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

Question

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!

Answer 1

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 $1 = p \cdot x(p,1) = p'\cdot x(p',1)$): $$ 2 - p' \cdot x(p,1) - p \cdot x(p',1) \le 0. \tag{1} $$

There are 3 cases to consider.

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 $p$).