Let $\alpha\in(-1/2,0)$ and $x\in (0,1)$. Define the function $$f(x) = \int_x^1 z^{\alpha-\frac{1}{2}} (z-x)^{\alpha}dz.$$

I have the feeling that $|f(x)-f(y)| \leq C|x-y|^{1/2}$ or any other power between 0 and 1. The (improper) integral is convergent so one should expect that the resulting function is continuous of some Hölder degree. Does anyone see a quick way to prove it?

Thank you for any hint!

  • $\begingroup$ There is a problem when $\alpha\le -1/4$: the limit as $x\to 0$ is infinite, since $\int_0^1 z^{\alpha-\frac{1}{2}} (z-0)^{\alpha}dz$ diverges. Did you mean $\alpha\in (-1/4,0)$? $\endgroup$ – user147263 Jan 13 '15 at 22:30
  • 1
    $\begingroup$ No, but I only look at positive values of $x$. If you want define $f$ on $[\varepsilon,1]$ for $\varepsilon>0$. The integrand is not continuous on $[\varepsilon,1]$ but maybe $f$ is. $\endgroup$ – mark Jan 13 '15 at 22:48

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