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Assume $\mu$ is a measure supported on a real finite interval $[a,b]$, and let $$p_\mu(z)=\int\log|z-t|d\mu(t),$$ denote the logarithmic potential associated to $\mu$. Are there (possibly simple) conditions on $\mu$ (or on its density $h$ if $d\mu=hdt$ is absolutely continuous w.r.t. the Lebesgue measure) that ensure that $p_\mu$ is (almost everywhere) differentiable as a function on $[a,b]$ ?

Are there classical references for such results ?

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I think there are necessary and sufficient conditions in the literature, but here is a simple sufficient condition : $d\mu=h\ dt$ with $h\in L^2([a,b],dt)$, because then the (distributional) derivative of $p_\mu$ is just the Hilbert transform $Hh$ of $h$, and $H$ maps $L^2$ to itself. Then $p_\mu$ is differentiable almost everywhere, as any absolutely continuous function $f$ (i.e. $f'\in L^1$).

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  • $\begingroup$ Thanks a lot ! Could it be extended to $h\in L^p$, $p>1$ ? At least, $H$ maps $L^p$ to $L^p$ and the last statement still works. $\endgroup$
    – user111
    Commented May 4, 2018 at 17:18
  • $\begingroup$ Very nice answer. I also looked at some other of your answers. Unfortunately there are not too many of us on Mathoverflow that support analysis. $\endgroup$
    – Piotr Hajlasz
    Commented May 5, 2018 at 3:09
  • $\begingroup$ @user111 If $H$ maps $L^p$ to itself (which I wasn't sure of :( ) then $h\in L^p (a,b)$ for some $p>1$ is enough. I know $H$ operates on Fourier transforms as multiplying with $i \sign(\xi)$ and that's straighforward for $L^2$... $\endgroup$
    – Jean Duchon
    Commented May 6, 2018 at 11:22
  • $\begingroup$ @PiotrHajlasz Thanks for your upvotes. Seems like we're a happy few to support analysis on MO, but that's OK to me. $\endgroup$
    – Jean Duchon
    Commented May 6, 2018 at 11:33

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