Page 47 - 36-1

NTU Management Review Vol. 36 No. 1 Apr. 2026

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We first prove that for all >0, <0, and 0 , Equation (E3) holds for all
0 if and only if
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� ( ) − ( )�� − ( )�
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� � � � �
− � ( ) − ( )�� − ( )� ≤0, (E4)
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� � � � �
for all 0 and 0 , as follows.
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The if part: Since >0 and <0, if Condition (E4) holds for all 0 and 0 ,
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then, for all >0, <0, and 0 , Equation (E3) holds for all 0 .
��
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0 such that
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The only if part: Suppose, by contradiction, that there exists
�� �
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∗
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� ( ) − ( )�� − ( )�
� � � �� � �
∗
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�
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− � ( ) − ( )�� − ( )� >0,
� � � �� � �
for all 0 . Due to continuity, the above equation holds for all − +
�� �� ��
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�� ⊂ 0 , where �� is positive and arbitrarily small. We define and as follows:
�
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= { �� − �� + �� ⊂ 0 | ( ) − ( )
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× − ( ) − ( ) − ( ) − ( ) > 0},
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∗
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� � � � � � �
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= { 0 �� − �� )∪( �� + �� | ( ) − ( )
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× − ( ) − � ( ) − ( )�� − ( )� ≤ 0}.
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In addition, we define and as follows:
�
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� � ∗ � � ∗ � ∗ �
= � �� ( ) − ( )�� − ( )� − � ( ) − ( )�� −
� � � � � � � � � �
( )�� ,
∗
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�
∗
= � �� ( ) − ( )�� − ( )� − � ( ) − ( )�� −
� � � � � � � � � �
( )�� .
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By the definitions of and , we know that >0 and ≤0. Then, consider an individual
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with utility �� defined as � �� ( − ( )) = 0 and
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− �� if
�
��
�� = � �
− if
0 otherwise,
��
where is a positive scalar that makes �� satisfy FOC (9) after integrating by parts. Because
��
=− < 0 and <0, we can define ( − ( )) = 0. Consequently,
��(� �)
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∗
��
�� �
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