vertex $v \in V(L(G)) \setminus
D$, implying that $D$ is a FD-set in $L(G)$, and so ${{\rm fd}}(L(G)) \le
|D| = \mH$. [$\Box$]{}

We remark that examples of graphs that possess spanning $r$-regular subgraphs, where $r > 0$ and where $H$ is not necessarily induced, are abundant. For example, regular graphs of even degree have a $2$-factor as do Hamiltonian graphs. Several interesting families of graphs possess a $1$-factor (or perfect matching), including regular bipartite graphs and connected claw-free graphs of even order. For further results about regular spanning graphs see for example [@Cs07; @HoVo].

Results
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Upper Bounds
------------

We first establish upper bounds on the fair domination of a graph in terms of its order. By Observation \[ob3\](a), ${{\rm fd}}(G) \le n$ with equality if $G = {{\overline{K}}}_n$. However this bound can be improved slightly if we restrict our attention to graphs without isolated vertices.

If $G$ is a graph of order $n \ge 3$ with $\delta(G) \ge 1$, then ${{\rm fd}}(G) \le n - 2$, and this bound is sharp. \[thm\_n2\]

**Proof.** We proceed by induction on $n \ge 3$. If $n \in
\{3,4\}$, it is a simple case to check that if $G$ is a graph 