I was reading the theory about the Riemann-Hilbert problem $\Phi^+(t)=G(t)\Phi^-(t)$ where $G(t)$ is a Holder continuous function on a closed curve $c$ with index $\operatorname{Ind}_cG(t)=0$. To solve this problem one takes logarithms on the above relation and reduces the problem to the additive R-H problem $\operatorname{Log}\Phi^+(t)-\operatorname{Log}\Phi^-(t)=\operatorname{Log}G(t)$ which using the Plemelj formulae has the solution given by the Cauchy Integral $$\operatorname{Log}\Phi(z)=\frac{1}{2\pi i}\int_{c} \frac{\operatorname{Log}G(t)}{t-z}\, dz.$$ But to use this formula the potential -here $\operatorname{Log}G(t)$- must satisfiy the Holder condition. The books that I have read about this justify this result using the fact that, since $\operatorname{Ind}_cG(t)=0$, the function $\operatorname{Log}G(t)$ is single-valued (which is true, no problem there) and thus Holder continuous.

So thats my question: How can we show that the function $\operatorname{Log}G(t)$ is H-continuous using the fact that $G(t)$ is H-continuous and $\operatorname{Ind}_cG(t)=0$?


Since the curve $G(t)$ does not pass through $0$, the infimum of $\|G\|$ is strictly positive. Call it $m$. The logarithm (locally defined) is Lipschitz with constant $1/m$, as long as we stay outside of the disk $D(0,m)$. Applying a Lipschitz function to a Hölder continuous function preserves Hölder continuity.

  • $\begingroup$ Can you show analytically that the logarithm is indeed Lipschitz for $\left|z\right|>m$? $\endgroup$ – peter Jun 12 '15 at 9:26
  • $\begingroup$ Yes, I can. Take the derivative. $\endgroup$ – user147263 Jun 12 '15 at 15:13
  • $\begingroup$ Based on the mean value theorem?That the derivative is bounded? I don't think that this is correct.Here $\operatorname{Log}$ is the complex logarithm. $\endgroup$ – peter Jun 12 '15 at 21:27
  • $\begingroup$ Well, you are free to think what you want. $\endgroup$ – user147263 Jun 12 '15 at 21:27

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