# a problem on functions of bounded variation

Which of the following statements are necessarily true ?
a. Any continuous function on [$0, 1$] is of bounded variation.
b. If $f : \mathbb{R} → \mathbb{R}$ is continuously differentiable, then its restriction to the interval [$−n, n$] is of bounded variation on that interval, for any positive integer $n$.
c. Any monotone function on [$0, 1$] is of bounded variation.

I know that (c) is true but not sure about the others. Can anybody help me for the other options?

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## 2 Answers

a) is false. Consider the function $$f(x) = \begin{cases} 0 & x=0 \\ x\cos\left(\frac{\pi}{x}\right) & x\neq 0\end{cases}$$ I will leave it to you to show that the function is not of bounded variation but is continous on $[0,\ 1]$.

b) is true. The function $f$ has continuous derivative $f'$ on every interval of the form $I_n = [-n,\ n]$. By the extreme value theorem, $|f'|$ attains a maximum $M$ on $I_n$. Take an arbitrary partition $$\mathcal{P}:\ -n = x_0 < x_1 < \cdots < x_m = n$$ of $I_n$. Then for each sub-interval $[x_{i-1},\ x_i]$, we have by the mean value theorem, $$f(x_i) - f(x_{i-1}) = f'(c_i)(x_i - x_{i-1})$$ for some $c_i \in [x_{i-1},\ x_i]$. This then implies \begin{align}\sum_{i=1}^m\left|f(x_i) - f(x_{i-1})\right| &= \sum_{i=1}^m\left|f(c_i)(x_i - x_{i-1})\right| \\ & \le \sum_{i=1}^m M(x_i - x_{i-1}) \\ & = 2Mn \end{align} This holds for any arbitrary partition so the function $f$ is of bounded variation on $I_n$.

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Hint

1. Consider $$f(x) := \begin{cases} x \cdot \sin \frac{1}{x} & x \in (0,1] \\ 0 & x=0 \end{cases}$$
2. Use the fact that the derivative $f'$ is bounded on $[-n,n]$ and apply mean value theorem.
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