# Proof for Marshallian Demand function

If you have a Marshallian demand function that is strictly convex, then it satisfies WARP. How to prove this?

• You mean strictly convex (Marshallian) demand, or strictly convex preferences?
– tdm
Commented Feb 19 at 7:46
• former case. Strictly convex Marshallian demand. Commented Feb 19 at 14:57
• Convex in which arguments? In prices? In income? In prices and income? Commented Feb 20 at 9:28

Let $$x(p,w)$$ be the demand at prices $$p$$ and income $$w$$.

Let $$x_0 = x(p_0,w)$$ and $$x_1 = x(p_1, w)$$. Note that $$p_0 x_0 = w = p_1 x_1$$

Assume that WARP is violated, so $$x_1 \ne x_0$$, $$p_0 x_0 \ge p_0 x_1 \text{ and } p_1 x_0 \le p_1 x_1.$$

Consider any $$\alpha \in (0,1)$$, define $$p_\alpha = \alpha p_0 + (1-\alpha) p_1$$ and consider $$x_\alpha = x(p_\alpha, w)$$ (which means that $$p_\alpha x_\alpha = w$$).

By convexity of the demand: $$p_\alpha(\alpha x_0 + (1-\alpha) x_1) > p_\alpha x_\alpha.$$

1. if $$p_\alpha x_0 \le p_\alpha x_1$$ then we have: $$p_\alpha x_1 > p_\alpha x_\alpha.$$ As such, \begin{align*} w = p_\alpha x_\alpha &< p_\alpha x_1,\\ &= \alpha p_0 x_1 + (1-\alpha) p_1 x_1,\\ &\le \alpha p_0 x_0 + (1-\alpha) w,\\ &= \alpha w + (1-\alpha) w = w, \end{align*} a contradiction.

2. If $$p_\alpha x_1 \le p_\alpha x_0$$, then: $$p_\alpha x_0 > p_\alpha x_\alpha.$$ In this case we have the contradiction: \begin{align*} w = p_\alpha x_\alpha &< p_\alpha x_0,\\ &= \alpha p_0 x_0 + (1-\alpha) p_1 x_0,\\ &\le \alpha w + (1-\alpha) p_1 x_1,\\ &= \alpha w + (1-\alpha) w = w. \end{align*}