12 3. PRELIMINARIES
of L−sf

for all s C with |s−s0| δ, and λs0 = λ. We will assume that h = hs0
is such that h = 1, where ν = ν−s0f

(cf. Theorem 3.1(a)(i)).
Set mf = min f and Mf = max f.
The following lemma is similar to Lemma 3.11 in Adachi-Sunada [AS].
Lemma 3.2. Let s = a + i b C. Under the above assumptions we have
(3.15) −Mf λs

∂a
λs −mf λs , s = a R , |a s0| δ .
Moreover, for any s C with |s s0| δ we have
(3.16) λs0
e−|f |∞ |s−s0|
|λs|≤ λs0
e|f |∞ |s−s0|
,
and
(3.17) |λs λ| 2 |f|∞ |s s0| λs0
e|f |∞ |s−s0|
.
Proof. Given x ΣC
+,
we have
σ(y)=x
e−sf (y)+ω(y)
hs(y) = λs hs(x) .
Differentiating this with respect to a and evaluating at a = s0 gives

σ(y)=x
f(y)
e−s0f (y)+ω(y)
h(y) +
σ(y)=x
e−s0f (y)+ω(y)
hs0 (y)
= λs0 h(x) + λ hs0 (x) ,
where λs0 =
(

∂a
λs
)
|s=s0
and hs0 =
(

∂a
hs
)
|s=s0
. Now integrating the above with
respect to ν implies

σ(y)=x
f(y)
e−s0f (y)+ω(y)
h(y) + L−s0f

(hs0 ) = λs0 + λ hs0 .
Since
L−s0f +ω(hs0 ) = λ hs0 ,
it follows that
λs0 =
σ(y)=x
f(y)
e−s0f (y)+ω(y)
h(y)
−mf
σ(y)=x
e−s0f (y)+ω(y)
h(y)
= −mf L−s0f
+g
h(y) = −mf λ .
The inequality −Mf λs λs0 is obtained in a similar way. Replacing s0 by any
a with |a s0| δ and using a similar argument one proves (3.15).
Thus, we have |

∂a
λs| |f|∞ λs for all s R, and therefore
(3.18) λs λs0
e|f |∞ |s−s0|
, s R .
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