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.

There is something about the $BV$ space

I have been putting it off for a while. But now that I can write Latex in the blogger I can talk about $BV$ spaces.
Solutions of some mathematical problems which have discontinuities along one-codimensional manifolds where the first distributional derivatives are measures prompt us to consider the $BV$ space. I will refer to Attouch et al's book: Variational Analysis in Sobolev and BV spaces.

Definition of $BV$ space: We say that a function $u:\Omega\rightarrow \mathbb{R}$ is a function of bounded variations i.e. $u\in BV \,$ if and only if it belongs to $L^1(\Omega)$ and its gradient $Du$ in the distributional sense is in $M(\Omega, \mathbb{R}^N)$.

The following statements are equivalent.
(i) $u\in BV(\Omega)$
(ii) $u \in L^1(\Omega)$ and $u_{x_i}\in M(\Omega)$ for all $i=1, 2, ..., N$.
(iii) $u \in L^1(\Omega)$ and $\vert u \vert_{BV}<\infty$, where $\vert u\vert_{BV}:=\sup \{\langle Du, \,\mathbf{\phi}\rangle: \mathbf{\phi} \in C_c(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}.$
(here $\langle Du, \,\mathbf{\phi}\rangle:=\sum_{i=1}^N \int_{\Omega}\, \mathbf{\phi}_i u_{x_i}$.)
(iv) $u \in L^1(\Omega)$ and $\vert u \vert_{BV}=\sup \{\int_{\Omega} u \,\mbox{ div } \mathbf{\phi}: \mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}<\infty$.

The implication (ii)$\implies $(iii) follows as $C_c(\Omega, \mathbb{R}^N)$ is dense in $C_0(\Omega, \mathbb{R}^N)$. The implication (ii)$\implies $(iii) follows as $C_c^{\infty}(\Omega, \mathbb{R}^N)$ is dense in $C_0(\Omega, \mathbb{R}^N)$ and $C_c(\Omega, \mathbb{R}^N)$.

Notations and facts:
$\Omega \subset \mathbb{R}^N$.
$C_0(\Omega, \mathbb{R}^N)$ is the space of all continuous functions that vanish at infinity. i.e. for a function $\phi\in C_0(\Omega, \mathbb{R}^N)$ the following is true: Given any $\epsilon > 0 $ there exists a compact set $K_{\epsilon} \subset \Omega$ such that $\sup_{x\in\Omega \backslash K_{\epsilon}}|\phi(x)|\leq \epsilon$.
The functions $\mathbf{\phi}$ in $C_0(\Omega, \mathbb{R}^N)$ are equipped with the uniform norm
$\Vert \phi \Vert_{\infty}:= \sup_{x \in \Omega}\{ |\phi(x)|\}$.
$C_c(\Omega, \mathbb{R}^N)$ is the subspace of $C_0(\Omega, \mathbb{R}^N)$ with a compact support in $\Omega$.
$C_b(\Omega,\mathbb{R}^N)$ is the subset of all bounded continuous functions from $\Omega$ into $\mathbb{R}^N$.

Density results:
$C_c(\Omega, \mathbb{R}^N)$ is dense in $C_0(\Omega, \mathbb{R}^N)$.
The space of infinitely differentiable and compactly supported functions $C_c^{\infty}(\Omega, \mathbb{R}^N)$ is dense is $C_0(\Omega, \mathbb{R}^N)$ and $C_c(\Omega, \mathbb{R}^N)$.

Duality results:
$M(\Omega, \mathbb{R}^N)$ denotes the space of all $\mathbb{R}^N$ valued Borel measures.
$M(\Omega, \mathbb{R}^N)$ is isomorphic to the product space $M^N(\Omega)$.
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 $C_c(\Omega, \mathbb{R}^N)$ and the dual norms $\Vert\cdot \Vert_{C_{0}^{'}(\Omega, \mathbb{R}^N)}$ and $\Vert \cdot \Vert_{C_{c}^{'}(\Omega, \mathbb{R}^N)}$ are equal to the total mass $|\cdot|(\Omega)$ :
$$|\mu|(\Omega)\equiv \int_{\Omega}|\mu|=\sup\{ \langle \mu, \,\mathbf{\phi}\rangle: \mathbf{\phi} \in C_0(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1 \} =\sup\{ \langle \mu, \,\mathbf{\phi}\rangle: \mathbf{\phi} \in C_c(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1 \}.$$ Here, $\langle \mu, \,\phi\rangle\equiv\mu(\phi)\equiv\int_{\Omega} \phi \, d\mu \equiv \sum_{i=1}^N\int_{\Omega} \phi_i d\mu_i.$

The relation between $BV$ and the Sobolev space $W^{1, 1}$:

According to Radon-Nykodym theorem there exists $\nabla u \in L^1(\Omega, \mathbb{R}^N)$ and measure $D_s u$ that is singular with respect to $\mathcal{L}^N|_{\Omega}$, the $N-$ dimensional Lebesgue measure restricted to $\Omega$, such that: $$Du=\nabla u \, \mathcal{L}^N|_{\Omega}+ D_s u.$$ Thus, $W^{1, 1}$ is a subspace of the $BV-$ space and for functions in $W^{1, 1}$ we can write $Du=\nabla u$.

Norm on the $BV$ space:

The $BV$ space is equipped with the norm $\Vert u \Vert_{BV}=\Vert u \Vert_{L^1}+\vert u\vert_{BV}$.
Equipped with this norm the space of $BV$ functions is complete. The completeness of the $BV$ space follows from the completeness of $L^1$ and lower semicontinuity property which is stated below.

Proposition about the lower semicontinuity property of $BV-$ seminorm:
If $\{u_n\}$ is a sequence in $BV(\Omega)$ with $\sup_n |u_n|_{BV}< \infty$ that converges strongly $u_n \rightarrow u$ in $L^1(\Omega)$ then $|u|_{BV}\leq\lim \inf_{n\rightarrow \infty} |u_n|_{BV}$ and $u\in BV(\Omega)$.
Proof:
We use the following definition of BV seminorm here: $u \in L^1(\Omega)$ and $\vert u \vert_{BV}=\sup \{\int_{\Omega} u \,\mbox{ div } \mathbf{\phi}: \mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}<\infty$.
Let $\phi\in C_c^1(\Omega, \mathbb{R}^N)$ with $\Vert \phi \Vert_{\infty}\leq 1$.
We first observe that
$$\lim_{n\rightarrow \infty}\int_{\Omega} u_n\, \mbox{ div } \phi =\int_{\Omega} u\, \mbox{ div } \phi. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
This follows as $u_n\rightarrow u$ in $L^1(\Omega)$ strongly, and that $\phi\in C_c^1(\Omega, \mathbb{R}^N)$. Indeed, $$ \big\vert \int_{\Omega} u_n\, \mbox{ div } \phi - u\, \mbox{ div } \phi\, \big\vert \leq \int_{\Omega} |u_n-u|\cdot |\mbox{ div } \phi | \rightarrow 0,\mbox{ as } n \rightarrow \infty.$$ Note that $ \int_{\Omega}u_n \mbox{ div } \phi \leq \sup\{\int_{\Omega} u_n \,\mbox{ div } \mathbf{\phi}: \mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}$, i.e. $$\int_{\Omega}u_n \mbox{ div } \phi \leq |u_n|_{BV}.$$ Taking $\lim\inf$ of both sides, we get $$\lim_{n\rightarrow \infty}\int_{\Omega}u_n \mbox{ div } \phi \leq \lim\inf_{n\rightarrow \infty} |u_n|_{BV}\,.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$ From (1) and (2) we get
$$\int_{\Omega}u \mbox{ div } \phi \leq \lim\inf_{n\rightarrow \infty} |u_n|_{BV}\,.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$
Taking supremum on the left side of (3) over all $\mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1$, we get $|u|_{BV}\leq\lim \inf_{n\rightarrow \infty} |u_n|_{BV}$. Moreover, if $|u_n|_{BV}<\infty$ for all $n$ we get $|u|_{BV}<\infty$, i.e. $u\in BV$.

The weak convergence of $u_n\rightharpoonup u$ in $BV\,$ :
We say that a sequence $u_n$ in $BV(\Omega)$ converges weakly to some $u$ in $BV(\Omega)$ (i.e. $u_n\rightharpoonup u$ in $BV(\Omega)$) if and only if the following two convergences hold:
(i) $u_n\rightarrow u$ strongly in $L^1(\Omega)$
(ii) $Du_n \rightharpoonup Du$ weakly in $M(\Omega, \mathbb(R)^N)$.

Proposition: If $\{u_n\}$ is a sequence in $BV(\Omega)$ with $\sup_n |u_n|_{BV}< \infty$ that converges strongly $u_n \rightarrow u$ in $L^1(\Omega)$ then $u_n\rightharpoonup u$ in $BV$.
Proof:
As $u_n\rightarrow u$ in $L^1$ we only have to show the weak convergence $Du_n\rightharpoonup Du$.

For $\phi\in C_c^{\infty}(\Omega, \mathbb{R}^N)$ we have that $\langle Du_n, \phi\rangle =-\int_{\Omega}u_n \mbox{ div }\phi\rightarrow -\int_{\Omega}u \mbox{ div }\phi =\langle Du, \phi\rangle$.

As result of the density of $C_c^{\infty}(\Omega, \mathbb{R}^N)$ in $C_0(\Omega, \mathbb{R}^N)$ and the boundedness of $|u_n|_{BV}$ we conclude that $Du_n\rightharpoonup Du$.

APPENDIX: Basic definitions

Let $X$ be some set, and $2^X$ symbolically represent its power set, the collection of all subsets of $X$.

Measure: A mapping $\mu:2^X\rightarrow [0, \infty]$ is called a measure on $X$ if
(i) $\mu(\emptyset)=0$ and
(ii) $\mu(A)\leq \sum_{k=1}^{\infty} \mu(A_k)$, whenever $A \subset \cup_{k=1}^{\infty} A_k$.

$\mu-$ measurable set: A set $A\subset X$ is $\mu-$ measurable if for each set $B\subset X$,
$\mu(B)=\mu(B\cap A)+\mu(B\setminus A)$.


$\sigma$-algebra: A subset $\mathcal{A} \subset 2^X$ is called a $\sigma$-algebra if it satisfies the following three properties:
(i) $\mathcal{A}$ contains the null set and the set $X$, i.e. $\emptyset, X \in \mathcal{A}$.
(ii) $\mathcal{A}$ is closed under complementation: $A \in \mathcal{A}$ implies $X\setminus A \in \mathcal{A}$.
(iii) $\mathcal{A}$ is closed under countable unions: If $A_k \in \mathcal{A}$ for $k=1, 2, \dots$, then so is the union, $\cup_{k=1}^{\infty}A_k $.

Borel $\sigma$-algebra of $\mathbb{R}^n$ is the smallest $\sigma$-algebra of $\mathbb{R}^n$ containing the open subsets of $\mathbb{R}^n$.

Regular measure: A measure $\mu$ on $X$ is regular if for each set $A\subset X$ there exists a $\mu-$ measurable set $B$ such that $A\subset B$ and $\mu(A)=\mu(B)$.

Borel set: A Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement.

Borel measure: A measure $\mu$ on $\mathbb{R}^n$ is called Borel measure if every Borel set is $\mu-$ measurable.

Borel regular measure: A measure $\mu$ on $\mathbb{R}^n$ is called Borel regular measure if $\mu$ is Borel and for each $A \subset \mathbb{R}^n$ there exists a Borel set $B$ such that $A\subset B$ and $\mu(A)=\mu(B)$.

Radon measure: A measure $\mu$ on $\mathbb{R}^n$ is called Radon measure if $\mu$ is Borel regular and $\mu(K)< \infty$ for each compact set $K \subset \mathbb{R}^n$.