Let $\rho\in L^1(R)$ be a given function. We associate it with $$U(x) = \frac{1}{2}\int_x^{+\infty}\rho(y)dy - \frac{1}{2}\int_{-\infty}^x\rho(y)dy.$$

Show that $x \mapsto U(x)$ is a continuous function.

My ideas:

Calculate $|U(x)-U(y)|$ and find an expression of type $C| x-y |$ using the mean value theorem for integrals.

  • $\begingroup$ It suffices to show "one of the integrals is continuous". You can use the uniform integrability of $|\rho|$ to do that. $\endgroup$ Dec 15 '20 at 14:31
  • $\begingroup$ See this. $\endgroup$ Dec 15 '20 at 14:47
  • $\begingroup$ I go to see the post that you give me $\endgroup$ Dec 15 '20 at 14:50

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