# Find $A \subset \mathbb{R}^n$ that minimizes its volume under the constraint $\int_A f = b$, where $f : \mathbb{R}^n \to \mathbb{R}$.

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 $$A$$.

Edited

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