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Let $$f(x) = x^2 \sin{\frac{1}{x^2}}$$ and $f(0) = 0$. $f$ is continuous and differentiable everywhere on $[0,1]$.

The hint of the problem suggests considering the indefinite integral of $f'$ over $[0,1]$.

However, $f'$ seems integrable on $[0,1]$, as I have calculated through Mathematica. I also found $\int_0^1 f'(x)dx = \sin{1}$.

So how can I solve this? Thanks in advance!

R. Feng
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    True, $\int_0^1 f'$ exists as an improper integral, but it's $\int_0^1|f'| $ you need to worry about. The latter integral does not converge, so $f$ is not AC. – zhw. May 23 '17 at 14:08
  • @zhw sir, how can I show that $\int_0^1 |f^{'}|dx$ does not converge? – R. Feng May 23 '17 at 15:02

2 Answers2

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Hint: The derivative $f'$ should be Lebesgue integrable, so $\int_0^1 |f'(t)|\, dt$ should be finite. Chopping this up into an infinite sum over intervals of monotonicity you realize that it is not finite.

KCd
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H. H. Rugh
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  • Thanks, how can I decompose the interval into an infinite sum over intervals of monotonicity? Does it mean $|f'|$ is monotonic on each of the intervals? – R. Feng May 23 '17 at 14:42
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    The easiest is to decompose into intervals of monotonicity of $\sin 1/x^2$, i.e. at the points where $1/x_n^2= \frac{\pi}{2}(2n+1)$, $n\geq 0$. But you may simply do as Chilango does in his answer, i.e. considering the variation of $f$ over a suitable partition (which more or less partitions into monotonicity intervals of $f'$). – H. H. Rugh May 23 '17 at 15:15
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For another approach, note that if we choose a partition $P=\left \{0, x_n,\cdots, x_1 \right \}$ where $x_n=\left ( \frac{n\pi}{2} \right )^{-1/2}, $ then $V(f,P_n)=\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k},$ which diverges, and therefore $f$ is not of bounded variation, hence can not be absolutely continuous.

Matematleta
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