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 $|Du_n|$.

Narrow convergence of measures and intermediate convergence of BV functions

Recall By the Riesz-Alexandroff representation theorem the dual of the space $C_0(\Omega,\mathbb{R}^N)$ (and thus of $C_c(\Omega,\mathbb{R}^N)$ can be isometrically identified with $M(\Omega,\mathbb{R}^N)$. i.e. any Borel measure $\mu$ is a bounded linear functional (continuous linear form if you like that) on $C_0(\Omega,\mathbb{R}^N)$ or $C_c(\Omega,\mathbb{R}^N)$. Thus, the weak convergence of a sequence of Borel measures $\mu_n$ is defined as follows.

Weak convergence of Borel measures: We say that a sequence of Borel measures $\mu_n$ converges weakly to $\mu$ i.e. $\mu_n\rightharpoonup \mu$ if $\langle\mu_n, \phi \rangle \rightarrow \langle \mu, \phi\rangle $ for all $\phi \in C_0(\Omega,\mathbb{R}^N)$.

Here, $\langle \mu, \,\phi\rangle\equiv\mu(\phi)\equiv\int_{\Omega} \phi \, d\mu \equiv \sum_{i=1}^N\int_{\Omega} \phi_i d\mu_i$.

Narrow convergence: A sequence $\{\mu_n\}$ in $M(\Omega, \mathbb{R}^N)$ narrowly converges to $\mu$ ($\mu_n\rightharpoonup \mu$ narrowly) in $M(\Omega, \mathbb{R}^N)$ if and only if $$\int_{\Omega} f\, d\mu_n \rightarrow \int_{\Omega} f d\mu,$$ for all $f \in C_b(\Omega, \mathbb{R}^N)$.

Narrow convergence is stronger than weak convergence.
If $\Omega=(0, 1)$ and $\mu_n=\delta_{\frac{1}{n}}$ then $\mu_n\rightharpoonup 0$ weakly. This is because for all $\phi \in C_0(\Omega)$, we have, $\mu_n(\phi)=\phi(\frac{1}{n})\rightarrow \phi(0)=0=\mu(\phi)$.

On the other hand, if $\phi\equiv1$ on $\Omega$, (note that $\phi \not\in C_0(\Omega)$ but $\phi \in C_b(\Omega)$), then we see that $\int_{\Omega} d\mu_n=\mu_n(\Omega)=1$ for all $n$. Essentially what happened with $\mu_n=\delta_{\frac{1}{n}}$ is that it converges weakly to $\mu=0$, but $\mu_n(\Omega)$, the size of $\Omega$ measured in $\mu_n$ does not converge to the size of $\Omega$ in measure $\mu(=0)$, that is $\mu_n\rightharpoonup \mu$ but $\mu_n(\Omega)\nrightarrow \mu(\Omega)$.

We have the following compactness result for the narrow topology.

PROKHOROV's sequential compactness theorem
If for a bounded subset $\mathcal{H}$ of $M^+(\Omega)$ it is true that for all $\epsilon$ there exists a compact subset $K_{\epsilon}\subset \Omega$ such that $\sup\{\mu(\Omega\ K_{\epsilon}):\mu\in \mathcal{H}\}\leq \epsilon$, then $\mathcal{H}$ is sequentially compact for the narrow topology.

The following equivalence proposition says that the convergence $\mu_n(\Omega)\rightarrow \mu(\Omega)$ is precisely what we need to ensure narrow convergence of measures.

Proposition
Let $\mu_n$ and $\mu$ be in $M^+(\Omega)$, non-negative Borel measures, then the following assertions are equivalent:
(i) $\mu_n\rightharpoonup \mu$ narrowly.
(ii)$\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu \rightharpoonup \mu$ weakly.

Proof
Note that as narrow convergence is stronger than weak convergence, it is clear that if $\mu_n\rightarrow \mu$ narrowly then $\mu \rightharpoonup \mu$. Also, $\mu_n(\Omega)\rightarrow \mu(\Omega)$ follows by taking $f\equiv 1$.

So, we need to only prove that if $\mu \rightharpoonup \mu$ weakly and $\mu_n(\Omega)\rightarrow \mu(\Omega)$ then $\mu_n\rightharpoonup \mu$ narrowly. That is, we need to show that for any function $f$ in $C_b(\Omega, \mathbb{R}^N)$ we have $\big| \int_{\Omega} fd\mu_n - \int_{\Omega} fd\mu \big|\rightarrow 0$. To do this we add and subtract $\int_{\Omega} f\,\phi d\mu_n$ and $\int_{\Omega} f\,\phi d\mu$ where $\phi$ is continuous function that is compactly supported in $\Omega$. The result follows as $f$ is bounded i.e. $\Vert f \Vert_{\infty}$ is a finite number and $\int_{\Omega} f\phi d\mu_n \rightarrow \int_{\Omega} f\phi d\mu$ due to compact support of $\phi$ in $\Omega . \hspace{25pt}\square$

More equivalence proposition
Let $\mu_n$ and $\mu$ be in $M^+(\Omega)$, non-negative Borel measures, then the following assertions are equivalent:
(i) $\mu_n\rightharpoonup \mu$ narrowly.
(ii) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(U)\leq \lim\inf_{n\rightarrow \infty} \mu_n(U)$ for all open subsets $U\subset \Omega$.
(iii) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(F)\geq \lim\inf_{n\rightarrow \infty} \mu_n(F)$ for all closed subsets $F\subset \Omega$.
(iv) $\mu_n(B)\rightarrow \mu(B)$ for all Borel subsets $B\subset\Omega.\,\,\hspace{25pt}\square$

Ok, so far so good. But images are not necessarily continuous functions. They might have jump discontinuities, i.e. edges. For this the following result due to Marle (Measure et probabilities) is helpful:

Proposition
Let $\mu_n$ and $\mu$ be non-negative Borel measures $M^+(\Omega)$, with $\mu_n\rightharpoonup \mu$ narrowly and let $f$ be a $\mu_n$ measurable bounded function for all $n$, such that it is discontinuous only on a set of null $\mu-$ measurable set then
$$\hspace{25pt}\int_{\Omega} f\, d\mu_n \rightarrow \int_{\Omega} f\, d\mu \hspace{25pt}\square$$

Now, that we have seen a notion of convergence for sequence of measures that is stronger than weak convergence, let's look at the a notion of convergence for sequence of functions in $BV$ space that is stronger than weak convergence.

Intermediate convergence
Let $\{u_n\}$ be a sequence in $BV(\Omega)$ and $u\in BV(\Omega)$. we say that $u_n\rightharpoonup u$ intermediately (or in the sense of intermediate convergence) if and only if
$$u_n\rightarrow u \mbox{ in } L^1(\Omega),$$
$$\hspace{25pt}\vert u_n\vert_{BV}\rightarrow \vert u\vert_{BV}\,\,\hspace{25pt}\square$$

Recall, from the previous post that for $\{u_n\}\in BV(\Omega)$ with bounded TV mass and $u_n\rightarrow u$ strongly in $L^1(\Omega)$, then we have $u_n\rightharpoonup u$ in $BV$, but $\vert u \vert_{BV}\leq \lim\inf_{n\rightarrow \infty}\vert u_n \vert_{BV}$ i.e. some of the mass is lost.

Ok, now I am hungry, so more about this tomorrow.