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Let $u(x)$ be a real function with the properties

  • $u(x)$ is continuous and non-decreasing in $x$
  • $u'(x)$ is non-increasing in $x$
  • $u(0)=0$
  • $u'(0)=1$.

In other words, $u(x)$ is a utility function.

In a rather old insurance paper the series on the left hand side of $$\sum_{k=0}^{\infty} \frac{\lambda^k}{k!}u(-\lambda u(-z)-zk)=0,$$ where $z>0$ and $\lambda>0$, is differentiated with respect to $\lambda$ rather non-chalantly and without further comment by differentiation of every summand, i.e differentiation and summation are exchanged. Is the uniform convergence of the series consisting of the derivatives of the summands, i.e $$\sum_{k=0}^{\infty} \left(\frac{\lambda^{k-1}}{(k-1)!}u(-\lambda u(-z)-zk)-\frac{\lambda^k}{k!}u'(-\lambda u(-z)-zk)u(-z)\right),$$ needed for this exchange to work really that easy to see? To be more specific, how would such a general series be approached?

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    differentiation is a linear operator and summation as well...as I see it, you can exchange it with no problem.2012-09-18
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    @Cristian Not true, Weierstrass M-test or showing series is cauchy is only thing that comes to mind. But I don't see it working unless $\lambda$ has some constraint.2012-09-18
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    @Mycroft does your original eq holds as an identity for $\lambda>0$?2012-09-18
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    @Cristian It does.2012-09-18
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    @Mycroft then, should the derivative of the lhs be equal to the derivative of the rhs implying that the sum of derivatives wrt $\lambda$ is zero?2012-09-18
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    @Cristian Well, yes the derivative of the LHS and the RHS should be equal, but this does not automatically imply that the sum of derivatives with respect to $\lambda$ is zero, as derivation can be performand summand-wise only once uniform convergence has been established.2012-09-18

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