# a) Find all the values of $\alpha$ such that $f'(0)$ exists. b) Find all the values of $\alpha$ such that $f$ is of bounded variation on $[0,1]$

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.

• 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? – user539887 Apr 24 '18 at 21:19
• for a) $f$ derivative is $$f^\prime(0) = \begin{cases} \alpha 0^{\alpha-1} \sin\left(\frac{1}{0^\beta}\right) - \frac{0^\alpha}{0^{\beta +1}} \cos\left(\frac{1}{0^\beta}\right) &\text{on }(0,1], \\ 0 & \text{if }x = 0. \end{cases}$$ – lomber Apr 24 '18 at 23:30
• @user539887 for b )$\gamma < 1$ – lomber Apr 24 '18 at 23:31
• but what is $\gamma$ here? – lomber Apr 24 '18 at 23:40
• 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 $\alpha-\beta-1$. – user539887 Apr 25 '18 at 6:23

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 $$\alpha>\beta$$.

• If $−1<\alpha−1<0$ or $-1< \alpha -\beta - 1<0$, then the integrands are unbounded and therefore not Riemann integrable. – David Feb 4 at 1:58

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.