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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 that $f_n$ is bounded in $BV_{loc}(\mathbb{R})$, so I have to show that $f_n$ is locally bounded in $L^1(\mathbb{R})$ and for any compact subset $K\subset \mathbb{R}$, it holds \begin{align*} \sup\limits_{\phi\in C_c^1(K),\\ \|\phi\|_{L^{\infty}(K)}\leq 1}\int\limits_{K} f_n\phi'\leq M<\infty, \end{align*} for all $n\in \mathbb{N}$. Since $|f_n|\leq 1$ it follows $\|f_n\|_{L^1(K)}\leq |K|$ for all $n\in \mathbb{N}$ and for all $K\subset \mathbb{R}$ compact. Let $\phi\in C_c^1(K)$, $\|\phi\|_{L^{\infty}(K)}\leq 1$. We have \begin{align*} \int\limits_{K} f_n(x) \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. Moreover I have $f_n$ is non-negative so that \begin{align}\label{1} \int\limits_{K} |f_n'(x)| dx =\int\limits_K f_n'(x) dx. \end{align} Since $|f_n|\leq 1$ and $f_n$ is monotonically increasing, I think that the derivative can't grow too fast. Is it possible to bound $\|f_n'\|_{L^1(K)}$ somehow ?

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    $\begingroup$ This is more suited for math.SE, please consider cross-posting there. As far as your question is concerned: your first integration by parts is not legitimate to start with, since the distributional derivative $f_n'$ can contain singular parts. But because $f_n'\geq 0$ in the sense of distributions you know it must be a measure, with moreover $\|f_n'\|_{TV}=1-0$ due to your "boundary conditions" at infinity. In particular your first integration by parts should read accordingly $\int \phi(x) df_n'(x)$, instead of using the absolutely continuous measure $f_n'(x) dx$ as you did $\endgroup$
    – leo monsaingeon
    Commented Apr 16, 2021 at 10:19
  • $\begingroup$ Thanks for your answer. I did ask it on math SE. Having $\|f_n'\|=1$ for all $n\in \mathbb{N}$ can I conclude that $f_n$ is bounded in $BV$ ? Maybe like this: Since $f_n'$ defines a measure, I can decompose it by Lebesgue's decomposition into two measures, the absolutely continuous part and the singular part. What I tried to compute is then just the absolutly continuous part ? $\endgroup$
    – user99432
    Commented Apr 16, 2021 at 10:35
  • $\begingroup$ yes, exactly, you tried to compute the ac part. But proving that the a.e.-well-defined pointwise derivative coincides with the absolutely continuous part of the distributional derivative (the measure) in the Lebesgue decomposition is not immediate. and indeed $\|f_n'\|_{TV}\leq 1$ gives that $f_n$ is BV, since by definition $\|f_n\|_{BV(K)}=\|f_n\|_{L^1(K)}+\|f_n'\|_{TV(K)}$ on any compact set $K$ and my "argument" proves that this quantity is finite $\endgroup$
    – leo monsaingeon
    Commented Apr 16, 2021 at 10:40
  • $\begingroup$ In the future if you crosspost to math.MO a question initially posted on math.SE that never received any answer then DO mention it in the MO post, and include the link. $\endgroup$
    – leo monsaingeon
    Commented Apr 16, 2021 at 10:42
  • $\begingroup$ Thank you very much! $\endgroup$
    – user99432
    Commented Apr 16, 2021 at 10:49

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