The eccentricity of a vertex $\epsilon (v) $ is the maximum of distances $d(v, u) $ over all other vertices $u$. The average distance of a vertex avgd$(v) $ is the average of all $d(v, u)$, more precisely $avgd(v) =\frac{\sum_{u\neq v} d(v, u)} {|V|-1}$
Now if the center of a graph is one single vertex $v$, that is, there is a least eccentric vertex $v$ , so $\epsilon (v) <\epsilon (u), u \neq v$, then avgd$(v) < \max(avgd(u)_{u \in V}) $ because this vertex $v$ can't simultaneously have the greatest average distance.
At least, that's what I assume. My intuition comes from a Graph where a small clique and a big clique are connected with one separate vertex. That vertex is then the least eccentric, but the small clique vertices have always a higher average distance. Anyone a proof or a counterexample?
Some of my thoughts about this conjecture :
I now realize that a graph consisting of one single vertex would be a trivial counterexample, so let's only consider simple connected graphs with more than one vertex.
I generated many graphs with a single vertex center and could indeed always find an $u$ with strictly greater average distance than the center. I wondered if we can also find such $u$ that is adjacent to the center, but that's not always the case.
Although it might sound intuitive or trivial that the most central point is not the point with the greatest average distance, it's not generally the case for any metric. It's not generally true for the Euclidean distance for example.
I thought it might be true because if there is singular center $ v$, then some $v*$ has a shortest path through the center for more than half of the other vertices. But this turned out to be false, I found graphs with a singular center but without such a $v*$
Counterexample. Take $6$ disjoint $2$-element vertex sets $V_i=\{x_i,y_i\}$ ($0\le i\le5$); draw $30$ edges making $V_i\cup V_j$ a clique for $i-j\equiv1\pmod6$; add a $13^\text{th}$ vertex $z$ and edges $y_0z$ and $y_3z$. You can easily verify that in this graph $$\epsilon(z)=2;$$ $$\epsilon(x_i)=\epsilon(y_i)=3;$$ $$\operatorname{avgd}(z)=\frac{22}{12};$$ $$\operatorname{avgd}(y_0)=\operatorname{avgd}(y_3)=\frac{19}{12};$$ and for vertices $v\notin\{z,y_0,y_3\}$, $$\operatorname{avgd}(v)=\frac{21}{12}.$$ Generalizing this construction, for integers $k\ge1$ and $n\ge2$ there is a graph of order $(4k+2)n+1$, radius $2$ and diameter $3$, with a unique vertex $z$ of eccentricity $2$ which has average distance $$\operatorname{avgd}(z)=\frac{2(4k+2)n-2}{(4k+2)n}=2-\frac1{(2k+1)n}$$ while $$\max_{v\ne z}\operatorname{avgd}(v)=\frac{(6k+4)n+1}{(4k+2)n}=\frac32+\frac{1+\frac1n}{4k+2}.$$ For this take $(4k+2)n$ vertices, divided into $4k+2$ "blobs" of $n$ vertices each; arrange the blobs in a circle; join each vertex to all other vertices in its own blob and the $k$ nearest blobs on either side; finally add a central vertex $z$ and join it to two vertices which are in diametrically opposite blobs.