Let $f : M \rightarrow \mathbb{R}$ be a continuous function on $M \subset \mathbb{R}^2$, where $M$ is compact (so $f$ is uniformly cont.). I have to show that $T : C(M) \rightarrow C(M) : Tf := \int_M \frac{1}{|t-\tau|} f(\tau) d\tau$ is bounded in the usual norm on $C(M)$, i.e. $\vert \vert f \vert \vert = \sup_{\tau \in M} f$.
If the kernel function was continuous, this would be child's play, since it would also be bounded and the rest follows easily enough. However, since $1/|x-y|$ has a singularity, I am unsure of what the proper angle of attack is. Can anyone help?
The point is that $1/|t|$ locally integrable on $\mathbb R^2$: using polar coordinates, for suitable $R$
$$ \int_M \frac{1}{|t-\tau|} \; dt \le \int_{0}^{2\pi} d\theta \int_0^R dr $$