Does $\hbox{pv-}\!\!\int_{-1}^1\!\frac{f(x)}{x}dx:=\lim_{\epsilon\to0}\int_{\epsilon\le|x|\le1}\!\frac{f(x)}{x}dx$ exist for all $f\in C[-1,1]$?

Question

My question concerns the existence of the principal value integral \begin{equation} \hbox{pv-}\!\!\!\int_{-1}^1\!\frac{f(x)}{x}\,\mathrm dx:=\lim_{\epsilon\to0}\int_{\epsilon\le|x|\le1}\!\frac{f(x)}{x}\,\mathrm dx \tag{1} \end{equation} for continuous functions $f\in C[-1,1]$. It is classically known that if we impose $C^{0,\alpha}$ Hölder continuity on $f$ for some $\alpha>0$, then this integral will exist. Furthermore, it is known that it is possible to define $$ Hf(x):=\hbox{pv-}\!\!\!\int_{\mathbb R}\frac{f(y)}{x-y}\,\mathrm{d}y $$ for a.e. $x\in\mathbb R$, with certain (weak) boundedness properties between Lebesgue spaces. However, since I know that $Hf$ need not be continuous for continuous $f$, it intuitively seems that it should be possible to construct some $f\in C[-1,1]$ s.t. (1) doesn't exist. Is this the case?

Answer 1

Consider the function

$$ f(x) = \frac{\operatorname{sign} x}{1 - \log |x|}, \qquad f(0) = 0. $$

Its graph looks like

$\hspace{10em}$

Since $f$ is odd, $f(x)/x$ is an even function and hence

$$ \int_{\epsilon \leq |x| \leq 1} \frac{f(x)}{x} \, dx = 2 \int_{\epsilon}^{1} \frac{dx}{x(1-\log x)} = 2\log(1-\log\epsilon) $$

which diverges as $\epsilon \to 0^+$.