14 3. PRELIMINARIES
Multiplying by ha(x), integrating with respect to νa = ν−af +g, and using the fact
that L−af +gq dνa = µa q dνa (see e.g. [PP]), gives
−i
ha(x)
ws(x)


σ(y)=x
f(y)
egs(y)
ws(y)⎠

dνa(x) =

∂b
αs .
Hence

∂b
αs


σ(y)=x
f(y)
egs(y)
ws(y)


ha(x) dνa(x)
|f|∞


σ(y)=x
eua(y)⎠

ha(x) dνa(x)
= |f|∞ ha(x) dνa(x) = |f|∞ .
Hence |αs 1| = |αs αa| |f|∞ |b|≤ |f|∞ |s s0|, and therefore
|λs λ| = |αs µa µs0 | |αs 1|· µa + |µa µs0 | 2 |f|∞ |s s0| λs0
e|f |∞ |s−s0|
.
This proves (3.17).
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