More on intermediate convergence

According to the vectorial version of the Riesz-Alexandroff representation theorem, the dual norm $\vert u \vert_{BV}=\Vert Du \Vert=\sup \{\int_{\Omega} u \,\mbox{ div } \mathbf{\phi}: \mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}$ is also the total mass $|Du|(\Omega)=\int_{\Omega}|Du|$ of the total variation $|Du|$ of the measure $Du$.

Moreover, from classical integration theory, the integral $\int_{\Omega}\phi \,d(Du)$ is well defined for all $Du$ integrable functions $\phi$ from $\Omega$ into $\mathbb{R}^N$, e.g. for functions in $C_b(\Omega, \mathbb{R}^N)$. For the same reasons, $\int_{\Omega}\phi\, d|Du|$ is well defined for all $|Du|$ integrable functions $\phi$, in particular for functions in $C_b(\Omega)$.

Thus, we can say that $|Du_n|\rightharpoonup |Du|$ narrowly in $M(\Omega)$ if $\int_{\Omega}\phi \, d|Du_n|\rightarrow \int_{\Omega}\phi\,d|Du|$, for all $\phi \in C_b(\Omega)$. If we choose $\phi\equiv 1$ on $\Omega$ then $\Vert Du_n \Vert \rightarrow \Vert Du \Vert$ i.e. $|u_n|_{BV}\rightarrow |u|_{BV}$.

Hence, $u_n\rightarrow u$ in $L^1(\Omega)$ and the narrow convergence of measures, $|Du_n|\rightharpoonup |Du|$ in $M(\Omega)$ implies the intermediate convergence, $u_n\rightharpoonup u$ (i.e. $u_n\rightarrow u$ in $L^1(\Omega)$ and $|u_n|_{BV}\rightarrow |u|_{BV}$).


We have the following proposition regarding the intermediate convergence.

Proposition
The following three assertions are equivalent
(i) $u_n\rightharpoonup u$ in the sense of intermediate convergence.
(ii) $u_n\rightharpoonup u$ in $BV(\Omega)$ and $\vert u_n \vert_{BV}\rightarrow \vert u \vert_{BV}$.
(iii) $u_n\rightarrow u$ strongly in $L^1(\Omega)$ and $\vert Du_n \vert\rightharpoonup \vert Du \vert$ narrowly in $M(\Omega)$.

Proof
The equivalence (i) to (ii) is a consequence of a proposition from the previous post. We have just seen above that (i) and (iii) are equivalent. We only need to prove that (ii) implies (iii).

Recall that $u_n\rightharpoonup u$ weakly in $BV(\Omega)$ means that
(a) $u_n\rightarrow u$ strongly in $L^1(\Omega)$ and
(b) the measures $Du_n\rightharpoonup Du$ weakly in $M(\Omega, \mathbb{R}^N)$.

Also, recall that for non-negative Borel measures $\mu_n$ and $\mu$ in $M(\Omega)$ the following statements are equivalent:
(c) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(U)\leq \lim\inf_{n\rightarrow\infty}\mu_n(U)$ for all open subsets $U\subset \Omega$.
(d) $\mu_n\rightharpoonup \mu$ narrowly in $M(\Omega)$.

Let, $\mu_n=|Du_n|$ and $\mu=|Du|$. We have been given that $|Du_n|(\Omega)\rightarrow |Du|(\Omega)$ i.e. we have $\mu_n(\Omega)\rightarrow \mu(\Omega)$. For showing the narrow convergence $\mu_n\rightharpoonup \mu$ we just need to prove the lower-semicontinuity property on $U\subset\Omega$,
i.e. we need to show that for $\mu(U)\leq \lim\inf_{n\rightarrow\infty}\mu_n(U)$ for all open subsets $U\subset \Omega$,
i.e. we need to show that for $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$.

To this effect we use the proposition from the previous blog which says that if $u_n\in BV(\Omega)$ with $|u_n|_{BV(\Omega)}$ bounded, with $u_n\rightarrow u$ in $L^1(\Omega)$, then we have the lower semicontinuity property i.e. $|Du|(\Omega)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(\Omega)$, and $u\in BV(\Omega)$.

Let $U\subset\Omega$ be any open subset. We have $u_n\in BV(\Omega)$, thus as $U\subset\Omega$, we have $u_n\in BV(U)$. Thus, we have $\sup_n |Du_n|(U)\leq \sup_n |Du_n|(\Omega)<\infty$.

The weak convergence $u_n\rightharpoonup u$ in $L^1(\Omega)$ implies the strong convergence $u_n\rightarrow u$ in $L^1(\Omega)$ and thus the strong convergence $u_n\rightarrow u$ in $L^1(U)$ follows as $U\subset \Omega$.

 Now that we have the conditions in the proposition: $u_n\in BV(U)$, $\sup_n |Du_n|(U)<\infty$, and $u_n\rightarrow u$ in $L^1(U)$ we have the lower semicontinuity property: $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$. 

As we have $|Du_n|(\Omega)\rightarrow |Du|(\Omega)$ and $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$, from the equivalence of (c) and (d) we get the narrow convergence: $|Du_n|\rightharpoonup |Du_n|$ narrowly in $M(\Omega). \,\,\hspace{25pt} \square$.

In conclusion, intermediate convergence implies narrow convergence of $|Du_n|$.