6 Gauge Freedom of Entropies on q-Gaussian Measures
141
Proof By the definition, we have that
d q,a ( p q (·; ξ 0 ), p q (·; ξ)) =
1
a
R
− q (x; ξ)
a ν q,a;ξ 0 (x)
=
1
a
R
− q (x; ξ)
a
− q (x; ξ 0 )
1−a χ q ( p q (x; ξ 0 ))dx
for ξ 0 , ξ ∈ R × q , which implies that
Ent q,1 ( p) = aEnt q,a ( p) = −
R
ln q ( p(x)) p(x)
q dx
=
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
−
R
p(x) − p(x)
q
1 − q
dx i f q > 1,
−
R
p(x) log( p(x))dx i f q = 1,
for p ∈ S q .
Recall that q = {σ > 0 | 1/(Z q σ ) < 1}. Since we observe that
lim
σ →∞
− q (x; 0, σ )
a
− ln q
1
Z q σ
=
⎧
⎪ ⎨
⎪ ⎩
∞ if a > 1,
1 if a = 1,
0 if a < 1, a = 0,
for x ∈ R, we apply the dominated convergence theorem a ≤ 1 and the monotone
convergence theorem for a > 1 to have
λD
(q,a)
( p, p q (·; 0; σ )) − D
(q,1)
( p, p q (·; 0; σ ))
− ln q
1
Z q σ
= −
λd q,a ( p, p) − d q,1 ( p, p)
− ln q
1
Z q σ
+
λd q,a ( p, p q (·; 0; σ )) − d (q,1) ( p, p q (·; 0; σ ))
− ln q
1
Z q σ
σ →∞
− −− →
⎧
⎪ ⎨
⎪ ⎩
λ · ∞ − M if a > 1,
(λ − 1)M if a = 1,
−M
if a < 1, a = 0,
for p ∈ S q and λ ∈ R, where we put 0 · ∞ := 0 and
M :=
R
χ q ( p(x))dx.
This constant M is obviously positive, and M is finite due to Lemma 5 in the next
section. This ensures that D
(q,a)
= λD
(q,1) for a = 1 and λ ∈ R.