5
$\begingroup$

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?

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

$\endgroup$
  • $\begingroup$ 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 Nov 1 '17 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 Nov 1 '17 at 7:33

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.