If you have a Marshallian demand function that is strictly convex, then it satisfies WARP. How to prove this?
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1$\begingroup$ You mean strictly convex (Marshallian) demand, or strictly convex preferences? $\endgroup$– tdmCommented Feb 19 at 7:46
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$\begingroup$ former case. Strictly convex Marshallian demand. $\endgroup$– babededeeptidoCommented Feb 19 at 14:57
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2$\begingroup$ Convex in which arguments? In prices? In income? In prices and income? $\endgroup$– BertrandCommented Feb 20 at 9:28
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1 Answer
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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: $$ p_\alpha(\alpha x_0 + (1-\alpha) x_1) > p_\alpha x_\alpha. $$
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.
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: $$ \begin{align*} w = p_\alpha x_\alpha &< p_\alpha x_0,\\ &= \alpha p_0 x_0 + (1-\alpha) p_1 x_0,\\ &\le \alpha w + (1-\alpha) p_1 x_1,\\ &= \alpha w + (1-\alpha) w = w. \end{align*} $$
