# Give bundles $x,y\in \mathbb R^n$, there must exist a budget $B\supset\{x,y\}$ and a demand $D(B)\in[x,y]$?

For a problem in revealed preference. Give bundles $$x,y\in \mathbb R^n$$, must there exist a budget $$B\supset\{x,y\}$$ and a demand $$D(B)\in[x,y]$$?

Intuitively, this mean that we have two bundles, and we want to determine a budget set $$B$$ including both $$x$$ and $$y$$, and the demand in $$B$$ is also in the interval $$[x,y]$$.

The budget set is determined by a price vector $$p\in\mathbb R^n$$.

Let $$U$$ be a $$C^1$$, monotone and quasi-concave utility function. Let $$\nabla U(z)$$ be the gradient of $$U$$ at the bundle $$z$$ (If we assume preferences are monotone then $$\nabla U(z) \gg 0)$$ for all $$z \in \mathbb{R}^n$$).

There are three cases to consider:

1. $$\nabla U(x) \cdot x \ge \nabla U(x) \cdot y$$. In this case, we can take the budget with price $$p = \nabla U(x)$$ and total income $$p \cdot x$$. This case corresponds to the setting where $$y$$ is below the hyperplane of the indifference curve at $$x$$. In this case $$x \in D(B)$$ and $$y \in B$$

2. $$\nabla U(y) \cdot y \ge \nabla U(y) \cdot x$$. In this case, we can take the budget with price $$p = \nabla U(y)$$ and total income $$p \cdot y$$. This case correponds to the setting where $$x$$ is below the hyperplane tangent to the indifference curve at $$y$$. In this case $$y \in D(B)$$ and $$x \in B$$.

3. Assume that case 1 and case 2 do not hold and consider the following problem $$\alpha^\ast = \max_{\alpha \in [0,1]} U(\alpha x + (1-\alpha) y).$$ The objective function is continuous and the constraint set $$[0,1]$$ is compact, so an optimal solution exists. Note that the derivative of the objective function is given by: $$\nabla U(\alpha x + (1-\alpha)y) \cdot (x - y).$$ First, let us exclude $$\alpha = 0$$ and $$\alpha = 1$$ as optimal solutions.

a) if $$\alpha = 0$$ is an optimal solution then the derivative of the objective function must be negative for $$\alpha = 0$$ (as it is on the boundary of the feasible set $$[0,1]$$): $$\nabla U(y)(x-y) \le 0,$$ But this gives $$\nabla U(y)\cdot y > \nabla U(y) \cdot x$$, which gives case 2.

b) if $$\alpha = 1$$ is an optimal solution, the derivative of the objective function must be positive at $$\alpha = 1$$: $$\nabla U(x)(x - y) \ge 0$$ But this gives $$\nabla U(x)\cdot x \ge \nabla U(x) \cdot y$$ which gives case 1.

Conclude that the optimal value of $$\alpha$$ must be strictly between zero and 1. Let $$z = \alpha x + (1-\alpha) y$$. As the solution is interior, we have that: $$\nabla U(z) \cdot (x - y) = 0.$$ In particular \begin{align*} \nabla U(z) \cdot z &= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\ &= \alpha \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot y,\\ &= \nabla U(z) \cdot y. \end{align*} and \begin{align*} \nabla U(z) \cdot z &= \nabla U(z)\cdot (\alpha x + (1-\alpha) y),\\ &= (\alpha-1) \nabla U(z)\cdot(x - y) + \nabla U(z) \cdot x,\\ &= \nabla U(z) \cdot x. \end{align*} As such, if we define the budget $$B$$ with price $$p = \nabla U(z)$$ and income $$p \cdot z$$, we have that $$x, y \in B$$ and that $$z \in D(B)$$ while $$z \in [x,y]$$.