Minimizing total variation with small norm in a compact set

Question

Take $\Omega\subseteq\mathbb R^n$ to be compact. Suppose for $t>0$ we define $f$ via the variational problem $$ \begin{array}{rl} \inf_{f\in\mathrm{BV}(\mathbb R^n)} & \mathrm{TV}[f]\\ \textrm{s.t.} & f\geq0\\ & f\equiv0\textrm{ on }\mathbb R^n\backslash\Omega\\ & \int_{\mathbb R^n} f=t. \end{array} $$ Is it possible to show that there exists a minimizer $f$ that satisfies $\sup_{x\in\mathbb R^n}f(x)\leq1$ for sufficiently small $t>0$?

My first attempt to do this in this post is obviously incorrect! This fact seems obviously true but I'm missing something.

Here, I am using the definition of total variation $$\mathrm{TV}(f):=\sup_{\|\phi\|_\infty\leq1}\int_{\mathbb R^n} f(x)\nabla\cdot\phi(x)\,dx.$$

UPDATE: It is clear that the above statement is true if and only if the $t=1$ solution admits a bounded function as its minimizer. But it seems to be nontrivial to show the latter.

Answer 1

Let $t=1$ and let $u$ be a minimizer. Let $m(r):=\mathcal{L}^n(\{u\geq r\})$, and suppose by contradiction $u$ is not bounded, that is $m(r)>0$ for every positive $r$. For positive $R$ define the truncation $u_R:=\inf\{u,R\}$ and then define the competitor $v_R=\tfrac{u_R}{\|u_R\|_1}$. By coarea $$TV(v_R)=\frac{1}{\|u_R\|_1}\int_0^R P(\{u\geq r\})dr.$$ By coarea again and by the isoperimetric inequality \begin{align} TV(u)&=\int_0^\infty P(\{u\geq r\})dr\\ &=\|u_R\|_1 TV(v_R)+\int_R^\infty P(\{u\geq r\})dr\\ &\geq \|u_R\|_1 TV(v_R)+c_n\int_{R}^\infty m(r)^{1-\frac{1}{n}}dr\\ &\geq \|u_R\|_1 TV(v_R)+\frac{c_n}{m(R)^\frac{1}{n}}\int_{R}^\infty m(r)dr\\ &=\|u_R\|_1 TV(v_R)+c_n\frac{1-\|u_R\|_1}{m(R)^\frac{1}{n}}. \end{align} Now from the minimality property of $u$ we obtain $$TV(v_R)\geq TV(u)\geq \|u_R\|_1 TV(v_R)+c_n\frac{1-\|u_R\|_1}{m(R)^\frac{1}{n}} $$ that is $$TV(v_R)\geq \frac{c_n}{m(R)^\frac{1}{n}}$$ (here we are dividing by $1-\|u_R\|_1$ and we are using the unboundedness assumption to make sure this is non-zero). But this is a contradiction because $m(R)$ goes to zero as $R\to\infty$, while $TV(v_R)$ stays bounded because $$TV(v_R)=\frac{1}{\|u_R\|_1}TV(u_R)\leq \frac{1}{\|u_R\|_1}TV(u)\leq 2 \,TV(u) $$ if we choose $R$ big enough so that $\|u_R\|_1\geq \tfrac12$. Therefore $u$ can not be unbounded.

Being more careful you can also find a quantitative bound on $\|u\|_\infty$ in terms of $t$.

Answer 2

Oops! This is an easy problem. The solution scales linearly with $t$, so obviously we can obtain $f$ as small as we want by scaling the $t=1$ case.

Sorry for posting something so straightforward!

EDIT: This is true only if a bounded solution exists.

Answer 3

The existence of a minimizer is easy, but I don't know how to prove boundness: take $B$ an open ball which contains the $\Omega$. It is easy to see that, since you are asking that $f$ vanishes outside $\Omega$, you can as well minimize the total variation over $B$. Take a minimizing sequence $f_n$. It is clear that $$ \sup TV(f_n) < \infty $$ So that, since $B$ is bounded smooth, you can extract a subseqence $f_{n_k} \to f$ in $L^1(B)$, with $f \in BV(B)$. The lower semicontinuity of the total variation gives: $$\liminf_k TV(f_{n_k}) \ge TV(f)$$ The first and the second conditons are easy to check.