Show boundedness of a sequence in $BV$

bounded-variationderivativesmonotone-functionsreal-analysistotal-variation

For any $n\in \mathbb{N}$ let $f_n:\mathbb{R}\to [0,1]$ be monotonically increasing and $\lim_{x\to -\infty} f_n(x)=0$ and $\lim_{x\to \infty} f_n(x)=1$. It follows $f_n$ is differentiable a.e..

I'm trying to show $f_n$ is bounded in $BV_{loc}(\mathbb{R})$, so I have to show that $f_n$ is locally bounded in $L^1$ and for any $K\subset \mathbb{R}$, it holds
\begin{align*}
\sup\limits_{\phi\in C_c^1(K),\\ \|\phi\|_{L^{\infty}(K)}\leq 1}\int\limits_{K} f_n \mathrm{div}\phi<\infty.
\end{align*}
Local boundedness in $L^1$ is already given since $|f_n(x)|\leq 1$ for all $x\in \mathbb{R}$ by using Hölder. Let $\phi\in C_c^1(K)$, $\|\phi\|\leq 1$. We have
\begin{align*}
\int\limits_{K} f_n(x) \mathrm{div}\phi(x)=-\int\limits_{K} f_n'(x)\phi(x) dx,
\end{align*}
since $f_n$ is differentiable a.e.. Then
\begin{align*}
\int\limits_{K} |f_n'(x)||\phi(x)|dx \leq \| f_n'\|_{L^1(K)} \| \phi\|_{L^{\infty}(K)} \leq \| f_n'\|_{L^1(K)}
\end{align*}
Now I have to bound the right-hand side. Unfortunately $f_n$ is not differentiable in the classical sense, so I can't just perform integration and use the fundamental theorem of integration to get this bounded. Since $f_n$ is non-negative
\begin{align*}
\int\limits_{K} f_n'(x) dx \leq f_n(\infty)-f_n(-\infty)=1.
\end{align*}
Can I use the last inequality ? Is there maybe an easier way to show that $f_n$ is bounded in $BV_{loc}(\mathbb{R})$?

Best Answer

I'm trying it this way: For any compact subset $K\subset \mathbb{R}$ we can find $a,b\in \mathbb{R}$: $K\subset [a,b]$. Let $P=(x_0,...,x_N)$ be a partition of $[a,b]$. By using that $f_n$ is monotonically increasing and $0\leq f_n\leq 1$ we have \begin{align*} \sum\limits_{k=0}^{N}|f_n(x_{k+1})-f_n(x_k)|=\sum\limits_{k=0}^N f_n(x_{k+1})-f_n(x_k)=f_n(b)-f_n(a)\leq 1 \end{align*} for every partition $P$ and any $n\in \mathbb{N}$. Hence $V_a^b(f_n)\leq 1$, so that $f_n\in BV([a,b])$ and thus $f_n\in BV(K)$. Since this is true for any compact $K$ we have now $f_n$ is bounded in $BV_{loc}(\mathbb{R})$. Can I do it like this ? If the answer is positive, does the above ansatz with the derivative also work ?

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