Do BV functions vanish at infinity?

Question

We were wondering if $BV$ functions in one dimension vanish at infinity. To detail the question, define for $u\in L^{1}(\mathbb{R})$ the $TV$ semi-norm by $$ |u|_{TV(\mathbb{R})}:=\sup_{\substack{\phi\in C^{1}_{\text{c}}(\mathbb{R})\\ \|\phi\|_{L^{\infty}(\mathbb{R})}\leq 1}}\int_{\mathbb{R}}\phi'(x)u(x)\,\mathrm{d} x. $$ With the norm $\|u\|_{BV(\mathbb{R})}:=\|u\|_{L^{1}(\mathbb{R})}+|u|_{TV(\mathbb{R})}$ we define $BV$ as $$ BV(\mathbb{R}):=\left\{u\in L^{1}(\mathbb{R}):\ |u|_{TV(\mathbb{R})}<\infty\right\}.$$ Then, $BV$ with the given norm is a Banach space. Do we have for $u\in BV(\mathbb{R})$ $$ \lim_{k\rightarrow\infty} \|u\|_{L^{\infty}((k,\infty))}=0? $$ We tried to use the typical approximation arguments, but could not get the result. Thus, is this result true? Does anyone have a hint on how to show or to disprove it?

In any case, thank you very much!

Alex

Answer 1

Let $u\in BV(\mathbf{R})$. Then $u$ admits a right continuous representative $\bar{u}$ such that the pointwise variation $$\mathrm{Var}(\bar{u}) := \sup\left\{\sum_{i=1}^{n}|\bar{u}(x_i)-\bar{u}(x_{i-1})|\right\},$$ where the supremum is taken over all finite partitions of $\mathbf{R}$, satisfies $\mathrm{Var}(\bar{u})=|u|_{TV(\mathbf{R})}$. In particular, $\bar{u}$ can be written as the difference of two bounded increasing functions. As a result, the limit of $\bar{u}(x)$ as $x\to\infty$ exists in $\mathbf{R}$. Since $u\in L^1(\mathbf{R})$ one can show (by a contradiction argument) that $\liminf_{x\to\infty}\bar{u}(x)=0$, and so we deduce $\lim_{x\to\infty}\bar{u}(x)=0$.

A good reference regarding BV functions (and where you can find proofs of all of the statements above) is Leoni's A First Course in Sobolev Spaces.