1]$. We get: $$ t\left(\sum_{i=1}^n \lambda_ix_i\right) + (1-t)\left(\sum_{j=1}^m \mu_jy_j\right) = \sum_{k=1}^{n+m} \xi_kz_k $$

where : $$ \xi_k = \begin{cases} t\lambda_k \quad \quad \ \ \quad \quad k \leq n \\ (1-t)\mu_{k-n} \quad k > n \end{cases} $$

and : $$ z_k = \begin{cases} x_k \quad \quad k \leq n \\ y_{k-n} \quad \ k > n \end{cases} $$

so all the $z_k \in A$ and $\sum_{k=1}^{m+n} \xi_k = \sum_{i=1}^n t\lambda_i + \sum_{j=1}^n (1-t)\mu_j = t + 1-t = 1$. $\xi_k \geq 0$ for all $k$ is clear.

Therefore , the set on the RHS is convex and contains $A$. In particular, therefore it contains $\mbox{conv}(A)$, completing the proof.

For part $c$, the example given in the link above by David suffices : I will reproduce and elaborate on it upon the request of OP.

For part $d$, just think of closure first. Fix $x_n \in \mbox{conv}(A), x_n \to x$, all we need to show is that $x \in A$.

Which seems very easy to see, and then when you look at the structure of $\mbox{conv}(A)$, realize is not a very easy task. The point is, each $x_n$ is an arbitrary finite convex combination of vectors from $A$, but finiteness still is not good enough : we'd like there to be an upper bound on the number of vectors in the linear combination, while keeping $\mbox{conv}(A)$ intact. This can be done in $\mathbb R^d$ at least:

Caratheodory Convexity Theorem : In $\mathbb R^n$, every vector in $\mbox{conv}(A)$ can be written as a convex combination of (no more than) $\color{blue}{n+1}$ vectors from $A$!

I'll adapt the proof from the source. Fix $x \in \mbox{conv}(A)$, and define the set $T = \{k : x$ can be written as a convex combination of $k$ elements of $A\}$. So $T$ is a non-empty subset of the natural numbers, and hence has a smallest element, say $k$. If we show $k \leq n+1$ we are done.

If not, then pick $x_1,...,x_k \in A$ and $\alpha_1,...,\alpha_k$ with $\sum_1^k \alpha_i = 1$ and $x = \sum_1^k \alpha_ix_i$. Since $k-1 > n$ , the set $\{x_i-x_1 : 2 \leq i \leq k\}$ has $k-1$ elements and hence is linearly dependent. Therefore, we can find $\lambda_1,...,\lambda_{k-1}$ not all zero such that $\sum_{1}^{k-1} \lambda_j (x_{j+1} - x_1) = 0$.

Let $C_1 = -\sum_{1}^{k-1} \lambda_i$ and $C_j = \lambda_{j-1}$ for $2 \leq j \leq k$, note that all the $C_i$ cannot be zero (one of $C_2,...,C_k$ is non-zero), and we have : $$ \sum_{i=1}^k C_i = C_1 + \sum_{j=2}^k C_j = - \sum \lambda_i + \sum\lambda_i = 0 $$

and : $$ \sum C_ix_i = \left(-\sum \lambda_i\right) x_1 + \sum \lambda_jx_j = \sum \lambda_j(x_j - x_1)= 0 $$

by choice of $\lambda_i$.

Let us assume WLOG that $C_j > 0$ for some $j$ (that is, we know one of them is non-zero, so we are assuming that this one is also positive : you can see how the argument changes if it is negative). Set $C = \min\left\{\frac{\alpha_i}{C_i} : C_i > 0\right\}$ (a non-empty set by choice) and let $\frac{\alpha_m}{C_m} = C > 0$. (That is, this is the $m$ for which the minimum is attained). We make some observations :

  • $\alpha_i - CC_i \geq 0$ for all $i$ and $\alpha_m - CC_m =0$ by assumption.

  • We have : $$ \sum_{i=1}^k (\alpha_i - CC_i) = \sum \alpha_i - C\sum C_i = 1-0 = 1 $$

  • Furthermore : $$ \sum_{i=1}^k (\alpha_i - CC_i)x_i = \sum \alpha_ix_i - C\sum C_ix_i = x-0 = x $$

Thus, we have written $x$ as a $k$-convex combination, but $\alpha_m-CC_m = 0$, so in fact the above is a $k-1$ convex combination of $x_i$ which remain elements of $A$. In other words, we have a contradiction since the above shows that $k-1\in T$, whose minimum was $k$.

This stemmed from the false assumption that $k > n-1$. Thus, as the conclusion desires, $k \leq n-1$.

Check out Helly's theorem as well!

But now with Caratheodory, what can we do? What we will do is show that $\mbox{conv}(A)$ is sequentially compact i.e. every sequence has a convergent subsequence within $\mbox{conv}(A)$. At least in $\mathbb R^d$ (it holds in more generality), compactness and sequential compactness are equivalent.

So pick a sequence $x_i \in \mbox{conv}(A)$. Write each $x_i$ as a $d+1$-convex combination, $x_i = \sum_{j=1}^{d+1} \lambda_{ji}y_{ji}$, in other words, $x_i$ is a convex combination of $y_{1i},y_{2i},...,y_{(d+1)i}$. We do a procedure which is quite common in functional analysis :

  • Note that $\lambda_{1i}$, the sequence of "first coefficients" of each convex combination, is an infinite subset of the compact $[0,1]$, hence has a convergent subsequence. Call the subsequence $\lambda_{n_k}$, and let $\lambda_{n_k} \to \lambda_1$ as $k \to \infty$ where $\lambda_1 \in [0,1]$.

  • Now, consider the sequence $y_{n_k}$, with $n_k$ as above. By sequential compactness, we get a further subsequence $y_{n_{k_l}}$ of this which converges in $A$, say $y_{n_{k_l}} \to y_1$ as $l \to \infty$. Note that $y_1 \in A$.

  • Further, note that $\lambda_{1n_{k_l}}$ being a subsequence of a convergent sequence also converges to $\lambda_1$.

  • Now, consider $\lambda_{2n_{k_l}}$, it has a convergent subsequence in $[0,1]$, some $y_{n_{k_{l_m}}} \to \lambda_2$. Note that $y_{n_{k_{l_m}}}$ is a subsequence of a convergent sequence and hence also goes to $y_1$.

  • Now proceed iteratively, and stop at $k$ (the stopping at finite time ensures that you are always left with a subsequence : after infinitely many transitions you cannot guarantee that there will be any subsequence left!)

At the end , you obtain $\lambda_i \in [0,1]$ and $y_i \in A$ for $1 \leq i \leq k$, such that there is a subsequence (which we index by $N$) such that $y_{iN} \to y_i$ and $\lambda_{iN} \to \lambda_i$ for all $1 \leq i \leq d+1$.

Now we just have to complete a few checks :

  • Each $\lambda_i \in [0,1]$.

  • We have $\sum_{i=1}^{d+1} \lambda_{iN} = 1$ for all $N$, so dragging $N \to \infty$ gives us $\sum_{i=1}^d \lambda_{i} = 1$.

  • We have $y_{Ni} \to y_i$ for all $i$ , hence by continuity of scalar multiplication we get $\lambda_{Ni}y_{Ni} \to \lambda_iy_i$ for all $i$. Taking the sum over $i$ gives $x_{N} \to \sum_{i=1}^n \lambda_iy_i$, a convex combination of vectors from $A$, hence belonging to $\mbox{conv}(A)$.

Which shows that $\mbox{conv}(A)$ is compact!

You might wonder if $(d)$ generalizes i.e. in spaces other than $\mathbb R^d$ can we get the same result ? Well, we have the following :

In a complete metric space $(X,d)$ with the topology induced by the metric, the convex hull(yes, this is the term for $\mbox{conv}$ !) of a compact set is totally bounded.

For clarification : a subset $S$ of a metric space is totally bounded if for each $\epsilon >0$ there are $x_1,...,x_N \in S$ ($N$ can vary with $\epsilon$) such that $S \subset \cup_1^N B(x_i,\epsilon)$. Or , for all very small values, $S$ can be covered by fintely many balls with radius that value.

This, combined with the general fact that (in a metric space) a set is compact if and only if it is closed and totally bounded, tells you :

The closed convex hull of a compact set is compact.

These are more difficult to prove, though, involving some other concepts.


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