The total variation of a continuous $f$ satisfies: $\lim_{\epsilon \to \infty} T_\epsilon ^1(f) = T_0^1(f)$.

Question

This is essentially a supremum and limits question.

I've encountered the following claim:

If $f$ is continuous on $[0,1]$, then $f$ satisfies: $\lim_{\epsilon \to 0^+} T_\epsilon ^1(f) = T_0^1(f)$.

where:

$T_a^b(f) := \sup\big\{\sum_{i=1}^n|f(x_i)-f(y_i)| \mid \{(x_i,y_i)\}_{i=1}^k \text{are a partition of [a,b]}\big\}$.

I realize that I have to use the formal definition of the limit to get around the fact that $\epsilon$ is a part of the partition's definition, so I tried using supremum definitions to get to something like the following:

Denoting the internal sum (i.e. the per-partition variation) by $V(f,P)$, we want to show that for every $\epsilon'>0$ there exists an $N>0$ such that for all $n>N$:

$$| \sup\big\{V(f,P) \mid \ P \text{ is a partition of } [0, 1]\big\} - \sup\big\{V(f,P) \mid \ P \text{ is a partition of } [\frac{1}{n}, 1]\big\} | <\epsilon'$$

but I'm not sure what to do with the supremums (my inuition is that somehow I can bound this by $|f(\frac{1}{k})-f(0)|$ where $k$ will be chosen such that $|\frac{1}{k}-0|\leq\delta$ and $\delta$ matches $\epsilon'$ from $f$'s continuity around $0$, so that we obtain $|f(\frac{1}{k})-f(0)| < \epsilon'$), but as I'm a bit rusty on handling supremums in such situations, I'm not sure how to move from the supremums inequality to this.

Edit:

I think the following comes close to solving it:

Let $P$ by a partition of $[\frac{1}{n},1]$.

Denote by $P^+$ the partition of $[0,1]$ such that its first two points are $\{0,\frac{1}{n}\}$ and the remaining are the points in $P$ after $x=\frac{1}{n}$.

Then we have:

$$\sup_{P}V(f,P^+)-\sup_PV(f,P)\leq \sup_P\big(V(f,P^+)-V(f,P)\big) \leq \sup_P\big(|f\big(\frac{1}{n}\big)-f(0)|\big)\leq\epsilon'$$

for $n$ sufficiently large, by continuity of $f$ around 0.

The first inequality follows from:

$$\sup(A+B)=\sup A+\sup B$$

where, in our case:

$$A:= V(f,P^+)-V(f,P)$$

$$B:=V(f,P)$$

It doesn't solve it though because the partitions on $[0,1]$ are limited to those of the type $P^+$.

Any advice?

Thanks!

Answer 1

Suppose that $f$ is continuous on $[0,1]$ and of bounded variation. (Note that there are continuous functions of unbounded variation such as $x \mapsto x \sin \frac{1}{x}$).

Since $f$ is continuous, given any $\eta > 0$ there is a $\delta > 0$ such that $0 < x < \delta $ implies $|f(x) - f(0)| < \eta/2$.

Since the total variation is a supremum of sums of absolute differences over all partitions, there exists a partition $0= x_0 < x_1 < \ldots < x_n = 1$ such that

$$\tag{1}T_0^1(f) \leqslant \sum_{k=1}^n |f(x_k) - f(x_{k-1})| + \eta/2$$

Suppose $0 < \epsilon <\hat{\delta} = \min(\delta,x_1)$. Since $\epsilon < x_1$ it follows from (1) that

$$\tag{2}T_0^1(f) \leqslant |f(x_1) - f(0)| + \sum_{k=2}^n |f(x_k) - f(x_{k-1})| + \eta/2 \\ \leqslant \underbrace{|f(\epsilon) - f(0)|}_{< \eta/2 \text{ since } \epsilon < \hat{\delta} \leqslant \delta} + \underbrace{|f(x_1) - f(\epsilon)| +\sum_{k=2}^n |f(x_k) - f(x_{k-1})|}_{\leqslant T_\epsilon^1(f) \text{ by definition of total variation on } [\epsilon,1]} + \eta/2 $$

Thus, for all $\epsilon$ such that $0 < \epsilon < \hat{\delta}$ we have $T_0^1(f) < \eta/2 +T_\epsilon^1(f) + \eta/2$, and, hence, $|T_0^1(f) - T_\epsilon^1(f)| < \eta$. This proves that

$$\lim_{\epsilon \to 0+}T_\epsilon^1(f) = T_0^1(f)$$