A simple graph's node is a (local) star graph center iff its average nearest neighbors degree equals 1

up:: Star Graph

The Average Nearest Neighbors’ Degree of a node $i$ (in a Simple Graph) measures how close $i$ is to a star graph’s center. In other words, $k_{nn}$ measures how the vicinity of $i$ compares to a star graph.

$i$ is a (local) star graph center $⟹$ $k_{nn}(i)=1$

Let $i$ be the center of a local star graph. Then this means that its neighbors connect exclusively to $i$, i.e. all their Node Degrees are equal to $1$. Thus, their sum is equal to $i$‘s degree (since this is a Simple Graph and undirected, i.e. no self-loops and at most one link between nodes).

$k_{nn}(i)=1⟹$ $i$ is a (local) star graph center

Let $k_{nn}(i)=1$. Then this means that $i$‘s Node Degree is not empty, i.e. it has neighbors. Thus, we can infer that

$$\sum _{<j>}k(j)=k(i)$$

But note that $i$‘s degree is equal to how many neighbors it has, i.e. $k(i)={\sum }_{<j>}1$.

Therefore

\sum_\limits{<j>} \left(k(j) - 1\right) = 0

Since we have that $k(j)\ge 1$ (since it has at least $i$ as its neighbor), then we have that $k(j)-1\ge 0$. Therefore, each term in this sum must be individually equal to zero.

Thus, each of $i$‘s neighbors connect exclusively to it. Thus, $i$ is the center of a local star graph.