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

+ω

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) .