Let $\Gamma$ be a smooth closed curve in the complex plane (for all practical purposes). Assume $f$ is a real-valued continuous function defined on $\Gamma$ and let $d\mu=fdm$, where $dm$ is the Lebesgue measure supported on the curve.

Is it fair to say that the logarithmic energy of $\mu$ is always non negative as long as the diameter of $\Gamma$ is small enough? More precisely $$\int_\Gamma\int_\Gamma\log\frac{1}{|z-w|}d\mu(z)d\mu(w)\geq0$$ provided $diam(\Gamma)<A$?


Yes, this is true if the supports of your measures are in the unit disk. All your other conditions are redundant: you do not need the curve, if you have one, it does not have to be smooth, and the measures do not need to have densities, and do not have to be positive.

The same is true also without the restriction on support if you additionally assume that $\mu(C)=0$.

Reference: Landkof, Inftoduction to modern potential theory.

See also Positivity of the Coulomb energy in two dimensions

  • $\begingroup$ A minus is missing in the question :) $\endgroup$
    – R W
    Apr 8 '15 at 3:03
  • $\begingroup$ I do not see any minus missing. I think my answer to the question as stated is correct. $\endgroup$ Apr 8 '15 at 3:06
  • $\begingroup$ Shouldn't one put minus in front of the logarithm??? Your answer is correct of course. $\endgroup$
    – R W
    Apr 8 '15 at 3:29
  • $\begingroup$ That minus sign is taken care of by having $\ln\frac{1}{|z-w|}$ as integrand. $\endgroup$
    – BigM
    Apr 8 '15 at 3:36
  • 1
    $\begingroup$ oups - sorry :) $\endgroup$
    – R W
    Apr 8 '15 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.