Let $I:=(0,1)$ be given. Let $u_0\in BV(I)$ and assume that $u_0$ is piece-wise constant. That is, $u_0'=0$ where $u_0'$ is the absolutely continuous part of weak derivative $Du_0$, where $|Du_0|$ denotes the total variation.
My question: let $u_1$ be the defined by $$ u_1:=\operatorname{argmin}_{u\in BV(I)}\int_I|u-u_0|^2dx+|Du|(I) $$ Then, do we have $u_1$ is piece-wise constant as well?
Update: the above statement is indeed hold. Now I am wondering the 2-dimensional case:
Let's assume that $\Omega:=(0,1)\times (0,1)$ and we define $\Omega_1,\ldots,\Omega_4$ be four sub-cubes with side length equal to $1/2$, i.e., $\Omega_1:=(0,1/2)\times(0,1/2)$ and so on. And $u_0$ is a piece-wise constant function over $\Omega$, i.e, it is constant in each $\Omega_i$ for $I=1,\ldots,4$. Then, would $u_1$, as defined above with $I=\Omega$, be a piece-wise constant too?