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

  • 1
    $\begingroup$ You mean strictly convex (Marshallian) demand, or strictly convex preferences? $\endgroup$
    – tdm
    Feb 19 at 7:46
  • $\begingroup$ former case. Strictly convex Marshallian demand. $\endgroup$ Feb 19 at 14:57
  • 2
    $\begingroup$ Convex in which arguments? In prices? In income? In prices and income? $\endgroup$
    – Bertrand
    Feb 20 at 9:28

1 Answer 1


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: $$ \begin{align*} w = p_\alpha x_\alpha &< p_\alpha x_0,\\ &= \alpha p_0 x_0 + (1-\alpha) p_1 x_0,\\ &\le \alpha w + (1-\alpha) p_1 x_1,\\ &= \alpha w + (1-\alpha) w = w. \end{align*} $$


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