The upper bound of $L^2$ norm of the minimizer in an minimizing problem.

Question

I am considering the following minimizing problem: $$ u_m:= \operatorname{argmin}_{u\in BV(\Omega)}\{ \frac{1}{2} \|u-u_0\|_{L^2}^2 + t |u|_{TV}\} $$ where $u_0\in BV(\Omega)\cap L^\infty(\Omega)$ and $t>0$ is a constant. $|\cdot|_{TV}$ denotes the total variation of $u$. ($\Omega\subset \mathbb R^2$ is open bounded smooth boundary)

I learned from the numerical simulation that $$ \|u_m\|_{L^2}\leq \|u_0\|_{L^2} $$ but I can not prove it analytically. Please advise!

Answer 1

I found three different ways to prove your inequality, but they are essentially the same. For convenience, let $t = 1$.

1. For $s \ge 0$, define $$f(s) = \frac12 \, \| s \, u -u_0\|_{L^2}^2 + |s \, u|_{TV}.$$ Then, $$0 = f'(1) = (u - u_0, u) + |u|_{TV}.$$

2. The optimality conditions of your problem yield $$0 \in u - u_0 + \partial |\cdot|_{TV}(u).$$ Here, $\partial |\cdot|_{TV}$ is the subdifferential of the total variation. Hence, $$|0|_{TV} \ge |u|_{TV} + (u_0 - u, 0-u).$$

3. Consider $u_0$ in your problem as a parameter. Then, the mapping $u_0 \mapsto u$ is the proximal point mapping of the TV-norm (w.r.t. the $L^2$-norm). Hence, it is Lipschitz (in $L^2$) with constant $1$. Since the solution with $u_0 = 0$ is $u = 0$, we have $\|u - 0\|_{L^2} \le \|u_0 - 0\|_{L^2}$.

Note that the strength of the results is decreasing, whereas the generality is increasing.