Real linear combinations of intervals Given intervals at $i\in\{0,1\}$ $I_i=[-a_i,a_i]$ where $0<a_0<a_1<1$ and a third interval $I=[-a,a]$ where $0<a<{1}$, when is there an $\alpha,\beta\in\Bbb R$ such that $\alpha I_0 +\beta I_1 =\{\alpha x +\beta y:x\in I_0, y\in I_1\}\subseteq[-a,a]$ with $\alpha+\beta=1$? 
That is if $x_0\in I_0, x_1\in I_1$,  $\alpha x_0 +\beta x_1\in I$ should hold true.
$-\infty<\alpha,\beta<\infty$ is range.
clear $a\in(a_0,a_1)$ is always possible even with $\alpha,\beta\in(0,1)$ by convex combination arguments.
QUESTIONS
1. If $a<a_0$ then does $\frac{1}2<a<a_0<a_1<1$ or  $0<a<a_0<a_1<\frac{1}2$ hold?
2. Is $a>a_1$ always possible?
 A: Part I
We first prove that

$$
\forall x_0 \in I_0, x_1 \in I_1, \ \alpha x_0 + \beta  x_1 \in I \tag{1}
$$
  is equivalent to
  $$
|\alpha|a_0 + |\beta|a_1 \leq a \tag{2}
$$

proof:


*

*First, prove if  (1) holds, then (2) holds.


Let $x_0 = $sign$(\alpha)a_0$, $x_1 =$sign$(\beta)a_1$, where the function sign$(x)$ returns the sign of $x$. Easy to observe that $x_0 \in I_0$ and $x_1 \in I_1$. Thus by (1), we have
$$
\alpha x_0 + \beta x_1 = |\alpha| a_0 + |\beta| a_1 \in I
$$
which implies
$$
|\alpha| a_0 + |\beta| a_1 \leq a
$$


*

*Next, prove if (2) holds, then (1) holds.


For $\forall x_0 \in I_0, x_1 \in I_1$, we can bound $\alpha x_0 + \beta x_1$ as follows:
$$
-(|\alpha| a_0 + |\beta| a_1) \leq \alpha x_0 + \beta x_1 \leq |\alpha| a_0 + |\beta| a_1
$$
Since (2) holds, thus we have
$$
-a \leq \alpha x_0 + \beta x_1 \leq a
$$
implying 
$$
\alpha x_0 + \beta x_1 \in I
$$
EDIT: A More Intuitive Proof
Observe that
$$\alpha I_0 = [\ -|\alpha|a_0,\ |\alpha|a_0\ ] \\
\beta I_1 = [\ -|\beta|a_1, \ |\beta|a_1\ ]
$$
Thus
$$
\alpha I_0 + \beta I_1 = [\ -(|\alpha|a_0 + |\beta|a_1),\ |\alpha|a_0 + |\beta|a_1\ ]
$$
implying
$$
\alpha I_0 + \beta I_1 \in I \Leftrightarrow |\alpha|a_0 + |\beta|a_1 \leq a
$$

Part II

According to the discussion in Part I, we only need to find when there exist $-\infty < \alpha, \beta <\infty$ with $\alpha + \beta = 1$ such that $|\alpha|a_0 + |\beta|a_1 \leq a$ holds. Substituting $\beta = 1 - \alpha$, we have
  $$
|\alpha|a_0 + |1 - \alpha|a_1 \leq a 
$$

We discuss for different cases below:


*

*Case 1. $a \geq a_1$
Let $\alpha = \frac{a + a_1}{a_0 + a_1} > 1$. Then
$$
|\alpha|a_0 + |1 - \alpha|a_1 = \alpha a_0 + (\alpha - 1) a_1 = \alpha (a_0 + a_1) - a_1 = a
$$
Therefore, when $a \geq a_1$, there exist $\alpha = \frac{a+a_1}{a_0 + a_1}$, $\beta = 1 - \alpha$ such that (1) holds.


*

*Case 2. $a < a_0$
We have
$$
|\alpha|a_0 + |1 - \alpha|a_1 > |\alpha|a + |1 - \alpha|a = (|\alpha| + |1 - \alpha|)a \geq a
$$
Here, the fact $|x + y| \leq |x| + |y|, \forall x, y\in\Re$ is used.
Therefore, it is impossible to find a solution in this case.


*

*Case 3. $a_0 \leq a < a_1$
Easy to show that $\alpha = \frac{a_1 - a}{a_1 - a_0}$ is a solution.


In summary, when $a \geq a_0$, there exist $\alpha, \beta \in \Re$ with $\alpha + \beta = 1$ such that (1) holds.

