I am trying to prove
if $A$ is convex and $x^*\in A$, then $D(A,x^*)=cone(A-x^*)$, where $D(A,x^*)=\{ d\in \mathbb{R^n}| \exists \delta >0$ such that $x^* +td \in A, \forall t \in (0,\delta) \}$ and $A-x^*=\{ a-x^*|a\in A\}$.
Definition of cone: $Cone(A) = \{ \lambda a\ | a\in A , \lambda \geq 0\}$
What I have tried so far:
I believe $cone(A-x^*)=\{ \lambda (a-x^*)| a\in A , \lambda \geq 0\}$
Since $A$ is convex, $A-x^*$ is also convex. Since $A-X^*$ is convex, $cone(A-x^*)$ is convex cone. So, $cone(A-x^*)=\{ \sum ^k _{i=1} \lambda_i x^{(i)}| k\in \mathbb{N}, \lambda_i \geq0, x^{(i)}\in A-x^*\}$
I am trying somehow to find a way to connect $\{ \sum ^k _{i=1} \lambda_i x^{(i)}| k\in \mathbb{N}, \lambda_i \geq0, x^{(i)}\in A-x^*\}$ and $D(A,x^*)$, but I am not sure....
Please help me.
To help you with the outline I've provided in my last comment, to prove $D(A,0) = \text{Cone}(A)$ when $A$ is convex and $0 \in A$, you need to prove two things:
The first is the harder the prove, and requires both that $A$ is convex and $0 \in A$. The second holds for any $A$.
To prove $\text{Cone}(A)\subseteq D(A,0)$, let $d \in \text{Cone}(A)$. Then there exist $\lambda > 0, a \in A$ with $d = \lambda a$. Let $\delta = \frac 1\lambda$. If $t \in (0,\delta)$, let $q = \frac t\delta = t\lambda$, so $q \in (0,1)$, and $td = \left(\frac q\lambda\right)\left(\lambda a\right) = qa$. But since $0, a \in A$, so is every point on the line segment connecting them. I.e., $td = qa \in A$. Therefore $d \in D(A, 0)$.