Is an NP-hard problem which is in NP is NP-complete?
Check the definitions of NP-completeness and NP-hardness. I suspect they use Karp reductions for one and Cook reductions for the other.

Then, the quote makes sense: it is indeed unknown if Cook reductions and Karp reductions are equivalent in this sense. This is related to the question whether co-NP = NP; Cook reductions can not separate the two but Karp reductions might. See here for some details.

However, the authors use different definitions than (most of) the rest of the CS world: the widely accepted standard is to use Karp reductions. I recommend you read our reference question (and get a better book).



To be fair, the authors want to talk about optimization problems, in which case they need a notion of (NP-)hardness that is not tied to decision problems. Karp reductions won't get them there directly. 

They picked a poor way out, arguably, by defining the notions differently from everybody else. There are cleaner (but more involved) ways, defining complexity classes for optimization problems rigorously, providing them with their own types of reduction, and tying them to the classic decision-problem classes using threshold languages.