I'm working on the following exercise:
Let $F = \{x \in \mathbb{R}^n \mid Ax = b, x > 0\}$ where $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$. Further, let $f: F \to \mathbb{R}$ be a differentiable function. We assume that the following problem has a solution $\overline{x}$.
$$\min_{x \in F} f(x)$$
Show that for all $u \in \ker(A)$ there is an $\epsilon_0$ such that for all $\epsilon \in [0,\epsilon_0]$ holds $$\overline{x} + u\epsilon \in F$$
Deduce that for all $u \in \ker(A)$ holds $\nabla f(\overline{x}) \cdot u \ge 0$
Deduce that there exists a $\lambda \in \mathbb{R}^n$ such that: $\nabla f(\overline{x}) = A^T\lambda$
I did 1) and 2) but I don't know what to do about 3). Could you help me with that?
We have strict equality to $0$ in 2. as the inequality has to hold for $u$ and $-u$. Thus $\nabla f\left(\bar{x}\right)$ has to be orthogonal on $\ker\, A$. We also have $\ker\, A\;\perp\;\mathtt{im}\, A^{T}$ with $\ker\, A\;\oplus\;\mathtt{im}\, A^{T}=\mathbb{R}^{n}$. Thus $\nabla f\left(\bar{x}\right)\in\mathtt{im}\, A^{T}$. qed