[see also: but, still, nevertheless]
For $j=1$ the operator is bounded, yet [= but] the integral (8) fails to be finite.
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.
In the remainder of this section, we study some properties of $K$, with the eventual aim (not realized yet) of describing $K$ directly using $G$.
Further research may yet explain the enigma.
Go to the list of words starting with: a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
y
z