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Let $ (\Omega, \Sigma, \mu) $ be a measure space and let $f$ be a real valued function on $\Omega $ such that $\mu (x :f(x) is finite for all t $\in \mathbb{R}$. Let the number $G>0$ be and given and defined a class of measurable functions $\Omega$ by $C= (g: 0\le g(x) \le 1 $ for all $x$ and $ \int g(x)\mu(dx)=G) $

Then the minimization problem I= $\inf_{g \in C} \int f(x)g(x) \mu(dx)$ is solved by $g(x)= \chi_{(f where $s=\sup(t: \mu((x :f(x) and $c\mu ((x :f(x)=s))=G-\mu ((x :f(x)

My question is, how can one show that g(x) given above is the minimizer.

Thanks

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    you are right, thanks Ben2012-11-24

2 Answers 2

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Let $h$ be any other member of $C$. To show that $g$ is a minimizer, you need to establish that $I(g) \leq I(h)$, which is equivalent to showing $ \int f(g-h) \leq 0. $

To show this, split the range of integration into the sub-level, sup-level, and level sets of $f$; i.e., $\{f, $\{f>s\}$, and $\{f=s\}$:

\begin{align} \int f (g-h) &= \int\limits_{\{fs\}} f (g - h) + \int\limits_{\{f=s\}} f (g - h) \\ &\leq s \int\limits_{\{f s\}} f h + \int\limits_{\{f=s\}} s (g-h) \\ &\leq s \int\limits_{\{f s\}} h + s \int\limits_{\{f=s\}} (g-h) \\ & \leq s \left( \int\limits_{\{fs\}} (g-h) + \int\limits_{\{f=s\}} (g-h) \right) \\ & = s \int(g - h) = s (G -G) = 0. \end{align}

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    @ Q. Huang: Rick's proof is totally correct. In the "saturation set" \{f we have by definition of $g=(...)$ that $g=1$, and $f$ is only allowed to take values $f\leq 1$ (everywhere) so indeed $g-f\geq 0$ there. As the rest of the proof, this really exploits the specific structure of the natural candidate $g$.2016-12-12
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The proof depends on the form of the minimizer $g$: $g(x) := \chi_{(f For any $h$ satisfying $0\le h\le1$ and $\int_\Omega h\,d\mu=G$, we compute \begin{equation*} \begin{split} \int_\Omega fg\,d\mu &= \int_{\{fs}, \end{split} \right. \\ &\Longrightarrow G=\int_\Omega h\,d\mu=\mu(\{f$G=\mu(\{f, the minimizer is unique (up to a set of measure zero); when $G=\mu(\{f\le s\})$, since $0\le h\le1$, the minimizer is also unique.

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    @jake In the answer of Rick there are some defects (see the comment under his answer). I have written down the right one.2016-05-01