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The first exercise from here is the following:

Let $x \in \mathcal{C}^{1\mbox{-var}} ([0,T], \mathbb{R}^d)$. Show that $x$ is the limit in $1$-variation of piecewise linear interpolations if and only if $x \in \mathcal{C}^{0,1\mbox{-var}} ([0,T], \mathbb{R}^d)$.

First, I'll clarify what everything means.

A function $x\in\mathcal{C}^{1\mbox{-var}} ([0,T], \mathbb{R}^d)$ is one such that the supremum over all partitions $\mathcal{P}=\{0=t_0,\ldots,t_n=T\}$ of $[0,T]$ of $\sum_k|x(t_{k+1})-x(t_k)|$ is bounded.

The space $\mathcal{C}^{0,1\mbox{-var}} ([0,T], \mathbb{R}^d)$ is the space of absolutely continuous functions on $[0,T]$. That is to say, if $x\in\mathcal{C}^{0,1\mbox{-var}} ([0,T], \mathbb{R}^d)$, then there exists a $y\in L^1[0,T]$ such that $x(t)=x(0)+\int_0^ty(s)\mathrm{d}s$ for all $t\in[0,T]$.

Here's my attempt at proving that $x$ absolutely continuous on $[0,T]$ implies it's the limit in $1$-variation of piecewise linear interpolations:

Start with some partition $\Pi=\{0=t_0,\ldots,t_m=T\}$, each at a distance $\delta$ from each other, and define the piecewise linear interpolation $$\psi(t) = x(t_i)+\frac{x(t_{i+1}+\delta)-x(t_i)}{\delta}(t-t_i)\qquad t\in[t_i,t_{i}+\delta).$$ We want to prove that, for any $\epsilon>0$, there is a $\delta$ such that $\|x(t)-\psi(t)\|_{1\mbox{-var}[0,T]}<\epsilon$. From the definition of $1$-var norm, this equals \begin{align*}&\sup_{\mathcal{P}[0,T]}\sum_k\left|x(t_{k+1})-\frac{x(t_i+\delta)-x(t_i)}{\delta}(t_{k+1}-t_i)-x(t_i)+\frac{x(t_i+\delta)-x(t_i)}{\delta}(t_{k}-t_i)\right|\\ \leq& \left|x(t_{k+1})-x(t_k)\right|+\left|\frac{x(t_i+\delta)-x(t_i)}{\delta}(t_{k+1}-t_i)\right|+\left|\frac{x(t_i+\delta)-x(t_i)}{\delta}(t_{k}-t_i)\right|. \end{align*} As $x$ is absolutely continuous, it's continuous, and as the domain is a compact set, it's uniformly continuous (Heine Cantor Theorem). So we can choose a $\delta$ such that $\left|x(t)-x(s)\right|<\frac{\epsilon}{3}$, for any $|t-s|<\delta$. So the above inequality is less than $\epsilon$, which is what we wanted to show.

How can I prove the other direction?

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1 Answer 1

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For simplicity we take a function $f: [0;1]\to \mathbb R$ which is not absolutely continuous (AC), we are going to prove that it is not the BV (bounded variation) limit of piecewise linear interpolations.

First, we need an other caracterisation of the absolute continuity : A function $g:[0;1]\to \mathbb R$ is AC if and only if for every $\varepsilon>0$ there exists some $\delta>0$ such that for every finite sequence $((a_n;b_n))_{n}$ of disjoint subintervals of $[0;1]$ we have : $\sum|b_n-a_n|<\delta \Rightarrow \sum |g(b_n)-g(a_n)|<\varepsilon$. You can see that this is a stronger condition than the uniform continuity (and also stronger than BV). The equivalence between the two definitions of the absolute continuity is the object of the "fundamental theorem of Lebesgue integral calculus".

Now take some $f$ not AC, there exists some $\varepsilon>0$ such that for every $\delta>0$ there exists a finite sequence of intervals $(a_n;b_n)$ such that $\sum|a_n-b_n|<\delta $ but $\sum |f(a_n)-f(b_n)|>\varepsilon$. The idea of the proof is that a non AC function has big variations over small sets whereas a piecewise linear function is Lipschitz and therefore cannot compensate these big variations on small sets. Now we take some points $0=t_0,\ldots, t_n=1$ and $h$ the piecewise linear approximation of $g$ associated with these points. Note that $h$ is Lpischitz, let's say it is $\lambda$-Lipschitz. Take some $\delta>0$ to be specified later, and take a finite sequence of intervals $((a_k;b_k))_k$ satisfying $\sum|a_k-b_k|<\delta $ and $\sum |f(a_k)-f(b_k)|>\varepsilon$. We can assume without loss of generality that the intervals are ordered : $0\leq a_1<b_1<a_2<b_2<\ldots<b_m\leq 1$. Now we look at the variation of $f-h$ associated with the points $(0;a_1;b_1;a_2;b_2;\ldots;b_m;1)=(c_0;\ldots;c_{2m+2})$, we have

$$\sum_i|(f-h)(c_{i+1})-(f-h)(c_i)|\geq \sum_k|(f-h)(a_k)-(f-h)(b_k)|$$

Since $h$ is $\lambda$-Lipschitz we have $\sum_k|h(a_k)-h(b_k)|\leq \lambda \delta$ and on the other hand $\sum_k|f(a_k)-f(b_k)|\geq \varepsilon$, it suffices to take $\delta<\varepsilon/2\lambda$ to have $$\sum_k|(f-h)(a_k)-(f-h)(b_k)|>\varepsilon/2$$ This in turn imply $\sum_i|(f-h)(c_{i+1})-(f-h)(c_i)|\geq \varepsilon/2$ and thus that the variation of $f-h$ is greater than $\varepsilon/2$, which is independant of $h$. This conclude the proof.

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