6
$\begingroup$

Suppose $\mu_j$ is a sequence of measures on $\mathbb{R}$. By the definition of weak convergence of measures, $\mu_j$ weak converges to $\mu$ means that for any bounded continuous function $f$, there holds that $$\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu.$$ My question is whether we can just check the equality $\int_{\mathbb{R}} f\mu_j \to \int_{\mathbb{R}} f\mu$ for all smooth functions $f$ with compact support.

I come across this problem while I try to prove the following problem:

If $\phi_j$ are a sequence of smooth convex functions defined on $\mathbb{R}$ with uniformly bounded second order derivative and $\phi_j$ converges to the convex function $\phi$ in $L^\infty(\mathbb{R})$, then $\phi_j''dx$ weakly converges to $\phi''dx$ as measures.

Can this be obtained just from the fact that the equality $$\int_{\mathbb{R}}f \phi_j''dx\to\int_{\mathbb{R}}f \phi''dx$$ holds for all smooth functions $f$ with compact support?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Wait a minute - how can a smooth convex function be in $L^\infty$? $\endgroup$
    – Nate Eldredge
    Commented Nov 1, 2017 at 9:05
  • $\begingroup$ Maybe he meant $\phi_j-\phi\to 0$ in $L^\infty(\mathbb{R})$. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 1, 2017 at 17:58

1 Answer 1

Reset to default
7
$\begingroup$

If the measures are probability measures, then yes you can; it's kind of a standard exercise.

The argument I've seen goes something like this: fix $\epsilon$ and choose a smooth compactly supported cutoff function $g$ with $0 \le g \le 1$ and $\int g\,d\mu \ge 1-\epsilon$ (possible by monotone convergence); let $K$ be the support of $g$. Then by assumption $\int g\,d\mu_j \to \int g\,d\mu \ge 1-\epsilon$ so we see that $\limsup_{j \to \infty} \mu_j(K) \ge 1-\epsilon$. In other words, $\{\mu_j\}$ is tight. Hence after passing to a subsequence, $\mu_j$ converges weakly to some measure $\nu$. Now we notice that $\int f\,d\mu = \int f\,d\nu$ for all smooth compactly supported $f$, and it follows from a monotone class type argument that $\mu=\nu$. Finally use the "double subsequence" trick to conclude that the original sequence $\mu_j$ also converges weakly to $\mu$.

If you don't assume they are probability measures then this can be false; let $\mu_j$ be a point mass at $j$ and let $\mu$ be the zero measure.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Are you assuming that $\mu$ is a probability measure? Otherwise, why couldn't you have $\mu_j$ a $\delta$-measure supported at $j$ and $\mu=0$. You would certainly have the correct convergence for all smooth functions with compact support (but not for bounded continuous functions) $\endgroup$
    – Anthony Quas
    Commented Nov 1, 2017 at 7:05
  • $\begingroup$ @AnthonyQuas: The question originally said "probability measure", and that's in the title, but apparently it was edited. You're right that the assertion is false without this assumption. $\endgroup$
    – Nate Eldredge
    Commented Nov 1, 2017 at 7:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .