Newtonian capacity: do we have $\operatorname{cap}(A\cup B)+\operatorname{cap}(A\cup C)-\operatorname{cap}(A)-\operatorname{cap}(A\cup B\cup C)>0$

Question

$\newcommand{\Cap}{\operatorname{cap}}$ For compact disjoint sets A,B,D each with positive Newtonian capacity do we have

$$\Cap(A\cup B)+\Cap(A\cup C)-\Cap(A)-\Cap(A\cup B\cup C)>0\text{ ?}$$

strictly positive.

Attempt

Not sure. For any two sets with positive capacity we have $\Cap(A_1)+\Cap(A_2)-\Cap(A_1 \cup A_2)>0$ (1); thus let $A_1=A\cup B$ and $A_1=A\cup C$ and we have

$$\Cap(A\cup B)+\Cap(A\cup C)-\Cap(A\cup B\cup C)>0.$$

More precisely in terms of potentials we have

\begin{align} & \Cap(A\cup B)+\Cap(A\cup C)-\Cap(A\cup B\cup C) \\[8pt] = {} & \int_{A\cup B}U_{A\cup C} \, d\mu_{A\cup B\cup C}+\int_{A\cup C}U_{A\cup B} \, d\mu_{A\cup B\cup C}>0 \end{align}

So we ask if the following is strictly positive:

$$\int_{A\cup B}U_{A\cup C} \, d\mu_{A\cup B\cup C} + \int_{A\cup C} U_{A\cup B} \, d\mu_{A\cup B\cup C}-\int_A U_A \, d\mu_A>0$$

Any ideas or references? Feel free to move the post around, I wasn't sure what tag to put.

Thank you

(1) pg. 402 http://books.google.ca/books?id=scOX7_JUp_kC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

Answer 1

Let $U_{A},\mu_{A}$ be the potential and equilibrium measure of set A (see (1)). Then the above expression is equal to

$$\int_{A\cup B}U_{A\cup C} \, d\mu_{A\cup B\cup C} + \int_{A\cup C} U_{A\cup B} \, d\mu_{A\cup B\cup \sigma B}-\int_{A} U_{A} \, d\mu_A\geq $$

using $A\subset A\cup B\Rightarrow U_{A}\leq U_{A\cup B}$

$$\geq \int_{A\cup B}U_{A} \, d\mu_{A\cup B\cup C} + \int_{A\cup C} U_{A} \, d\mu_{A\cup B\cup \sigma B}-\int_{A} U_{A} \, d\mu_A =$$

by $\Delta U_{A}=\mu_{A}$ and Green's theorem we have

$\int U_{A}d\mu_{A\cup B}=-\left \langle U_{A},\Delta U_{A\cup B} \right \rangle=\int U_{A\cup B}d\mu_{A}$ ; and so

$=\int_{A\cup B}U_{A\cup B\cup C} \, d\mu_{A} + \int_{A\cup C} U_{A\cup B\cup C} \, d\mu_{A}-\int_{A} U_{A} \, d\mu_A>0$