content_image/320231.jpg content_image/320229.jpg content_image/320230.jpg content_image/320228.jpg content_image/320226.jpg
is a holomorphic function of z on$_{C}$. Consequently, d ( x ; µ ) is a continuous function of x on$_{R}$. 

Proof: It can be shown that ∫ ∞ −∞ φ ( y − z ) log φ ( y − z )d y is a constant, thus a holomorphic function of z on$_{C}$. Therefore, it remains to prove that 

is a holomorphic function of z on$_{C}$. 

First, by Jensen’s inequality, we have 

<img src='content_image/320231.jpg'>

<img src='content_image/320229.jpg'>

where a = − E$_{µ}$ { X } and b = 1 2 $^{(}$E$_{µ}$ $^{{}$X 2 $^{}}$+ log(2 π ) ) are real numbers due to the fact that µ ∈ Λ( q ) . Thus, p$_{Y}$ ( y ; µ ) ∈ [ e − 1 $_{2}$y $^{2}$− ay − $^{b}$, 1] , i.e., 

As a result, we have 

<img src='content_image/320230.jpg'>

<img src='content_image/320228.jpg'>

which is integrable. (Here Re ( z ) and Im ( z ) represent the real and imaginary parts of z , respectively.) It follows that ξ ( z ) given by (24) exists for any µ ∈ Λ( q ) and z ∈$_{C}$ . 

Suppose U is an open and bounded subset of $_{C}$. There exists an r > 0 such that | Re ( z ) | ≤ r and | Im ( z ) | ≤ r for all z ∈ U . It is easy to check that 

<img src='content_image/320226.jpg'>

