Shepard's Lemma

* If Hicksian demand is single-valued, it coincides with the derivative of the expenditure function.
* Thm: if the utility function u(\cdot)is continuous and Hicksian demand h(p,u) single-valued (function), then the expenditure function e(p,u) is differentiable in price at (p,u) with \frac{\partial e(p,u)}{\partial p_i} = h_i(p,u).
* Intuition: if the price of a good increases by a very (infinitesimally) small amount (\varepsilon, say $0.01), then the unique optimal consumption bundle now costs \varepsilon h_i(p,u)more.
* This is basically the envelope theorem intuition, and the proof for Shepard's lemma accordingly uses the envelope theorem.
== Proof for the Differentiable Case ==
The expenditure function e(p,u)
 is the value function of the constrained optimization problem (expenditure minimization problem) characterized by the Lagrangian:
\mathcal{L} (\mathbf{x}, \lambda; \mathbf{p}, u) = \mathbf{p} \mathbf{x} + \lambda(u - U(\mathbf{x}))
By the envelope theorem, the derivatives of the expenditure function with respect to the price of a good are: 
\frac{\partial e(\mathbf{p},u)}{\partial p_i} = \frac{\partial \mathcal{L}}{\partial p_i} = x^h_i
Where x_i^h
 is the minimizer. In other words/by definition of the Hicksian demand function: x^h_i = h_i(\mathbf{p},u)
. Hence:
\frac{\partial e(\mathbf{p},u)}{\partial p_i} = \frac{\partial \mathcal{L}}{\partial p_i} = h_i(\mathbf{p},u)