# Mumford and Shah functional

## Introduction

Let $\Omega$ be a bounded set in $\mathbb{R}^2$, $K$ be the closed set in $\Omega$

$u_0 \in L^{\infty}(\Omega)$ be the raw image

$u\in H^1(\Omega\setminus K)$ be the processed image, $u$ may have jump discontinuities on $K$
$$
F^{MS}(u,K) = \displaystyle\int_{\Omega\setminus K} (u-u_0)^2dx+\int_{\Omega\setminus K}|\nabla u|^2dx+\mathcal{H}(K)
$$
Define $K$ as $S_u$ the set of jump discontinuities of $u$

$F^{MS}(u,K)$ becomes
$$
F(u) = \displaystyle\int_{\Omega\setminus S_u} (u-u_0)^2dx+\int_{\Omega\setminus S_u}|\nabla u|^2dx+\mathcal{H}^1(S_u)\
$$
The goal is to find $u^{*} = \inf_{u\in SBV(\Omega)\cap L^{\infty}(\Omega)} F(u)$

---
### BV space

Consider that for $u \in L^{1}(\Omega)$,

The space of bounded variation, $BV(\Omega) = \left\{u\in L^{1}(\Omega): \sup\left\{\displaystyle\int _{\Omega}u \mbox{div}\phi : C^{\infty}_c(\Omega), \hspace{5pt} ||\phi||_{L^{\infty}}\leq 1\right\}<\infty\right\}$

---
### SBV space

the measure $Du$ can be decomposed into two part (Lebesgue measure decomposition)

$Du = D^{a.c}u+D^{s}u$

and $Du^{a.c}$ is the measure absolutely continuous with respect to $dx$ ($Du^{a.c}<<dx$)

if $dx(E) = 0, D^{a.c}u(E) = 0$

$Du^s$ is the singular part with respect to $dx$

for existing $A, B$ disjoint and union be $\Omega$ , where $dx(A)=0, D^su(B)=0$

Moreover $D^su$ can further be decomposed into $D^ju+D^cu$ where $D^ju$ is the jump part, $D^c u$ is the Cantor part

Consider that if $u\in BV(\Omega)$, the measure gradient $Du$ can be decomposed into
$$
Du = D^{a.c}u+D^ju+D^cu
$$
if $u\in SBV(\Omega)$, $D^cu(B) =0$ for all $B$ such that $\mathcal{H}^1(B)<\infty$

the total variation of $Du$ in $SBV$ space,
$$
||Du||(\Omega) =\displaystyle\int_{\Omega}|\nabla u|dx+\int_{S_u}|u^+-u^-|d\mathcal{H}
$$
---
### Compactness of SBV

First we have $u_n\in SBV(\Omega)$ be a sequence of functions

1. $|u_n|\leq C$ for some $C<\infty$ a.e
2. $\displaystyle\int_{\Omega} |\nabla u_n|^2dx+\mathcal{H}^1(S_{u_n}) \leq C$

Then,

1. $\exists \left\{u_{n_k}\right\}\subset\left\{u_n\right\}$ $u_{n_k}\rightarrow u \in SBV$ a.e
2. $\nabla u_{n_k} \rightharpoonup\nabla u \in L^2(\Omega)^2$
3. $\displaystyle\mathcal{H}(S_u)\leq\liminf_{n_k\rightarrow\infty}\mathcal{H}^1(S_{n_k})$
---
### Lemma

---
### Lemma

#### Every Bounded Sequence in a $L^p$, $1<p<\infty$ Has a Weak Convergent Subsequence

consider that $\left\{u_n\right\}$ be a bounded sequence in $L^2$ space

Since $L^2$ is separable, given a sequence $\left\{\phi_n\right\}$ of functionals which are bounded in the dual space of $L^2$

Consider that $\phi_n(u_1)$ which is bounded in $\mathbb{R}$ therefore, we have a convergent subsequence, $\phi_{s(1,n)}(x_1)$ that converge to a real number $a_1$

Then we have a subsequence $\phi_{s(2,n)}(x_2)$ that converge to a real number $a_2$

Continuing the process, we can choose $n_k = s(k,k)$ such that $\phi_{s(k,k)}(x_m)$ that converge to $a_m$

Next, we want to show that $\phi_{n_k}\rightarrow\phi$
$$
|\phi_{n_k}(x)-\phi_{n_l}(x)|=|\phi_{n_k}(x)-\phi_{n_k}(x_N)+\phi_{n_k}(x_N)-\phi_{n_l}(x_N)+\phi_{n_l}(x_N)-\phi_{n_l}(x)|\\
\leq |\phi_{n_k}(x)-\phi_{n_k}(x_N)|+|\phi_{n_k}(x_N)-\phi_{n_l}(x_N)|+|\phi_{n_l}(x_N)-\phi_{n_l}(x)|\\
\leq 2M\epsilon+|\phi_{n_k}(x_N)-\phi_{n_l}(x_N)|
$$
Therefore the sequence is cauchy, which will converge to some function $\phi$ in $L^2$. This shows the weak * converge.

To sum up, if the space is separable and there is a sequence that is bounded above, there exists a weakly * convergence subsequence.

By the discussion above,

---
## Existence of minimizer of $F(u)$

A functional admits it minima if and only if is obey the following conditions:

1. infimum should exists
2. $F$ is coercive
3. $F$ is lower semi-continuous

The first condition prevent the functional has no infimum

The second and the third are the key of the existence of minimizer.

---
### Coercivity

$V = SBV(\Omega)\cap L^{\infty}(\Omega)$

If $F$ is coercive, meaning that $\displaystyle\lim_{||u||_V\rightarrow\infty}F(u) = +\infty$

This condition is to give a uniform bound on the minimizing sequence $\left\{u_n\right\}_{n=1}^{\infty}$.

By the reflexivity of the space $V$, one can obtain a sebsequence $\left\{u_{n_k}\right\}_{k=1}^{\infty}\subset \left\{u_n\right\}_{n=1}^{\infty}$, such that $u_{n_k} \rightharpoonup \overline{u}$ in $V$.

---
### Lower semi-continuous

The lower semi-continuous admits that
$$
F(\overline{u})\leq \liminf_{k\rightarrow\infty}F(u_{n_k})
$$
Then $\overline{u}$ is the minimizer of $F(u)$

Proof of the Existence:

1. Take $u=0$, since $u_0\in L^{\infty}(\Omega)$ therefore, the functional is finite.

2. [Find a convergence subsequence, l.s.c]

Consider for $\left\{u_n\right\}_{n=1}^{\infty}\subset V$ be a sequence such that $\displaystyle\lim_{n\rightarrow\infty}F(u_n) = \inf_{u\in V} F(u)$.

WLOG, $||u_n||_{\infty}\leq||g||_{\infty}$ , $\displaystyle \sup_{u_n\in V} F(u_n) < C$ for some $C<\infty$.

Since $F(u_n)<C$, and $g\in L^{\infty}$, therefore $u_n\in L^2$ , and $\displaystyle\int_{\Omega\setminus S_u}|\nabla u_n|^2dx+\mathcal{H}^1(S_{u_n})<C$

By the compactness of SBV space, there exists a subsequence $u_{n_k}\rightarrow u\in SBV$ $a.e$ and $u_{n_k}\rightarrow u$ in $L^1$, $\nabla u_{n_k}$ converge weakly to $\nabla u$ in $L^2(\Omega)^2$

Since $u_{u_k}$ is in $L^2$ and bounded by $g$, we can obtain that $u_{n_k}$ is converge weakly in $L^2$.

therefore we deduce that $\displaystyle F(u) \leq\liminf_{n_k\rightarrow \infty} F(u_{n_k}) = \inf F$

## Ambrosio-Torelli Approximation (Convergence)

The functional $F(u)$ can be written in approximation form by Ambrosio-Torelli.
$$
AT_\epsilon(u,v) = \left\{\begin{array}{l}
\displaystyle\int_{\Omega}(u-g)^2dx+\beta\int_\Omega v^2|\nabla u|^2 +\int_\Omega(\epsilon|\nabla v|^2+\frac{(v-1)^2}{4\epsilon})dx\hspace{3pt}(u,v)\in H^1(\Omega)^2 v\in\left[0,1\right].\\
+\infty\hspace{5pt} o.w
\end{array}\right.
$$
We aim to construct a sequence of function $\left(u_\epsilon, v_\epsilon\right)$ that converge to $\left(u, 1\right)$ such that the sequence $AT_\epsilon(u_\epsilon, v_\epsilon)\rightarrow F(u)$

That is we are willing to show that the final term will converge to $\mathcal{H}^1(S_u)$


The final term can be written as
$$
\displaystyle \int_{A_\epsilon}(\epsilon|\nabla v_\epsilon| ^2+\frac{(v_\epsilon-1)^2}{4\epsilon})dx+ \int_{B_\epsilon}(\epsilon|\nabla v_\epsilon|^2+\frac{(v_\epsilon-1)^2}{4\epsilon})dx+ \int_{\Omega \setminus(A_\epsilon\cup B_\epsilon)}(\epsilon|\nabla v_\epsilon|^2+\frac{(v_\epsilon-1)^2}{4\epsilon})dx
$$
for $v_\epsilon$ be small on $A_\epsilon$ and $1-w(\epsilon)$ outside $A_\epsilon\cup B_\epsilon$

Therefore the equation above is equivalent to
$$
\displaystyle \int_{A_\epsilon}\frac{1}{4\epsilon}dx+ \int_{B_\epsilon}\left(\epsilon|\nabla v_\epsilon|^2+\frac{(v_\epsilon-1)^2}{4\epsilon}\right)dx+ \int_{\Omega \setminus(A_\epsilon\cup B_\epsilon)}\left(\frac{w(\epsilon)^2}{4\epsilon}\right)dx
$$
For choosing $w(\epsilon)$ and the bandwidth of $A_\epsilon$ such that the first and third term will vanish.

Then the question will become $\displaystyle\lim_{\epsilon\rightarrow0}\int_{B_\epsilon}\left(\epsilon|\nabla v_\epsilon|^2+\frac{(v_\epsilon-1)^2}{4\epsilon}\right)dx$

Define that $\tau(x) = d(x,S_u)$

Define a real value function $\sigma_\epsilon$ such that $v_\epsilon(x) = \sigma_\epsilon(\tau(x))$ with $|\nabla \tau| = 1$

Rewrite the equation, we have
$$
\displaystyle\int_{B_\epsilon}\left(\epsilon\sigma_\epsilon'(\tau(x))^2+\frac{\left(\sigma_\epsilon(\tau(x))-1\right)^2}{4\epsilon}\right)dx
$$
Set $t = \tau(x)$ then the integration is thought as integrate through the set $\left\{x: \tau(x) = t\right\}$

Modify the integral we have
$$
\displaystyle\int_{\eta_\epsilon}^{\gamma_\epsilon}\left(\epsilon\sigma_\epsilon'(t)^2+\frac{\left(\sigma_\epsilon(t)-1\right)^2}{4\epsilon}\right) \mathcal{H}\left(\left\{x: \tau(x) = t\right\}\right)dt
$$
Since our problem is to minimize the functional, so that we set
$$
\left\{\begin{array}{l}\sigma_\epsilon'(t) = \displaystyle \frac{\sigma_\epsilon(t)-1}{2\epsilon}\\ \sigma_\epsilon(\eta(\epsilon)) = 0\end{array}\right.
$$


therefore, the region becomes $\displaystyle\frac{1}{2\epsilon}\int_{\eta_\epsilon}^{\gamma_\epsilon}e^{\left(\frac{\eta_\epsilon-t}{\epsilon}\right)}\mathcal{H(\left\{x:\tau(x) = t\right\})}dt$

Since $\mathcal{H}\left(\left\{x: \tau(x) = t\right\}\right)$ is a monotone decreasing function w.r.t $t$ ( 一圈一圈縮小)

by the Mean value theorem,

$\exists t_0$ such that
$$
\displaystyle\frac{1}{2\epsilon}\int_{\eta_\epsilon}^{\gamma_\epsilon}e^{\left(\frac{\eta_\epsilon-t}{\epsilon}\right)}\mathcal{H(\left\{x:\tau(x) = t\right\})}dt \\
= \frac{\mathcal{H(\left\{x:\tau(x) = t_0\right\})}}{2\epsilon}\int_{\eta_\epsilon}^{\gamma_\epsilon}e^{\left(\frac{\eta_\epsilon-t}{\epsilon}\right)}dt\\
=\frac{\mathcal{H(\left\{x:\tau(x) = t_0\right\})}}{2}\left(1-e^{\left(\frac{\eta_\epsilon-\gamma_\epsilon}{\epsilon}\right)}\right)
$$
take $\gamma_\epsilon$ such that $\displaystyle \frac{\gamma_\epsilon}{\epsilon} \rightarrow \infty$

then it converge to $\mathcal{H}^1(S_u)$

To sum up,

We show the convergence of a sequence $AT_\epsilon(u_\epsilon, v_\epsilon)$ converge to $F(u)$ as $\left(u_\epsilon, v_\epsilon\right)\rightarrow(u,1)$

## Existence of minimizer

### Coercivity

given the pertubation

### Lower semi-continuous

By the lemma above

## What's left?

solution

steepest descent method