Check if $F$ is epimorphism For given polymonial $p \in \mathbb R [x]_2$ we have 
$$f_p^*: \mathbb R[x]_2 \rightarrow \mathbb R$$

$$f_p^*(q) = p(-1)q(1) + p(0)q(0) + p(1)q(-1) $$
where $q\in \mathbb R [x]_2$
We have also  $$F: \mathbb R[x]_2 \rightarrow (\mathbb R[x]_2 )*$$ which is given with formula:
$$F(p) = f_p^* $$
where $p \in \mathbb R [x]_2$
Check if $F$ is epimorphism
My try
Ok, it seems to be simple, I have just to show that:
$$F(\alpha t + \beta w) = \alpha F(t) + \beta F(w) $$
so I have
$$\forall q. \alpha f_t^* (q) + \beta f_w^*(q) = f_{\alpha t+ \beta w}^*(q) $$
I want to go from right to left but how to use polymonial $\alpha t+ \beta w$?
 A: For linearity, all you have to see is that for all real $x$ and $y$, $(\alpha t + \beta \omega)(x).q(y) = (\alpha t(x) + \beta \omega(x)).q(y) = \alpha t(x)q(y) + \beta \omega(x)q(y)$. Since you sum this behavior three time with $(-1,1)$, $(0,0)$ and $(1,-1)$ for $x$ and $y$, linearity is pretty straight-forward.
For surjectivity all you have to check if that given a linear form $g \in \mathbb{R}[x]_2$ there exists a polynomial $p$ of degree 2 such that $g = F(p)$
The thing to see here is that a polynomial of degree 2 can be determined given three different points and their values (it is Lagrange interpolation).
Consider $g \in \mathbb{R}[x]_2^*$. All you have to do is determined its values on a base, for example $\{1, x, x^2\}$ and it will be uniquely defined since $g$ is a linear form. By evaluation on this base you get three reals $g(1) = \alpha$, $g(x) = \beta$, $g(x^2) = \gamma$.
It gives you the linear system:
$p(-1) + p(0) + p(1) = \alpha$
$p(-1) - p(-1) = \beta$
$p(-1) + p(1) = \gamma$
I let you convince yourself that this linear system is always resoluble (its determinant is -2)
It gives you a the values of p for $0$, $1$, $-1$ which allows you, by Lagrange interpolation, to determine $p$, which is what you wanted to prove surjectivity.
