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I have a question on the proof of lemma 3.3: Let $g\in L_0$ satisfy $u \circ g = b$, $z\in Y$ satisfy $J(z)=0$ where $J$ satisfy lemma 3.2, and let $f=\alpha g + (1-\alpha)z$. Then $u\circ f = \alpha u\circ g + (1-\alpha) u \circ z$. It is then stated that $\alpha u\circ g + (1-\alpha) u \circ z = \alpha b$. This implies that $u \circ z = 0$. Why is this the case?

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    $\begingroup$ This question could be improved to be easier to answer and be more of a reference to others by linking to the underlying paper and explaining the notation, particularly what that circular operator does. $\endgroup$ – BKay Feb 9 '16 at 16:06
  • $\begingroup$ It would be helpful as a reference for this question (and in understanding the below answer) if you stated what Lemma 3.2 is. $\endgroup$ – Kitsune Cavalry Mar 15 '16 at 17:36
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Notice, $z$ is in $Y$, and so it is a lottery. Lemma 3.1 uses the standard vNM techniques to identify $u: Y \to \mathbb{R}$. Now, $J$ is defined as a function from $L_0$ to $\mathbb{R}$, so the statement $J(z) = 0$ is via the natural identification between $Y$ and constant acts, $L_c$.

Condition (ii) of Lemma 3.2 says that $J(y) = u(y)$ for all $y \in L_c \cong Y$. So, $u(z) = J(z) = 0$, or, in the other notation, $u \circ z = 0$.

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