[$≠$ verify; see also: obey]
Here $a$ and $b$ are chosen to satisfy (2). [Not: “to verify (2)”]
This forces $f$ to satisfy (6).
The function of Lemma 2 can be made to satisfy (6).
The operator $P$ satisfies essentially the same inequality as $F$ does.
However, $A$ fails to satisfy (3).
Then $F$ need not satisfy (2). [= $F$ does not necessarily satisfy (2).]
The products $F_iG_i$ are very close to satisfying (1).
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