Suppose $\mu_j$ is a sequence of probability 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?

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?