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This is a qualifying exam problem from Indiana University.

Prove or provide a counterexample to the following statement:

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function, then there exists a real number $L$ such that

$$\lim_{\epsilon \rightarrow 0} \int_{\epsilon \leq|x|\leq1} \frac{f(x)}{x}dx=L$$

End of question.

I have shown that if f is differentiable at $0$ then is the statement is true, but I am having a hard time to find a counterexample with $f$ merely continuous or give a proof.

Is the statement true? Or can someone provide counterexample. Thanks.

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Note that f(x)=1 is not a counterexample as the PV is 0. – KWO Oct 28 '12 at 14:06
Yes, since the PV exists for all constants, you can as well assume $f(0)=0$. And in that case it is equivalent to the question whether $\lim\limits_{\epsilon \to 0} \int_\epsilon^1 \frac{f(x)}{x} \, dx$ exists and is finite. (A positive answer to this is clearly a positive answer to the question, and a counterexample can be extended to an odd function and will give a counterexample to the original question.) – Lukas Geyer Oct 28 '12 at 14:17
up vote 3 down vote accepted

This limit won't exist in general, a counterexample is $f(x) = \frac{1}{1+|\ln x|}$ for $x > 0$, $f(x)=0$ for $x\le0$, in which case the limit is $\infty$.

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Thanks, that's right. Appreciate it. – KWO Oct 28 '12 at 14:28

To find a counterexample, it suffices to consider an odd function $f$ only. Rewriting the integral as

$$ 2 \int_{\epsilon}^{1} \frac{f(x)}{x} \, dx = 2 \int_{0}^{\log(1/\epsilon)} f(e^{-t}) \, dt,$$

it suffices to find a continuous function $g(t) = f(e^{-t})$ defined on $t \geq 0$ such that $\lim_{t\to\infty} g(t) = 0$ and $\int_{0}^{\infty} g(t) \, dt = \infty$. A possible choice is

$$g(t) = \frac{1}{1+t},$$

corresponding to

$$f(x) = \frac{\mathrm{sign}(x)}{1+\left|\log x\right|}.$$

p.s. I think I'm late.

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Thanks, a bit late :) – KWO Oct 28 '12 at 14:28
Thanks again, appreciate it. – KWO Oct 28 '12 at 14:29

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