[see also: but, still, nevertheless
For j=1 the operator is bounded, yet the integral (8) fails to be finite. [= but the integral]
Then F is strictly increasing and yet has zero derivative on a dense set.
In the next section we introduce yet another formulation of the problem.
The as yet unproved conjecture of Newman is that......
On the other hand, as yet, we have not taken advantage of the basic property enjoyed by S: it is a simplex.
The conjecture has not been proved yet.
Further research may yet explain the enigma.