# Prove: The law of demand holds if WA, Walras' law, homogeneity of degree 0, and homogeneity of degree 1 in wealth hold for Walrasian demand functions

## Problem

I am asked to prove the following result (MWG Exercise 2.F.5):

The law of demand always holds if the walrasian demand function $$x(\mathbf{p},w)$$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

## My Attempt So Far

This is what I have so far:

Suppose that the Walrasian demand function $$x(\mathbf{p},w)$$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $$w$$. We want to prove that, for any price change from $$\mathbf{p}$$ to $$\mathbf{p}'$$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $$w$$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $$\mathbf{p}$$ and $$\mathbf{p}'$$. We consider the following cases:

Case 1: Suppose that $$x(\mathbf{p},1) = x(\mathbf{p}',1)$$. Then the inequality $$(2)$$ always holds with $$(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$$.

From now on, suppose that $$x(\mathbf{p},1) \neq x(\mathbf{p}',1)$$.

Case 2: Suppose that $$\mathbf{p}' \cdot x(\mathbf{p},1) > 1$$ and $$\mathbf{p} \cdot x(\mathbf{p}',1) > 1$$. Notice that, by Walras' law, $$1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Here is where I got stuck:

Case 3: Suppose that $$\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$$ and $$\mathbf{p} \cdot x(\mathbf{p}',1) > 1$$.

Case 4: Suppose that $$\mathbf{p}' \cdot x(\mathbf{p},1) > 1$$ and $$\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$$.

## My Question

Could someone please help me derive the inequality $$(2)$$ for Case 3 and Case 4 above? We could basically just consider Case 3. I want to prove that, if $$\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$$ and $$\mathbf{p} \cdot x(\mathbf{p}',1) > 1$$, then \begin{align*} \mathbf{p}' \cdot x(\mathbf{p},1) + \mathbf{p} \cdot x(\mathbf{p}',1) \geq 2. \end{align*} Thanks a lot in advance!

We need to show that: $$(p - p')\cdot(x(p,1) - x(p', 1)) \le 0.$$ Note that if $$x(p,1) = x(p', 1)$$ then this is obviously satisfied, so assume that $$x(p,1) \ne x(p', 1)$$.
Note that the condition is equivalent to (using $$1 = p \cdot x(p,1) = p'\cdot x(p',1)$$): $$2 - p' \cdot x(p,1) - p \cdot x(p',1) \le 0. \tag{1}$$
1. If $$1 \le p \cdot x(p',1)$$ and $$1 \le p' \cdot x(p,1)$$ then we can add the two inqualities together to obtain condition $$(1)$$.
2. If $$p \cdot x(p,1) = 1 \ge p \cdot x(p', 1)$$ then let $$z \ge 1$$ be such that $$1 = p \cdot x(p', z)$$. (here $$z = 1/(p \cdot x(p', 1))$$. Then WARP requires that $$p' \cdot x(p', z) = z \le p' \cdot x(p,1)$$. We have the two conditions: $$1 = p \cdot x(p', z) \text{ and } z \le p' \cdot x(p,1).$$ Adding both together gives: $$1 + z - p \cdot x(p', z) - p' \cdot x(p,1) \le 0.$$ Using $$x(p', z) = x(p', 1) z$$ this can be rewritten as: $$1 + z\underbrace{(1 - p \cdot x(p',1))}_{\ge 0} - p' \cdot x(p,1) \le 0.$$ As $$z \ge 1$$, we get, $$2 - p \cdot x(p', 1) - p' \cdot x(p,1) \le 0.$$ This gives condition $$(1)$$
3. If $$p' x(p', 1) = 1 \ge p' x(p,1)$$ we get a similar reasoning as in point $$2$$ before (exchanging $$p'$$ and $$p$$).