The $BV$ minimizing problem depends on a moving domain.

Question

Let $v\in BV_{loc}(\mathbb R^2)$ be given. Define a moving domain $Q_s$ by $$ Q_s:=(s,1+s)\times(0,1). $$ where $0

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

Hence, we have $\|u_s\|_{BV((0,2)\times(0,1))}$ is uniformly bounded (just extend $u_s$ from $Q_s$ to $(0,2)\times (0,1)$ by letting $u_s=0$ outside $Q_s$). Then there exists $\bar u$ such that $u_s\to \bar u$ weakly in $BV$.

My question: do we have $\bar u = u_0$ ($u_0$ is defined by using $(1)$ with $s=0$)? If not, would it help if I assume $v\in C^\infty$?

Thank you!

Answer 1

Yes. The reason is that the map $s\mapsto u_s$ is continuous in $L^2$ norm (even Hölder continuous, with exponent $1/2$).

Let $F(s)$ be the minimal value of the functional we're minimizing, and let $v_s$ be the restriction of $v$ to the moving domain. Note that $v_s$ and $v_t$ differ by at most $C|s-t|$ in the $L^2$ norm.

Consider two values $s,t$ with minimizers $u_s, u_t$. Let $u=(u_s+u_t)/2$. By the parallelogram law, for any function $g\in L^2$ we have $$ \|u-g\|_2^2 \le \frac{1}{2}(\|u_s-g\|_2^2 + \|u_t-g\|_2^2) - \frac{1}{4}\|u_s-u_t\|_2^2 \tag{1}$$ (The parallelogram has vertices $g$, $u_s$, $u_t$, and center point $u$.) Additionally, $$|u|_{TV}\le \frac12(|u_t|_{TV} + |u_s|_{TV})\tag2$$ by the triangle inequality.

The inequalities (1) and (2) imply that if $u_t-u_s$ has large $L^2$ norm, the average $u$ becomes preferable to either of them for the purpose of minimization, contradicting the minimizing properties of $u_t, u_s$.

To verify this, use (1) with $g=v_s$, then with $g=v_t$, and add the results:
$$ (\|u-v_s\|_2^2 + |u|_{TV}) + (\|u-v_t\|_2^2 + |u|_{TV}) \le \frac12(\|u_t-v_s\|_2^2 + \|u_t-v_t\|_2^2) + |u_t|_{TV} + \frac12(\|u_s-v_s\|_2^2 + \|u_s-v_t\|_2^2 ) + |u_s|_{TV} - \frac{1}{4}\|u_s-u_t\|^2 $$ Since $v_t$ and $v_s$ are close, the right hand side is at most $$F(t)+F(s)+ C\|v_t-v_s\|_2 - \frac{1}{4}\|u_s-u_t\|_2^2 $$ But the left hand is at least $F(t)+F(s)$. Hence, $\|u_s-u_t\|_2^2 \le C\|v_t-v_s\|_2$ as claimed.