[Looking for Confirmation](X,G) is a duality, then (X,G1) is also one iff G1 is dense in G Edit: I noticed, that my original post has not had any replies, therefore I wrote the problem again (together with my attemps, which are also already more detailed that 2 days ago) and structured it better. Still, I would apprechiate any help! 
This is a problem from my functional analysis lecture, chapter locally convex vector spaces:

$(F,G)$ is a duality, 

which means $G \subseteq F^*$ with $F^*$ beeing the algebraic dual space of F and $G$ operates seratating on $F$, which means:
$ \forall a,b \in F, a \ne b: \exists f \in G: \langle a,f \rangle \ne \langle b,f\rangle $
(In this notation $\langle a,f \rangle$ means $f$ applied to $a$.)

Now I want to show, that for any $G_1 \subseteq G: (F,G_1)$ is a duality if and only if $G_1$ is dense in $G$ with respect to the weak star topology $\sigma (G,F)$ 

which is the weakest topology such that all $ x \in F$ act continuous on $G$. As $G_1$ is already a subset of $G \subseteq F^*$, the dualitiy comes to effect exactly when $G_1$ acts separating on F:
$ \forall a,b \in F, a \ne b: \exists g \in G_1: \langle a,g \rangle \ne \langle b,g\rangle$
At first, I will show: $\mathbf{G_1}$ weak star dense in $\mathbf{G}$ $\mathbf{\implies}$ $\mathbf{G_1}$ separating on $\mathbf{F}$
We know $ \forall a,b \in F, a \ne b: \exists f \in G: \langle a,f \rangle \ne \langle b,f\rangle$
Density: $\exists (f_n)_{n \in \mathbb{N}}$ series in $G_1$: $f_n \stackrel{\ast}{\rightharpoonup} f$ which means $\forall x \in F: \langle x,f_n \rangle \rightarrow \langle x,f \rangle$.
Now $\langle x, \cdot \rangle: F^* \rightarrow \mathbb{C}: f \mapsto \langle x, f \rangle$ is continuous, which means in the two series:
$\{ \langle a, f_1 \rangle, \langle a, f_2 \rangle, ... \} \subseteq \mathbb{C}, \langle a, f_n \rangle \rightarrow \langle a, f \rangle$
$\{ \langle b, f_1 \rangle, \langle b, f_2 \rangle, ... \} \subseteq \mathbb{C}, \langle b, f_n \rangle \rightarrow \langle b, f \rangle$
with $\langle a, f \rangle \ne \langle b, f \rangle$ there must be some $N \in \mathbb{N}: \forall n \ge N : \langle a, f_n \rangle \ne \langle b, f_n \rangle$.
So I found a $f_N \in G_1: \langle a, f_N \rangle \ne \langle b, f_N \rangle$ and as $a,b$ were arbitrary elements of $F$ I have shown that $G_1$ acts separating on $F$.
Question: Is that proof correct? I'm still not sure about the part where I assume an $f_N$ such that the inequality come to effect.
Now I come to $\mathbf{G_1}$ separating on $\mathbf{F}$$\mathbf{\implies}$  $\mathbf{G_1}$ weak star dense in $\mathbf{G}$ 
For $f\in G$, the sets
$U_f(x_1,\ldots,x_n,\epsilon) := \{g\in G; |  \langle x_j, f \rangle - \langle x_j, g \rangle| < \epsilon, j=1,\ldots,n\}$,
with $x_1,\ldots,x_n\in F, n \in {\mathbb N}, \epsilon > 0$, are weak star open sets in $G$ and define a neighbourhood basis of $f$.
I want to show, that any $U_f$ contains an element of $G_1$, but... that's the point where I'm stuck! Another fact, which might be useful, even though I don't know yet if I need to use it:
$G_1 \subseteq G \subseteq F^*$, so $(F,G_1)$ is a duality already and therefore we can assume, that $F \subseteq G_1^*$ and $F$ acts separating on $G_1$.
I would appreciate any help or hint, I'm thinking about that proof for nearly two weeks now and I would really like to know how it works :D thanks already!
 A: Okay, I think I got something and it sounds good to me. I would still be thankful, if somebody checked it if it really works that way, not only the part in this answer but also the other direction which is in the original post.

Now I come to $\mathbf{G_1}$ separating on $\mathbf{F}$$\mathbf{\implies}$  $\mathbf{G_1}$ weak star dense in $\mathbf{G}$ 

Assume $G_1$ is not dense in $G$, then $\overline G_1$ is a strict linear subspace of $G$.
$(F,G_1)$ is a duality, therefore any $a \in F$ defines via $\langle a, \cdot \rangle: G_1 \rightarrow \mathbb{C}$ a linear functional on $G_1$. 
Now define a seminorm an $G$ via the minkowski functional $\mu_{G_1}(f) = \{ t > 0 : f \in \frac{1}{t} G_1 \}, f \in G$.
Choose $a \in F$ such that $ \lvert \langle a, \cdot \rangle \rvert \vert_{\overline G_1} = 0$, then $ \lvert \langle a, \cdot \rangle \rvert \vert_{\overline G_1} \le \mu(\cdot)_{G_1}\vert_{\overline G_1}$.
Then, thanks to Hahn-Banach, there exists an $ A \in F : \langle a, \cdot \rangle \vert_{\overline G_1} = \langle A, \cdot \rangle \vert_{\overline G_1}$ and $\langle A, \cdot \rangle \vert_{ G} \le \mu(\cdot) \vert_{G}$. Especially, $\langle A, \cdot \rangle \vert_{ G} $ may be $ > 0$.
So in general: $\langle A, \cdot \rangle \vert_{ G} \ne \langle b, \cdot \rangle \vert_{ G}$ with $b :=  0$. But then, because $G_1$ is also separating, $\langle A, \cdot \rangle \vert_{ G_1} = \langle a, \cdot \rangle \vert_{ G_1} \ne \langle b, \cdot \rangle \vert_{ G}$, which is not the case, as they both are $=0$. So we have a contradiction and therefore $G_1$ must be dense in $G$.
