Timeline for Does Newtonian capacity increase strictly when mass is spread?
Current License: CC BY-SA 3.0
12 events
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| Jan 19, 2015 at 21:58 | history | edited | Thomas Kojar | CC BY-SA 3.0 |
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| Jan 19, 2015 at 13:46 | comment | added | Alexandre Eremenko | @TKM: I agree with Christian's last message. The measure cannot be supported by A or B because you said that they are both of positive capacity. | |
| Jan 19, 2015 at 3:52 | comment | added | Christian Remling | @TKM: I don't think this is interesting enough for an answer. Just write $\int\int \ldots$ as a sum of four parts, corresponding to $A$ or $B$ as the region of integration, and look at what happens to this if you now shift $B$ and that part of the measure (this will not be the new equilibrium measure, but it gives a lower bound on the capacity). | |
| Jan 19, 2015 at 2:47 | comment | added | Thomas Kojar | @Remling could you add more detail and put it as an answer? Eg. why capacity would stay the same if it was supported on A or B? | |
| Jan 19, 2015 at 1:59 | comment | added | Christian Remling | 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. | |
| Jan 19, 2015 at 1:56 | comment | added | LSpice |
@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).
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| Jan 19, 2015 at 1:41 | history | edited | Thomas Kojar | CC BY-SA 3.0 |
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| Jan 19, 2015 at 1:06 | comment | added | Christian Remling | 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. | |
| Jan 18, 2015 at 23:11 | comment | added | Thomas Kojar | The map that sents $A\sqcup (B+rv)$ to $A\sqcup B$ decreases distances. The $\sqcup$ is standard notation for disjoint union. | |
| Jan 18, 2015 at 23:10 | history | edited | Thomas Kojar | CC BY-SA 3.0 |
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| Jan 18, 2015 at 22:07 | comment | added | Alexandre Eremenko | 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? | |
| Jan 18, 2015 at 21:44 | history | asked | Thomas Kojar | CC BY-SA 3.0 |