= Solution
For <secant domination for expected utility derivatives>, at each sample value $x$ set $g_x(\theta)=U(\theta x)$. This is a differentiable concave function, regardless of the sign of $x$. Fix $\theta_0$ and $\delta>0$. For $\theta\in[\theta_0-\delta,\theta_0+\delta]$, concavity bounds its derivative between the two outer secant slopes:
$$
\frac{g_x(\theta_0+2\delta)-g_x(\theta_0+\delta)}\delta\leq g'_x(\theta)\leq\frac{g_x(\theta_0-\delta)-g_x(\theta_0-2\delta)}\delta.
$$
Because $U<0$, finiteness of $F$ means $\mathbb E|U(\eta X)|<\infty$ at each of these four endpoints. Their absolute values divided by $\delta$ therefore give a common integrable bound on $|g'_X(\theta)|$. Difference quotients satisfy the same bound by the <mean value theorem>. The <dominated convergence theorem> permits differentiation under the <expectation>, and pointwise continuity of $XU'(\theta X)$ with the same bound gives continuity of the derivative. \b[Hence $\boxed{F'(\theta)=\mathbb E[XU'(\theta X)]}$ and $F\in C^1(\mathbb R)$.] This argument proves absolute <integrability> of the displayed derivative; no unjustified differentiation assumption is needed.
Back to article page
