Let $\Gamma=(V,E)$ where $V=\{v_1,v_2,v_3,v_4,v_5,v_6\}$ and $E=\{\{v_1,v_2\}, \{v_2,v_3\},\{v_2,v_4\}, \{v_3,v_5\}, \{v_4,v_6\} \}$. Let $f: V \to \mathbb{R}, f(v_j)=(-1)^j.$ Define $$(\Delta f)(x) = \begin{cases} \sum_{\{x,y\} \in E(x)} (f(x)-f(y)), E(x)\ne\emptyset \\ 0, E(x) = \emptyset\end{cases},$$ where $E(x)=\{\{x,y\} \in E \mid y \in V\}$. Compute the value of $\Delta f$ at each vertex. Show that $\Delta : F(V) \to F(V)$, $f\mapsto \Delta f$ is a linear transformation, where $F(V) = f:V \to \mathbb{R}.$
Wouldn't $E(x)$ be here exactly the same as $E$? If so I would have that $$\Delta f(v_1) = 0 \\ \Delta f(v_2) = (-1)^1-(-1)^2 = -2\\ \Delta f(v_3) =(-1)^2-(-1)^3 = 2 \\ \Delta f(v_4) =(-1)^2-(-1)^4 = 0 \\ \Delta f(v_5)=(-1)^3-(-1)^5 = 1 \\ \Delta f(v_6)=(-1)^4-(-1)^6 = 0$$
I'm not entirely sure I understood this corrrectly. Any feedback/hints would be appreciated.
$E(x)$ is defined in the text of the exercise to be the set of edges with one vertex being $x$. Judging from your notation for the set of edges $E$ (to be precise: from writing edges as sets $\{u,v\}$ and not ordered tuples $(u,v)$) I guess we are dealing with undirected graphs. So in your example we get $$\begin{align*} E(v_1)&=\{\{v_1,v_2\}\}\\ E(v_2)&=\{\{v_2,v_1\}, \{v_2,v_3\},\{v_2,v_4\}\}\\ E(v_3)&=\{\{v_3,v_2\}, \{v_3,v_5\}\}\\ &... \end{align*}$$ Hence we get by definition of $\Delta f$ $$\begin{align*} \Delta f (v_1) &= (-1)^1 - (-1)^2 = -2\\ \Delta f(v_2) &= ((-1)^2-(-1)^1) + ((-1)^2 - (-1)^3) + ((-1)^2 - (-1)^4)\\&= 2 + 2 + 0 = 4\\ \Delta f(v_3) &= ((-1)^3 - (-1)^2) + ((-1)^3 - (-1)^5) = -2 + 0 = -2\\ &... \end{align*}$$