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Let $f\colon[0,1] \rightarrow \mathbb R$ (all real numbers) be defined by $\displaystyle f(x) = x \sin \left(\frac{\pi}{2x}\right)$ if $0 \lt x \le 1$ and $f(x) = 0$ if $x=0$.

Determine whether $f$ is of bounded variation or not.

I've been struggling with this problem and looking for any help. Thanks!

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At the points where $\sin\left( \frac{\pi}{2x}\right)=\pm 1$, what are the values of $x\sin\left( \frac{\pi}{2x}\right)$. Note that these points won't be the max and min, but they will still tell you something useful about the variation of $f$. –  Aaron Mar 17 '13 at 0:13
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Check example 4. It is related to your problem. –  Mhenni Benghorbal Mar 17 '13 at 0:13
    
@Mhenni Benghorbal,+1 for pointing out the example. –  juniven Mar 17 '13 at 0:18
    
@jun: Thanks for the comment. Glad to assist. –  Mhenni Benghorbal Mar 17 '13 at 1:20

1 Answer 1

Fix an integer $N$, and define $x_n:=\frac 1{4n+1}$, $y_n:=\frac 1{4n+3}$. Consider the partition $$0<y_N<x_N<y_{N-1}<x_{N-1}<\dots<y_1<x_1<1.$$ The sum $\sum_k|f(t_k)-f(t_{k-1})|$ for this partition behaves approxitively as $C\log N$, where $C$ is a constant independent of $N$.

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