# Duality of Cost Functions
## Intro
Varian(1992, P.84) states that if a differentiable function $\phi (w,y)$ statisfies four criteria, then it is the cost function of the technology $V^*(y) = \{x \ge 0|w \cdot x \ge \phi(w,y), \text{ for all } w \ge 0\}$.

The four criteria are,

1. $\phi (tw, y) = t\phi(w, y) \text{ for all } t \ge 0$.

2. $\phi (w, y) \ge 0 \text{ for } w\ge 0 \text{ and } y \ge 0$.

3. $\phi (w', y) \ge \phi (w, y) \text{ for } w' \ge w$.

4. $\phi (w, y)$ is concave in $w$.

However, Varian did not use the second and the third criteria in the proof. A natural question is, are these criteria redundent?

The usage of the second criterion is ruling out negative cost, which implies free lunch of the technology. Since the construction of $V^*(y)$ rules out negative vectors, the second criterion is required.

Although the third criterion is the immediately result of cost-minimizing behavior, it may be unnecessary. In fact, just modifying Varian's proof, we would find the redundancy of the third criterion.

## Redundancy
**Claim**
Let $w$ be a vector and $y$ be a scalar in $\mathbb{R}^n$. If a function $\phi (w,y): w \times y \to \mathbb{R}$ statisfy the following two criteria, then $\phi(w,y)$ is non-increasing in $w$.

1. $\phi (tw, y) = t\phi(w, y) \text{ for all } t \ge 0$.

4. $\phi (w, y)$ is concave in $w$.

**Proof**
Given a $w$, we define

$$x(w,y) = (\frac{\partial \phi(w, y)}{\partial w_1}, ..., \frac{\partial \phi(w, y)}{\partial w_n})$$

and note that since $\phi(w, y)$ is H.D.1 in $w$, Euler's law implies that

$$\phi(w,y) = \sum_{i=1}^n w_i \frac{\partial \phi(w, y)}{\partial w_i} = w \cdot x(w,y) $$

By the concavity of $\phi(w, y)$ in $w$, we have

$$\phi(w,y) \le \phi(w',y) + D\phi(w', y)(w - w') \text{ for all } w $$

Applying Euler's law, we have lemma 1

$$\phi(w, y) \le w' \cdot x(w',y) + x(w',y) (w -w') = w \cdot x(w',y) \text{ for all } w $$

Suppose $\phi(w,y)$ is not non-drceasing in $w$, i.e., there exists two vector $w'' \ge w'$ and a scalar $y' \ge 0$ such that

$$\phi (w'', y') < \phi (w', y')$$

Since $w'' \ge w'$, we have

$$w' \cdot x(w'',y') \le w'' \cdot x(w'',y') = \phi (w'', y')$$

Combing it with lemma 1, we have a contracdition

$$w' \cdot x(w'',y') \le \phi (w'', y') <\phi(w',y')\le w' \cdot x(w'',y')$$

Hence, $\phi(w,y)$ must be non-dreasing in $w$.

## Reference
Varian, H. R., & Varian, H. R. (1992). Microeconomic analysis (Vol. 3). New York: Norton.