5
$\begingroup$

Let $\{ \sigma_n \}$ be a sequence of positive measures on the complex unit circle $\mathbb{T}$ with its borel sets, and Suppose that $\{ \sigma_n \}$ converges weakly to $\sigma$ which is also such a measure. Suppose that $\mu$ is another positive measure on $\mathbb{T}$ such that $\sigma_n\ll\mu$ for every $n$. Does this imply that $\sigma\ll\mu$? If so, can we tell anything about $d\sigma/d\mu$?

$\endgroup$
  • $\begingroup$ You can use \ll for $\ll$. $\endgroup$ – martini Sep 19 '12 at 12:17
  • $\begingroup$ In fact, the usual way for physicists to talk about "the delta function" is to approximate this singular measure as the weak limit of a sequence of absolutely continuous measures. $\endgroup$ – GEdgar Sep 19 '12 at 13:14
6
$\begingroup$

No, it doesn't. Let $\lambda$ be the arglength measure and $\phi_n \ge 0$ a continuous function on $\mathbb T$ with $\int_{\mathbb T} \phi_n\, d\lambda = 1$ and $\phi_n(x) = 0$ if $|x-1| \ge \frac 1n$. Then for each continuous function $f\colon \mathbb T \to \mathbb R$ we have $\int_{\mathbb T} f\phi_n d\lambda \to f(1)$, that is $\phi_n \lambda \to \delta_1$ weakly. But $\delta_1$ is not $\lambda$-continuous.

$\endgroup$
  • $\begingroup$ What if all the functions $d\sigma_n/d\mu$ are non-negative and uniformly bounded (there is some $M>0$ such that for every $n$ we have $d\sigma_n/d\mu <M$)? $\endgroup$ – user25640 Sep 20 '12 at 21:32

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.