> This is clearly the case for f(n) = 1, g(n) = lg(n), hence the constant function is O(lg(n)).

It is not, because constant function has no logarithmic behavior. To distinguish that is the whole point of big-O notation. When you tell someone "this is a O(logn) operation" they do expect log performance.

Yes O(c) is strictly under O(n) too, but so what? We can just as happily declare most functions double-exponential in complexity, but what use is that? It would be one of those formally correct but practically useless definitions.