Suppose I have a compact measurable set $\Omega\subseteq\mathbb R^n$ and a sequence of functions $\{f_k\}_{k=1}^\infty\subset C^\infty(\mathbb R^n)$ with the following set of properties:
- $f_k\geq0$
- $\int_{\mathbb R^n} f_k\rightarrow0$ as $k\rightarrow\infty$
- $f_k|_{\mathbb R^n\backslash\Omega}\equiv0$
- $\mathrm{TV}(f_k)\rightarrow0$ as $k\rightarrow\infty$
- $f_k(x_k)>1$ for some $x_k\in\Omega$
Is it possible to prove that there is a contradiction?
In particular, the vanishing total variation (property 4) "wants" $f_k$ to decay slowly away from $x_k$ , but properties 1, 2, and 3 together imply that $f_k$ should be close to zero; somehow these seem to contradict.
Here, I am using the definition of total variation $$\mathrm{TV}(f):=\sup_{\|\phi\|_\infty\leq1}\int_{\mathbb R^n} f(x)\nabla\cdot\phi(x)\,dx.$$
Best Answer
Unless $n=1$, the total variation does not constrain pointwise values. There is no contradiction.
A sequence with all these properties is $f_k(x)=f(kx)$ where $f$ is nonnegative, smooth and compactly supported with $f(0)>1$. Indeed, $TV(f_k)=k^{1-n}TV(f)\to 0$ via a change of variable.