Let $f\colon \mathbb{R}\to \mathbb{R}, x\mapsto -\ln|x|$ and $g\colon \mathbb{R}\to \mathbb{R}, x \mapsto 1/\sqrt{x^2-1}\mathbf{1}_{[-1,1]}(x)$, where $\mathbf{1}_{S}(x)=1$ if $x\in S$ and $\mathbf{1}_{S}(x)=0$, otherwise. Consider the convolution: $$ [f\ast g](y) = \int_{-\infty}^\infty f(x-y)g(x)\mathrm{d}x = -\int_{-1}^1 \frac{\ln|x-y|}{\sqrt{x^2-1}} \mathrm{d}x $$
How to prove (or disprove) that $[f\ast g](y)$ is a continuous function for $y\in[-1,1]$?
N.B. I know that the convolution of an integrable function with a bounded function is continuous. In my case, however, the two functions $f$ and $g$ are both unbounded. I tried to compute the convolution integral with some symbolic math software, but I got some pretty messy results. Any help/comment/suggestion is welcome. Thanks a lot!
Theorem: If $f\in L^p$ and $g\in L^q$ with $\frac{1}{p}+\frac{1}{q}=1$, then $f*g$ is continuous.
Proof: Take $y\in\Bbb R$ and a sequence $h \to 0$. Then by Hölder's inequality $$ |(f*g)(y+h)-(f*g)(y)| = \left|\int_{\Bbb R} \left(f(y+h-x)-f(y-x)\right) g(x)\, \mathrm d x\right| \\ \leq \left(\int_{\Bbb R}|f_y(x-h)-f_y(x)|^p\,\mathrm d x\right)^{1/p}\, \|g\|_{L^q} $$ where $f_y(x) = f(y-x)$. By the continuity of the translation operator in $L^p$ (see e.g. Brezis, Functional Analysis, Sobolev spaces and PDE, Lemma $4.3$), the $L^p$ integral above converges to $0$ when $h\to 0$, proving the result.
Application: In your case, notice that if $y\in[-1,1]$, since $x\in[-1,1]$ in the integral of the convolution, then $x-y\in[-2,2]$ and so $f*g = \tilde{f}*g$ where $\tilde{f} = f$ on $[-2,2]$ and $\tilde{f} = 0$ on $\Bbb R\setminus [-2,2]$.
Now, notice that $\tilde{f}\in L^p$ for any $p\in[1,\infty)$, and $g\in L^q$ for any $q < 2$ (because $g(x) \sim 1/\sqrt{2\,(|x|-1)}$ when $|x|\to 1^-$). So for example $$ \tilde{f}\in L^3 \text{ and } g\in L^{3/2} $$ and you can use the above theorem to obtain that $\tilde{f}*g = f*g$ is continuous.