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Thomas Kojar
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Does Newtonian capacity increase strictly when mass is spread?

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, $dist(A,B)<dist(A,B+rv)$. 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 $\mu_{A\sqcup B}\geq \mu_{A\sqcup (B+rv)}$.

Thomas Kojar
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