Editing Expenditure function (section)

=== Proofs ===

(1) As in the above proposition, note that 

<math>e(\lambda p,u)=\min_{x\in\mathbb{R}^n_+ :u(x)\geq u}</math> <math>\lambda p\cdot x=\lambda \min_{x\in\mathbb{R}^n_+ :u(x)\geq  u}</math> <math>p\cdot x=\lambda e(p,u)</math>

(2) Continue on the domain  <math>e</math>: <math>\textbf R_{++}^N*\textbf R\rightarrow \textbf R
</math>  

(3) Let <math>p^\prime>p</math> and suppose <math>x \in h(p^\prime,u)</math>. Then <math>u(h)\geq u</math>, and <math>e(p^\prime,u)=p^\prime\cdot x\geq p \cdot x</math> . It follows immediately that <math>e(p,u)\leq e(p^\prime,u)</math>. 

For the second statement, suppose to the contrary that for some <math>u^\prime > u</math>, <math>e(p,u^\prime)\leq e(p,u)</math> Than, for some <math>x \in h(p,u)</math>, <math>u(x)=u^\prime>u</math>, which contradicts the "no excess utility" conclusion of the previous proposition 

(4) Let <math>t \in(0,1)</math> and suppose <math>x \in h(tp+(1-t)p^\prime)</math>. Then, <math>p \cdot x\geq e(p,u)</math> and <math>p^\prime \cdot x\geq e(p^\prime,u)</math>, so <math>e(tp+(1-t)p^\prime,u)=(tp+(1-t)p^\prime)\cdot x\geq</math><math>te(p,u)+(1-t)e(p^\prime,u)</math>. 

(5) <math>\frac{\delta(p^0,u^0)}{\delta p_i}=x^h_i(p^0,u^0)
</math>