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Take any real-valued function $u$ taking values in $[0,1]$. Fix any CDF on $\mathbb{R}$, say $F_0$. Consider a maximization problem $\max_{F \in \mathcal{F}} \int^1_0 u(x) dF(x)$ where $\mathscr{F}$ is the set of distributions $F$ such that $F_0$ is a mean-preserving spread of $F$.

Can we say the following?: if we multiply a convex function $h$ to $u$, then $f^*$ chosen under $u\cdot h$ is more dispersed than the distribution $f^{**}$ chosen under $u$.

I know adding a convex function $h$ to $u$ makes the solution more dispersive, but I'm curious whether the similar thing holds about multiplication.

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    Your question is very unclear. "I know adding a convex function..." - can you provide the source of that knowledge? – 2017-02-16
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    The difference between $\int u(x) + h(x) dF(x)$ and $\int u(x) dF(x)$, which is $\int h(x)dF(x)$, increases when $F$ is replaced by another $F'$ which is a mean-preserving spread of $F$, which follows from Jensen's inequality.This implies that I choose more dispersed distribution under $u + h$ than under $u$, which follows from, say, http://faculty.insead.edu/vanzandt/teaching/CompStatics.pdf That sentence is not really a part of the question, btw. – 2017-02-17
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    And what are $f^*$, $f^{**}$? Maximizers? – 2017-02-18
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    Exactly, $f^*$ is a maximizer chosen under $u\cdot h$ and similarly for $f^{**}$. – 2017-02-18

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I do not think the property you are asking for is true. In fact, take $u(x)=1-x$ and $h(x)=x$. Both are convex, but their product is not. Or take $u(x)=(1-x)^2$ and $h(x)=\frac{1}{u(x)}$. Then both functions are convex but their product is constant. Given that, $f^{*}$ chosen under $u\cdot h$ can be anything.

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    I see! I should have been more careful about phrasing my question, but I would like to know more unambiguous counter example in which a maximizer of each problem is unique. (We can still choose a maximizer appropriately to make my statement --- multiplying a convex function makes me choose more dispersed distribution --- true. ) – 2017-02-17
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    @tarou2, one may take two strictly convex functions such that their product is concave. Then there are unique maximizers both for $u$ (it is $F_0$), and for $u\cdot h$ (it is $\delta_{\mu}$, where $\mu$ is the mean of $F_0$). So not just the maximizer of $u\cdot h$ isn't more dispersed in general, but it can even be the least dispersed element of $\mathcal F$. – 2017-02-18
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    Thank you, that makes sense. Since a product of non-negative convex functions is not necessarily convex, we should be able to construct an example. Would you happen to have an example where a product of non-negative convex functions turns out to be concave, which I could not discover? – 2017-02-19