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For $\alpha$, $\beta$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{\alpha}\sin(1/x^{\beta})$ for $0 < x \leq 1$ and $f(x) = 0$ for $x=0$. Show that if $\alpha > \beta$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $\alpha \leq \beta$, then $f$ is not of bounded variation on $[0, 1]$.

Proof: Let $x,y>0$. Then, $f(x) - f(y) = \int^x_y f'(t) dt$. Then, there is some partition $V(f, P)$ = $\sum^n_{k=1}|f(x_{k+1}) - f(x_k)|$ = $\sum^n_{k=1}|\int_{x_k}^{x_{k+1}}f'(t)dt|$ $\leq$ $\sum^n_{k=1}\int_{x_k}^{x_{k+1}}|f'(t)|dt$ $\leq$ $\int_{0}^{1}|f'(t)|dt$ < $\infty$.

If $\int_{0}^{1}|f'|$ < $\infty$, then $f \in BV[0, 1]$.

Since if $f$ is increasing on $[a, b]$, $\int^a_b f'(t) dt \leq f(a)-f(b)$, would it be helpful to break up into increasing\decreasing intervals?

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Which part of the statement do you want to show exactly? – Cantor Dec 9 '12 at 17:10
I want to show both, but I guess I was wanting to show it would be of bounded variation first. – Jake Casey Dec 12 '12 at 3:06

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