The BV minimizing problem depends on a moving weighted parameter

Question

Let $v\in L^2(Q)$ be given, where $Q:=(0,1)\times(0,1)$ is an unit square. Define a sequence of parameter function $\alpha_s$ by $$ \alpha_s(x):= \begin{cases} 1&\text{ if }x\in(1/2+s,1)\times(0,1)\\ 2&\text{ if }x\in(0,1/2+s)\times(0,1) \end{cases} $$ where $0

Define $$ u_s:=\operatorname{argmin}\{\|u-v\|_{L^2(Q)}^2+|\alpha_su|_{TV(Q)}:\,\,u\in BV(Q)\}\tag 1 $$ where $BV$ denotes the bounded variation space and $TV$ denotes the total variation seminorm.

My question: do we have $u_s\to u_0$ in $L^1$ as $s\to 0$? ($u_0$ is defined by letting $s=0$ in $(1)$)

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I am also wondering what if I change $(1)$ by replacing $BV$ with the Ambrosio-Tortorelli functional, i.e.,

\begin{multline} (u_s,z_s):=\operatorname{argmin}\{\|u-v\|_{L^2(Q)}^2+\int_Q |\nabla u|^2(z^2+1)\alpha_sdx+\\ \int_Q[|\nabla z|^2+(1-z)^2]\alpha_s dx:\,\,u,z\in W^{1,2}(Q)\}\tag 2 \end{multline} Then, do we have $(u_s,z_s)\to (u_0,z_0)$ in $L^1$? or even weakly in $W^{1,2}$?

Thank you!

Answer 1

For every $u\in BV(Q)$ define $w:=\alpha_{s}u$. Then $w\in BV(Q)$ (this can be seen by looking at the weak derivatives of $w$) and so we can write $$ \Vert u-v\Vert_{L^{2}}+|\alpha_{s}u|_{TV}=\Vert\alpha_{s}^{-1}w-v\Vert_{L^{2}% }+|w|_{TV}. $$ Conversely, given $w\in BV(Q)$, if we define $u:=\alpha_{s}^{-1}w$ we have that $u\in BV(Q)$ and $$ \Vert\alpha_{s}^{-1}w-v\Vert_{L^{2}}+|w|_{TV}=\Vert u-v\Vert_{L^{2}} +|\alpha_{s}u|_{TV}. $$ Thus, $$ \inf\{\Vert u-v\Vert_{L^{2}}+|\alpha_{s}u|_{TV}:\,u\in BV(Q)\}=\inf \{\Vert\alpha_{s}^{-1}w-v\Vert_{L^{2}}+|w|_{TV}:\,w\in BV(Q)\}. $$ The latter problem is much simpler to deal with. Let now $w_{s}$ be a minimizer for the latter problem (do you need a proof of this?). By taking $w=0$ you have that $$ \Vert\alpha_{s}^{-1}w_{s}-v\Vert_{L^{2}}+|w_{s}|_{TV}\leq\Vert v\Vert_{L^{2} }=M $$ for every $s$. Since $\Vert w_{s}\Vert_{L^{2}}^{2}\leq2\Vert\alpha_{s} ^{-1}w_{s}\Vert_{L^{2}}\leq2\Vert\alpha_{s}^{-1}w_{s}-v\Vert_{L^{2}}+2\Vert v\Vert_{L^{2}}\leq4M$, it follows that the sequence $\{w_{s}\}_{s}$ is bounded in $BV(Q)$ and so by the Rellich-Kondrachov theorem it admits a subsequence $\{w_{s_{n}}\}$ which converges in $L^{q}(Q)$ for every $1\leq q<2$ to a function $z\in BV(Q)$ with $z\in L^{2}(Q)$. By selecting a further subsequence you can assume that $w_{s_{n}}$ converges pointwise $\mathcal{L}^{2}$ a.e. in $Q$. By the lower semicontinuity of the total variation seminorm and Fatou's lemma you have that $$ m_{0}:=\Vert\alpha_{0}^{-1}z-v\Vert_{L^{2}}+|z|_{TV}\leq\liminf_{n\rightarrow \infty}(\Vert\alpha_{s_{n}}^{-1}w_{s_{n}}-v\Vert_{L^{2}}+|w_{s_{n}}|_{TV}). $$ On the other hand, if $w\in BV(Q)$, then since $w_{s_{n}}$ is a minimizer $$ \Vert\alpha_{s_{n}}^{-1}w_{s_{n}}-v\Vert_{L^{2}}+|w_{s_{n}}|_{TV}\leq \Vert\alpha_{s_{n}}^{-1}w-v\Vert_{L^{2}}+|w|_{TV}% $$ and so letting $n\rightarrow\infty$, you get \begin{align*} \Vert\alpha_{0}^{-1}z-v\Vert_{L^{2}}+|z|_{TV} & \leq\liminf_{n\rightarrow \infty}(\Vert\alpha_{s_{n}}^{-1}w_{s_{n}}-v\Vert_{L^{2}}+|w_{s_{n}}|_{TV})\\ & \leq\liminf_{n\rightarrow\infty}\Vert\alpha_{s_{n}}^{-1}w-v\Vert_{L^{2}% }+|w|_{TV}=\Vert\alpha_{0}^{-1}w-v\Vert_{L^{2}}+|w|_{TV}, \end{align*} where the last equality follows by the Lebesgue dominated convergence theorem. This shows that $z$ is a minimizer of the problem for $s=0$. Now if $w_{0}$ is another minimizer of the problem with $s=0$, then taking $\frac{1}{2}% z+\frac{1}{2}w_{0}$, by the convexity of the norms \begin{align*} & \Vert\alpha_{0}^{-1}(\tfrac{1}{2}z+\tfrac{1}{2}w_{0})-v\Vert_{L^{2}} +|\tfrac{1}{2}z+\tfrac{1}{2}w_{0}|_{TV}\\ & \leq\tfrac{1}{2}\Vert\alpha_{0}^{-1}z-v\Vert_{L^{2}}+\tfrac{1}{2} |z|_{TV}+\tfrac{1}{2}\Vert\alpha_{0}^{-1}w_{0}-v\Vert_{L^{2}}+\tfrac{1} {2}|w_{0}|_{TV}=m_{0}. \end{align*} Since the $L^{2}$ norm is strictly convex, we have that $z=w_{0}$. Hence, the full sequence must converge to the same function $z$, which is a minimizer of the problem with $s=0$.