# bounded variation

• Dec 24th 2009, 07:19 PM
Kat-M
bounded variation
Let $f(x)$ be defined by $f(0)=0$ and $f(x)=x^2 sin(1/x)$ for $x \neq 0$. Prove that $f$ is of bounded variation on [-1,1].
• Dec 24th 2009, 08:03 PM
tonio
Quote:

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

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

Tonio
• Dec 24th 2009, 08:10 PM
Kat-M
Quote:

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

Tonio

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

Is is correct?
• Dec 24th 2009, 08:38 PM
tonio
Quote:

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

Is is correct?

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

Tonio