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Given any interval ${[a,b]}$, define the total variation ${\|F\|_{TV([a,b])}}$ of ${F}$ on ${[a,b]}$ as $$ \displaystyle \|F\|_{TV([a,b])} := \sup_{a \leq x_0 < \ldots < x_n \leq b} \sum_{i=1}^n |F(x_i) - F(x_{i+1})|. $$ Let $F:[a,b]\to\mathbb{R}$ be a continuous function.

Can one conclude that $$\|F\|_{TV([a,b])}=\lim_{\epsilon\to 0+}\|F\|_{TV([a+\epsilon,b])}?$$

If $F$ is absolutely continuous than this can by done by noting that $$ \|F\|_{TV([c,d])}=\int_c^d|F'(x)|\ dx. $$ What can one say in the general case?

  • Have you tried exploring the canonical examples with infinite total variation? – Alex R. Jan 04 '17 at 00:10
  • @AlexR. In that case, I assume that one also has $\lim_{\epsilon\to0+}|F|_{TV([a+\epsilon,b])}=\infty$ and thus the equality is also true? –  Jan 04 '17 at 00:13

1 Answers1

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I decided to make my comment an answer since it appears to be long.

Observe that $\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon,b])}\le\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a,b])}$ since if $P$ is a partition of $[a+\epsilon,b]$

then

$$\sum_{i=1}^{n}\left\lvert F(x_{i})-F(x_{i-1})\right\rvert \le\left\lvert F(a+\epsilon)-F(a)\right\rvert+\sum_{i=1}^{n}\left\lvert F(x_{i})-F(x_{i-1})\right\rvert\le\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a,b])}$$

Taking a supremum over partitions gives the described result. Thus, taking a limit in $\epsilon$ we obtain

$$\limsup_{\epsilon\to0^{+}}\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon,b])}\le\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a,b])}$$

Observe that through a similar proof we may obtain:

$$\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon_{1},b])}\le\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon_{2},b])}$$

when $\epsilon_{1}>\epsilon_{2}$. Now take a partition $P=\left\{a=x_{0}<\ldots<x_{n}=b\right\}$. By continuity of $F$ at $x=a$ we can find $\delta>0$ so that $\left\lvert F(x)-F(a)\right\rvert<\eta$ if $\left\lvert x-a\right\rvert<\delta$. By perhaps refining our partition and increasing the variation we may assume $\left\lvert x_{1}-a\right\rvert<\delta$. So:

$$\sum_{i=1}^{n}\left\lvert F(x_{i})-F(x_{i-1})\right\rvert\le\eta+\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([x_{1},b])}\le\eta+\liminf_{\epsilon\to0^{+}}\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon,b])}$$

So $$\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a,b])}\le\eta+\liminf_{\epsilon\to0^{+}}\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon,b])}$$

So $$\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a,b])}\le\liminf_{\epsilon\to0^{+}}\left\lvert\left\lvert F\right\rvert\right\rvert_{TV([a+\epsilon,b])}$$

Note that if you remove continuity of the function at $a$ then the limit is not true. For example, consider $f:[0,1]\to\mathbb{R}$ defined by $f(x)=\chi_{\left\{0\right\}}(x)$. Then $\left\lvert\left\lvert f\right\rvert\right\rvert_{TV([\epsilon,1])}=0$ for $\epsilon>0$ but $\left\lvert\left\lvert f\right\rvert\right\vert_{TV([0,1])}=1$.

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