Unbounded Hilbert Transform of $C_{c}(\mathbb{R})$ Function I want to give an example of a continuous, compactly supported function (denoted $C_{c}(\mathbb{R})$) $f$, such that the distribution $H(f)$, where $H$ is the Hilbert transform, does not coincide with a continuous function. I.e., there does not a continuous function $g$ such that
$$\langle{Hf,\phi}\rangle=\langle{g,\phi}\rangle,\quad\forall \phi\in\mathcal{S}(\mathbb{R})$$
My attempt was the function
$$f(y):=\dfrac{\text{sgn}(y)}{\log\left|y\right|}\phi(y)$$
where $\phi$ is a smooth cutoff function which avoids the singularity at $y=\pm 1$.
It's easy to show that the principal value integral giving $H$ diverges to $\infty$ in modulus at the origin, as there is no cancellation. However, I could not rigorously (at least by an argument I found convincing) show that
$$\lim_{\delta\rightarrow 0}\int_{\left|y\right|\geq\delta}\dfrac{\text{sgn}(x-y)\phi(x-y)}{y\log\left|x-y\right|}dy$$
blows up as $x\rightarrow 0$. Does anyone have any suggestions for showing this? Alternatively, is there a simpler way to find such an example? I was thinking perhaps stacked copies of trapezoid functions converging in distribution to the characteristic function $\chi_{[-1,1]}$ might work. But I didn't calculate the Hilbert transform of a trapezoid function.

You're on the right track, but let's make it easier. Let $f(y) = |1/\ln y\,|(1-y)^2, 0 < y <1$ and $f=0$ elsewhere. For $x<0, f(x+y)$ has support in $(0,\infty).$ Thus for such $x, f(x+y)/y \ge 0.$ As $x\to 0^-, f(x+y)/y \to f(y)/y = $ pointwise a.e. So by Fatou's Lemma,
$$\infty = \int f(y)/y\,dy \le \liminf_{x\to 0^-}\int f(x+y)/y\,dy.$$
This gives the desired blow up. Now that's not quite the Hilbert transform there at the end, but a simple modification will give what you need.
