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I am reading Giovanni Leoni's A First Course in Sobolev Spaces and trying to solve the following problem:

Let $u : [0, 1]\rightarrow\mathbb{R}$ be defined by $$u(x) = \begin{cases}x^{-c}\sin(x^d), & 0 < x\leq 1,\\0, & x= 0,\end{cases}$$ where $c, d > 0$. For which $c, d$ is $u\in BV([0, 1])$?

The original problem states that $c$ and $d$ can be any real numbers, but I have been able to deal with the other cases. Could someone please give me feedback about whether my solution is correct?

My attempt:

If $d < c$, then the function is not bounded and hence cannot be of bounded variation. Now I show that it if $d\geq c$, then $u\in BV([0, 1])$.

For $x > 0$ we have $$u'(x) = dx^{d-c-1}\cos(x^d)-cx^{-c-1}\sin(x^d).$$ If $d > c$ then using the fact that $|\cos(y)|\leq 1$ and $|\sin(y)|\leq y$ for $y\geq 0$ we have $$|u'(x)|\leq (d+c)x^{d-c-1}.$$ If $d = c$ then using the fact that $\sin(y)\geq y-\frac{y^3}{3}$ for $y\geq 0$ we have \begin{align}u'(x)\leq \frac{d}{x}\Big(1-\frac{1}{x^d}\Big(x^d-\frac{x^{3d}}{6}\Big)\Big) = \frac{d}{6}x^{2d-1}.\end{align} Similarly, using the fact that $\cos(y)\geq 1-\frac{y^2}{2}$ for $y\geq 0$ we have $$-u'(x)\leq \frac{d}{x}\Big(\frac{x^d}{x^d}-\Big(1-\frac{x^{2d}}{2}\Big)\Big) = \frac{d}{2}x^{2d-1}.$$ Thus, for $d = c$ we have $$|u'(x)|\leq \frac{d}{2}x^{2d-1}.$$ If $0 = x_{0} < x_{1} < \cdots < x_{N}$ is a partition, then \begin{align}\sum_{i = 1}^{N}|u(x_{i})-u(x_{i-1})| &\leq \Big|\frac{\sin(x_1^d)}{x_1^d}\Big|+\sum_{i = 2}^{N}\int_{x_{i-1}}^{x_{i}}|u'(x_{i})|\:\text{d}x\\&\leq \begin{cases}1+\frac{d+c}{d-c}(x_{N}^{d-c}-x_{1}^{d-c}), & d > c,\\1+\frac{1}{4}(x_{N}^{2d}-x_{1}^{2d}), & d = c,\end{cases}\end{align} which shows that $u\in BV([0, 1])$.

Is this the correct approach? Any help is appreciated. Thank you.

Karthik Kannan
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1 Answers1

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Your approach seems fine. You can also use this representation $$u(x) = x^{d-c}\frac{\sin(x^d)}{x^d}, \quad x> 0\,,\quad \quad (*)$$ for an alternative approach. As you observed for $d<c$, $(*)$ gives that the function is unbounded on $[0,1]$. Now, notice that since $\sin$ is analytic on $\mathbb{R}$, it follow that $$\frac{\sin(x^d)}{x^d} = \displaystyle \sum_{n\ge 0}\frac{x^{2nd}}{(2n+1)!}, \quad x>0$$ is also. Thus, in the case $d\ge c$, $u$ is bounded in absolute value on $[0,1]$ and has an integrable derivative on $(0,1]$, which gives the claim.

Pantelis Tassopoulos
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