TITLE: Is the matrix filled with the areas of pairwise intersections of disks in a plane always positive semidefinite?
QUESTION [5 upvotes]: Consider disks $s_1, \cdots, s_n$ in the plane and let $a_{ij}$ be the area of $s_i\cap s_j$. Is it true that for any real numbers $x_1,\cdots, x_n$ we have 
$$ \sum_{i,j=1}^n x_ix_j a_{ij} \geq 0$$

Equivalent formulation: one can put $a_{ij}$ into a matrix $A$ and ask whether it is positive semidefinite. For $n=2$ this is true since 
$$a_{12}^2\le \min(a_{11},a_{22})^2 \le a_{11}a_{22} $$

REPLY [11 votes]: The question above has an affirmative answer. More generally, we have the following result:
Theorem: Suppose $s_1, \dots, s_n$ are measurable sets in the plane $\mathbf{R}^2$ with finite area. Let $a_{i,j}$ be the area of $s_i \cap s_j$. Then
$$
\sum_{i,j=1}^n x_i x_j a_{i,j} \ge 0
$$
for all real numbers $x_1, \dots, x_n$.
Proof: In the inner product space $L^2(\mathbf{R}^2)$ (with the usual area measure), let $f_i$ be the characteristic function of $s_i$. Thus
$$
a_{i,j} = \int_{s_i \cap s_j} 1 = \int_{\mathbf{R}^2} f_i f_j = \langle f_i, f_j \rangle.
$$
Now suppose $x_1, \dots, x_n$ are real numbers. Let $f = x_1 f_1+ \dots + x_n f_n$. Then
\begin{align*} 
\sum_{i,j=1}^n x_i x_j a_{i,j} &= \sum_{i,j=1}^n x_i x_j \langle f_i, f_j \rangle\\[6pt]
&= \Bigl\langle \sum_{i=1}^n x_i f_i, \sum_{j=1}^n x_j f_j \Bigr\rangle\\[6pt]
&=\langle f, f \rangle \\[6pt]
&\ge 0,
\end{align*}
as desired.