[see also: difficult, complex, complicated, involved, intricate]
This is the hard part of Jones's theorem.
The theorem indicates that arbitrary multipliers are much harder to handle than those in $M(A)$.
The calculation of $M(f)$ is usually no harder than the calculation of $N(f)$.
The cases $p=1$ and $p=2$ will be the ones of interest to us, but the general case is no harder to prove.
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