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We start with two disjoint compact sets A and B with positive capacities. Then, we translate B s.t. $B+rv$ is disjoint from A and B and ,more importantly, $dist(x,y)<dist(x,y+rv)$ for all $x\in A$ and $y\in B$. Does this imply

$$Cap(A\sqcup B)<Cap(A\sqcup (B+rv))?$$

We have $\leq $ because contraction maps decrease capacity (Landkoff in chapter "metric properties of capacities"). The map that sents $A\sqcup B+rv$ to $A\sqcup B$ decreases distances.

I will post as I find things.

We have $Cap(A)=[inf_{\mu(A)=1}\int_{A}\int_{A}\frac{1}{|x-y|^{d-2}}d\mu(x) d\mu(y)]^{-1}$. Call the infimum measure $\mu_{A}$ (equilibrium measure).

So the diffulty is in showing that $\mu_{A\sqcup B}\geq \mu_{A\sqcup (B+rv)}$.

Why does there exist measure $\mu'_{A\cup B+rv}$ s.t. $\int\int_{A\cup B} \frac{1}{|x-y|} d\mu_{A\cup B}>\int\int_{A\cup B+rv} \frac{1}{|x-y|} d\mu'_{A\cup B+rv}$?

If we take any measure supported on $A$ and $B+rv$, it is not obvious to me that that the above inequality is true.

If we set $\mu'_{A\cup B+rv}(A):=\mu_{A\cup B}(A)$, how can we shift the measure while keeping the integral lower or equal?

Maybe just define $\mu'_{A\cup B+rv}(B+rv):=\mu_{A\cup B}(B)$

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  • $\begingroup$ I did not understand how do you use the Landkof result. Why does there exist a contraction which sends $A$ to $A$ and $B$ to $B+rv$? (I do not think this is true without further restrictions.) And what does your strange sign like $\Pi$ upside down nean? Is this the union? $\endgroup$
    – Alexandre Eremenko
    Commented Jan 18, 2015 at 22:07
  • $\begingroup$ The map that sents $A\sqcup (B+rv)$ to $A\sqcup B$ decreases distances. The $\sqcup$ is standard notation for disjoint union. $\endgroup$
    – Thomas Kojar
    Commented Jan 18, 2015 at 23:11
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    $\begingroup$ I also don't see why your map decreases distances. Your condition that $\textrm{dist}(A,B)$ goes up after translating $B$ does not prevent individual points from $B$ from getting closer to a fixed point in $A$ after translation. $\endgroup$
    – Christian Remling
    Commented Jan 19, 2015 at 1:06
  • $\begingroup$ @AlexandreEremenko, as a minor curiosity, note that the similarity to an upside-down $\Pi$ is not a coincidence; $\prod$ (\prod) right-side up is a product, whereas $\coprod$ (\coprod) upside down is a coproduct (set-theoretic, in this case). $\endgroup$
    – LSpice
    Commented Jan 19, 2015 at 1:56
  • $\begingroup$ Now the claim seems clear. Since you can just use the same equilibrium measure as before to estimate the $\inf$ (the $B$ part suitably translated), the capacity could only stay the same if the equilibrium measure was supported by $A$ or $B$, but this is impossible because the potential is constant q.e. on $A\cup B$ and $A,B$ have positive capacity. $\endgroup$
    – Christian Remling
    Commented Jan 19, 2015 at 1:59

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