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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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    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.)2012-10-28

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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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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.