Problem: I'm stuck showing the opposite direction. I wanna show that $N(x̄,Ω)=∂δΩ(x̄)$ subdifferential of normal cone, where $N(x̄,Ω)=\{g \in Rn | \langle g, z-x̄\rangle ≤ 0, ∀ z ∈ Ω\}$.
My attempt:
First direction $∂δΩ(x̄) ⊂ N(x̄,Ω)$:
Let's take $g ∈ ∂δΩ(x̄)$ then
$$\langle g, z-x̄\rangle ≤ δΩ(z) - δΩ(x̄), ∀z ∈ Rn \,\text{ and }\, δΩ(x̄)=0\, \text{ since }\, x̄ ∈ Ω.$$ Hence
$$\langle g, z-x̄ \rangle ≤ δΩ(z), ∀z ∈ Rn$$. Also $∀z ∈ Rn ⇒ ∀ z ∈ Ω ⇒ δΩ(z)=0.$ Therefore
$$\langle g, z-x̄\rangle ≤ 0 , ∀ z ∈ Ω ⇒ g ∈ N(x̄,Ω).$$ In opposite direction we have to show that $N(x̄,Ω) ⊂ ∂δΩ(x̄)$.
My problem is starting here because I don't know how I can start to prove other side. If you have any idea please help me.
You should learn how to type math formulas. Also, please clearly define all the used notation. For example, not all readers are familiar with the notation $\delta_{\Omega}$ for the indicator function of a set $\Omega$.
Back to the question. The results follow directly from the definitions of the normal cone and the subdiferential. Just write down the subdiferential using the definition and simplify it: \begin{align} \partial \delta_{\Omega}(x) &= \left\{g\in\mathbb{R}^n\mid \langle g, z - x \rangle \le \delta_{\Omega}(z) - \delta_{\Omega}(x) \ \forall z\in\mathbb{R}^n\right\}\\ &= \begin{cases} \emptyset &\mbox{if }x\not\in\Omega\\ \left\{g\in\mathbb{R}^n\mid \langle g, z - x \rangle \le \delta_{\Omega}(z) \ \forall z\in\mathbb{R}^n\right\} &\mbox{if }x\in\Omega \end{cases}\\ &= \begin{cases} \emptyset &\mbox{if }x\not\in\Omega\\ \left\{g\in\mathbb{R}^n\mid \langle g, z - x \rangle \le \delta_{\Omega}(z)\ \forall z\in\Omega\right\} &\mbox{if }x\in\Omega \end{cases} \end{align} which is precisely the definition of the normal cone $N_{\Omega}(x)$.