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This is a follow-up to a previous question.

Consider a function $[a,b]\to\mathbb{R}$. For $f$ being of bounded variation (BV), it is necessarily true that one must have

(1) $f$ is differentiable almost everywhere in $[a,b]$;

(2) $\int_a^b|f'(x)|\ dx<\infty$.

However, (1) and (2) are not sufficient to imply that $f$ is BV due to existence of counterexamples. One can see that it is because there might be too many "bad points" in (1): if $f$ is differentiable everywhere on $[a,b]$ and (2) is true, then $f$ is BV.

I would like to know how bad the exceptional set where $f$ is not differentiable in (1) can be. Consider

(1a) $f$ is differentiable everywhere in $[a,b]$ except at one point;
(1b) $f$ is differentiable everywhere in $[a,b]$ except at finitely many points;
(1c) is differentiable everywhere in $[a,b]$ except at countably many points;

Which one of them together with (2) can imply that $f$ is BV?

  • 0
    If you don't require $f$ to be continuous, [Daniel Fischer's comment](https://math.stackexchange.com/questions/2080737/if-f-is-differentiable-almost-everywhere-and-f-is-abolutely-integrable-the#comment4275733_2080737) gives a counterexample to 1c). – 2017-01-02
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    For (1a), under the assumption of continuous $f$, I think you can argue as follows, provided I'm not mistaken: Let $P=\left\{a=x_{0}<\ldots$c$. Up to considering a refinement of $P$ we may assume $c=x_{j}$ for some $j$. Then $$\sum_{i=1}^{n}\left\lvert f(x_{i})-f(x_{i-1})\right\rvert=\sum_{i=1,i\neq j,j+1}^{n}\left\lvert f(x_{i})-f(x_{i-1})\right\rvert+\left\lvert f(x_{j})-f(x_{j-1})\right\rvert+\left\lvert f(x_{j+1})-f(x_{j})\right\rvert$$ – 2017-01-03
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    Observe that $f$ is differentiable everywhere on $[a,x_{j-1}]$ and $[x_{j+1},b]$ and $f'$ is integrable so the sum is bounded by $\int_{a}^{b}\left\lvert f'(x)\right\rvert dx$ whereas the other terms are bounded by $4\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{\infty}([a,b])}$. The partition was arbitrary so the total variation is finite. A similar argument should hold for (1b). – 2017-01-03

1 Answers 1

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If I'm not mistaken it appears as though all of the statements you suggest imply bounded variation (provided we assume $f$ is continuous). For (1a) combined with (2) we can argue as follows:

Let $P=\left\{a=x_{0}

\begin{align*}\sum_{i=1}^{n}\left\lvert f(x_{i})-f(x_{i-1})\right\rvert&=\sum_{i=1,\,i\neq j,j+1}^{n}\left\lvert f(x_{i})-f(x_{i-1})\right\rvert+\left\lvert f(x_{j})-f(x_{j-1})\right\rvert+\left\lvert f(x_{j+1})-f(x_{j})\right\rvert\\ &=\sum_{i=1,\,i\neq j,j+1}^{n}\left\lvert \int_{[x_{i-1},x_{j}]}f'(t)dt\right\rvert+\left\lvert f(x_{j})-f(x_{j-1})\right\rvert+\left\lvert f(x_{j+1})-f(x_{j})\right\rvert\\ &\le\sum_{i=1,\,i\neq j,j+1}^{n}\int_{[x_{i-1},x_{i}]}\left\lvert f'(t)\right\rvert dt+4\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{\infty}\left([a,b]\right)}\\ &\le\int_{a}^{b}\left\lvert f'(t)\right\rvert dt+4\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{\infty}\left([a,b]\right)} \end{align*}

where I have used that $f$ is differentiable everywhere on $[a,x_{j-1}]$ and $[x_{j+1},b]$ and $f'$ is integrable on these sets. Since $P$ was arbitrary we see that $f$ is of bounded variation. A similar proof should hold for (1b) combined with (2). For (1c) combined with (2) we have to appeal to the Henstock-Kurzweil integral. We require the following Theorem:

Let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $[a,b]\setminus E$ where $E$ is a countable set. Then $f'$ is Henstock-Kurzweil integrable on $[a,b]$ and $$\int_{a}^{b}f'(x)\,dx=f(b)-f(a)$$

With these Theorem we have for a partition $P=\left\{a=x_{0}<\ldots

\begin{align*} \sum_{i=1}^{n}\left\lvert f(x_{i})-f(x_{i-1})\right\rvert&=\sum_{i=1}^{n}\left\lvert\int_{[x_{i-1},x_{i}]}^{HK}f'(t)\,dt\right\rvert=\sum_{i=1}^{n}\left\lvert\int_{[x_{i-1},x_{i}]}^{L}f'(t)\,dt\right\rvert\\ &\le\sum_{i=1}^{n}\int_{[x_{i-1},x_{i}]}^{L}\left\lvert f'(t)\right\rvert dt=\int_{[a,b]}^{L}\left\lvert f'(t)\right\rvert dt\\ \end{align*}

where I have used the superscripts $HK$ and $L$ on the integrals to denote Henstock-Kurzweil and Lebesgue respectively. Since the partition was arbitrary we see that $f$ has bounded variation. Note by a more careful estimate for (1a) and (1b) with (2) (or perhaps appealing again to the Henstock-Kurzweil integral) we can obtain the better upper bound $\int_{a}^{b}\left\lvert f'(t)\right\rvert dt$ by using the continuity of the function.