NP-hardness does not imply lower bound, strictly speaking? A problem is NP-hard iff every NP problem can be polynomially-time reduced to it.


Hardness is often intuitively explained as a lower bound. But it isn't, strictly speaking. For the sake of the argument, assume P=NP. Now the definition becomes:

A problem is P-hard iff every P problem can be polynomially-time reduced to it.


Because the above definition uses the polynomial-time reduction, the overall running time is polynomial (reduction + solving the resulting problem), no matter how easy is the resulting problem. Hence we could get an absurd result: a problem, which runs in constant time (hence lower bound is constant), is P-hard.

The definition makes total sense if we assume that NP>P. Do people assume NP>P?

The same question arises for the definition of PSPACE-hardness, where book authors use polynomial-time reduction, rather than something strictly easier than PSPACE.

I guess the answer to this question is simply "yes", sorry for the rant.
Yes, if P = NP then any decision problem is NP hard as long as there is one problem instance where the answer is "YES" and one where the answer is "NO". 

Just check the definition of NP-hard (and check why problems where every instance has the answer YES or every instance has the answer NO are not NP-hard by that definition).