1. ## bounded variation

Let $\displaystyle f(x)$ be defined by $\displaystyle f(0)=0$ and $\displaystyle f(x)=x^2 sin(1/x)$ for $\displaystyle x \neq 0$. Prove that $\displaystyle f$ is of bounded variation on [-1,1].

2. Originally Posted by Kat-M
Let $\displaystyle f(x)$ be defined by $\displaystyle f(0)=0$ and $\displaystyle f(x)=x^2 sin(1/x)$ for $\displaystyle x \neq 0$. Prove that $\displaystyle f$ is of bounded variation on [-1,1].

$\displaystyle f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3$ on $\displaystyle [-1,1]$, so using the mean value theorem for differentiable functions, $\displaystyle |f(x)-f(y)|\le 3|x-y|$ . Now take any partition of the given interval and from the above you get at once that f has bounded variation there.

Tonio

3. Originally Posted by tonio
$\displaystyle f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3$ on $\displaystyle [-1,1]$, so using the mean value theorem for differentiable functions, $\displaystyle |f(x)-f(y)|\le 3|x-y|$ . Now take any partition of the given interval and from the above you get at once that f has bounded variation there.

Tonio
$\displaystyle \Sigma_1^n |f(x_i)-f(x_{i-1})| \leq 3\Sigma_1^n |x_i -x_{i-1}|=3*2=6<\infty$.

Is is correct?

4. Originally Posted by Kat-M
$\displaystyle \Sigma_1^n |f(x_i)-f(x_{i-1})| \leq 3\Sigma_1^n |x_i -x_{i-1}|=3*2=6<\infty$.

Is is correct?

Yup...it'd be a nice exercise for you to prove that in fact $\displaystyle x^p\sin 1\slash x$ has bounded variation iff $\displaystyle p>1$ ...

Tonio