Graphs with Minimum Degree Three are Totally Dominated by Half of the Vertices

Let = () be a graph without isolated vertices. We say that ⊂ is a if every vertex has a neighbor in . Let () denote the minimum cardinality of a total dominating set of . We prove that if is a graph of order and minimum degree at least 3 then () ≤ /2. If equality holds then must be 3-regular and ≡ 0 mod 4. For every ≡ 0 mod 4 it is known that there are 3-regular graphs for which equality holds, hence the result is sharp. This solves the conjecture of Favaron et al.[4] and improves all previously known upper bounds