# Does the range of convex hull of a set equal to the range of that set?

I'm trying to construct convex hulls of figures, and this question came up to my mind. Literally what I want to know, is that if the statement is true in Banach spaces.

My idea about this is that when I want to make some figure convex, I just connect the parts where it is "concave" with the line. And if so, than intuitively the range of body itself and the range of its convex hull should be the same.

Thank you. And together with this, if one tell me any useful technique for constructing convex hull by means of support function, I would be very grateful.

• What is the range of a set and the range of a convex hull? What do you mean by constructing a set from the support function? – copper.hat Dec 29 '16 at 5:35
• Litterally, a range of a set $A$, is the following $R(A) = sup\{d(x,y)| x,y \in A \}$. Where $sup$ is for supremum (not for a support), and $d$ is some metric. For simplicity we can take Euclidean distance. Or in the case of Banach spaces metric induced from norm. – kolobokish Dec 29 '16 at 13:41
• Sometimes called diameter. – kolobokish Dec 29 '16 at 13:48
• I found a partial answer to my question here math.stackexchange.com/questions/1836008/… – kolobokish Dec 29 '16 at 13:48

It is straightforward to see that if $A \subset B$ then $\operatorname{diam} A \le \operatorname{diam} B$, hence $\operatorname{diam} C \le \operatorname{diam} (\operatorname{co} C)$.

If $x,y \in \operatorname{co} C$, then there are $\alpha,\beta$ with $\sum_i \alpha_i = \sum_j \beta_j = 1$ and $\alpha_i \ge 0, \beta_j \ge 0$, and $u_i,v_j \in C$ such that $x = \sum_i \alpha_i u_i, y = \sum_j \beta_j v_j$. (All of the sums are over a finite range.)

Then $x-y = \sum_i \sum_j \alpha_i \beta_j (u_i-v_j)$ and so $\|x-y\| \le \sum_i \sum_j \alpha_i \beta_j \|u_i-v_j\| \le \operatorname{diam} C$ and so $\operatorname{diam} (\operatorname{co} C) \le \operatorname{diam} C$.

This is true in any normed space.

It is not clear what you mean by constructing the set from a support function.

One can always define $C = \{ x | \langle h, x \rangle \le \sigma(h) \}$, where $\sigma$ is the support function. Then $C$ is convex and $\sup_{x \in C} \langle h, x \rangle = \sigma(h)$.

Elaboration:

Note that $\sum_i \alpha_i = \sum_j \beta_j =1$. \begin{eqnarray} x-y &=& \sum_i \alpha_i u_i - \sum_j \beta_j v_j \\ &=& \sum_i \alpha_i (u_i - \sum_j \beta_j v_j) \\ &=& \sum_i \alpha_i \sum_j \beta_j(u_i - v_j) \\ &=& \sum_i \sum_j \alpha_i \beta_j(u_i - v_j) \end{eqnarray}

• I don't understand the part of the proof, when you write $x-y = \sum_{i}\sum_{j}\alpha_{i}\beta_{j}(u_{i}-v_{j})$. – kolobokish Dec 30 '16 at 1:19
• I added an elaboration. – copper.hat Dec 30 '16 at 1:27