On structure of graphs with average degree equal to maximum average degree

Question

For a simple graph $G$, the $\text{average-degree}(G)=|E(G)|/|V(G)|$ and
the maximum average degree $\text{mad}(G)=\max\{\text{average-degree}(H)\colon H \text{ is a subgraph of } G\}$.

If $\text{average-degree}(G)=\text{mad}(G)$, what can we say about the structure of $G$?

Clearly, $\text{average-degree}(G)=\text{mad}(G)$ for every regular graph $G$. I suppose the graph being regular is not a necessary condition.

Context: Both parameters are equal for extremal graphs in some graph coloring problems.

Answer 1

This question is a bit open-ended, but here are some observations:

Some non-regular graphs can have this property, for example all trees have this property.

For graph coloring, this property seems to have limited use. In particular, there is a reduction from any graph $G$ to a graph $G'$ that satisfies this property and $\chi(G') = \chi(G)+1$, where $\chi(G)$ denotes the chromatic number of $G$. The reduction is as follows:

Denote $n = |V(G)|$. Construct $G'$ by adding $n$ new vertices, each connected to every vertex of $V(G)$ but not to each other. Now, $\chi(G') = \chi(G)+1$ because the new vertices need an additional color but only one additional color suffices. I claim that $G'$ has the average degree property. Let $G'[X]$ be the induced subgraph of $G'$ with the highest average degree. I claim that $G'[X] = G'$. Let $A$ denote the original vertices $V(G)$ and $B$ denote the added vertices. Now, the average degree of $G'[X]$ is strictly less than $|A \cap X|$, because each vertex in $B \cap X$ contributes $|A \cap X|$ edges, and after counting edges adjacent to $B \cap X$, the vertices in $A \cap X$ contribute strictly less than $|A \cap X|$ edges. Now, if $X \cap B \neq B$, we could add a vertex in $B$ to $X$, contributing $|A \cap X|$ new edges, and therefore increasing average-degree. Therefore $X$ must contain $B$. Suppose $X \cap A \neq A$. Now, as $B \subseteq X$, adding a vertex of $A$ to $X$ contributes at least $|B| > |A \cap X|$ new edges, and therefore increases the average degree. Therefore $X$ must contain both $A$ and $B$, so $G'[X] = G'$.