How to show $f(x) = \begin{cases} \frac{\sin(x)}{x}, & \text{when } x \neq 0,\\ 0, & \text{when } x = 0. \end{cases}$ is of bounded variation?

Question

Prove or disprove: $f(x) = \begin{cases} \frac{\sin(x)}{x}, & \text{when } x \neq 0,\\ 0, & \text{when } x = 0. \end{cases}$ is of bounded variation on $[0,1]$ and $\sup_{p \in P} V(f,p) =2$.

My try:

Let $p=\{0,x_{1},x_{2},...,x_{n-1},1\}$

$$ \begin{split} V(f,p) &= \sum_{k=1}^{n}|f(x_{k})-f(x_{k-1})| \\ &= \left|\frac{\sin(x_{1})}{x_{1}}\right| + \left|\frac{\sin(x_{2})}{x_{2}}-\frac{\sin(x_{1})}{x_{1}}\right| + \ldots \\ &+ \left|\frac{\sin(x_{n-1})}{x_{n-1}}-\frac{\sin(x_{n-2})}{x_{n-2}}\right| + \left|\sin(1)-\frac{\sin(x_{n-1})}{x_{n-1}}\right| \end{split} $$

I tried the definition of variation but I was not capable of reaching the result — Math Admiral, Jul 20 '23 at 14:16
could you update your questions with what you tried and how it did not work? would be much easier to guide you that way — gt6989b, Jul 20 '23 at 14:17
I tried to make the variation bounded by $2 $ but I am not capable of doing it — Math Admiral, Jul 20 '23 at 14:24
You can use that if $f$ is $C^1$, then $V_a^b(f)=\int_a^b\lvert f'(x)\rvert~\mathrm{d}x$. See this for the details — Lorago, Jul 20 '23 at 14:33
@Lorago But the derivative of my function is non elementary function so can't be integrated easily. — Math Admiral, Jul 20 '23 at 14:39
It bothers me that the function is being defined to have a discontinuity at $x=0$, since $\lim_{x\to 0} \sin x/x = 1$. Would you mind checking that the problem statement did not become garbled in its presentation here? That said, I'm voting to leave open. — hardmath, Jul 25 '23 at 16:20

score 2 · Accepted Answer · answered Nov 26 '24 at 03:12

For reals $a<b$, $a,b\in [0,1]$ let $V[a,b]$ be the variation of the function $f$ at the segment $[a,b]$. Then $V[0,1]=\sup_{x\in (0,1]} f(x)+V[x,1]$. But when $x\in (0,1]$ then $f'(x)=\frac{x\cos x-\sin x}{x^2}\le 0$, because $x\le \tan x$. So the function $f$ is nonincreasing at $(0,1)$ and so $V[x,1]=f(x)-f(1)$. Therefore $$V[0,1]=\sup_{x\in (0,1]} 2f(x)-f(1)=2-\sin 1.$$

How to show $f(x) = \begin{cases} \frac{\sin(x)}{x}, & \text{when } x \neq 0,\\ 0, & \text{when } x = 0. \end{cases}$ is of bounded variation?

1 Answers1