# Examples about bounded variation function

Problem. Give an example of a function with bounded variation $$f:[0,1] →\mathbb R$$ with $$f'$$ integrable in $$[0,1]$$, and such that the function $$g:[0,1]→\mathbb R$$ defined by $$g(x)=f(x)-f(0)-\int_0^xf'(s)\,\mathrm ds$$ vanishes at the points $$\displaystyle x_n = \sum_{j=1}^n\frac{1}{2^j}$$, $$g$$ is positive on $$(x_n,x_{n+1})$$ if $$n$$ is odd and is negative if $$n$$ is even.

This example does not seem intuitive for me. How can I try to build this function? Is there any strategy for this?

In addition to the assumptions, the unique other restriction that I saw clearly: $$\int_0^xf'(s)\,\mathrm ds ≠ f(x) - f(0).$$ The only non-trivial function that I know that satisfies this is the Cantor-Lebesgue function, but this function does not work for this problem. So, I tried to start with a function that was not absolutely continuous, but I could not get an example.

$$\def\aeeq{\stackrel{\mathrm{a.e.}}{=}}\def\d{\mathrm{d}}$$Denote by $$C(x)$$ the Cantor function on $$[0, 1]$$ and define$$h(x) = \begin{cases} \dfrac{1}{2} C(2x); & 0 \leqslant x \leqslant \dfrac{1}{2}\\ \dfrac{1}{2} C(2(1 - x)); & \dfrac{1}{2} < x \leqslant 1 \end{cases},$$ then $$h$$ is continuous, $$h(0) = h(1) = 0$$, $$0< h(x) \leqslant 1$$ for $$0 < x < 1$$, $$h' \aeeq 0$$, and the variation of $$h$$ on $$[0, 1]$$ is$$V_{0, 1}(h(x)) = \frac{1}{2} V_{0, \frac{1}{2}}(C(2x)) + \frac{1}{2} V_{0, \frac{1}{2}}(C(2(1 - x))) = \frac{1}{2} V_{0, 1}(C(x)) + \frac{1}{2} V_{0, 1}(C(x)) = 1.$$
Now, denote $$x_0 = 0$$ and take$$f(x) = \sum_{k = 1}^∞ (-1)^k (x_k - x_{k - 1}) h\left( \frac{x - x_{k - 1}}{x_k - x_{k - 1}} \right) I_{[x_{k - 1}, x_k)}(x),$$ where $$I_A$$ is the indicator function. Note that $$f(x)$$ is continuous at $$x = 1$$ since $$x_n - x_{n - 1} → 0\ (n → ∞)$$ and $$0 \leqslant h(x) \leqslant 1$$ for $$0 \leqslant x \leqslant 1$$, thus $$f$$ is continuous on $$[0, 1]$$. The variation of $$f$$ on $$[0, 1]$$ is\begin{align*} V_{0, 1}(f(x)) &= \sum_{k = 1}^∞ (x_k - x_{k - 1}) V_{x_{k - 1}, x_k} \left( h\left( \frac{x - x_{k - 1}}{x_k - x_{k - 1}} \right) \right)\\ &= \sum_{k = 1}^∞ (x_k - x_{k - 1}) V_{0, 1}(h(x)) = \sum_{k = 1}^∞ (x_k - x_{k - 1}) = \lim_{n → ∞} x_n = 1. \end{align*} Also, $$f' \aeeq 0$$. Thus,$$g(x) = f(x) - f(0) - \int_0^x f'(s) \,\d s = f(x). \quad \forall x \in [0, 1]$$ For any $$n \geqslant 1$$, $$g(x_n) = f(x_n) = (-1)^{n + 1} (x_{n + 1} - x_n) h(0) = 0$$. If $$n$$ is odd, then$$g(x)\Bigr|_{(x_n, x_{n + 1})} = f(x) \Bigr|_{(x_n, x_{n + 1})} = (-1)^{n + 1} (x_{n + 1} - x_n) h\left( \frac{x - x_n}{x_{n + 1} - x_n} \right) > 0.$$ If $$n$$ is even, then$$g(x)\Bigr|_{(x_n, x_{n + 1})} = f(x) \Bigr|_{(x_n, x_{n + 1})} = (-1)^{n + 1} (x_{n + 1} - x_n) h\left( \frac{x - x_n}{x_{n + 1} - x_n} \right) < 0.$$ Therefore, this $$f$$ satisfies all the requirements.
• Awesome! I would just like to clarify one point. I dont know if I got why $f$ is continuous in $[0,1]$. – Lucas Corrêa Oct 21 '18 at 19:55
• @LucasCorrêa $h(0)=h(1)=1$ implies that $f$ is continuous on each $[x_n,x_{n+1}]$, and it has been proved that $f$ is also continuous at $x=1$. – Saad Oct 22 '18 at 0:26