I read on the Wikipedia page that bounded variation (BV) functions have only jump-type discontinuities. Why is that? Suppose at some $a\in\mathbb{R}$, the limit $\lim_{x\rightarrow a^+}f(x)$ doesn't exist (or is infinite). Why would such a function $f$ not be BV?

  • A BV function can be expressed as sum of two monotone functions (each with different direction of growth). So the problem reduces to showing that a monotone function can only have jump kind of discontinuity. See the answer math.stackexchange.com/a/581672 for more details. – Paramanand Singh Nov 28 '13 at 17:07
up vote 4 down vote accepted

If $f$ is unbounded near $a$, it can clearly not have bounded variation, since

$$V(f,[a,b]) \geqslant \lvert f(x)-f(a)\rvert$$

for all $x \in (a,b]$.

If $-\infty < m = \liminf\limits_{x \searrow a} f(x) < \limsup\limits_{x\searrow a} f(x) = M < +\infty$, let $\varepsilon = (M-m)/3$. Then we can find sequences $x_n \searrow a$ and $y_n \searrow a$ with $x_n > y_n > x_{n+1}$ and $f(y_n) < m+\varepsilon$, $f(x_n) > M-\varepsilon$ for all $n$, so

$$V(f,[a,b]) \geqslant \sum_{n=1}^N \lvert f(x_n) - f(y_n)\rvert \geqslant N\cdot \varepsilon.$$

Letting $N\to \infty$, we see that then $V(f,[a,b]) = +\infty$.

  • This is cool as you have directly shown how non-existence of left and right limits interferes with bounded variation. As mentioned in my comment to the question I normally reduce any BV function to monotone function. So essentially this proof of yours is very simple and smart.(+1) – Paramanand Singh Nov 28 '13 at 17:10

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