Prove that $\operatorname{p.v.}(k\;*f)$ does not exist if $k(x)=|x|^{-n+i\gamma}$ and $f\in\mathcal{C}_c^1$ I put a bounty only because I need quickly a solution, NOT because I know it's difficult - maybe it is, maybe not. I'm trying to do it, but without results. If I get some "intermediate result" (something I think it could be useful) I will write it here immediately.
Consider the function $k(x)=|x|^{-n+i\gamma}$ defined on $\Bbb R^n$, where $\gamma$ is any nonzero real number and $i$ is the imaginary unit.
$k$ is homogeneous of degree $-n+i\gamma$.
I have to show that
$$
\operatorname{p.v.}(k\;*f)(x):=\lim_{\epsilon\to0^+}
\int_{\{|t|>\epsilon\}}|t|^{-n+i\gamma}f(x-t)\;dt
%(k\chi_{\{|x|>\epsilon\}})*f
$$
does NOT exists for $f\in\mathcal C_c^1(\Bbb R^n)$ (i.e. $\mathcal{C}^1$ functions with compact support), but we can find a sequence $\epsilon_j\to0$ such that
$$
\lim_{j\to+\infty}\int_{\{|t|>\epsilon_j\}}|t|^{-n+i\gamma}f(x-t)\;dt
%(k\chi_{\{|x|>\epsilon_j\}})*f
$$
exists for such $f\;$.
We exclude the trivial case $f\equiv 0$.
I don't know where to start, can someone help me please?
 A: To analyse the behaviour as $\varepsilon \to 0$, it is useful to write $f(x-t) = f(x) + \bigl(f(x-t) - f(x)\bigr)$. Since neither of these two terms approaches $0$ at $\infty$, we can only use that decomposition on sets of finite measure, so we split $k$ into two parts, $k = k_1 + k_2$, where $k_1 = \chi_{\{ t : \lvert t\rvert \leqslant 1\}}\cdot k$. Then $k_2$ is locally integrable on $\mathbb{R}^n$, whence
$$k_2 \ast f \colon x \mapsto \int_{\mathbb{R}^n} k_2(t)f(x-t)\,dt$$
is a continuous function on $\mathbb{R}^n$.
For $0 < \varepsilon < 1$, we have
$$\int_{\lvert t\rvert > \varepsilon} k(t)f(x-t)\,dt = (k_2\ast f)(x) + \int_{\varepsilon < \lvert t\rvert \leqslant 1} k(t)f(x-t)\,dt$$
and writing $f(x-t) = f(x) + \bigl(f(x-t) - f(x)\bigr)$, we need to analyse the behaviour of
$$\int_{\varepsilon < \lvert t\rvert \leqslant 1} k(t)\bigl(f(x-t) - f(x)\bigr)\,dt\tag{$\ast$}$$
and
$$f(x)\cdot\int_{\varepsilon < \lvert t\rvert \leqslant 1} k(t)\,dt \tag{$\ast\ast$}$$
as $\varepsilon \to 0$. The assumptions on $f$ are more than sufficient to conclude that $(\ast)$ has a limit as $\varepsilon \to 0$. Since $f\in \mathcal{C}_c^1(\mathbb{R}^n)$, all partial derivatives of $f$ are continuous functions with compact support, hence bounded, and that implies that $f$ is Lipschitz continuous, $\lvert f(x) - f(y)\rvert \leqslant L_f \cdot \lvert x-y\rvert$. It would suffice that $f$ be $\alpha$-Hölder continuous for any $\alpha > 0$ to see that $(\ast)$ has a limit, for then
$$\bigl\lvert k(t)\cdot \bigl(f(x-t) - f(x)\bigr)\bigr\rvert = \lvert t\rvert^{-n}\cdot \lvert f(x-t) - f(x)\rvert \leqslant C\cdot \lvert t\rvert^{-n}\cdot \lvert (x-t)-x\rvert^{\alpha} = C\cdot \lvert t\rvert^{\alpha-n},$$
so $k(t)\cdot \bigl(f(x-t)-f(x)\bigr)$ is Lebesgue-integrable over the unit ball $\{ t : \lvert t\rvert \leqslant 1\}$.
It remains to see how $({\ast}\ast)$ behaves as $\varepsilon \to 0$. We can evaluate that integral explicitly using spherical coordinates:
\begin{align}
\int_{\varepsilon < \lvert t\rvert \leqslant 1} \lvert t\rvert^{-n + i\gamma}\,dt
&= \omega_{n-1}\cdot \int_{\varepsilon}^1 r^{-n+i\gamma}r^{n-1}\,dr \\
&= \omega_{n-1}\cdot \int_{\varepsilon}^1 r^{-1+i\gamma}\,dr \tag{$r = e^{\rho}$}\\
&= \omega_{n-1} \cdot \int_{\log \varepsilon}^0 e^{i\gamma \rho}\,d\rho\\
&= \frac{\omega_{n-1}}{i\gamma} \bigl(1 - e^{i\gamma \log \varepsilon}\bigr),
\end{align}
where $\omega_{n-1}$ is the $(n-1)$-dimensional volume of the unit sphere $\{ t : \lvert t\rvert = 1\}$.
As $\varepsilon \to 0$, the exponential $e^{i\gamma \log \varepsilon}$ traverses the whole unit circle in $\mathbb{C}$ infinitely often, so
$$\lim_{\varepsilon \to 0} \int_{\lvert t\rvert > \varepsilon} k(t)f(x-t)\,dt$$
exists only for $x$ with $f(x) = 0$. But for every $z\in \mathbb{C}$ with $\lvert z\rvert = 1$ we can find sequences $\varepsilon_j \to 0$ such that $e^{i\gamma \log \varepsilon_j} \to z$ (we can even choose the sequence so that $e^{i\gamma \log \varepsilon_j} = z$ for all $j$), and for such sequences we have
$$\lim_{j \to \infty} \int\limits_{\lvert t\rvert > \varepsilon_j} k(t) f(x-t)\,dt = \frac{\omega_{n-1}(1-z)}{i\gamma}f(x) + \int\limits_{\lvert t\rvert \leqslant 1} k(t)\bigl(f(x-t) - f(x)\bigr)\,dt + \int\limits_{\lvert t\rvert > 1} k(t)f(x-t)\,dt.$$
