3. PRELIMINARIES 13
Now assume that s = a + ib C, |s s0| δ. It follows from Theorem 3.1(b)
that |λs| is equal to the spectral radius µa of the operator L−af +g. Using (3.18)
with s replaced by a and λs by µa, one gets
|λs| = µa λs0
e|f |∞ |a−s0|
λs0
e|f |∞ |s−s0|
for all s C with |s s0| δ. Similarly, |λs|≥ λs0 e−|f |∞ |s−s0|.
To prove (3.17), following the proof of Theorem 4.5 in [PP], consider
ua = (−af + ω) log ha σ + log ha log µa , gs = ua i b f ,
where ha is a positive eigenfunction of L−af

corresponding to µa with ha dνa =
1, νa = ν−af +ω. It then follows that Lua 1 = 1 and Lgs ws = αs ws for some αs C
and ws Fθ(ΣC)
+
with |αs| = 1 and |ws(x)| = 1 for all x ΣC
+.
Moreover,
Lgs q = αs ws Lua (q/ws) for any q Fθ(ΣC)
+
and λs = αs µa. Also, a direct
calculation shows that
(3.19) Lua (q) =
1
µaha
L−af
+g
(haq) .
It follows from perturbation theory that αs and ws depend analytically on s.
Fix a R for a moment and differentiate (for a given x ΣC
+
, as in part (a) above)
the relation
σ(y)=x
egs(y)
ws(y) = αs ws(x)
with respect to b to get
−i
σ(y)=x
f(y)
egs(y)
ws(y) +
σ(y)=x
egs(y)

∂b
ws (y)
=

∂b
αs ws(x) + αs

∂b
ws (x) .
That is
−i
σ(y)=x
f(y)
egs(y)
ws(y) + Lgs

∂b
ws (x) =

∂b
αs ws(x) + αs

∂b
ws (x) ,
which is equivalent to
−i
σ(y)=x
f(y)
egs(y)
ws(y) + αs ws(x)Lua
1
ws

∂b
ws (x)
=

∂b
αs ws(x) + αs

∂b
ws (x) .
Using (3.19), the latter implies

i
ws(x)
σ(y)=x
f(y)
egs(y)
ws(y) +
αs
µaha(x)
L−af
+g
ha
ws

∂b
ws (x)
=

∂b
αs +
αs
ws(x)

∂b
ws (x) .
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