Riemann surface
matlab

Clearly the weak axiom implics that there exists $w>0$ such that for every $\mathrm{p}, \mathrm{p}^{\prime}$, and $\mathrm{w}^{\prime}$, if $\mathrm{p} \cdot x\left(\mathrm{p}^{\prime}, \mathrm{w}^{\prime}\right) \leq \mathrm{w}$ and $x\left(\mathrm{p}^{\prime}, \mathrm{w}^{\prime}\right) \neq x(\mathrm{p}, \mathrm{w})$, then $\mathrm{p}^{\prime} \cdot x(\mathrm{p}, \mathrm{w})>$ w.

Conversely, suppose that such a $\mathrm{w}>0$ exists and that $\mathrm{p} \cdot x\left(\mathrm{p}^{\prime}, \mathrm{w}^{\prime}\right) \leq \mathrm{w}$ and $x\left(\mathrm{p}^{\prime}, \mathrm{w}^{\prime}\right) \neq x(\mathrm{p}, \mathrm{w}) . \quad$ Let $\alpha=\mathrm{w}^{\prime} / \mathrm{w}$. Then $x\left(\mathrm{p}^{\prime}, \mathrm{w}^{\prime}\right)=x\left(\mathrm{p}^{\prime}, \alpha \mathrm{w}\right)=x\left(\alpha^{-1} \mathrm{p}^{\prime}, \mathrm{w}\right)$ by the homogeneity assumption, and $p \cdot x\left(\alpha^{-1} \mathrm{p}^{\prime}, w\right) \leq w$ and $x\left(\alpha^{-1} p^{\prime}, w\right) \neq x(p, w)$. But this implies that $\left(\alpha^{-1} p^{\prime}\right) \cdot x(p, w)>w$, or, equivalently, $p^{\prime} \cdot x(p, w)>\alpha w=w^{\prime}$. Thus the weak axiom holds.

2.F.7 By Propositions 2.E.2 and 2.E.3,
$$\mathrm{p} \cdot \mathrm{S}(\mathrm{p}, \mathrm{w})=\mathrm{p} \cdot \mathrm{D}{\mathrm{p}} x(\mathrm{p}, \mathrm{w})+\mathrm{p} \cdot \mathrm{D}{\mathrm{w}} x(\mathrm{p}, \mathrm{w}) x(\mathrm{p}, \mathrm{w})^{\mathrm{T}}=\mathrm{p} \cdot \mathrm{D}{\mathrm{p}} x(\mathrm{p}, \mathrm{w})+x(\mathrm{p}, \mathrm{w})^{\mathrm{T}}=0$$ By Proposition 2.E.1 and Walras’ law, $$\mathrm{s}(\mathrm{p}, w) p=\mathrm{D}{\mathrm{p}} x(\mathrm{p}, \mathrm{w}) \mathrm{p}+\mathrm{D}{\mathrm{w}} x(\mathrm{p}, w) x(\mathrm{p}, \mathrm{w}){\mathrm{p}}^{\mathrm{T}}=\mathrm{D}{\mathrm{p}} x(\mathrm{p}, \mathrm{w}) \mathrm{p}+\mathrm{D}{\mathrm{w}} x(\mathrm{p}, \mathrm{w}) \mathrm{w}=0 .$$

\end{prob}

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