# Where does this proof use the fact that the consumption level is positive?

Consider the following problem.

Now consider the following theorem and proof. My question is, where is it used in the theorem that $c^\star + \alpha D^T \theta \ge 0$? That is, why is that important? What goes wrong if it is negative?

Source: Duffie, Dynamic Arbitrage Pricing Theory

The problem naturally restricts the set of feasible consumption vectors to $\mathbb{R}_+^S$ (in a sense, it's hard to eat -1 apples). A strictly positive vector $\mathbf{c}^*$ implies an interior solution which gives you clean, uncomplicated optimality conditions.
• Let $\mathbf{\Delta}$ be some arbitrary vector in $\mathbb{R}^S$. Since vector $\mathbf{c}^*$ is strictly positive, the vector $\mathbf{c}^* + \alpha \mathbf{\Delta}$ is also strictly positive (i.e. feasible consumption) for $\alpha$ in some small enough neighborhood $[-k, k]$.
• Let $\boldsymbol{\theta} \in \mathbb{R}^N$ be an arbitrary vector such that $\boldsymbol{\theta} \cdot q$ = 0. This means $\boldsymbol{\theta}$ is a zero cost portfolio and it's affordable to move consumption in the direction $\mathbf{\Delta} = D' \boldsymbol{\theta}$.
The basic idea is that if $\mathbf{c} = \mathbf{c}^*$ solves $\max_{\mathbf{c} \in X} U(c)$ then for any zero cost portfolio $\boldsymbol{\theta}$, we have that $\alpha = 0$ solves $\max_{\alpha \in \mathbb{R}} U\left(c^* + \alpha D' \boldsymbol{\theta} \right)$. Take the derivative with respect to $\alpha$ and you get an optimality condition which leads you to a state price vector.