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]
Now (3) is clear. As for (4), it 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 $C^1$ for (3) to be well defined.
Computing $f(y)$ can be done by enumerating $A(y)$ and testing each element for membership in $C$.
Therefore, the system (5) has a solution of the sought-for type.
[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 $x\in X$” if you mean: since $x\in X$. Also, avoid starting a sentence with a For in this sense.]
[Do not use for followed by an ing-form to indicate the purpose of one's actions: instead of: “For showing that $f$ is bounded, we observe”, write: To show that $f$ is bounded, we observe. However, you can use for to indicate the `purpose' of a thing: This provides a method for recognizing pure injective modules.]
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