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For any positive real numbers $ \alpha $ and $\beta$, define

$f(x) = \begin{cases} x^{\alpha} \sin\frac{1}{x^\beta} && \text{if $x \in (0,1]$,}\\ 0 && \text{if $x = 0$}\end {cases}$

a) For a given $\beta > 0$ , find all the values of $\alpha$ such that $f'(0)$ exists.

b) For given $\beta >0,$ find all the values of $ \alpha$ such that $f$ is of bounded variation on $[0,1]$

My ANSWER: For $ a) $ $\alpha \ge \beta$ then $f$ is bounded variation . now $ f'(0) $will exists if $\alpha \ge \beta \ge 0$

For $b)$ same condition as for $(a)$

Edit answer :$f$ has derivative $$\displaystyle f^\prime(x) = \begin{cases} \alpha x^{\alpha-1} \sin\left(\dfrac{1}{x^\beta}\right) - \dfrac{x^\alpha}{x^{\beta +1}} \cos\left(\dfrac{1}{x^\beta}\right) &\text{on }(0,1], \\\\ 0 & \text{if }x = 0. \end{cases}$$ Hence $$\vert f^\prime(x) \vert \le \alpha x^{\alpha-1} + x^{\alpha - \beta -1}$$ The integrals $\int_0^1 x^{\alpha-1}dx$ and $\int_0^1 x^{\alpha - \beta -1}dx$ both converge for $1 < \alpha < 1 +\beta $. Hence $$V_0^1(f) \le \int_0^1 \vert f^\prime(x) \vert dx$$ and $f$ is of bounded variation on $[0,1]$ as the RHS of the inequality is finite.

Is my answer correct ??? or incorrect ?? Please rectify it.

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    $\begingroup$ Some hints: for a): apply the definition of derivative at zero; for b): for which $\gamma \in \mathbb{R}$ is the integral $$\int\limits_{0}^{1} x^{\gamma}\, dx$$ convergent? Also, are you sure that your formula for the derivative on $(0, 1]$ correct? $\endgroup$
    – user539887
    Commented Apr 24, 2018 at 21:19
  • $\begingroup$ for a) $f$ derivative is $$f^\prime(0) = \begin{cases} \alpha 0^{\alpha-1} \sin\left(\frac{1}{0^\beta}\right) - \frac{0^\alpha}{0^{\beta +1}} \cos\left(\frac{1}{0^\beta}\right) &\text{on }(0,1], \\ 0 & \text{if }x = 0. \end{cases}$$ $\endgroup$
    – jasmine
    Commented Apr 24, 2018 at 23:30
  • $\begingroup$ @user539887 for b )$\gamma < 1$ $\endgroup$
    – jasmine
    Commented Apr 24, 2018 at 23:31
  • $\begingroup$ but what is $\gamma$ here? $\endgroup$
    – jasmine
    Commented Apr 24, 2018 at 23:40
  • $\begingroup$ I repeat: for a), apply the definition of derivative (not use the formula that have no application here; by the way, that formula is still wrong). $\gamma$ is any real number, for which you can substitute $\alpha-1$ and $\alpha-\beta-1$. $\endgroup$
    – user539887
    Commented Apr 25, 2018 at 6:23

2 Answers 2

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For the first part $f'(0)$ exists if $\displaystyle\lim_{h\rightarrow 0^+}\dfrac{f(0+h)-f(0)}{h}$ exists. Without loss of generality fix $\beta>0$, we have the limit $$\displaystyle\lim_{h\rightarrow 0^+}\dfrac{(0+h)^{\alpha}\sin\Big(\dfrac{1}{(0+h)^{\beta}}\Big)-0}{h}=\lim_{h\rightarrow 0^+}h^{\alpha-1}\sin\Big(\dfrac{1}{h^{\beta}}\Big) .$$ Using squeeze theorem we see the limit exists if $\alpha\geq \beta+1$.

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For the second part we calcualte $f'(x)$.

$$ f'(x)=\begin{align}\begin{cases}\alpha x^{\alpha-1}\sin\Big(\dfrac{1}{x^{\beta}}\Big)-\dfrac{\beta x^{\alpha}}{x^{\beta+1}}\cos\Big(\dfrac{1}{x^{\beta}}\Big), &x\neq0\\0, &x=0\end{cases}\end{align}$$ If total variation is finite we say $f$ is of bounded variation. Also if $f$ is differentiable and its derivative is Riemann-integrable, its total variation is given by

$${\displaystyle V_{a}^{b}(f)=\int _{a}^{b}|f'(x)|\,\mathrm {d} x.} .$$

$$V_0^1(f)=\int_0^1|f'(x)|dx\leq\alpha\int_0^1x^{\alpha-1}dx+\beta\int_0^1x^{\alpha-\beta-1}dx=1+\dfrac{\alpha}{\alpha-\beta} $$ (The above integrals Riemann integrable only if $\alpha-1>-1$ and $\alpha-\beta-1>-1$). Thus the total variation $V_0^1(f)$ is finite if $\alpha>0$ and $\alpha>\beta$.

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  • $\begingroup$ If $−1<\alpha−1<0$ or $-1< \alpha -\beta - 1<0$, then the integrands are unbounded and therefore not Riemann integrable. $\endgroup$
    – David
    Commented Feb 4, 2019 at 1:58
  • $\begingroup$ Can you show why $f'(0) = 0$. $\endgroup$
    – user0
    Commented Aug 5, 2021 at 21:17
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As explained in Yadati Kiran's answer, if $\alpha > \beta > 0$, then $f'$ is integrable. By the Fundamental Theorem of Calculus, we have $$ \int_y^x f' = f(x) - f(y) $$ for any $x > y > 0$. Taking the limit as $y \to 0$, we obtain $$ f(x) = \int_0^x f' $$ by continuity of integration. Now let $P = \{x_0, \cdots, x_k\}$ be a partition of $[0,1]$. We then have \begin{align*} V(f, P) &= \sum_{i=1}^k |f(x_i) - f(x_{i-1})| \\ &= \sum_{i=1}^k \left|\int_0^{x_i} f' - \int_0^{x_{i-1}} f'\right| \\ &= \sum_{i=1}^k \left|\int_{x_i}^{x_{i-1}}f'\right| \\&\leq \sum_{i=1}^k \int_{x_{i-1}}^{x_i}|f'| \\ &= \int_0^1 |f'| \\ &< \infty \end{align*} Since $P$ was arbitrary, $f$ is of bounded variation.

Now suppose $\beta \geq \alpha > 0$. Fix an index $n$ and consider the following partition: $$ P_n = \left\{0, \left(\frac{2}{2 n \pi}\right)^{1 / \beta}, \left(\frac{2}{(2n - 1) \pi}\right)^{1 / \beta}, \cdots, \left(\frac{2}{\pi}\right)^{1 / \beta}, 1 \right\} $$ It can be shown \begin{align*} V(f, P_n) &= c \sum_{k=0}^{n - 1} \frac{1}{(1 + 2k)^{\alpha / \beta}} + \sin(1) - c \\ &\geq \frac{c}{2^{\alpha/\beta}} \sum_{k=1}^{n} \frac{1}{k^{\alpha / \beta}} + \sin(1) - c \end{align*} where $$ c = 2 \left(\frac{2}{\pi}\right)^{\alpha / \beta} $$ Since $0 < \alpha / \beta \leq 1$, the series on the right-hand side diverges. Therefore $f$ is not of bounded variation.

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