4
$\begingroup$

If $\mu$ is a complex measure on a $\sigma$-algebra $M$, show that every set $E \in M$ has a subset $A$ for which $$|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$$ The suggestion is as follows: Put $d\mu=e^{i\theta}d|\mu|$. Let $A_\alpha$ be the subset of $E$ where $cos(\theta - \alpha) > 0$. Show that $$Re[e^{-i\alpha}\mu(A_\alpha)]=\int_Ecos^+(\theta-\alpha) d|\mu|,$$ and integrate with respect to $\alpha$.

Show, by an example, that $1/\pi$ is the best constant in this inequality.

Thanks for the help.

$\endgroup$
1
$\begingroup$

From the definitions in the hint, we have:

$$ e^{-i\alpha}\mu(A_\alpha) = \int_{E} \chi_{A_\alpha} e^{i(\theta - \alpha)}\,d|\mu| $$

Thus:

$$ \left|\mu(A_\alpha)\right| \ge \Re\left(e^{-i\alpha}\mu(A_\alpha)\right) = \Re\left(\int_{E} \chi_{A_\alpha} e^{i(\theta - \alpha)}\,d|\mu|\right) = \int_E \cos^+(\theta-\alpha)\,d|\mu| $$

Choose $\alpha = \alpha_0$ to maximize the integral on the RHS. The value at $a_0$ is at least as large as the average:

$$ \left|\mu(A_{\alpha_0})\right| \ge \frac{1}{2\pi} \int_{-\pi}^\pi \int_E \cos^+(\theta-\alpha)\,d|\mu|d\alpha $$

Use Fubini's theorem to swap integral signs on the RHS and obtain:

$$ \left|\mu(A_{\alpha_0})\right| \ge \int_E \frac{1}{2\pi} \int_{-\pi}^\pi \cos^+(\theta-\alpha)\,d\alpha d|\mu|= \frac{1}{2\pi}\int_E 2 \,d|\mu| = \frac{1}{\pi}|\mu|(E) $$


To show that this is the best possible constant, consider the unit circle equipped with $d\mu = e^{i\theta} \, dm$ where $m$ is the Lebesgue measure. For any measurable set $E$ and $A \subset E$, pick $\alpha$ so that $e^{-i\alpha}\mu(A) \ge 0$ (always possible). I'll let you show that $\left|\mu(A)\right| \le \frac{1}{\pi}|\mu|(E)$.

$\endgroup$
  • $\begingroup$ @CHG No problem. I also added an example to show that $1/\pi$ is the best possible constant. $\endgroup$ – Ayman Hourieh Mar 23 '13 at 17:09

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.