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title: The Duality Theorem?
tags: [discussion]

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tags: discussion
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# The Duality Theorem?




## Statement
MWG P.64
**Definition 3.F.1:** For any nonempty closed set $K \subset \mathbb{R}^L$, the support function of $K$ is defined for any $p \in \mathbb{R}^L$ to be

$$\mu_K(p)=\text{Infimum} \{p \cdot x: x \in K\}.$$

MWG P.66
**Proposition 3.F.1:** (The Duality Theorem). Let $K$ be a nonempty closed set, and let $\mu_K(\cdot)$ be its support function. Then there is a uniqure $\bar{x} \in K$ such that $\bar{p} \cdot \bar{x} = \mu_K(\bar{p})$ if and only if $\mu_k(\cdot)$ is differentiable at $\bar{p}$. Moreover, in this case, 

$$\nabla \mu_K(\bar{p})=\bar{x}.$$

## Counterexample
Consider the case of $\mathbb{R}^2$. Let $A = \{(x,y): y=-x - \frac{1}{x}\text{ and }x \le -1\}$ , $\bar{x} = (0,0)$ and $K = A \cup \{\bar{x}\}$. 

Clearly, $K$ is a nonempty closed set.

If $\bar{p} = (1,1)$, for any point $x' \in A$, we have $\bar{p} \cdot x' = -\frac{1}{x}>0$. Hence $\bar{p} \cdot \bar{x}=0$ is the infimum of $\mu_K(\bar{p})$, and  $\bar{x}$ is the  uniqure element in $K$ such that $\bar{p} \cdot \bar{x} = \mu_K(\bar{p})$. 

By the duality theorem, $\mu_K(\bar{p})$ is differentiable at $\bar{p}$ and $\nabla \mu_K(\bar{p})=\bar{x}.$, i.e. 

$$\lim_{h\to 0} \frac{\|\mu_K(\bar{p}+h) - \mu_K(\bar{p}) - \bar{x} \cdot h\|}{\|h\|}=0$$

Since $\bar{x} = (0,0)$,

$$\lim_{h\to 0} \frac{\|\mu_K(\bar{p}+h) - 0 -  0 \|}{\|h\|}=0$$

Let $h=(t,-t)$ where $t>0$. for any point $x' \in A$, we have $(\bar{p}+h) \cdot x' = 2tx-\frac{1+t}{x}$ and $(\bar{p}+h) \cdot \bar{x} =0$.

Therefore, for small enough $t>0$, $\mu_K(\bar{p}+h) \to -\infty$, and $\lim_{h\to 0} \frac{\|\mu_K(\bar{p}+h) \|}{\|h\|} \neq 0$.