Theorem about convex sets

Question

I want to prove a theorem (the link is after the whole text here) and in order to do that I need to prove three preliminary statements. I tried to prove them all but I'm already stuck in the first, and this one seems to be necessary to prove the others. I hope someone here can help me on this, thanks.

We have two compact convex subsets $A_1, A_2$ of $\mathbb{R}^n$, also assume that the origin is in the interior of both subsets. Denote $E_1$ for the linear hull of $A_1$, same thing for $E_2$. Furthermore, for any $x_0\in A_1$, denote $C_{x_0} = \{y\in E_2: \ \forall x\in A_1, \ \langle y, x_0-x\rangle \geq 0\}$. Finally, consider $\lambda_1,\lambda_2 >0$ two arbitrary and fixed real numbers and consider the set $A = \lambda_1A_1+\lambda_2A_2$. The three steps to prove the theorem are:

i) Show that for any $a\in A$ there are $x\in A_1$ and $y\in C_x\cap A_2$ such that $a = \lambda _1 x+ \lambda_2 y$.

ii) This decomposition is unique.

iii) The map $a\mapsto (x,y)$ described above is Lipschitz and smooth almost everywhere.

My attempt: Ok, for the first one I know than $a$ can be written as $a=\lambda_1 x + \lambda_2 y$ for some $x\in A_1, y\in A_2$, but it's possible that $y\notin C_x\cap A_2$. My idea was to consider elements like $ty$, $t\in\mathbb{R}$, which are elements in $C_x$, and knowing that $0\in A$, we could try to take smalls values for $t$ to get elements in $A_2$, one of them will fits the properties desired. But the way to show this is nothing clear.

About the third item, to show that the map is smooth almost everywhere, I can use a result which says that if a function is Lipschitz in an open subset of $\mathbb{R}^n$, then this function is smooth almost everywhere. Therefore, I want to show there is some convenient open subset containing $A$. I just don't know how to show this.

By the way, this is for proving this theorem, the part which says One can show that f is a homogeneous polynomial of degree n, therefore it can be written as...

Thank you!

I think it's false as written. If $a$ is in the interior of $A_1$ then $C_a$ is empty: take $x=a+\varepsilon y\in A_1$; then $\langle y,a-x\rangle < 0$. Maybe you want this just for $a$ in the boundary of $A$? — , Jun 08 '14 at 18:00
How do you know that $x=a+\varepsilon y\in A_1$? We know that $a=\lambda_1 x+ \lambda_2 y$ for some $x\in A_1, y\in A_2$, the best we can do is to right $x=\frac{a}{\lambda_1}-\frac{\lambda_2 y}{\lambda_1}$. — Integral, Jun 08 '14 at 18:13
My $x$ is the one in the defining condition of $C_a$, but yours is the one in the decomposition $a=\lambda_1x+\lambda_2y$. They're not related. — , Jun 08 '14 at 18:19
It's a bit confusing. Could you keep the notation $a\in A, x\in A_1, y\in A_2$. So you are saying that don't matter which $a\in A$ and $y\in E_2$ I get, there will at least one $x\in A_1$ not satisfying $\langle y, x-a \rangle \geq 0$ ? — Integral, Jun 08 '14 at 18:38
@Steven Taschuk I made a mistake when writing the problem. The sets are not $C_a$, but $C_x$. I hope this change is enough to solve the problem. — Integral, Jun 08 '14 at 19:55
When you say the word "interior", are you using it in the sense of the topology induced from $\mathbb{R}^n$? Because I don't see why you bother defining $E_1$ and $E_2$, since both will equal to $\mathbb{R}^n$. But if you allow $E_1$ and $E_2$ to be proper subspaces of $\mathbb{R}^n$, then the statement (i) is clearly false unless the two subspaces are mutually orthogonal. — Willie Wong, Jun 10 '14 at 13:26
Don't have the time to write a full proof now, but I think what you should do is to first define $B = \lambda_1^{-1}(a - \lambda_2 A_2)$, and let $C = A_1\cap B$ which is still a compact set. Let $x\in C$ be the minimiser of $d(a,C)$. The associated $x,y$ should be the unique decomposition you need in the first two steps. — Willie Wong, Jun 10 '14 at 13:59
@Integral To be clear.... do you mean linear hull or convex hull? — Squirtle, Jun 16 '14 at 20:04

Theorem about convex sets

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