Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ (\Omega, \Sigma, \mu) $ be a measure space and let $f$ be a real valued function on $\Omega $ such that $\mu (x :f(x)<t) $ 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 $g(x)= \chi_{(f<s)}(x) + c\chi_{(f=s)}(x)$ where $s=\sup(t: \mu((x :f(x)<t)) \le G) $ and $c\mu ((x :f(x)=s))=G-\mu ((x :f(x)<s))$

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


share|cite|improve this question
I think you have made a typo in your question.The $s$ should be $sup\{t:μ\{x:f(x)<t\}\le G\}$ and $μ(x:f(x)<t)$ is finite for all $t$. – Ben Nov 24 '12 at 16:04
you are right, thanks Ben – jake Nov 24 '12 at 17:30
up vote 6 down vote accepted

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., $\{f<s\}$, $\{f>s\}$, and $\{f=s\}$:

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

share|cite|improve this answer
thanks rick, 15char – jake Nov 28 '12 at 3:44
How did you assert that $\int_{\{f<s\}}f(g-h)\le s\int_{\{f<s\}}(g-h)$? Is it known that $g-h\ge0$ on the set $\{f<s\}$? I don't think it is right... – Q. Huang May 1 at 15:05

The proof depends on the form of the minimizer $g$: $$g(x) := \chi_{(f<s)}(x) + c\chi_{(f=s)}(x).$$ 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_{\{f<s\}}f\,d\mu + c\int_{\{f=s\}} f\,d\mu \\ &= \int_{\{f<s\}}f\,d\mu + cs\mu(\{f=s\}) \\ &= \int_{\{f<s\}}f\,d\mu + s\big(G-\mu(\{f<s\})\big) \\ &= \int_{\{f<s\}}(f-s)\,d\mu + sG \\ &= \int_{\{f<s\}}(f-s)\,d\mu + s\int_\Omega h\,d\mu \\ &= \int_{\{f<s\}}(f-s)\,d\mu + \int_\Omega(s-f)h\,d\mu + \int_\Omega fh\,d\mu\\ &= \int_{\{f<s\}}(f-s)(1-h)\,d\mu + \int_{\{f\ge s\}}(s-f)h\,d\mu + \int_\Omega fh\,d\mu \\ &\le 0 + 0 + \int_\Omega fh\,d\mu \\ &= \int_\Omega fh\,d\mu. \end{split} \end{equation*} The equality above holds if and only if \begin{equation*}\begin{split} \left\{ \begin{split} \int_{\{f<s\}}(f-s)(1-h)\,d\mu &= 0, \\ \int_{\{f\ge s\}}(s-f)h\,d\mu &= 0, \end{split} \right. &\iff \left\{ \begin{split} h=1 &\quad\text{in } {f<s}, \\ h=0 &\quad\text{in } {f>s}, \end{split} \right. \\ &\Longrightarrow G=\int_\Omega h\,d\mu=\mu(\{f<s\})+\int_{\{f=s\}} h\,d\mu. \end{split}\end{equation*} Hence, when $G=\mu(\{f<s\})$, 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.

share|cite|improve this answer
@jake In the answer of Rick there are some defects (see the comment under his answer). I have written down the right one. – Q. Huang May 1 at 15:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.