Positive linear combinations of intervals

Question

Given two intervals at $i\in\{0,1\}$ $I_i=[-a_i,a_i]$ where $0<a_0<a_1=1-a_0<1$ and a third interval $I=[-a,a]$ where $0<a<\frac{1}2$, when is there an $\alpha,\beta\in\Bbb R$ such that $\alpha I_0 +\beta I_1\subseteq[-a,a]$ with $\alpha+\beta=1$, $\alpha,\beta>0$? That is if $x_0\in I_0, x_1\in I_1$, $\alpha x_0 +\beta x_1\in I$ should hold true.

Answer 1

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

proof:

• We first prove that if (1) holds, then (2) holds.

Let $x_0 = a_0$ and $x_1 = a_1$. Easy to see that $x_0 \in I_0$ and $x_1 \in I_1$, thus by (1), we have $$ \alpha x_0 + \beta x_1 = \alpha a_0 + \beta a_1 \leq a $$

• Next, we prove if (2) holds, then (1) holds.

If $x_0 \in I_0$ and $x_1 \in I_1$, then we can bound $\alpha x_0 + \beta x_1$ as follows: $$ -a \leq -(\alpha a_0 + \beta a_1) \leq \alpha x_0 + \beta x_1 \leq \alpha a_0 + \beta a_1 \leq a $$ thus $\alpha x_0 + \beta x_1 \in I$.

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According to the discussion above, we only need to find when there is $0 < \alpha, \beta < 1$ with $\alpha + \beta = 1$ such that $$ \alpha a_0 + \beta a_1 \leq a \tag{3} $$

Substituting $\beta = 1 - \alpha$ into (3), we get $$ \alpha a_0 + (1 - \alpha) a_1 \leq a $$

• When $a_1 > a > a_0$, let $\alpha = \frac{a_1 - a}{a_1 - a_0}$, thus $1 > \alpha > 0$. We have $$ \alpha a_0 + (1 - \alpha)a_1 = a_1 + \alpha(a_0 - a_1) = a $$

Thus when $a_1 > a > a_0$, there exist $\alpha = \frac{a_1-a}{a_1 - a_0}$ and $\beta = 1- \alpha$ satisfying the requirement.

• When $a \geq a_1$, this case is impossible since $a_1 > 0.5$ and $a < 0.5$.

• When $a \leq a_0$, we have $$ \alpha a_0 + (1 - \alpha)a_1 > \alpha a + (1 - \alpha)a = a $$

Thus when $a \leq a_0$, no solution exists.

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In summary, there exist $0 < \alpha, \beta < 1$ with $\alpha + \beta = 1$ satisfying the requirement in question iff $a > a_0$.