Let $\Omega\subset \mathbb R^2$ be a open bounded smooth boundary domain. Let $u_0\in L^\infty(\Omega)$ be given. We define, for $\alpha>0$, that $$ u_\alpha:=\operatorname{argmin}_{u\in BV(\Omega)}\int_\Omega|u-u_0|^2dx+\alpha TV(u) $$ Then since the total variation seminorm is strictly convex, we have $u_\alpha\in BV\cap L^\infty$ is uniquely defined.
My question: I somehow remembered that the following convergence hold: $$ u_\alpha\to \frac1{|\Omega|}\int_\Omega u_0\,dx, \text{ as }\alpha\to\infty $$ i.e., $u_\alpha$ goes to the average of $u_0$ as $\alpha\to \infty$. I remembered that was just a one line proof but now I can not find any reference which provide such result. tried to proof it but so far I have no luck. Any helps? Thank you!
Let $\bar u_0$ denote the average of $u_0$. Using the constant function $\bar u_0$, the fact that $u_\alpha$ is the unique minimizer implies
$$ \tag{*} \int |u_\alpha - u_0|^2 dx + \alpha TV(u_\alpha) < \int |u_0 - \bar u_0|^2 dx$$ for all $\alpha$. Since the first term is non-negative and the RHS is a constant independent of $\alpha$, in particular we have $0\le TV(u_\alpha) < C / \alpha$ and thus $TV(u_\alpha)\to 0$. By the compactness theorem for functions of bounded variation this implies there is a subsequence converging in $L^1$ to a limit $u_\infty$ with zero variation; i.e. a constant.
In fact we can improve this convergence to $L^2$, since (as you can hopefully prove easily) $\| u_\alpha \|_\infty \le \| u_0 \|_\infty$ and thus $u_\alpha - u_0$ is uniformly bounded.
From (*) we also get
$$ \int |u_\alpha - u_0|^2 dx < \int |u_0 - \bar u_0|^2 dx$$
which along with the $L^2$ convergence $u_\alpha \to u_\infty$ implies
$$ \int |u_\infty - u_0|^2 dx \le \int|u_0 - \bar u_0|^2 dx.$$
Since $\int |u_0 - c|^2 dx$ is minimized (for constant $c$) at $c = \bar u_0$ this implies $u_\infty = \bar u_0$, so we know $u_\alpha \to \bar u_0$ on a subsequence. Since the limit is independent of the subsequence, we in fact get convergence on the whole sequence as desired.