
periphery of a graph
Let (eccentricity of v)$\displaystyle e(v)=Max_{u \in V(G)} d(u,v)$ and $\displaystyle diam(G)= Max_{v \in V(G)} e(v)$, where V(G) is the set of vertices of G and d(u,v) is the distance between u and v. Let (periphery of G) Pe(G) be the graph induced by the vertices of G that have eccentricity equivalent to diam(G). Now suppose that the graph H has n vertices. Prove that H is Pe(G) if and only if $\displaystyle \Delta(H) \le n2$. ($\displaystyle \Delta(H)$ is the maximum degree in H)

Re: periphery of a graph
You need an additional constraint/premise of some kind. It's not true as stated.
$\displaystyle \text{If } K_n \text{ is the complete graph on } n \text{ verticies } (n \ge 2), \text{ then}$
$\displaystyle diam(K_n) = 1, \text{ and } e(v) = 1 = diam(K_n) \ \forall \ v \in V(K_n),$
$\displaystyle \text{and so } Pe(K_n) = K_n.$
$\displaystyle \text{But } \Delta(K_n) = n1 = V(K_n)  1,$
$\displaystyle \text{which contradicts this problem's claim that}$
$\displaystyle H = Pe(K_n) (= K_n) \Rightarrow \Delta(H) \le V(H)  2.$

Re: periphery of a graph
You're right. Assume that no vertex in H has eccentricity 1.

Re: periphery of a graph
$\displaystyle \text{Let } G \text{ be a connected graph such that } diam(G) > 1 \text{ (i.e connected, not complete, having 3+ verticies.)}$
$\displaystyle \text{ASSUME } \exists v \in H = Pe(G) \ni deg_H(v) = V(H)1.$
$\displaystyle \text{Since } v \in Pe(G), \exists w \in G \ni e(v) = dist(v, w) = diam(G).$
$\displaystyle \text{Since } diam(G) = dist(v, w) \le e(w) \le diam(G),$
$\displaystyle \text{have that } e(w) = diam(G), \text{ and so } w \in H.$
$\displaystyle \text{But then } v, w \in H, \text{ and so since } deg_H(v) = V(H)1,$
$\displaystyle \text{have that } (v,w) \in E(H).$
$\displaystyle \text{Since } H \text{ is an induced subgraph of } G, \text{ have that } (v,w) \in E(G).$
$\displaystyle \text{But } (v,w) \in E(G) \Rightarrow dist(v, w) = 1 \Rightarrow diam(G) = dist(v, w) = 1, \text{ contrary to the premises.}$
$\displaystyle \text{Thus the initial assumption led to a contradiction.}$
$\displaystyle \text{Thus } x \in H \Rightarrow deg_H(x) \le V(H)2.$
$\displaystyle \text{Therefore } \Delta(Pe(G)) \le V(Pe(G))2.$

Re: periphery of a graph

Re: periphery of a graph
$\displaystyle \text{Let } H \text{ be a graph such that } \Delta(H) < V(H)1 \text{ and } H > 1.$
$\displaystyle \text{Then for all } v \in H \text{ have that } deg_H(v) < V(H)1,$
$\displaystyle \text{and so if } v \in H \text{ there exists } w \in H \text{ such that } (v, w) \notin E(H).$
$\displaystyle \text{Define } G \text{ by } V(G) = V(H) \cup \{ x \}, E(G) = E(H) \cup \{ (x,v)  v \in V(H) \}.$
$\displaystyle \text{Then for all } v \in H \subset G, dist_G(v, x) = 1 \text{ since } (v, x) \in E(G).$
$\displaystyle \text{Thus } e_G(x) = 1.$
$\displaystyle \text{From before, if } v \in H \subset G, \text{ there exists } w \in H \text{ such that } (v,w) \notin E(H),$
$\displaystyle \text{and so by the definition of } E(G), \text{ also have that } (v,w) \notin E(G).$
$\displaystyle \text{So for all } v \in H \subset G, \text{ this proves that } e_G(v) > 1, \text{ since } dist_G(v, w) > 1.$
$\displaystyle \text{Now, for all } v, w \in H \subset G, \{(v, x), (x, w) \} \subset E(G),$
$\displaystyle \text{and so, for all } v, w \in H \subset G, \ dist_G(v, w) \le 2.$
$\displaystyle \text{But from that it follows that, for all } v \in H \subset G, e_G(v) \le 2.$
$\displaystyle \text{(That conclusion also requires also remembering that } dist(v, x) = 1.)$
$\displaystyle \text{So have shown that, for all } v \in H \subset G, 1 < e_G(v) \le 2.$
$\displaystyle \text{Therefore, if } v \in G, v \ne x, \text{ then } e_G(v) = 2. \text{ Also, } e_G(x) = 1.$
$\displaystyle \text{That proves that } diam(G) = 2, \text{ and so also that } Pe(G) = H,$
$\displaystyle \text{since } Pe(G) = \ < \{ v \in V(G)  e_G(v) = diam(G) = 2 \} > \ = \ < H > \ = H.$