# Bathtub principle proof

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.

Thanks

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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

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}

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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.

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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. – Q. Huang May 1 at 15:29