Given $f : \mathbb{R}^n \to \mathbb{R}$, smooth and $\int_{\mathbb{R}^n} f= 1$, I would like to find the area $A \subset \mathbb{R}^n$ that minimizes its volume under the constraint $\int_A f = b$, where $b$ : constant in $(0,1)$.

My attempt

By the method of Lagrange multiplier, define

$$ g = V(A) + \lambda \left( \int_A f - b \right) $$ where $V(A)$ denotes the volume of the area $A$.

My claim is that, the optimal $A$ is the area that satisfies

$$ A = \{a \in \mathbb{R}^n : f(a) \ge \xi \} $$

where $\xi$ is a quantity defined by $b$.

My claim is based on the following reasoning under a very strong condition. In the case of $n=1$, and $f$ : unimodal, $A$ should be in the form of $A = [a_1, a_2]$ that contains the mode. And we have

$$ \begin{aligned} \frac{\partial g}{\partial a_1} &= -1 - \lambda f(a_1) = 0 \\ \frac{\partial g}{\partial a_2} &= 1 + \lambda f(a_2) = 0 \end{aligned} $$

thus $f(a_1) = f(a_2)$ and second derivative gives concaveness by unimodality, so $A$ is optimal.

But I can't think of what form $A$ should take when $n \ge 2$ even in unimodal case, to prove my claim.

Also, since my calculus seems weak, so I wonder if there is any method to take derivative wrt $A$.


Assume that one condition has been added : same conditions else, but such that $A$ is unique. But still cannot proceed.


A minimizing set $A$ need not be unique. Consider for example $$ f(x) = \begin{cases} 1 \quad \text{if}\; x \in [0,1] \\ 0 \quad \text{otherwise} \end{cases} $$ By construction $\int_\mathbb{R} f = 1$. Then all sets $A \subset [0,1]$ having measure equal to $b$ are solutions. That's a lot of solutions, certainly uncountably many.

  • $\begingroup$ Thx, what if in the case such $A$ is unique? $\endgroup$ – Moreblue Oct 3 '18 at 2:53
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    $\begingroup$ Are you asking under what conditions the solution in unique? $\endgroup$ – Hans Engler Oct 4 '18 at 1:50
  • $\begingroup$ To be precise, if $f$ is such that there should not be infinitely many $A$, could things change? $\endgroup$ – Moreblue Oct 4 '18 at 4:10
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    $\begingroup$ Maybe there is no such $f$. Please think about concrete examples first before asking such general questions. $\endgroup$ – Hans Engler Oct 4 '18 at 12:36

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