1
We write z=(x,y) for the common point of A and B.
We first consider the M/G/1 queue, where M (for “Markov'') means that......
For D a smooth domain, the following are equivalent.
For m not an integer, the norm can be defined by interpolation.
For (ii), consider...... [= To prove (ii), consider]
As for (4), this is an immediate consequence of Lemma 6. [= Concerning (4)]
Thus F is integrable for the product measure.
Then for such a map to exist, we must have H(M)=0.
The problem with this approach is that V has to be C1 for (3) to be well defined.
Computing f(y) can be done by enumerating A(y) and testing each element for membership in C.
This observation prompted the author to look for a more constructive solution.
Therefore, the system (5) has a solution of the sought-for type.
2
[see also: because] We must have Lf=0, for otherwise we can replace f by f-Lf. [= because otherwise]
It turns out that it suffices to show that A=1, for if this is proved, the preceding remark shows that......
[This use of “for'' sometimes leads to confusion; e.g., never write: “for x∊ X'' if you mean: since x∊ X. Also, avoid starting a sentence with a “For'' in this sense.]