Rudin-Osher-Fatemi image denoising model

Today we will finally touch the subject of the celebrated image denoising model proposed by Rudin, Osher and Fatemi in 1992, ( L. Rudin, S. Osher, E. Fatemi, "Nonlinear total variation based noise removal algorithms", Physica D, vol. 60, pp. 259–268, 1992.)

This model is motivated by the following problem in image restoration:
Given a noisy image $f:\Omega\rightarrow \mathbb{R}$, where $\Omega$ is bounded open subset of $\mathbb{R}^2$, we need to obtain a clean image $u$.

For solving this problem they proposed the following minimization problem:
$$\mbox{ minimize } \int_{\Omega}|\nabla u|\\\mbox{ subject to } \int_{\Omega} u=\int_{\Omega}f \, \hspace{10 pt}\mbox{ and } \hspace{10 pt}\int_{\Omega} |f-u|^2 = \sigma^2.$$ where $\sigma^2$ is the estimated global variance of the additive noise. This can be also formulated in terms of soft constraint minimization:
$$u_{\lambda}=\arg\min_u \{\,|u|_{BV(\Omega)} + \lambda \Vert f-u\Vert_{L^2(\Omega)}^2\}.$$ where $|u|_{BV(\Omega)}$ is the BV-seminorm.
In general, this belongs to the class of minimization problems of the following type
$$u_{\lambda}=\arg\min_u \{\,J(u) + H(f, u) \}$$ where $J$ is a convex nonnegative regularization functional and the fitting functional $H$ is convex nonnegative with respect to $u$ for fixed $f$. Due to convexity there exists a minimizer $u_{\lambda}$ such that '0' is in the sub-differential of $J(u) + H(f, u)$ at $u_{\lambda}$ i.e.
$$0\in\partial_u J(u_{\lambda})+\partial_u H(u_{\lambda}, f).$$
(here subdifferential $\partial_u J$ at $u_{\lambda}$ is the set $\partial_u J(u_{\lambda}):=\{v:J(u)\geq J(u_{\lambda})+\langle v, u-u_{\lambda} \rangle \}$. )

If $H(u, f)=\lambda\Vert f-u\Vert_{L^2(\Omega)}^2$, then $\partial_u H(u, f)=2\lambda (u-f)$. Moreover if $H(u, f)=\lambda\Vert f-Ku\Vert_{L^2(\Omega)}^2$, where $K:L^2(\Omega)\rightarrow \mathcal{H}$, $\mathcal{H}$ being some Hilbert space, then $\partial_u H(u, f)=2\lambda K^*(Ku-f)$, where $K^*$ denotes an adjoint of $K$.

Formally, we can write $\partial J(u)=-\mbox{div}\Big( \frac{\nabla u}{|\nabla u|}\Big)$, and we can write the Euler-Lagrange differential equation for the Rudin Osher Fatemi model as follows:
$$ 0=-\mbox{div}\Big( \frac{\nabla u}{|\nabla u|}\Big)+2\lambda(u-f)$$ or equivalently,
$$ u=f+\frac{1}{2\lambda}\mbox{div}\Big( \frac{\nabla u}{|\nabla u|}\Big)$$
We impose Neumann boundary condition $\nabla u\cdot\nu=0$ in the boundary $\partial \Omega$, this ensures the condition on the mean: $\int_{\Omega}u=\int_{\Omega}f$.

Dependence on the scaling parameter $\lambda$:

The parameter $\lambda$ determines the denoising level. If $\lambda$ is small, then it is equivalent to penalizing the fidelity term more and the minimizer $u_{\lambda}$ is very close to the original image. On the other hand, for large $\lambda$ the minimizer is a cartoon representation of the image $f$.



We perform a few experiments on the image of Lenna:

Original image, Lena.

For $\lambda=0.8$ and the image in the range [0, 255] we get the following result:

The denoised image $u_{\lambda}$ for $\lambda=0.8$

We see above that for this $\lambda=0.8$ the denoising is minimal.


For $\lambda=0.05$ and the image in the range [0, 255] we get the following result:

The denoised image $u_{\lambda}$ for $\lambda=0.05$.

Here we see the details are gone, but we still see a lot of structure. Let's see what happens if we take $\lambda=0.01$, with image in [0, 255].

The denoised image $u_{\lambda}$ for $\lambda=0.01$.

We here see that we get only a cartoon image of Lena. In other words we get only large-scale structures in the image of Lena.

Properties of the ROF denoising:
Let us look at some properties of the ROF model now.

1. The $L^2$-norm of the minimizer $u_{\lambda}$ is bounded, and the bound is independent of $\lambda$.

This nice property can be proven (Nezzar) by realizing that $u_{\lambda}$ is a minimizer of the ROF functional, thus for any $w\in BV(\Omega)$ we have:
$$|u_{\lambda}|_{BV(\Omega)}+\lambda \Vert f-u_{\lambda}\Vert_{L^2(\Omega)}^2\leq |w|_{BV(\Omega)}+\lambda \Vert f-w\Vert_{L^2(\Omega)}^2.$$This is true even if we choose $w \equiv 0$, which gives us:$$|u_{\lambda}|_{BV(\Omega)}+\lambda \Vert f-u_{\lambda}\Vert_{L^2(\Omega)}^2\leq \lambda \Vert f\Vert_{L^2(\Omega)}^2.$$As $0\leq |u_{\lambda}|_{BV(\Omega)}$ we have:$$\Vert f-u_{\lambda}\Vert_{L^2(\Omega)}^2\leq \Vert f\Vert_{L^2(\Omega)}^2.$$By expanding the $L^2$ term and using Cauchy-Schwartz inequality we get $$\int_{\Omega} f^2-2fu_{\lambda}+u_{\lambda}^2 \leq \int_{\Omega}f^2,\\
\int_{\Omega} u_{\lambda}^2 \leq \int_{\Omega}2fu_{\lambda} \leq 2 \Vert u_{\lambda}\Vert_{L^2(\Omega)}\Vert f \Vert_{L^2(\Omega)},\\
\Vert u_{\lambda} \Vert_{L^2(\Omega)}\leq 2 \Vert f \Vert_{L^2(\Omega)}.$$

2. The average of the minimizer $u_{\lambda}$ is the same as the average of the original image $f$.

To prove this it suffices to prove that $\int_{\Omega}u_{\lambda}=\int_{\Omega}f.$ This follows from integrating the Euler-Lagrange differential equation and Neumann boundary condition.

Now let us discuss properties of the ROF minimization proven by Yves Meyer. Let's define a few things first.

Definition: Let $G$ denote the space as follows $G=\{v=\mbox{div } \bar{g}: \bar{g}=(g_1, g_2); \mbox{ with }g_1, g_2 \in L^{\infty}(\mathbb{R}^2)\}.$
The star-norm $\Vert v \Vert_*$ of $v\in G$ is defined as follows:
$$\Vert v \Vert_* =\inf \{\Vert |\bar{g}|_{l^2}\Vert_{L^{\infty}}:v=\mbox{div } \bar{g}\}$$

3. If $v\in L^2(\mathbb{R}^2)$, then
$$ \int f\, v \leq |f|_{BV(\Omega)} \Vert v\Vert_{*}.$$
This property is true for $f\in W^{1, 1}(\mathbb{R}^2)$. Then we can use the density of $W^{1, 1}(\mathbb{R}^2)$ in $BV(\mathbb{R}^2)$.

4. Meyer: If $\Vert f \Vert_* \geq \frac{1}{2\lambda}$, then the Rudin-Osher-Fatemi decomposition $f=u_{\lambda}+v_{\lambda}$ is characterized by the following two conditions:$$ \Vert v_{\lambda} \Vert_* = \frac{1}{2\lambda} \mbox{ and } \int u_{\lambda}v_{\lambda}=\frac{1}{2\lambda} |u_{\lambda}|_{BV}.$$
Moreover, if $\Vert f \Vert_* \leq \frac{1}{2\lambda}$, then $u_{\lambda}=0$ and $v_{\lambda}=f$.

To prove this assertion, consider that if $u_{\lambda}$ is the minimizer of the ROF energy then $u_{\lambda}+\epsilon h$ is not! for a constant $\epsilon$ and a function $h$. This gives $|\int vh|\leq \frac{1}{2\lambda}|h|_{BV}$. Secondly, taking $h=u_{\lambda}$ we obtain equality.

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 $|Du_n|$.

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 $K \subset \mathbb{R}^n$.