# Bounded variation condition if $f(0)=0$ and $f(x) =x^{\alpha}\sin(x^{-\beta})$ for $x\in(0,1]$

Let $f:[0,1] \rightarrow \Bbb R$ with $f(x) = \begin{cases} x^{\alpha}\sin(x^{-\beta})\;\; \text{if}\; x \neq 0 \\ f(x) =0 \;\;\text{if}\; x =0\end{cases}$

Find out for which value of$\;\alpha, \beta$ $f(x)$ would become of bounded variation on [0,1].

Question: one person advised for this problem set, "Since $f(x)$ is locally absolutely continuous, a necessary and sufficient condition is $f'\in L^1$

Which Set does $L^1$ represent? and what is the meaning of "absolutely continuous"?

• @JeanMarie edited. – Daschin May 26 '17 at 13:26
• No \begin{cases}...\end{cases} in titles, please. – Did May 26 '17 at 15:42
• If you haven't studied the Lebesgue Integral or absolute continuity, it's probably better just to do the exercise from scratch. – Matematleta May 26 '17 at 15:46

If $\alpha>\beta$, then $f$ is $C^1$ on $(0,1)$, so for any partition $P=\left \{ 0,x_1,\cdots,x_{n-2},1 \right \}$, there are $c_i\in (x_k,x_{k-1})$ such that

$V(f,P)=\sum |f(x_k)-f(x_{k-1})|=\sum |f'(c_k)|(x_k-x_{k-1})\le \\ (\alpha+\beta)\sum (x_k-x_{k-1})=\alpha+\beta<\infty .$

Since $P$ is arbitrary, it follows that $f$ is of bounded variation whenever $\alpha>\beta.$

On the other hand, if $\alpha\le \beta,$ then choosing $P_n=\left \{0, x_n,\cdots, x_1 \right \}$ where $x_n=\left ( \frac{n\pi}{2} \right )^{-1/\beta},$ then $V(f,P_n)=\sum_0^n \left ( \frac{k\pi}{2} \right )^{-\alpha/\beta},$ which diverges as $n\to \infty.$

Thus, if $\alpha\le \beta,\ f$ is not of bounded variation.

• how is $\sum |f(x_k)-f(x_{k-1})|=\sum |f'(c_k)|(x_k-fx_{k-1})$ derived? – Daschin May 27 '17 at 0:56
• Mean Value Theorem. – Matematleta May 27 '17 at 3:36
• So isn't it $x_{k-1}$ not $fx_{k-1}$? – Daschin May 27 '17 at 7:22
• What I understand about notation $C^1$ means there $\exists$ 1-th continuous deratives for the function. If my understadning is correct, how could one know that $f$ is $C^1$ when $\alpha\gt\beta$? – Daschin May 27 '17 at 7:34
• Yes it's a typo. Thanks – Matematleta Jun 7 '17 at 0:10

Hint:

If $\alpha < 0$ and $\beta \geq 0$ then $f$ is not bounded in $[0,1]$, hence it cannot be $BV$.

Let us consider the case $\alpha \geq 0$. Since $f$ is continuously differentiable in $[\epsilon, 1]$ for every $\epsilon\in (0,1)$, its variation in $[\epsilon, 1]$ is given by $$\text{T.V.}(f, [\epsilon, 1]) = \int_\epsilon^1 |f'(x)| \, dx.$$ In particular, the function $f$ is in BV if $$\lim_{\epsilon\to 0+} \int_\epsilon^1 |f'(x)|\, dx$$ is finite, i.e. if the improper integral $\int_0^1|f'(x)|\, dx$ is convergent.