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].

please help me on this one.

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- Dec 24th 2009, 06:19 PMKat-Mbounded 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].

please help me on this one. - Dec 24th 2009, 07:03 PMtonio

$\displaystyle f'(x)=2x\sin 1\slash x-\cos 1\slash x \Longrightarrow |f'(x)|\le 2+1=3$ 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 - Dec 24th 2009, 07:10 PMKat-M
- Dec 24th 2009, 07:38 PMtonio