Suppose the following:
- $f$ and $f_m$ ($m\in\mathbb N$) are real-valued functions on $[0,1]$;
- $f(0)=f_m(0)=0=f(1)=f_m(1)$ for each $m\in\mathbb N$;
- $f$ and $f_m$ ($m\in\mathbb N$) all share the same Lipschitz constant $\kappa>0$, meaning that $|f(y)-f(x)|\leq\kappa|y-x|$ and $|f_m(y)-f_m(x)|\leq\kappa|y-x|$ for each $x,y\in[0,1]$ and $m\in\mathbb N$; and
- $\lim_{m\to\infty}\sup_{x\in[0,1]}|f_m(x)-f(x)|=0$ (that is, we have uniform convergence).
I am trying to either prove or disprove the following (the derivatives exist almost everywhere by Lipschitz and hence absolute continuity): $$\int_0^1|f'(t)|\,\mathrm dt\leq\liminf_{m\to\infty}\int_0^1|f'_m(t)|\,\mathrm dt.$$
It’s quite easy to come up with examples showing that the inequality can actually be strict, but I am wondering whether it always holds in the first place, or whether there is a neat counterexample.
Any feedback is greatly appreciated.
Here is a solution. Take $g\in\{f,f_1,f_2,\ldots\}$ to be a generic function satisfying the premises. Based on Lipschitz, and hence absolute, continuity, one can derive (with a little work) that $$\|g\|=\int_0^1|g'(t)|\,\mathrm dt,$$ where $$\|g\|\equiv\sup\left\{\sum_{i=1}^k|g(a_i)-g(a_{i-1})|\right\},$$ and the supremum is taken over finite partitions $0\equiv a_0<a_1<\cdots<a_{k-1}<a_k\equiv 1$, $k\in\mathbb N$. Hence, the problem has been transformed to comparing total variations.
Now, consider any such partition. One then has, by uniform and hence pointwise convergence, that \begin{align*} \sum_{i=1}^k|f(a_i)-f(a_{i-1})|=&\;\lim_{m\to\infty}\left\{\sum_{i=1}^k|f_m(a_i)-f_m(a_{i-1})|\right\}\\=&\;\liminf_{m\to\infty}\left\{\sum_{i=1}^k|f_m(a_i)-f_m(a_{i-1})|\right\}\\ \leq&\;\liminf_{m\to\infty}\|f_m\|. \end{align*} Since this holds for any arbitrary partition, one has $$\|f\|\leq\liminf_{m\to\infty}\|f_m\|,$$ as desired.